So all numbers below 2*10^60 has been checked, and all of them eventually reached 1. That means that all larger numbers only need to reach ANY of the previously tested numbers to prove that that number too eventually will reach 1. This is not a solution for the problem but it can speed up the testing of numbers (which I'm sure that most mathematicians have already thought of)
plus it basically rolls down hill from there if we apply the same logic to all the evens that equal and even that equals that range of numbers and so on
Yeah this was one of my first assignments in an algorithms paper at university. You use the previous numbers so you don't have to keep calculating every number again.
I wanted to check the comments to see interesting things, what we get : -Comments saying "how is that a problem". -People who think they can solve the problem (and do what every mathematician failed to do in the last century) in a 2 lines comment on youtube. I guess the real mystery is why I expected something from the comments at the first place...
Well, isn’t this easy? It’s just that this pattern always ends up hitting a number in the pattern: y=2^x. And once it hits a number in that pattern it has to go all the way down to 1 and then you know what happens after that. It just depends on wether this pattern has a starting point that does not hit the equation: y=2^x. Please point out if I’m wrong but I think the conclusion is that this pattern always hits a y value in the equation: y=2^x.
Yes, it´s believe to hit a 2^x. But how do you prove it. You have said how you can prove this whole thing (so if you prove that every number reaches a number expressed by 2^x, then you have the proof), but you do not have the proof. You just have a shortcut.
The whole problem is just that the sequence is arbitrary. Just for illustration. You have a number n for which the cycle ends after 25 steps, then for n+1 it takes 400 steps and for n+2 it only takes 20 steps. So, there is no real pattern.
to anyone that says that: "i don't see the problem" the problem is to prove that n will at some point will hit the cycle 4->2->1 and if you say that is will hit 2^x you also have to prove why is does
ATABA I did. The reason it always ends back up at the sequence 4, 2, and 1 are multiple factors. First, I will explain why the numbers will always end at 4, and then 2. If you notice, no matter what number you start with, you always end up with far more evens than there are odds. This is because of the algorithms that have been chosen. Every other number divided by 2 will give you another even or an odd. And if it’s even, you divide again, (remember that most numbers in these sequences are even), so therefore you will always go down, bit by bit, and all even numbers can be divided by the numbers in the sequence at the end: 4, and 2. And of course 2/2 will always give you the 1 in the sequence. Now to further explain: when you get an odd, you must multiple that number by 3, (an odd number), and then add 1 to the whole. This will always leave you with an even number. (Again, the even numbers will always bring you back down eventually, and back to 4, 2, and then 1). If you take any odd number, (remember that the first number doesn’t matter, but the last, so therefore keep in mind there’s only actually 10 whole numbers you have to worry about), like 3 for instance, or even 23, or 43, you then multiply by 3. So, 3x3= 9, and now you add the 1 to get you to the even number. 9+1=10. Or, 23x3= 69+1= 70. Or, 43x3= 129+1= 130. Let’s try 7. 7x3= 21+1= 22. Or, 17x3= 51+1=52. Starting to see a pattern here? They always end up at the same number (referring to the last number on the right), and then always end up with an even. And since you’re only multiplying every once in a while to get , and you’re dividing evens most of the time throughout the sequence, you will always end the sequence with 4, 2, and then 1. It is simple really. I figured this out by viewing the sequence a couple times and within maybe 30 minutes? Also, a lot of times (if not every time), you will send back up with a number that does already end the sequence back down to 4, 2, 1. As you can see in the case of 11 and its sequence, and then 29 and its sequence which eventually leads to 11 and it's own sequence which leads to the infamous 4, 2, 1. And if you look at all the sequences shown in this video, you can see a pattern, not with the sequences themselves persay (even though there somewhat is), but with the percentage of how many times the odds are compared to all the numbers in the sequence. If you count all the numbers including the 4, 2, 1, and either exclude or include the number you start at, odd shows up in a range of 25% - 37% of the time. Usually around 33% of the time. And if you were to count all the odds and compare to all theevents, with or without counting the starting number and the 4, 2, 1, you'll end up with the odds only being 25% - 37% of theevents, again with 33% being the usual percentage. Also, also, if you take the starting number and find how many times the designated number goes into it, a strange and certain pattern exists again. (Odds only). For example, let’s look at the starting number 11. So, this is an odd number which will mean you have to multiply by 3 and add 1. If you find out how many times 3 goes into 11, (3.67 times), and you take those 3.67 times and divide it by the starting number 11, (3.67/11), you will get 33%. Basically giving an obvious glimpse at the approximate percentage the odds will show in the sequence. Let’s try 29. 3 goes into 29, 9.67 times. 9.67/29= 33%. Let’s do 8,199. 8,199/3= 2,733. This divided by 8,199= 33%. I’m basically pointing out that this should be obvious with every odd number divided by 3, and then taking the answer and dividing the original odd number to get 33%. This problem is simple yet not, because the “patterns” and answers are right in front of our eyes, just hidden in plain sight. And to add to this, if you do the evens only. The starting number is even, so we’ll do 10. You must divide by 2, which will give you the amount of times 2 goes into the starting number. 2 goes into 10, 5 times. And then 5/10 is 50%. Basically this is alluding to the fact that after every time you divide, every other time you will end up at an even or an odd. 50% of the time you will end up at and even. And like I said, after you do the 3n+1, 100% of the time you will end up at an even. So approximately 33% (from previous paragraph) of the numbers in any sequence from this conjecture will be an odd number (this isn’t a rule of thumb, but a better understanding), and the rest will be evens. And since the majority of the numbers will end on an even, and the even you have to divide compared to only multiplying about 33% of the time, you will always end back up at 4, then 2, then 1. Like I said, 4 and 2 go into all even numbers (although 4 obviously don’t go into 2), and 2/2 will always equal 1. This is why all numbers you start with (all sequences) will end up back at 4, 2, and 1. Oh, and it don’t matter what numbers you do these to. The starting numbers or any even number in the sequence or just period. You will always get that 50% chance that you’ll end up with an even or odd. 20/2=10 (Even). 10/2=5 (Odd). 12/2=6 (Even). 14/2=7 (Odd). 16/2=8 (Even). Etc, etc. 50%. Same with doing that with odds. It doesn’t matter what number; starting or in the sequence or just period. You’ll always have an answer that’s even (after adding 1), and in the sequence with these equations of sorts, you’ll have about 33% of the sequence be odd and the rest be even. Again, this isn’t a set rule, but more of an answer that just gives you a rounded reasoning for the problem. I don’t have too high of a vocabulary to better and fully explain this. Hopefully people can understand what I’m referring to and how I got the answers I did. I didn’t show the math too much, but if you do it yourself you’ll see. I can’t exactly show a picture of the math notes in my journal.
Having given this 3 minutes of thought, If N is ODD, it will ALWAYS produce an EVEN number.However, an EVEN number will produce an EVEN number half the time, and an ODD number half the time. When an ODD number produces an even, it does so with x 3 the original value, plus the 1 to make it EVEN. However, an Even number will ALWAYS divide by 2 once, and 1/2 the time divide by 4, and 1/4 of the time divide by 8, and so one. There will be sequences of GROWTH ( * 3 / 2 * 3 / 2 ), however not possibly forever. There are simply *MORE* times when a 2nd or 3rd or 4th division by 2 will collapse the total, no matter how large it gets. Any power of 2 becomes an irreversible *Zip-Line to collapse". This does not predict the # of steps, but rather argues that in the end it MUST devolve into the trivial cycle 4 -> 2 -> 1 because that *happens* to be the minimal set where ( # * 3 + 1 ) produces a power of 2. Perhaps then one might wish to examine *how many* values, multiplied by 3 then add 1, result in a power of 2, or have LCDs with Powers of 2 > 2. I remember asking friends would they buy an investment that doubled every year, but in the off years would lose 51% of it's value. Most said *Heck YES!* thinking that a 100% gain must ALWAYS be better than a 51% loss. After all, 51 is less than 100. But they fail to comprehend that a 51% loss, even after 1 two-year cycle, results in a smaller total amount that will be doubled the 3rd year. ANY Loss > 50% must eventually erode the bi-annual 100% gain to 0. My $0.02
+Texas75023 Try to change the problem to 3n-1, and insert 5. Your argument can be used in that problem, but it has a counterexample. It's possible than 3n+1 also has a loop, but isn't found yet.
+Anonymous71475 Changing the problem, is ... well... changing the problem. However, I understand the consideration of Symmetry. I tried (at random) to find another example. The first alternative I tried, was 8795. And VOILA! it had a different cycle ! At iteration 63 arrives at *122*. At iteration 81 arrives again, at *122*. A Cycle of 18 operations. At iteration 99 again, arives at *122*. So a definite cycle. And checking further across the first 100,000 initial numbers, the "subtraction" form of the problem apparently has 2 cycles, based on the eventual occurrance of certain numbers ( 2, 5, 7, 10, 14, 17, 20, 25, 34, 37, 41, 50, 55, 61, 68, 74, 82, 91, 110, *122*, 136, 164, 182, 272 ). Now, I initially *did not check* for lengths of cycle, or alternate entry points, for simplicity sake. But then I thought to examine the cycle members. For example [ 5, 7, 10, 14, 20 ] form one cycle-set (not cycle ordered) which could be entered anywhere in the cycle, resulting in the same repeating member set. The remaing values (excluding "2") form a second complete cycle of 18 steps. Entry into the cycle at ANY of the 18 numbers, results in the inescable looping. ) Running the first 1,000,000 initial seeds, these same 24 cycle elements eventually appear. Expanding that to the first 3,000,000 initial values, *SURPRISE!* I still get the same 24 elements. Why these? I suspect if one would simply examine each number, these elements constitute 24 "traps", much like Powers of 2 in the "addition" or ( 3 n + 1 ) form. Only I do not see any such other "trap" values in the ( 3 n + 1 ) domain. Again, my $0.02
Texas75023 Still... there is a possibility of undiscovered loop for 3n+1... Even if the smallest member of the loop has more digits than the atoms in the universe, the conjecture would be wrong. So... the way you did it won't solve the problem at hand... To solve it, you need to either: 1. Prove that a loop or a number that goes to infinity can't exist. Personally speaking I don't think the latter case can exist, but I can't prove it. 2. Give a counter example. This maybe is the easiest one if IT EXISTS, anyone without a background of mathematics can do it with AMAZING LUCK, if IT EXISTS. 3. Prove that there must be a counter example somehow. Either this or number one has to be done if all computable numbers have been checked.
***** I agree with your assertions. Of the three cases, Counter-example (2) seems to be the easiest when one exists. Proof (1) to Inifinity is always the harder challenge to perform with mathematical rigor. Proof of the existence of counter-example is less frequently used, but necessary to consider as an approach to disprove an assertion. The 3n+1 problem is interesting, and perhaps the 3n-1 problem provides insights that will be the key to the proof. I think an understanding of the 2 demonstrated cycles in 3n-1 might point to the nature of the behavior, but still not provide either a path to a counter-example of 3n+1, or a proof that said counter-example even exists.
I made my own program and calculated numbers 1-20,000,000 (20 Million) I determined that the number that took the most operations was 15,733,191 It took 704 turns to get to 1 I also calculated that for numbers 1-20,000,000 each number averages at 162 operations, pretty cool
I tried to reach 1-100.000.000 using mat.h, but the program crashed and so I can't run it in my computer, could I send you the script for you to use it in YOUR computer and send me the results later?
So you can find the sequences in excel: Start with a column in a sequence (it could be 30, it could be 300), in column A, so that A1=1, A2=2. This is your reference column. In cell b1, put this formula: =IFERROR(IF(A1=1,"",IF(ISODD(A1),(A1*3)+1,(A1/2))),"") This determines if cell A1 =1. If it is, it ends the sequence, and creates a clear cell. The clear cells are important for visual reasons. If its not 1, it determines whether that cell is odd or even, and makes the calculation based on the problem (a1*3)+1 for odd, or (a1/2) for even. Now fill that down for all your numbered rows, and fill right for all your chosen columns as far as you want. You end with all the sequences of the numbers for your specified rows. (30 or 300! Whatever you chose) Then you see some really interesting properties. In some cases, like for the number 31, the next number is 94. 94 is the 6th number in the sequence of 27, so from then on, both 31 and 27 are exactly the same. The reason that these high sequence numbers dont go off into infinity, is because you end up with a sequence of numbers within that sequence that are even, which halves each time. For example, in the number 31, the 84th number and onward are 976, 488, 244, 122 and then 61. You could then argue that those numbers (27 and 31) become exactly the same as the sequence 61 produces (and even the preceding numbers). So as you can see, the sequence drops dramatically because of the number of even numbers in a row. Because 31 and 27 are the same after 6 numbers (in 27), that sequence also falls into 27, and the reduction occurs quickly. Each number will create a sequence that falls within the sequence of another number. So no infinity. Edit: Check out these sequences from 3377, 3378, and 3379. They align at 1426, and follow the same sequence, making these 3 numbers exactly the same as the sequence 1426 makes. They also have exactly the same amount of numbers in their sequence: 3377 3378 3379 10132 1689 10138 5066 5068 5069 2533 2534 15208 7600 1267 7604 3800 3802 3802 1900 1901 1901 950 5704 5704 475 2852 2852 1426 1426 1426 713 713 713 2140 2140 2140 1070 1070 1070 535 535 535 1606 1606 1606 803 803 803 2410 2410 2410 1205 1205 1205 3616 3616 3616 1808 1808 1808 904 904 904 452 452 452 226 226 226 113 113 113 340 340 340 170 170 170 85 85 85 256 256 256 128 128 128 64 64 64 32 32 32 16 16 16 8 8 8 4 4 4 2 2 2 1 1 1 Edit 2: Maybe the question should be "Would a (virtually) infinitely high number create a finite number of numbers in a sequence?" The numbers will always decrease to 4, 2, 1... but how high does a number have to start to end up, not with the numbers going off into infinity, but having an infinite number of numbers in a sequence?
Or if you think that 0 is actually even or odd, www.google.com/search?sourceid=chrome-psyapi2&ion=1&espv=2&ie=UTF-8&q=is%200%20even%20or%20odd&oq=Is%200%20Even%20or%20odd&rlz=1C1CHZL_enUS697US698&aqs=chrome.0.0l6.4272j0j7 SO that proves that 0 is even, so then divide by 2, you get 0 again, and again so you never get to one.
Yes, there are theorems that are impossible to prove. That is the essence of Godel's Theorem. However, no one knows which theorems are the impossible ones. Moreover, almost no mathematicians worry about it. But, some have wondered whether the 3x+1 problem is one of those unprovable results ...
For those not trained in mathematics, think of it as a puzzle. Can you find a starting number that does NOT eventually reach 1? How do you know that it will even get below where it started? For mathematicians, this problem is fascinating, because while it is so easy to explain, nobody can prove that every number eventually reaches 1. The process is deterministic (there is no randomness in how the next number is generated), but after many iterations it ACTS as if it is random. The larger of implication is that probability theory may be able to say something profound about deterministic processes.
Tipping Point Math Yeah, thinking about it and reading what you said it does sounds fascinating or interesting, so from what I understand it's just acts like a puzzle and that's it?
How can something be random? If everything inside the universe is governed by fundamental mathematical laws, it is possible to predict everything with 100% certainty. Just because we haven't discovered the algorithms behind something doesn't mean it's random, it's just unpredictable until we've discovered how it works.
Wow guys, thank you so much for pointing out logically in two sentences why it always goes to 4-2-1. You should tell this to the mathematicians working on this problem, they will be so glad to hear that a bunch of random people on the internet finally saw and pointed out what they haven't been able to see for so many decades! Guys, yes, we all know why it goes the way it goes, but just because you can explain it doesn't mean it is mathematically proven.
Your first two sentences (ignoring "easy") do not imply the third; Consider the function that sends all odd numbers that leave a remainder of 1 upon division by 4 to 1,000,000 and all other odd numbers to 1,000,002.
Don't ya think if it was so "easy" that you can come up with the "proof" (what you've written is hardly a cohesive prove) then mathematicians wouldn't be sitting on this problem for decades? Or are you so much smarter than all of them?
Several of the comments ask about replacing 3x+1 with 3x+d where d is an odd number. It is conjectured that for a fixed d, there are at most a finite number of cycles and all numbers eventually reach one of these cycles. For fun, look at the case with d=-1, that is, the 3x-1 problem. Can you find three different cycles?
jim blonde Jim what is the text showing, I can't understand what it represents? Could you put it in a readable form ? i.e. literally putting it as text please
Wich numbers do not add up to 1? And even though i do not understand the problem of the problem beeing there, how would you like to get it solved? Change the formula? Then u change the cause of the problem?
No shit, you're taking X (10, in this case) subtracting X (10) and adding one. It works for negatives too because a negative minus a negative becomes addition, which everyone should know. So it's 0 + 1 every time. This is what? 5th grade/year (elementary/primary) school knowledge?
Another interesting thing about this problem is, that if you graph the number of steps taken to get to 1 for enough numbers (e.g. 1 to 10^5), it starts to look like a root-function.
What about the imaginary unit? --- Let's back up a bit... Let us pretend we can do this with imaginary numbers... now, how would you visualize such thing? Well, we use a number line for the reals, and we can plot the complex in a plane... What if we color the numbers depending on the number of iterations they take to reach 1? Done: 3:17 Note: Black = goes to infinity --- What? Didn't they say they all go to 1? Well, all INTEGERS - as far as we can tell - go to 1. Not all reals tho, much less all complex... but, is 0.5 even or odd? What about i? Is i even or odd? Well, first we need to extend the function f(x) = x/2 : (x = 0 mod 2), f(x) = x*3+1 : (x = 1 mod 2) to something that would work for any complex number. We call this an Analytic Continuation. It is identical to the described function for integers, and it is guaranteed to be continous. To be honest, I don't know how to do that... yet, the paper "The 3n+l-Problem and Holomorphic Dynamics" has this: f(z) = (1/2)*z + (1/2)*(1-cos(pi*z))*(z+(1/2)) + (1/pi)*((1/2)-cos(pi*z))*sin(pi*z)+h(z)*sin(pi*z)^2 where h is an arbitrary holomorphic function. Ok, this is not exactly a continuation of f(x) = x/2 : (x = 0 mod 2), f(x) = x*3+1 : (x = 1 mod 2) instead it is a continuation of f(x) = x/2 : (x = 0 mod 2), f(x) = (x*3+1)/2 : (x = 1 mod 2) - that is, it is doing two steps for the odd numbers. You can try on Wolfram|Alpha, for example: (1/2)*z + (1/2)*(1-cos(pi*z))*(z+(1/2)) + (1/pi)*((1/2)-cos(pi*z))*sin(pi*z)+h(z)*sin(pi*z)^2 where z = 5 It should yield 8, because (5*3+1)/2 =8. The interesting thing is that you can plug any complex number you fancy... so let's see... (1/2)*z + (1/2)*(1-cos(pi*z))*(z+(1/2)) + (1/pi)*((1/2)-cos(pi*z))*sin(pi*z)+h(z)*sin(pi*z)^2 where z = i It yields... -h(i)*sinh(pi)^2+(1/2)*((1+4i)-(1+2i)*cosh(pi))-(i*sinh(pi)*(2cosh(pi)-1))/(2pi) And there you go! :S
Video: The problem is: how can we prove this algorithm will eventually yield infinite sequence 4, 2, 1? Half the comments section: What is the problem? What is the question? The answer is 0. This is obvious wth.
Josh Brady well if someone proved that say, all numbers do eventually end up on 4 2 1, maybe someone would be able to find an application for it. Maybe in pseudorandom number generation... maybe in encryption... I don't know just doing armchair speculation.
Math also makes a lot of unnecessary problems for itself.. this "problem" isn't actually a problem... nobody can find a number that won't lead to this same pattern.. and number starting from 1 all the way into the hundreds of billions have been tested.. I'm sure billions more by now and they still can't find one that doesn't lead to one.... so what does it teach us if 4568296368262527494826254 ends at...2... what did we learn?... but that wouldn't happen anyway because cut that number in half and the resulting number has already been tested... almost seems like a waste of resources to build an expensive ass computer just to run this program... just creating more problems
Yes, I came to the same conclusion. Eventually a series of x/2 and 3x+1 will result in a power of 2. So how do you prove that a power of 2 will always eventually result?
I'll prove that 1+1=56. heres my theory. Heres my example. When adding an odd number by 1, it will be even. If it happens to be 56, it will quickly prove my theory. If it isnt, multiply it by 0 and add 1 to get it to 1. Keep adding 1 and repeating the process until it becomes 56. There. Im a 9th grader by the way ...some people really have no idea what a proof is.
That's not the part that needed figuring out. The part that confuses everybody is the fact that it takes a seemingly random amount of steps to get the starting number down to one for each different starting number
I'm no mathematician but I think the problem is: so far, the rate of decrease (n/2) eventually outstrips the rate of increase (3n+1). The question is, can it be proved that this will always be the case.
Multiplying an odd number by 3 gets you another odd number, then 1 is added making it even again, allowing it to be halved, this will reach a different number each time until it gets to a number that can reach 16, 32, 64, etc. There is no pattern, just a law of operation.
Ding Ding! The way it's presented makes it sounds like an equation to solve. Let me try to make up some nonsense too umm... Start with any whole number, if it's greater than 10 subtract 5, if it's less than 11 add 3 you get a repeating 6, 9, 12, 7, 10, 13, 8, 11...
how do you know it'll always eventually culminate in some power of 2? how do you know it will not just infinitely produce even numbers but never an even number that's a power of 2?
Here's my take: 3 times any odd number stays odd. (try it) Add one and you get an even number. Because of this you will never get two odd numbers in a row. However, you can get many even numbers in a row (44->22, 48->24). In fact - ~half the even numbers you get will stay even. Due to this, inevitably the number will go lower and lower until it "locks on" to the pattern that ends in 4-2-1.
It's good intuition for understanding what's generally going on, but doesn't solve the problem. How do we know that there isn't a number that has even numbers in a row so rarely that it overpowers the halving?
It's a pattern - and with patterns we assume they go on forever. It's the same as asking "Sure, 3,6,9,12.. is a pattern, but how do we know there isn't a number really high that still adds 3 but isnt divisible?" its implied
@Miroslav Mandic Woosh is the sound that is made when a joke goes over your head. A lot of times, the response to someone taking a joke seriously is simply, "Woosh."
for people who didn't get it the unsolved problem here is that we couldn't prove that *every existing number* reaches that cycle after calculations, we reached pretty high numbers but there's still a chance that there's some large number we didn't reach that doesn't end in that cycle.
After review of the problem it is clear every number will reach 4 2 1. There is never a number that will not reach it and it is due to the (+1) When a number reaches an ODD value it will triple and +1 making it even. Its mostly a game of how many even numbers can you divide by 2 before hitting the next odd value. Your never going to find a number that will triple +1 and divide by 2 and break the cycle. The +1 will change the pattern of any number that has the potential to break the cycle. Is there a pattern or another mathematical explanation for the results of each number? Not really, each number is unique, the ones that can divide the most by 2 will always be the key changers in the values. Until it eventually reaches 4, 2, 1. You will never end up with 2 odd values after each other and you can skip half of the odd values by simply multiplying the first odd value+.333 * 1.5 rounding up. will always = skip over a few values.
I agree that any odd number will become even with one application of the function. I agree that an even number will divide by 2 until reaching an odd number. I can't follow the rest of the argument; how do you get from here to all numbers eventually reach 1?
In mathematics you're not allowed to say "These numbers will do exactly this" without offering proof, so simply saying "As long as the even number divides into a multiple of 2 it will divide all the way down to 1" doesn't solve the problem. I'm pretty much sure that after reaching 5x2^60 mathematicians are pretyt certain that it works with any number but till now there is still no proof written down. That's what this video is about. Kind regards, Meta Custom Computers
Someone please correct me if Im wrong but I see a very clear/definitive solution to this. First, let's say you start with an even number. You will divide by 2 until you either reach 1 as your first odd number and you will get the 4, 2, 1 cycle. Or, you will reach an odd number before 1. then you multiply it by 3 and add 1. this will always result in a even number again because an odd number x 3 is always odd, then once you add 1 it will be even. Basically what this means is that it is impossible to have 3n+1 twice in a row but is not impossible to have n/2 twice in a row. This means that as our base number grows larger due to a repeating series of 3n+1, n 2, 3n+1, n/2.... the statistical probability of getting more n/2 twice (or more) in a row grows, at some point becoming nearly impossible to avoid. When this happens it will more than counteract the 3n+1 and will continually happen until the series converges to 1 because the probability of dividing will always be higher than multiplying.
Note that 3 (the factor by which you multiply) is larger than 2 (the factor by which you divide). Therefore, it is possible to divide by 2 more times than you multiply by 3 and still end up with a number larger than the number that you started with.
+supergreatsuper no if you divide a number by 2 twice, then you are dividing it by 4 which will shrink number (compared to operation of multiplying it by 3)
If you divide by 2 three times and only multiply by 3 two times, you end up with a number larger than the number you started with. Therefore, "it is possible to divide by 2 more times than you multiply by 3 and still end up with a number larger than the number that you started with" as claimed.
The problem is that there is no exception, regardless of the number, and that there is no real pattern correlated to any of the other starting numbers. For people who were confused.
But the system always ends with 16,8,4,2,1, and then, 4,2,1 repeating. Mathematically isn't that the only outcome, when a number reaches such close amounts to 0 it has to eventually falls into a pattern. Doesn't that mean that a number, no matter how big or small, can be divided by 2 if even, and multiplied by 3 then have 1 added to that if odd? Why isn't that a solution?
There is clear pattern dude, I'm going to prove that soon) For example, I found that in average ratio between odd numbers and even numbers in those sequences is 1 to 2, thats why /2 prevails over *3. But a lot of work is still needed, won't disclose anything until figure that out)
+Neon Lights Also, the reason for that is obvious, because 3*n+1 ALWAYS leads to even number as a result, but /2 could result in even number, in fact, half of even numbers could be divided more than once by 2. That is why we have this pattern
3 times an odd number is always an odd. Adding 1 makes the number even which means it will be divided by two. If the last digit in the number is a 2, 6, or 0 the number will be odd repeating the cycle until 1. Dividing a whole number by 2 will always be a positive integer and since 1 is odd 3(1)+1=4 which then causes the cycle to repeat
You only have to check all possible combinations of 1 , 2 , 4 , 8 , 16 , 32 ,64etc. to get the shortest possible count or , all other numbers are poluted , this is clean halving time i.e. 64^32 / 2*191 = 1 , 64^32 each time divided by 2 takes you 191 steps to get 64^32 to 1 OR from 64 - 32 - 16 - 8 - 4 - 2 - 1 = 1 , six subtractions from 64 to 1 , reminds me of bits and bytes All combinations from these number to the power of (mix them) will result in 'low' steps to get to the one , tho
Oh fuck off. I'm going into engineering, but pure mathematics great too. It's kinda like an exploratory mission into the system that we live in [5]. And engineers couldn't do what the do without the underlying math behind it.
Have you ever heard of the Fibonacci Sequence? 1 1 2 3 5 8 13 21 34... you get it by adding the previous two numbers in the sequence 1+1=2, 1+2=3, 2+3=5 and so on. Just looking at it, it might seem to be just folly but it appears everywhere, from snail shells to plant growth to the shape of the galaxy. Maybe the solution to this problem will become part of the foundation of the next paradigm shift in physics and in 100 years every engineer will have to study it in university. Or maybe the proof techniques invented to solve this problem will be applied to other "impossible" problems. Or maybe this problem is truly impossible to prove which will have powerful implications on the field of mathematics. Or maybe the solution of this problem is completely inconsequential and unimportant The fact of the matter is that we don't know. Engineers - for all their importance to society - work firmly within the bounds of human knowledge, and thusly for all the work they do it is immediately obvious how it is beneficial. Mathematicians (and scientists as well) work outside the bounds of human knowledge and don't have this privilege of knowing beforehand what is "important".
Wait.. how is this a problem? 3n+ 1 (n being uneven) always results in an even number you'll eventually hit a factor of 2 resulting in more divisions than multiplications.
But 3>2, so it's possible to divide more times than you multiply and still end up with a larger number than you started with. How do you know there are more "enough" divisions?
Depending on your definition of "chain", I agree that you will always reach a chain of divisions. However, how do you know that these chains will be long enough, and that there will be enough of them?
your example is trivial, as you purposefully chose a power of two. But then again, you are claiming to understand the solution to a problem that has never been cracked by the most brilliant minds for the past century or so; so clearly I fell to a troll. Stupid me.
most comments prove that math is being taught the wrong way... Guys come in, the problem is to prove it, the fact that every number we have plugged goes to 1 doesn't mean it will happen with all numbers.
Actually that is 100% logical: Every odd number when tripled gives an odd number, add 1, it's even, you divide it by 2, if its even, keep dividing, if it is odd, get it even again, you will keep doing so until you either progressively reach 4, or until you reach a power of 2, it may near an infinity of tries, but at the end , it works out, the difference in number of tries is related to how far away is that number from the nearest power of 2
The only reason I can think for this is because every time you use an odd number, it will go to an even number. As an even number is every other number, the probability of you getting a chain of them is 1/4, 1/8 etc... Due to this, it is unavoidable that you will get a chain of numbers that will continuously divide by 2. As the 2nd or 3rd division is less effective, you are in effect multiplying it by 1/4. Which, if you were to multiply by 3 and add 1, you would get 3/4 + 1, which, unless the number is 4, this produces a smaller number. Due to probability, every number will eventually reach a point where it will hit a chain of divisions, causing the number to decrease by a factor of 2. This is what results in all numbers reaching the chain.
+Tipping Point Math Thanks for the support! +Callum Chiverton I can't really tell if you're claiming you have a proof or not so I'll go ahead and assume you are just in case. I agree that any number will eventually be divided by 2 twice in a row. How do you know this will be enough to counteract the gains due to the multiplications by 3?
supergreatsuper See, that's the thing. There is almost no way to guarantee that this can happen, the only thing that can make it "certain" is the probability of you getting a longer chain with the more times you cycle through the calculation. You could even go on long enough and, based on probability, you could get 10000 divisions in a row. There is just no way to prove every single number with just probability, as this applies to everything in life
2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 1 2 3 4 5 6 7 8 1 2 3 4 1 2 Using this as a base example, we can see that as you go up in numbers, that every 2 numbers you can divide by 2 (obviously). Out of these, numbers it produces, you have another 1/2 chance that it will divide. This means you get a 1/4 chance that you can counteract the gains of the multiplication by 3. If we were to take this probability into account and say that it would happen EVERY 2nd or 4th or 8th division. Keeping n as the same number throughout, but applying the logic n/2 3n/2 + 1 9n/2 + 1/2 9n/4 + 1/4 27n/4 +5/4 27n/8 +5/8 81n/8 +13/16 81n/16+13/32 81n/32+13/64 81n/64+13/128 This could go on for nearly forever. However, it proves that with time, the probability of the division occurring more than once can counteract the multiplication by 3. I wouldn't be surprised if I messed up here somehow, I rushed this, but for me this is proof, even though it isn't hard mathematical proof, which can only come from having a calculation with n that then results in the amount of steps needed to get it to 1. For me, however, this is proof because you can go to infinity, with each number you use having a chance to divide more than once. This means that you can theoretically go to an almost unlimited number with almost unlimited steps, however, the more steps you perform the higher the chance there is for you to get chains of divisions, so a near unlimited number of steps in my eyes will eventually bring a number of divisions that will result in 1
Suppose just a few numbers were to form a loop or a chain that did not include 1 (that is, most numbers reach 1, but a few (still infinitely many) don't). Would this affect the overall probability? The next question is: How strong must the definition of "few" be to not affect the overall probability? Could it be almost all numbers?
an odd number multiplied by three is always an odd number. an odd number + 1 is always an even number. therefore, following this formula, if you have an odd number the next number is guaranteed to be an even number. HOWEVER, when you have an even number there is no guarantee. you can have an odd number, or an even number. when you get an even number from an even number, the number you get is typically lower than the odd number you got there from. over time, this formula harbors a disproportionate amount of even numbers, and therefore, a disproportionate amount of division. this leads to the unavoidable cycle of 4 > 2 > 1. am i missing anything here? tl;dr you're always more likely to get an even number, so more division
well more divisions isnt the odd thing causing the repeating pattern like with 28 14 7 22 11 34 17 52 26 13 40 the numbers are going from high to low there whats odd is that eventually one of those odd numbers when multipled by 3 +1 is a power of 2 so therefore you keep dividing by 2 until you get one 256 128 64 32 16 8 4 2 1 then the pattern starts thats why 8192 had so few steps because its a power of 2 and goes straight into the pattern
How does continuously adding 1 make any odd number eventually get to a power of 2? Don't forget that you are doing lots of operations between the adding 1.
+supergreatsuper because the additional one forces an even number. Take any odd number and apply this. It will come out even. Then division. If it's a multiple of 4 it's gonna come out even. Then divide again. It doesn't necessarily have to become a power of 2 to get to smaller numbers, but at that point there are more multiples of two. 2,4,8,16,32,64,128. Get in this range and the multiplication is more likely to hit a power of two than in higher numbers on your next multiplication. Once you hit it, it's inevitable it will reach 1.
*I thought its when your girlfriend seems mad at you and you ask her if everything is ok and she says yes and is still clearly mad at you specifically...*
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Well of course you're right! I mean it's not like as if much more talented and experienced mathematicians worked on this problem! What I'm trying to say is - the answer is not so easy!
I don't see why it couldn't end any other way (unless you're using the word "end" literally, as in the sequence stops. Which technically it does not stop at 1 either - it just goes back to 4, 2, 1, ...).
On the contrary, it is the geniuses that look for the simple answer. In fact, mathematicians in general look for the elegant solutions (e.g. proofs without words).
Not a mathematician by any stretch, but this seems very intuitive. Half of numbers are odd, half are even. So 50/50 chance of performing each operation. However the 3n+1 always yields an even number which will be halved so the step should really be described as (3n+1)/2. The n/2 can yield an even or odd number with equal likelihood so that single calculation can be considered a complete step. So the equally likely options are multiply by 1.5 or by .5 (you cN ignore the +.5 because that just rounds the number up to the next number which may be odd or even and just invokes the above method anyway. So, to simplify if I have a 50/50 chance of multiplying by 1.5 or by .5, the expected impact is to multiply by .75 over each set of 2 goes, so the odds are stacked towards a downward direction. Sure you can find numbers that will have a 'hot streak' of being multiplied by 1.5 several times im a row, but eventually the methods weighting towards 1 will kick in.
Well, if it ever reaches a power of two, it gets caught in the cycle of 4-2-1, as it keeps halving until it reaches one. Multiplying an odd whole number by three and adding one will always produce an even number, which will let you get loser to small numbers. As you get smaller numbers, you get a higher probability of the 3x+1 function producing a power of two. But, to get to the 4-2-1 cycle, you HAVE to reach a power of two. There is no other method of reaching one. For example, to get one, you know there are only two formulas that can produce each number: 3x+1 and x/2. For one of these two produce one, the previous number would either have to be two or zero. It can't be zero, as that would require starting at zero or a negative number, so we know you have to reach two first. If we do the same to two, we must either get four or one third. Again, we know the sequence will never produce any number other than a positive integer, so it can't be one third. We must first reach four. To reach four, we must either have eight or one. We know that to get one, we must first get four. So, getting one will not give us a solid method of reaching the cycle. We must first get eight. Since 8-1, or seven, is not divisible by three, we know the only way to reach eight is by getting twice its value, which is 16. To get one, you HAVE to reach 16. Now, this is where the methodology begins to split. Both methods (3x+1 and x/2) will produce a positive integer for the answer. Having the first expression equal 16 gives us five as our previous number, while the second expression gives us 32. Five gives us ten, ten gives us 20 and 3, 3 gives us six, and twenty gives us forty. If we continue this on forever, we will eventually include every single whole number in our pattern. We've already got one through six in our pattern, plus many other greater numbers. It always loops back.
supergreatsuper The rate of expansion for the set is too rapid for it to not include every number. The rate just keeps getting faster as you come across specific numbers that work for both 3n+1 and n/2, like 16.
It's the infinite monkeys, writing on infinite typewriters quip. You may not understand, or comprehend, that every number will arise, it might sound unlikely every number will pop up even once, but in an infinite sequence, every number will show up an infinite number of times.
It is common usage of the word "problem" to denote (from the second definition from Google) "an inquiry starting from given conditions to investigate or demonstrate a fact, result, or law."
huh... you learn something new everyday. Thanks. but it's really mind bugging that the same words have different meanings. When someone tells me they have a problem I assume that it is something that is making their life harder in some way.
According to Wiktionary, the "good" connotation is outdated (see en.wiktionary.org/wiki/egregious) However, there are such words and they are called Janus words (see en.wikipedia.org/wiki/Auto-antonym) (Yeah I just looked this all up)
This series always occurs because the pattern can cause up to four divisions backwards but only one multiplication forwards at a time this leads to statistically more divisions backwards in the series which leads to the enevitable approach of the 8-4-2-1 at this point the pattern because infinitely repeating simply because the numbers cannot escape the pattern
You made the mistake of calling it inevitable without any proof of that. Also, statistically speaking, airplanes tend to land on airports. Somehow, 2 of them still managed to land inside the twin towers.
+kek u wot m9 HUEHUEHUEHUEHUEHUEHUEHUEHUEHUEHUEHUE I would agree with you but in terms of airplanes there are a lot more variables to consider that could cause statistic variation, but in this the only variable is what we are in fact measuring, it would be like saying an airplane might not land because it in fact is not an airplane
okay, stupid here, why is this a problem? watched the video 3 times and dont understand where the problem situation lies. I was taught that every even number can be divided by two and if you add two odd numbers you get a even number. So it makes sense they get in this 4-2-1 circle in the end. There is no problem. That is just proofing the rule.
The problem is in showing that every starting number eventually iterates to 1. As noted in the video, the number 27 wanders a lot before eventually reaching 1.
Tipping Point Math and even if it would go on in another way or only would proof the law of simple mathematics. That's wrong with humanity, they want to see a problem where no problem is on the columns they created
Supposedly by your rules, there is no problem, because you're only looking at the conditions. The actual problem is proving that there are no exceptions to the conditions. If you're going to assume that there are no exceptions to all rules, then that's a pretty naive and ignorant way to live. It's like seeing one of those erroneous headlines nowadays and immediately believing it.
Konohura well, there are exceptions in patterns but not in rules. If you dont follow a rule in math its plain wrong. If its a exception in a pattern, this exception is created by a rule. Comparing this to believing in headlines is not an appropiate analogy.
Once a result becomes a power of 2, the pattern is unavoidable. That's the pattern. The proof required is for any given positive number m, what its probability of avoiding being a power of 2. Any even number that is not a power of 2 is the product of 2 and an odd number and it will thereby avoid the pattern for at least one more round as the result will be odd and the function 3n+1 will be applied. Another way to look at the issue is for any given odd number greater than 1, what is the probability that 3n+1 is not a power of 2?
+Bryon Lape After looking at the comments, I do want to be clear. I am not stating the probabilities are the proof, but the start of the idea. I've put many different numbers into the calculator. I am amazed how many reach a result of 5 and then fall into 16->8->4->2->1.
+Bryon Lape The probability that 3n+1 is a power of 2 is 0. More precisely, the density (natural density, but replace the set of natural numbers with the set of positive odd integers) of the set of odd numbers for which 3n+1 is a power of 2 is 0. This is not hard to prove, and I encourage you to try and prove this.
+supergreatsuper First I want to be clear that Bryon's assertion does not solve the conjecture as the assumption does not prove the recursion will not spiral out of control (get rid of or try substituting another number for 3 and you can easily find examples that will). But, I want to address 3n+1's ability to evaluate to a power of 2. If n is a sum of sequential powers of 4 starting from 4^0 then 3n+1 will evaluate to a power of 4 and, by default, a power of 4 is a power of 2. This is easy though a bit long to prove. First remember the following rule of summation: now replace i=0, c=1 and r=4 we get the sum of powers of 4 from 4^0 to 4^j simplifying we get multiply top and bottom by -1 and flipping around we get now, lets substitute this for n and put it into we get The 3s cancel -1+1=0 so, when n=sum of powers of 4 we get: Which is a power of 4 which is also a power of 2. So, it is not impossible that 3n+1 can not evaluate to a power of 2. Try it out with sums of powers of 4 (the powers of 4 are 1, 4, 16, 64, 256, . . . ) I'll start the list of sums so there is no confusion: 1, 5, 21, 85, 341, 1365, 5461, 21845, . . . Q.E.D. :)
+Robert Dostal Darn thing go rid off all the formulas.... Will it take html? (hard to write out sigmas any other way) [tried, nope] From Google docs? (they don't make great but will do) k=ijcrk =c(ri-rj+1)/(1-r) nope, that doesn't work either.
+Robert Dostal Did you write that using LaTeX? I can read LaTeX code, so it would be just find if you wrote out the source code. If not, what did you use? There is a chance I will be able to read that as well. At any rate, I said that the probability that it becomes a power of 2 is 0, not that it is impossible (actually, because a uniform probability distribution is impossible on the set of natural numbers, I used natural density instead of probability). What this means is if you "pick a random odd natural number," then the probability that it goes to a power of 2 in one step tends to 0 as the numbers get arbitrarily large. I interpreted Bryon's claim is "all numbers eventually reach a power of 2," which would clearly solve the conjecture.
The solution is simple. 3n+1 always equals an even number. If n is odd, it WILL be followed by an even result. There cannot be 2 odd numbers following one after the other. More than 2/3 of all n values will be even. Odd numbers increase the value to 3/1(plus 1). Even numbers decrease the value to 1/4 because you will always end up halving twice. half of a half is a quarter. 1/4 of 3 = 3/4 25% of 300% = 75% The math is forcing the value of n towards 1. When it hits an odd value, it jumps up to an even value and starts going down a different sequence. The math is designed to search for the correct sequence to reach 1. The only way that this sequence will not reach {4,2,1,...} is if every even result is an even number that is made up of two separate odd numbers, or if the sequence repeats a multi-digit n value (odd or even) more than once.
This would be a great way to stump a class of third graders. Bonus points: tell them that if any of them can find a number that doesn't hit 4-2-1, you'll order the whole class pizza. xD And then if any of them come up with a cool insight about the problem, order pizza anyway.
Many people have tried to use induction, but it has not worked. Indeed, this problem seems infuriating because it resists every approach. Of course many newcomers to the problem may try induction or modular arithmetic, but all of the low-hanging fruit was picked long ago.
this video doesn't really explain why its a problem. The actual problem that mathematicians are trying to solve is to repeat the process indefinitely without invoking the Collatz Conjecture. It's an impossible problem because it was designed to take advantage of how we define even and odd. If anything, the Collatz Conjecture is a problem of language not mathematics.
if I start with an even no., the next no. can either be an odd no. or an even no. if the no. I'd even, the process continues and if the even no is a power of 2, the end will have a series of 4,2,1. and if initially the no. was odd, multiplying 3 and adding 1 makes it even again putting it back into the loop. so no matter which no. you start with you will eventually reach a no. that is a power of 2.
unless the no. is a power of 2, in which case the answer will lead to the series(4,2,1) the no. will eventually become a power of 2 since only then will it stop. the stopping point is the series.
According to Google (and personal experience), a problem is "an inquiry starting from given conditions to investigate or demonstrate a fact, result, or law." Surely you'll agree the Collatz Conjecture fits that description?
the problem is, why does every number end with 4, 2, 1? there doesn't seem to be any pattern in which it should, yet it does every single time with any number you choose to start with, as if there should be a pattern in which it should end this way, so why does it?? that's the problem
No the problem is "does every number end at 1?". It is fairly obvious why numbers that do end at 1 do so, namely the pattern eventually ends on some power of two, which then reduces all the way to 1. Every number that has been tried has ended at one, but that is not a proof.
Ah, well if the problem is "Does every number end at one?", then I'll accept that as a valid problem. There is no limit to the number of numbers you can try this with, so we will never know. That sounds like a genuine problem. Thanks for the input :)
In both results, integer/2 or integerX3+1, more possible outcomes that are even exist than that are odd. Because an even result requires a halving, eventually the result is going to degrade to 1. It's even number survivor bias and the result is a continual down counting.
The point is to reach a power of 2 which will then become 1. This means that only powers of 2 that are of the form 3x+1 where x is odd will work. That would be 4, 16, 64, 256, 1024, and so on. These are the powers of 4. So numbers that reach a power of 4 will work. The numbers that do this are 1, after which you've already finished, 5, 21, 85, 341, etc. This is a sequence with a common difference of powers of 4. These numbers are only reachable by dividing by 2, from 2, 10, 42, 170, 682, etc. Thus, only numbers that reach 4, 7, 20, 31, 84, 127, 340, 511, 1364, 2047, etc. will work, plus those from earlier lists: 2, 4, 5, 7, 10, 16, 20, 21, 31, 42, 64, 84, 85, 127, 170, 256, 340, 341, 511, 682, 1024, 1364, 2047, etc. Keep increasing the size of this list. If you expected me to prove it here, I'm sorry.
This is a shitty problem. Its more like "You dont have a formular." Everyone with brain knows how to solve it but it is not like a formular.. If it hits y=2^x than its shrinks to 4 2 1 And because infinity its 100% possible.. -> Solved :I
The reason it always ends back up at the sequence 4, 2, and 1 are multiple factors. First, I will explain why the numbers will always end at 4, and then 2. If you notice, no matter what number you start with, you always end up with far more evens than there are odds. This is because of the algorithms that have been chosen. Every other number divided by 2 will give you another even or an odd. And if it’s even, you divide again, (remember that most numbers in these sequences are even), so therefore you will always go down, bit by bit, and all even numbers can be divided by the numbers in the sequence at the end: 4, and 2. And of course 2/2 will always give you the 1 in the sequence. Now to further explain: when you get an odd, you must multiple that number by 3, (an odd number), and then add 1 to the whole. This will always leave you with an even number. (Again, the even numbers will always bring you back down eventually, and back to 4, 2, and then 1). If you take any odd number, (remember that the first number doesn’t matter, but the last, so therefore keep in mind there’s only actually 10 whole numbers you have to worry about), like 3 for instance, or even 23, or 43, you then multiply by 3. So, 3x3= 9, and now you add the 1 to get you to the even number. 9+1=10. Or, 23x3= 69+1= 70. Or, 43x3= 129+1= 130. Let’s try 7. 7x3= 21+1= 22. Or, 17x3= 51+1=52. Starting to see a pattern here? They always end up at the same number (referring to the last number on the right), and then always end up with an even. And since you’re only multiplying every once in a while to get , and you’re dividing evens most of the time throughout the sequence, you will always end the sequence with 4, 2, and then 1. It is simple really. I figured this out by viewing the sequence a couple times and within maybe 30 minutes? Also, a lot of times (if not every time), you will send back up with a number that does already end the sequence back down to 4, 2, 1. As you can see in the case of 11 and its sequence, and then 29 and its sequence which eventually leads to 11 and it's own sequence which leads to the infamous 4, 2, 1. And if you look at all the sequences shown in this video, you can see a pattern, not with the sequences themselves persay (even though there somewhat is), but with the percentage of how many times the odds are compared to all the numbers in the sequence. If you count all the numbers including the 4, 2, 1, and either exclude or include the number you start at, odd shows up in a range of 25% - 37% of the time. Usually around 33% of the time. And if you were to count all the odds and compare to all theevents, with or without counting the starting number and the 4, 2, 1, you'll end up with the odds only being 25% - 37% of theevents, again with 33% being the usual percentage. Also, also, if you take the starting number and find how many times the designated number goes into it, a strange and certain pattern exists again. (Odds only). For example, let’s look at the starting number 11. So, this is an odd number which will mean you have to multiply by 3 and add 1. If you find out how many times 3 goes into 11, (3.67 times), and you take those 3.67 times and divide it by the starting number 11, (3.67/11), you will get 33%. Basically giving an obvious glimpse at the approximate percentage the odds will show in the sequence. Let’s try 29. 3 goes into 29, 9.67 times. 9.67/29= 33%. Let’s do 8,199. 8,199/3= 2,733. This divided by 8,199= 33%. I’m basically pointing out that this should be obvious with every odd number divided by 3, and then taking the answer and dividing the original odd number to get 33%. This problem is simple yet not, because the “patterns” and answers are right in front of our eyes, just hidden in plain sight. And to add to this, if you do the evens only. The starting number is even, so we’ll do 10. You must divide by 2, which will give you the amount of times 2 goes into the starting number. 2 goes into 10, 5 times. And then 5/10 is 50%. Basically this is alluding to the fact that after every time you divide, every other time you will end up at an even or an odd. 50% of the time you will end up at and even. And like I said, after you do the 3n+1, 100% of the time you will end up at an even. So approximately 33% (from previous paragraph) of the numbers in any sequence from this conjecture will be an odd number (this isn’t a rule of thumb, but a better understanding), and the rest will be evens. And since the majority of the numbers will end on an even, and the even you have to divide compared to only multiplying about 33% of the time, you will always end back up at 4, then 2, then 1. Like I said, 4 and 2 go into all even numbers (although 4 obviously don’t go into 2), and 2/2 will always equal 1. This is why all numbers you start with (all sequences) will end up back at 4, 2, and 1. Oh, and it don’t matter what numbers you do these to. The starting numbers or any even number in the sequence or just period. You will always get that 50% chance that you’ll end up with an even or odd. 20/2=10 (Even). 10/2=5 (Odd). 12/2=6 (Even). 14/2=7 (Odd). 16/2=8 (Even). Etc, etc. 50%. Same with doing that with odds. It doesn’t matter what number; starting or in the sequence or just period. You’ll always have an answer that’s even (after adding 1), and in the sequence with these equations of sorts, you’ll have about 33% of the sequence be odd and the rest be even. Again, this isn’t a set rule, but more of an answer that just gives you a rounded reasoning for the problem. I don’t have too high of a vocabulary to better and fully explain this. Hopefully people can understand what I’m referring to and how I got the answers I did. I didn’t show the math too much, but if you do it yourself you’ll see. I can’t exactly show a picture of the math notes in my journal.
Unfortunately, while what you said constitutes ideas for a potential proof, it does not solve the problem at all. The real question is whether there exists any loop patern other than 4 -> 2 -> 1. Note that it does not have to be of size 3, but at least size 2 (although size 2 is impossible). There needs to only exist one to provide a counter-example to Collatz conjecture.
@@codycast well I don’t know how to actually properly “proof” conjectures. But I would like to work with someone who does. I know for a fact I’m right, so I can really help someone who actually knows how to proof math equations and conjectures. I looked up 1 video on how to do so and I immediately just went to sleep bc of how complicated it was 😅
@@alexandrezeddam7817 which one does not exist. In this conjecture the other digits do not matter whatsoever. The only thing that matters is the last digit (since this is based on odd or even). Odd and even numbers are actually just 1 digit, and will always end up to another ONE digit. The other numbers/digits do not matter at all. That’s the whole point of my argument on this matter and why I postulate no one or no computer will ever find a different ending sequence.
I'm not sure what the big question is here. All it seems this problem is likely asking is how likely it is for this equation to get several even numbers in a row? I mean, any even number is divided by two. and anytime you get an odd number the 3n+1 ensures you get an even number 100% of the time. So then you just need several even numbers in a row to ensure that the division overtakes the multiplication right? I mean, as unscientific as it might sound, but shouldn't the law of large numbers kinda answer that? I mean there is literally an infinite amount of numbers, so either the law of large numbers ensures that you would eventually reach the pattern, like it should, or we accept that we just can't ever test it because it is literally infinite. As a side note, I could be wrong here, but a pattern almost identical to this should technically also exist for 5n+3 right? It would likely exponentially increase the amount of times it would take to reach the pattern, but it should still work. Edit: Actually I guess a simpler way to look at it would be, instead of trying to get several even numbers in a row, all you are really doing is trying to hit a power of 2. That's also part of the idea behind the 3n+1 and 5n+3. As long as even numbers are divided by 2 then you can create a repeating pattern with odd numbers. Basically find an equation for odd numbers, that when applied to the number one ends in a power of 2. You also need to make sure that both numbers in the equation are odd numbers(Such as the 3 and 1, or 5 and 3 in my example), so that way all odd numbers will end up even. And then once again in the end, the law of large numbers should ensure you eventually hit a power of 2. Anyone else have thoughts on this, or did I miss something and look stupid? Or did I look stupid anyway lol
THe problem is that when N is odd, we will always get an even number afterwards by using 3n+1. Hence if we start the problem with an odd number, that will be the only odd number in the sequence. If we start by an even number, n might be an odd number as the 2nd number of the sequence but never odd again. Therefore, no matter how big the number is, everything will always keep getting divided by 2 as there will no longer be any odd number ever in the sequences. That is my theory at least :)
Kevin Nguyen Mmm, are you saying that no number divided by 2 can be odd again? What about 42,18, and 6, just to name a few. Sorry, but I think I just got confused on what you were saying.
huh. That is true. I only tested out a few and they didnt turn out to be odd for like a 2nd time within a sequence. Then again, I did think about it for like a minute :) it seems that I were wrong haha. Thank you man.
There is no positive number that doesn't reach 1, because there are infinite positive numbers and once the number hits a power of 2, it is able to hit 1. Also, the chance for higher numbers to reach a power of 2 declines by half each number in the 2 squared chain.
+Zara auto that will only work as a counter example if one actually exists which is what the computers are currently trying to do. If no numbers exist that do not end in 4, 2, 1 then your challenge becomes impossible.
+Oliver Foggin why WOULD it ever NOT end 4 2 1? You perform a simple conditional check to decide which function to use, if it's even, you half, if it's odd, you increase the size of the number but make it even. You then feed that number back into the algortihm and repeat until it reaches the end cyclical pattern. The ONLY variable in this algorithm is the starting number. If the starting number is not a power of two, and thus can't be collapsed in minimal steps, you perform arbitrary steps on it, until, eventually, it WILL hit a power of two number, and collapse down. There is no problem or challenge to be found here, but it's hard to see a pattern because the algorithm is arbitrary. 3n+1 could literally ANY other equation that ensures an even number... such as n+1. Eventually this pattern will also always end cyclically (although as 2, 1... 2, 1 in this case.). This thing is stupid and I don't even know why it is here nor why people can't wrap their heads around it. Eventually you hit a power of two, so the n/2 will collapse to 1, which will be the first odd number in sequence once n becomes a power of two, at which point you once again start from whatever results from the OTHER equation when n is 1, and then collapse back down. As long as you turn the odd n into any even number, and don't have ANY other variables in this process, this will ALWAYS result in an eventual repeating pattern. Can I get my nobel prize in mathematics now please?
WilliumBobCole that's what the entire video is about. What you have done is claimed to have solved the problem... that many many maths geniuses have not been able to solve with multiple super computers. Yeah... right.
+WilliumBobCole But you're making the assumption that you *do* eventually hit a power of 2, which is essentially what you are trying to prove in the first place. Based on intuition, this may be true, although being the genius you are, you have excluded this portion of the proof, due to it being too simple. So no, you don't get a nobel prize quite yet, unless you include the proof of reaching a power of 2, so we inferior humans can understand.
***** what I did was explain how and why the end sequence is the same. The video never explained what the problem WAS, so to me it seemed like they were saying that maybe there is a number somewhere that breaks the pattern... but I explained that no, that will never be the case, and simplified the algorithm since the 3n is completely unnecessary and arbitrary. No problem was presented here FOR maths geniuses or super computers TO solve, just an infinitely large pool of numbers to run through this simple algorithm for no reason. If there WAS a number that broke the pattern of reaching a cyclical loop, then there would be the problem of working out WHY, and are there MORE like it, but no such number CAN exist because we can prove that logically it will always become a power of two after a varying amount of steps... so I don't understand your point, and again, the point for this video's existence... unless you are suggesting the problem is why some numbers require more steps than others before 'completion'?...
Suppose you're trying to solve a problem, but you only have so many proof techniques. It is completely possible that no combination of the proof techniques you have already will solve the problem; you might need to come up with hundreds of new techniques to solve it, which is too hard to do if you're only working on this problem. So one way of coming up with new proof techniques is making new problems that may be easier than the original problem. They might only require one or two new techniques to solve, which is more feasible. Eventually, you might get all of the techniques you need to solve the original problem. Along the way, you might have created some problems that you still can't solve. These can also be used as milestones: as you keep doing more math, every once in a while you will solve one of those unsolved problems, which can sort of "keep track" of how far mathematics has come.
Once you reach a power of two, then the process is trivial as you keep halving to 1. Since 3n+1 will always produce an even number from an odd number you keep reducing the number of odd factors in a number. What’s left to prove is that indefinitely repeating f(y) = (y-1)/3 covers the whole natural numbers for y in 2**b, b in (1,2,3,...)
F4T4L 405 That's not a valid explanation, you don't have prove that after dividing any even number by two, you don't just immediatly get an odd number again, then divide by 2, get an odd number again, etc. If a number would go like that it would continuesly become bigger, seeing as 3n+1increases the number by more than n/2 can decrease it.
But the number doesn't get smaller always, for example, here I ran a script which uses that logic, and with starting N 99 it becomes 298 at next step. $ python solver.py 99 N is 99 N is 298 N is 149 N is 448 N is 224 N is 112 N is 56 N is 28 N is 14 N is 7 N is 22 N is 11 N is 34 N is 17 N is 52 N is 26 N is 13 N is 40 N is 20 N is 10 N is 5 N is 16 N is 8 N is 4 N is 2 N has reached 1
Yes...10 billion...that's a big cycle that we can't really solve the question by hand. As someone said in comments, it won't work for 5n+1, so that makes it even more challenging. What about, say 7n+1? Since 7 is another Mersenne Prime like 3, if this holds good for 7n+1, I guess it wouldn't remain as hard as it seems.
For the modified problems with 5n+1, 7n+1, 9n+1, etc, the conjecture is that there are at most a finite number of cycles but "most" starting points run away to infinity.
Tipping Point Math No, I mean if 7n+1 also reaches to 1 in the cycle. The problem here is that after about 5 steps, it becomes too extensive for humans. Though if all numbers of form 2^x-1 follow the exact same pattern for mn+1, ie. leading to 1, then perhaps it can have a simpler way to solve.
no it wont work with 5n+1. but if you look closly. 1 --> 2 ---> 4 thats byte logic. if we use the same concept, it works. 0n + 1 = 1 1n + 1 = 1, 2 3n + 1 = 1, 2, 4 6n + 2 = 1, 2, 4, 8 12n + 4 = 1, 2, 4, 8, 16 (still missing the formula to explain how to get 0n and 1n, but they work)
Now I wonder, what does this problem look like in binary? The even rule would simply shift everything to the right. The odd rule would be multiplying by 11 (3 in binary) and adding 1. Now what?
This is an interesting approach. Think of it as follows: - If there are any zeros on the right, eliminate them (this is the division by 2 for even numbers) - with no zeros on the right, add a zero on the right then add this to the unshifted number (this is multiplying by 3). Now add 1. This will produce at least one zero on the right. Remove all of those zeros and repeat this process.
I wrote a program to do this in java, printing the number in each iteration in binary and decimal. I noticed that the 3n + 1 step tends to slowly turn numbers into alternating 1s and 0s. Multiplying that by 3 turns it into a string of 1s, and adding 1 turns it all into a power of two, which then divides down to 1. I think the next question is, how does 3n + 1 achieve this?
+Tipping Point Math I just realized your graph look wrong. The powers of two should be incrementing with a slope of one, but your 8 value appears to be less than your 4 value.
Why not reverse the problem ? Considering the main tree that is made by the powers of 2 & the branches with their predecessors ... The question would be demonstrated if we provide a necessary condition to reach all integers comprized between 2^n & 2^(n+ 1) .
Instead of checking each number 1,2,3,4,... etc .. & apply the rule : divide by 2 when even or multiply by 3 & add 1 if odd , we can reverse the problem & consider the main tree of the powers of 2 which obviously leads to the final loop 4,2,1 for you always divide by 2 in this case . Then for each power of 2 , namely 2,4,8,16,32,... we check their predecessors . For example 16 has two predecessors : 32 & 5 , we have multiplied by two & substracted 1 & divided by three to reach there , each predecessor has a maximum of two predecessors & we keep on going like this , reaching number after number . The theorem would be demonstrated if we provide a necessary condition to fill all gaps between two consecutive powers of 2 . Is this clear enough ?
Take the normal way : 5 is odd so it gives 3x5 +1 =16 ; 32 is even so it gives 32/2=16 . 5 & 32 are the two predecessors ( or previous values ) leading to 16 . Taken the reversed way : 16x2=32 , (16-1)/3=5 . We keep on going like this for each new value we get ; for ex. 5x2=10 ; (5-1)/3=(4/3) in this case the result is not an integer so we reject it . Any number "n" has one or two previous values depending on the cases . The idea with this method is to see if we can find a necessary condition to fill all the gaps between two given values , in this case 2^n & 2^(n+1) . Of course these are only some personal ideas to deal with the problem , i didn't pretend i had solved it . The advantage with reversing the problem is that instead of getting a single image for each value you can have two sometime ...
It’s simple. If we multiply an odd number by 3 and add 1, we will always get an even number. If we divide by 2, we have a chance of either getting an even or an odd number. This means that overall it is more likely that we will get an even number as an outcome or performing one of the two operations. Because we are decreasing in number when we divide by two, and we are dividing by 2 more than multiplying by 3, we will eventually reach a lower number. We only need to look at the ratios of even to odd numbers in the 1’s place that can come as a result of specific 1’s place values to see this. 0 has a ratio of 1:1 (will either have a 0 or 5 1 has a ratio of 1:0 (4) 2 has a ratio of 1:1 (6 or 1) 3 has a ratio of 1:0 (0) 4 has a ratio of 1:1 (2 or 7) 5 has a a ratio of 1:0 (6) 6 has a ratio of 1:1 (8 or 3) 7 has a ratio of 1:0 (2) 8 has a ratio of 1:1 (4 or 9) 9 has a ratio of 1:0 (8) That’s an overall ratio of 10:5. So a rough estimate could say that for ever 15 we multiply by, we divide by 20, or for every 3 we multiply by, we divide by 4. This, of course, is not exactly true if we have some of the crazy numbers, the ratios will be a bit different just by the nature of probability, but overall there should always be more even results, unless there is some weird starting number I am missing. If that is the case, tell me, please. Also I realize this probably is not entirely a valid proof, but maybe it helps someone to get an exact proof with the right mathematical terms, or maybe it’s just worthless, idk.
Multiplication is repeated addition therefore 3x = x + x + x +1 the size of x can really be ignored because you are creating 3 instances of x which can be considered as 3 then adding 1 to create an even number. An even number will eventually reduce to multiples of 2. Divide 2 by 2 and get 1 and we have the pattern: 4 from 3 + 1, 2 from reducing multiples of 2, and 1 which is the result of 2/2. 4 - 2 - 1.
that music makes even math sound dumb
HAH!! "EVEN" HAHAHAHA
*grunts*
Your picture is badass
The Elmafia zfZ
Thanks Elmo!
😘
Officer: Is there a problem?
Me: *Shows this video*.
Can't thumbs up on cell phone. But your comment made me laugh aloud 😂
+Naif Shaikh try pressing and holding to get a small menu, there should be a "like" button there.
+282FUN Nope, press and hold on Android RUclips gives just Reply and Delete options.
Naif Shaikh Welp.
+yumguy47 😂😂😂
After hearing "... INVENTED THE PROBLEM..." I realised what the humankind is all about :).
Bots
what do you mean?
He means humankind creates a lot of problems but dont/don't know solve it, can be applied to situations other than math, ig
me: 6969696969696969696969696969
generator: this is the seventh time this input has been submitted
So all numbers below 2*10^60 has been checked, and all of them eventually reached 1.
That means that all larger numbers only need to reach ANY of the previously tested numbers to prove that that number too eventually will reach 1.
This is not a solution for the problem but it can speed up the testing of numbers (which I'm sure that most mathematicians have already thought of)
Yes, this is an important aspect of any search algorithm for this problem.
so basically that rolls out all even numbers that are < 2x (if x is 2*10^60) and we only need to do the odd numbers when in that range
Nice
plus it basically rolls down hill from there if we apply the same logic to all the evens that equal and even that equals that range of numbers and so on
Yeah this was one of my first assignments in an algorithms paper at university. You use the previous numbers so you don't have to keep calculating every number again.
Wow surely everyone in the comment know math more than mathematician mentioned in the video
So true haha
False.
didn't you know that everyone in the comments section suddenly has a PhD in math?
xD
everyone in the comment section are professors xD
I've edited the comment... Now it'll be complete unexplainable section .
it equals fish you uncultured swine
NO OT EQUALS 5.000000000000000...
Hello no, 2 + 2 = 22
ahhh I see you've talked with Big Brother
Sydrahh MPS ihavetheproofthattwoplustwoisfive
what?.. i was listenning the background music whole time
I wanted to check the comments to see interesting things, what we get :
-Comments saying "how is that a problem".
-People who think they can solve the problem (and do what every mathematician failed to do in the last century) in a 2 lines comment on youtube.
I guess the real mystery is why I expected something from the comments at the first place...
Magima Thossa this is proof that 99% percent of the population is retarded
Exactly my thoughts... people think too highly of themselves in today's day and age.
"The problem with the world is that the intelligent people are full of doubt while the stupid ones are full of confidence." -- Charles Bukowski
The problem is that math proofs aren't taught in school.
jonathan haroun, i think the same
...I only came for the thumbnail...
+Eduayn Norahs me too
yeah XD
appel?
+Theseus Dude I wasn't even trying lol
What about decimals
+The Nerdy Cats ??? I completely forgot what the video was about
I dont see where is the problem...
Explaining why every starting number reaches the cycle.
'Prove that with these rules all starting numbers will eventually reach 1'
Well, isn’t this easy? It’s just that this pattern always ends up hitting a number in the pattern: y=2^x. And once it hits a number in that pattern it has to go all the way down to 1 and then you know what happens after that. It just depends on wether this pattern has a starting point that does not hit the equation: y=2^x. Please point out if I’m wrong but I think the conclusion is that this pattern always hits a y value in the equation: y=2^x.
Yes, it´s believe to hit a 2^x. But how do you prove it. You have said how you can prove this whole thing (so if you prove that every number reaches a number expressed by 2^x, then you have the proof), but you do not have the proof. You just have a shortcut.
The whole problem is just that the sequence is arbitrary. Just for illustration. You have a number n for which the cycle ends after 25 steps, then for n+1 it takes 400 steps and for n+2 it only takes 20 steps. So, there is no real pattern.
to anyone that says that: "i don't see the problem" the problem is to prove that n will at some point will hit the cycle 4->2->1 and if you say that is will hit 2^x you also have to prove why is does
Thanks!
No problem 😁
I still don't see why this is a problem.
@@citymoose lol
ATABA I did.
The reason it always ends back up at the sequence 4, 2, and 1 are multiple factors.
First, I will explain why the numbers will always end at 4, and then 2. If you notice, no matter what number you start with, you always end up with far more evens than there are odds. This is because of the algorithms that have been chosen. Every other number divided by 2 will give you another even or an odd. And if it’s even, you divide again, (remember that most numbers in these sequences are even), so therefore you will always go down, bit by bit, and all even numbers can be divided by the numbers in the sequence at the end: 4, and 2. And of course 2/2 will always give you the 1 in the sequence.
Now to further explain: when you get an odd, you must multiple that number by 3, (an odd number), and then add 1 to the whole. This will always leave you with an even number. (Again, the even numbers will always bring you back down eventually, and back to 4, 2, and then 1). If you take any odd number, (remember that the first number doesn’t matter, but the last, so therefore keep in mind there’s only actually 10 whole numbers you have to worry about), like 3 for instance, or even 23, or 43, you then multiply by 3. So, 3x3= 9, and now you add the 1 to get you to the even number. 9+1=10. Or, 23x3= 69+1= 70. Or, 43x3= 129+1= 130. Let’s try 7. 7x3= 21+1= 22. Or, 17x3= 51+1=52. Starting to see a pattern here? They always end up at the same number (referring to the last number on the right), and then always end up with an even.
And since you’re only multiplying every once in a while to get , and you’re dividing evens most of the time throughout the sequence, you will always end the sequence with 4, 2, and then 1. It is simple really. I figured this out by viewing the sequence a couple times and within maybe 30 minutes?
Also, a lot of times (if not every time), you will send back up with a number that does already end the sequence back down to 4, 2, 1. As you can see in the case of 11 and its sequence, and then 29 and its sequence which eventually leads to 11 and it's own sequence which leads to the infamous 4, 2, 1.
And if you look at all the sequences shown in this video, you can see a pattern, not with the sequences themselves persay (even though there somewhat is), but with the percentage of how many times the odds are compared to all the numbers in the sequence. If you count all the numbers including the 4, 2, 1, and either exclude or include the number you start at, odd shows up in a range of 25% - 37% of the time. Usually around 33% of the time.
And if you were to count all the odds and compare to all theevents, with or without counting the starting number and the 4, 2, 1, you'll end up with the odds only being 25% - 37% of theevents, again with 33% being the usual percentage.
Also, also, if you take the starting number and find how many times the designated number goes into it, a strange and certain pattern exists again. (Odds only).
For example, let’s look at the starting number 11. So, this is an odd number which will mean you have to multiply by 3 and add 1. If you find out how many times 3 goes into 11, (3.67 times), and you take those 3.67 times and divide it by the starting number 11, (3.67/11), you will get 33%. Basically giving an obvious glimpse at the approximate percentage the odds will show in the sequence.
Let’s try 29. 3 goes into 29, 9.67 times. 9.67/29= 33%.
Let’s do 8,199. 8,199/3= 2,733. This divided by 8,199= 33%.
I’m basically pointing out that this should be obvious with every odd number divided by 3, and then taking the answer and dividing the original odd number to get 33%. This problem is simple yet not, because the “patterns” and answers are right in front of our eyes, just hidden in plain sight.
And to add to this, if you do the evens only. The starting number is even, so we’ll do 10. You must divide by 2, which will give you the amount of times 2 goes into the starting number. 2 goes into 10, 5 times. And then 5/10 is 50%. Basically this is alluding to the fact that after every time you divide, every other time you will end up at an even or an odd. 50% of the time you will end up at and even. And like I said, after you do the 3n+1, 100% of the time you will end up at an even. So approximately 33% (from previous paragraph) of the numbers in any sequence from this conjecture will be an odd number (this isn’t a rule of thumb, but a better understanding), and the rest will be evens.
And since the majority of the numbers will end on an even, and the even you have to divide compared to only multiplying about 33% of the time, you will always end back up at 4, then 2, then 1. Like I said, 4 and 2 go into all even numbers (although 4 obviously don’t go into 2), and 2/2 will always equal 1. This is why all numbers you start with (all sequences) will end up back at 4, 2, and 1.
Oh, and it don’t matter what numbers you do these to. The starting numbers or any even number in the sequence or just period. You will always get that 50% chance that you’ll end up with an even or odd. 20/2=10 (Even). 10/2=5 (Odd). 12/2=6 (Even). 14/2=7 (Odd). 16/2=8 (Even). Etc, etc. 50%.
Same with doing that with odds. It doesn’t matter what number; starting or in the sequence or just period. You’ll always have an answer that’s even (after adding 1), and in the sequence with these equations of sorts, you’ll have about 33% of the sequence be odd and the rest be even. Again, this isn’t a set rule, but more of an answer that just gives you a rounded reasoning for the problem.
I don’t have too high of a vocabulary to better and fully explain this. Hopefully people can understand what I’m referring to and how I got the answers I did. I didn’t show the math too much, but if you do it yourself you’ll see. I can’t exactly show a picture of the math notes in my journal.
The problem isn't that it happens, it's that we can't prove that it happens.
Ridwaan Mahomed I can prove it
The problem is that to see this statement I have to read comments instead of watching the video
Woah! What's your Proof? You actually can prove it??
Having given this 3 minutes of thought, If N is ODD, it will ALWAYS produce an EVEN number.However, an EVEN number will produce an EVEN number half the time, and an ODD number half the time.
When an ODD number produces an even, it does so with x 3 the original value, plus the 1 to make it EVEN. However, an Even number will ALWAYS divide by 2 once, and 1/2 the time divide by 4, and 1/4 of the time divide by 8, and so one.
There will be sequences of GROWTH ( * 3 / 2 * 3 / 2 ), however not possibly forever. There are simply *MORE* times when a 2nd or 3rd or 4th division by 2 will collapse the total, no matter how large it gets. Any power of 2 becomes an irreversible *Zip-Line to collapse". This does not predict the # of steps, but rather argues that in the end it MUST devolve into the trivial cycle 4 -> 2 -> 1 because that *happens* to be the minimal set where ( # * 3 + 1 ) produces a power of 2.
Perhaps then one might wish to examine *how many* values, multiplied by 3 then add 1, result in a power of 2, or have LCDs with Powers of 2 > 2.
I remember asking friends would they buy an investment that doubled every year, but in the off years would lose 51% of it's value. Most said *Heck YES!* thinking that a 100% gain must ALWAYS be better than a 51% loss. After all, 51 is less than 100. But they fail to comprehend that a 51% loss, even after 1 two-year cycle, results in a smaller total amount that will be doubled the 3rd year. ANY Loss > 50% must eventually erode the bi-annual 100% gain to 0.
My $0.02
+Texas75023 Try to change the problem to 3n-1, and insert 5. Your argument can be used in that problem, but it has a counterexample. It's possible than 3n+1 also has a loop, but isn't found yet.
+Anonymous71475 Changing the problem, is ... well... changing the problem. However, I understand the consideration of Symmetry.
I tried (at random) to find another example. The first alternative I tried, was 8795. And VOILA! it had a different cycle !
At iteration 63 arrives at *122*.
At iteration 81 arrives again, at *122*. A Cycle of 18 operations.
At iteration 99 again, arives at *122*. So a definite cycle.
And checking further across the first 100,000 initial numbers, the "subtraction" form of the problem apparently has 2 cycles, based on the eventual occurrance of certain numbers ( 2, 5, 7, 10, 14, 17, 20, 25, 34, 37, 41, 50, 55, 61, 68, 74, 82, 91, 110, *122*, 136, 164, 182, 272 ). Now, I initially *did not check* for lengths of cycle, or alternate entry points, for simplicity sake. But then I thought to examine the cycle members. For example [ 5, 7, 10, 14, 20 ] form one cycle-set (not cycle ordered) which could be entered anywhere in the cycle, resulting in the same repeating member set. The remaing values (excluding "2") form a second complete cycle of 18 steps. Entry into the cycle at ANY of the 18 numbers, results in the inescable looping. )
Running the first 1,000,000 initial seeds, these same 24 cycle elements eventually appear.
Expanding that to the first 3,000,000 initial values, *SURPRISE!* I still get the same 24 elements.
Why these? I suspect if one would simply examine each number, these elements constitute 24 "traps", much like Powers of 2 in the "addition" or ( 3 n + 1 ) form. Only I do not see any such other "trap" values in the ( 3 n + 1 ) domain.
Again, my $0.02
Texas75023
Still... there is a possibility of undiscovered loop for 3n+1... Even if the smallest member of the loop has more digits than the atoms in the universe, the conjecture would be wrong. So... the way you did it won't solve the problem at hand... To solve it, you need to either:
1. Prove that a loop or a number that goes to infinity can't exist. Personally speaking I don't think the latter case can exist, but I can't prove it.
2. Give a counter example. This maybe is the easiest one if IT EXISTS, anyone without a background of mathematics can do it with AMAZING LUCK, if IT EXISTS.
3. Prove that there must be a counter example somehow. Either this or number one has to be done if all computable numbers have been checked.
*****
I agree with your assertions. Of the three cases, Counter-example (2) seems to be the easiest when one exists.
Proof (1) to Inifinity is always the harder challenge to perform with mathematical rigor.
Proof of the existence of counter-example is less frequently used, but necessary to consider as an approach to disprove an assertion.
The 3n+1 problem is interesting, and perhaps the 3n-1 problem provides insights that will be the key to the proof. I think an understanding of the 2 demonstrated cycles in 3n-1 might point to the nature of the behavior, but still not provide either a path to a counter-example of 3n+1, or a proof that said counter-example even exists.
+Texas75023 Something to note: the 3n-1 problem is the same as the 3n+1 problem for negative integers.
I made my own program and calculated numbers 1-20,000,000 (20 Million)
I determined that the number that took the most operations was 15,733,191
It took 704 turns to get to 1
I also calculated that for numbers 1-20,000,000 each number averages at 162 operations, pretty cool
Exploration is indeed cool.
heh nerd
I tried to reach 1-100.000.000 using mat.h, but the program crashed and so I can't run it in my computer, could I send you the script for you to use it in YOUR computer and send me the results later?
+Luis Parra for sure the only limitation In my program was that Java couldent process more then 20 mil, send me a pastebin link
So you can find the sequences in excel:
Start with a column in a sequence (it could be 30, it could be 300), in column A, so that A1=1, A2=2. This is your reference column. In cell b1, put this formula:
=IFERROR(IF(A1=1,"",IF(ISODD(A1),(A1*3)+1,(A1/2))),"")
This determines if cell A1 =1. If it is, it ends the sequence, and creates a clear cell. The clear cells are important for visual reasons. If its not 1, it determines whether that cell is odd or even, and makes the calculation based on the problem (a1*3)+1 for odd, or (a1/2) for even.
Now fill that down for all your numbered rows, and fill right for all your chosen columns as far as you want.
You end with all the sequences of the numbers for your specified rows. (30 or 300! Whatever you chose)
Then you see some really interesting properties. In some cases, like for the number 31, the next number is 94. 94 is the 6th number in the sequence of 27, so from then on, both 31 and 27 are exactly the same.
The reason that these high sequence numbers dont go off into infinity, is because you end up with a sequence of numbers within that sequence that are even, which halves each time. For example, in the number 31, the 84th number and onward are 976, 488, 244, 122 and then 61. You could then argue that those numbers (27 and 31) become exactly the same as the sequence 61 produces (and even the preceding numbers). So as you can see, the sequence drops dramatically because of the number of even numbers in a row. Because 31 and 27 are the same after 6 numbers (in 27), that sequence also falls into 27, and the reduction occurs quickly.
Each number will create a sequence that falls within the sequence of another number.
So no infinity.
Edit: Check out these sequences from 3377, 3378, and 3379. They align at 1426, and follow the same sequence, making these 3 numbers exactly the same as the sequence 1426 makes. They also have exactly the same amount of numbers in their sequence:
3377 3378 3379
10132 1689 10138
5066 5068 5069
2533 2534 15208
7600 1267 7604
3800 3802 3802
1900 1901 1901
950 5704 5704
475 2852 2852
1426 1426 1426
713 713 713
2140 2140 2140
1070 1070 1070
535 535 535
1606 1606 1606
803 803 803
2410 2410 2410
1205 1205 1205
3616 3616 3616
1808 1808 1808
904 904 904
452 452 452
226 226 226
113 113 113
340 340 340
170 170 170
85 85 85
256 256 256
128 128 128
64 64 64
32 32 32
16 16 16
8 8 8
4 4 4
2 2 2
1 1 1
Edit 2: Maybe the question should be "Would a (virtually) infinitely high number create a finite number of numbers in a sequence?" The numbers will always decrease to 4, 2, 1... but how high does a number have to start to end up, not with the numbers going off into infinity, but having an infinite number of numbers in a sequence?
*Just start with 0, its not even or odd So i win*
Or if you think that 0 is actually even or odd, www.google.com/search?sourceid=chrome-psyapi2&ion=1&espv=2&ie=UTF-8&q=is%200%20even%20or%20odd&oq=Is%200%20Even%20or%20odd&rlz=1C1CHZL_enUS697US698&aqs=chrome.0.0l6.4272j0j7
SO that proves that 0 is even, so then divide by 2, you get 0 again, and again so you never get to one.
So unless *SOMEHOW* someone proves that 0 is and odd number, then 0 will never turn into 1 by ties rules
And then after i type all of this i realize that he said it has to be a positive number and 0 is not positive or negative
;_; RIP ME!!!
+Vixen you just lost an argument to yourself, wow
+Vixen I like your thought process. I usually have little debates with myself, just like you did. Gg
2:00 **Immediately checking 5 · 2^60 + 1**
"who invented the problem" ... and get famous for
i know what i'm going to do next
I'll help you.
Prove that pi does not contain the Bee Movie script.
pangit mo
anak ko nya ako si naruto at pinakalas sa lahat
Was wondering... What if the conjecture is correct but there is no way to prove it? Could it be possible to prove something is impossible to prove?
Yes, there are theorems that are impossible to prove. That is the essence of Godel's Theorem. However, no one knows which theorems are the impossible ones. Moreover, almost no mathematicians worry about it. But, some have wondered whether the 3x+1 problem is one of those unprovable results ...
BubuSnow93 shut up bojack
and that is what faith is... ha..
en.m.wikipedia.org/wiki/List_of_statements_independent_of_ZFC
You’re welcome
undecidable proposition
So the real problem is Collatz for creating the problem?
the creation of new math problems isn't a problem. its how maths advances
Just simply put infinity as a number.
The problem is done.
Say 123 for GF
Infinity is not a number, its more of a concept.
Say 123 for GF you cant cuz it's not a number.you can define it's limit when n closes off to infinity which is pretty easy
Use pie, Pie solves all problems.
I'm on break from school and I'm on YT learning math what am I doing!
anthony battista studying math
EXPLOOSION! !!! Lool
anthony battista Ypure being a loser... With me cuz I am in weekend
your imitating me XD my spring break just started today
Same. I've spent the whole day watching math videos. Spring break is wild.
Tipping Point Math: "What can you see in this problem?"
Me: "It's a waste of time"
For those not trained in mathematics, think of it as a puzzle. Can you find a starting number that does NOT eventually reach 1? How do you know that it will even get below where it started?
For mathematicians, this problem is fascinating, because while it is so easy to explain, nobody can prove that every number eventually reaches 1. The process is deterministic (there is no randomness in how the next number is generated), but after many iterations it ACTS as if it is random. The larger of implication is that probability theory may be able to say something profound about deterministic processes.
Tipping Point Math
Yeah, thinking about it and reading what you said it does sounds fascinating or interesting, so from what I understand it's just acts like a puzzle and that's it?
Tipping Point Math I found a number that wont lead to 1: 0, does this count? Cause it is true
+Theredghost 0 is not positive (see 0:28)
How can something be random? If everything inside the universe is governed by fundamental mathematical laws, it is possible to predict everything with 100% certainty. Just because we haven't discovered the algorithms behind something doesn't mean it's random, it's just unpredictable until we've discovered how it works.
I could easily write down the proof to this but i really need to go feed my cat.
Can you give us the proof now Lord Alafin
@@yoshimitsu8643 Cat's gotta be fed, all we can do is wait.
Its such an easy problem, i could write the answer right now but i gotta wash the dishes.
Are you guys serious? 😂 😂
U haven't done feeding it yet
wacthing comments... some of the best mathematicians that world has ever known can't reach the answer, but youtube community did it again :)
No one in the comments have the explanation so far...
Kuro Neko Are you sure? All i see is idiots saying WHAT IS THE PROBLEM THOO?
that's true as well, ppl nowdays are advanced, they dont even see the problem :D
Kuro Neko which is not advanced,thats showing that people are getting stupid,no one has gotten the answer,the person who made they have to
They refuse the answer, because it's beneath their level of mathematical understanding.
Wow guys, thank you so much for pointing out logically in two sentences why it always goes to 4-2-1. You should tell this to the mathematicians working on this problem, they will be so glad to hear that a bunch of random people on the internet finally saw and pointed out what they haven't been able to see for so many decades! Guys, yes, we all know why it goes the way it goes, but just because you can explain it doesn't mean it is mathematically proven.
Every even number can be halved. 3x+1 on an odd number will always be an even number. Any real number will always produce the same end pattern.
I think the problem arises from predicting the number of steps required for any given number.
Your first two sentences (ignoring "easy") do not imply the third; Consider the function that sends all odd numbers that leave a remainder of 1 upon division by 4 to 1,000,000 and all other odd numbers to 1,000,002.
And proving that evary number does indeed converge to 1
Don't ya think if it was so "easy" that you can come up with the "proof" (what you've written is hardly a cohesive prove) then mathematicians wouldn't be sitting on this problem for decades? Or are you so much smarter than all of them?
Says the user with the username "hellowutlol".
When will this be patched?
Update will come.. not immediately but eventually..
@@rajeev_kumar Second coming of Jesus, or Antichrist
@@urorazbojnik5678 Nope, I'll just patch the Sim, Lol.
Several of the comments ask about replacing 3x+1 with 3x+d where d is an odd number. It is conjectured that for a fixed d, there are at most a finite number of cycles and all numbers eventually reach one of these cycles. For fun, look at the case with d=-1, that is, the 3x-1 problem. Can you find three different cycles?
I made this real quick www.khanacademy.org/cs/collatz-conjecture/5186778061864960 so, ummm yeah
jim blonde Jim what is the text showing, I can't understand what it represents? Could you put it in a readable form ? i.e. literally putting it as text please
how's that? *shows new version*
Wich numbers do not add up to 1? And even though i do not understand the problem of the problem beeing there, how would you like to get it solved? Change the formula? Then u change the cause of the problem?
+Tipping Point Math I found a 2,1 cycle
I know anotherone:
x-x+1, it will always gets to one.
Theredghost wtf you're right
Theredghost im gonna test it with 1 Billion numbers by hand just to be sure
No shit, you're taking X (10, in this case) subtracting X (10) and adding one. It works for negatives too because a negative minus a negative becomes addition, which everyone should know.
So it's 0 + 1 every time. This is what? 5th grade/year (elementary/primary) school knowledge?
LKCshow100 i would have never guessed wow thanks
ShavdezProductions it does -12 - -12 = 0 because - - = +
Another interesting thing about this problem is, that if you graph the number of steps taken to get to 1 for enough numbers (e.g. 1 to 10^5), it starts to look like a root-function.
Jannis what about i?
What do you mean by "i"? Imaginary numbers?
You can even check the scatterplot of this in oeis.org/A006577/graph. It really looks like a root function.
What about the imaginary unit?
---
Let's back up a bit...
Let us pretend we can do this with imaginary numbers... now, how would you visualize such thing?
Well, we use a number line for the reals, and we can plot the complex in a plane... What if we color the numbers depending on the number of iterations they take to reach 1?
Done: 3:17
Note: Black = goes to infinity
---
What? Didn't they say they all go to 1? Well, all INTEGERS - as far as we can tell - go to 1. Not all reals tho, much less all complex... but, is 0.5 even or odd? What about i? Is i even or odd?
Well, first we need to extend the function f(x) = x/2 : (x = 0 mod 2), f(x) = x*3+1 : (x = 1 mod 2) to something that would work for any complex number. We call this an Analytic Continuation. It is identical to the described function for integers, and it is guaranteed to be continous.
To be honest, I don't know how to do that... yet, the paper "The 3n+l-Problem and Holomorphic Dynamics" has this:
f(z) = (1/2)*z + (1/2)*(1-cos(pi*z))*(z+(1/2)) + (1/pi)*((1/2)-cos(pi*z))*sin(pi*z)+h(z)*sin(pi*z)^2
where h is an arbitrary holomorphic function.
Ok, this is not exactly a continuation of f(x) = x/2 : (x = 0 mod 2), f(x) = x*3+1 : (x = 1 mod 2) instead it is a continuation of f(x) = x/2 : (x = 0 mod 2), f(x) = (x*3+1)/2 : (x = 1 mod 2) - that is, it is doing two steps for the odd numbers.
You can try on Wolfram|Alpha, for example:
(1/2)*z + (1/2)*(1-cos(pi*z))*(z+(1/2)) + (1/pi)*((1/2)-cos(pi*z))*sin(pi*z)+h(z)*sin(pi*z)^2 where z = 5
It should yield 8, because (5*3+1)/2 =8.
The interesting thing is that you can plug any complex number you fancy... so let's see...
(1/2)*z + (1/2)*(1-cos(pi*z))*(z+(1/2)) + (1/pi)*((1/2)-cos(pi*z))*sin(pi*z)+h(z)*sin(pi*z)^2 where z = i
It yields...
-h(i)*sinh(pi)^2+(1/2)*((1+4i)-(1+2i)*cosh(pi))-(i*sinh(pi)*(2cosh(pi)-1))/(2pi)
And there you go! :S
Yeah I believe it was proven that the process tends to look like the graph y=x^.84
Video: The problem is: how can we prove this algorithm will eventually yield infinite sequence 4, 2, 1?
Half the comments section: What is the problem? What is the question? The answer is 0. This is obvious wth.
Most people do not know what mathematics is. When you ask them, sometimes you are lucky just to get a remedial background in arithmetic
Zero is not even or odd so it would just stay zero
@@iBlaze1232 Nah, 0 is even.
Random Dude hmmm
Maybe it's 0 but cuz the number is even so..
0 is a symbol that is absolute. Nothings without it yet nothing has it. 1D+2D=3D but 0 is the element with no ending or beginning.
I honestly can't see how this is a problem.
OK there is now a lion at my front door, clearly it is a problem, I'm a fool
Is there a big number you can pick that doesn't go to 4->2->1? Nobody knows.
+Nifty Fingers but that's not a problem... does the pattern of 4,2,1 have any significance to anything?
Josh Brady well if someone proved that say, all numbers do eventually end up on 4 2 1, maybe someone would be able to find an application for it. Maybe in pseudorandom number generation... maybe in encryption... I don't know just doing armchair speculation.
Math also makes a lot of unnecessary problems for itself.. this "problem" isn't actually a problem... nobody can find a number that won't lead to this same pattern.. and number starting from 1 all the way into the hundreds of billions have been tested.. I'm sure billions more by now and they still can't find one that doesn't lead to one.... so what does it teach us if 4568296368262527494826254 ends at...2... what did we learn?... but that wouldn't happen anyway because cut that number in half and the resulting number has already been tested... almost seems like a waste of resources to build an expensive ass computer just to run this program... just creating more problems
Pretty sure every number is just gonna bounce around randomly until it hits one of those numbers that easily collapses into 1
yeah but "pretty sure" isn't a proof, which is what they're looking for
Yes, I came to the same conclusion. Eventually a series of x/2 and 3x+1 will result in a power of 2. So how do you prove that a power of 2 will always eventually result?
I'll prove that 1+1=56. heres my theory. Heres my example. When adding an odd number by 1, it will be even. If it happens to be 56, it will quickly prove my theory. If it isnt, multiply it by 0 and add 1 to get it to 1. Keep adding 1 and repeating the process until it becomes 56. There. Im a 9th grader by the way
...some people really have no idea what a proof is.
ReaSeBy lol, just about to point that
ReaSeBy but you never reach 56. You stay at one.
I FIGURED IT OUT:
This equation forces the numbered to end up to the smallest even number that when divided by 2 it isn't a prime number.
Btw there is no pattern. It just keeps going until it reaches 4
+NinjaBallista that's a patern
That's not the part that needed figuring out. The part that confuses everybody is the fact that it takes a seemingly random amount of steps to get the starting number down to one for each different starting number
+NinjaBallista math wiz over here
+Tysoreny It does need to be figured out and has been. The part that confuses everybody now is proving it.
I'm no mathematician but I think the problem is: so far, the rate of decrease (n/2) eventually outstrips the rate of increase (3n+1). The question is, can it be proved that this will always be the case.
Multiplying an odd number by 3 gets you another odd number, then 1 is added making it even again, allowing it to be halved, this will reach a different number each time until it gets to a number that can reach 16, 32, 64, etc.
There is no pattern, just a law of operation.
Ding Ding! The way it's presented makes it sounds like an equation to solve. Let me try to make up some nonsense too
umm... Start with any whole number, if it's greater than 10 subtract 5, if it's less than 11 add 3
you get a repeating 6, 9, 12, 7, 10, 13, 8, 11...
Vyor Generic Last Name you're a genius
That's a fine and dandy explanation, but can you prove it? Proving that is the problem being explained in the video.
how do you know it'll always eventually culminate in some power of 2? how do you know it will not just infinitely produce even numbers but never an even number that's a power of 2?
Everyone knows that... The problem is proving the thing. What you said is equivalent to the information we get from the tests.
Here's my take:
3 times any odd number stays odd. (try it) Add one and you get an even number. Because of this you will never get two odd numbers in a row. However, you can get many even numbers in a row (44->22, 48->24). In fact - ~half the even numbers you get will stay even. Due to this, inevitably the number will go lower and lower until it "locks on" to the pattern that ends in 4-2-1.
Tim S well go tell all the mathematicians... except wait, can we prove that all even numbers times three plus one will always give a even number?
An even number multiplied by itself any number of times stays even. Add one and it becomes odd. But I'm not sure how that's related to anything.
i came to the same conclusion :|
It's good intuition for understanding what's generally going on, but doesn't solve the problem. How do we know that there isn't a number that has even numbers in a row so rarely that it overpowers the halving?
It's a pattern - and with patterns we assume they go on forever. It's the same as asking "Sure, 3,6,9,12.. is a pattern, but how do we know there isn't a number really high that still adds 3 but isnt divisible?" its implied
I got one even easier: n-n always = 0. No matter what number you use. Checkmate eggheads!
Harvard wants to: know your location
That's not a problem.
@Miroslav Mandic n is a variable. A variable stands for any number. So, what they're saying is that any number minus itself is 0
Damn it I got wooshed
@Miroslav Mandic Woosh is the sound that is made when a joke goes over your head.
A lot of times, the response to someone taking a joke seriously is simply, "Woosh."
for people who didn't get it the unsolved problem here is that we couldn't prove that *every existing number* reaches that cycle after calculations, we reached pretty high numbers but there's still a chance that there's some large number we didn't reach that doesn't end in that cycle.
After review of the problem it is clear every number will reach 4 2 1. There is never a number that will not reach it and it is due to the (+1) When a number reaches an ODD value it will triple and +1 making it even. Its mostly a game of how many even numbers can you divide by 2 before hitting the next odd value. Your never going to find a number that will triple +1 and divide by 2 and break the cycle. The +1 will change the pattern of any number that has the potential to break the cycle. Is there a pattern or another mathematical explanation for the results of each number? Not really, each number is unique, the ones that can divide the most by 2 will always be the key changers in the values. Until it eventually reaches 4, 2, 1. You will never end up with 2 odd values after each other and you can skip half of the odd values by simply multiplying the first odd value+.333 * 1.5 rounding up. will always = skip over a few values.
I agree that any odd number will become even with one application of the function. I agree that an even number will divide by 2 until reaching an odd number. I can't follow the rest of the argument; how do you get from here to all numbers eventually reach 1?
As long as the even number divides into a multiple of 2 it will divide all the way down to 1
In mathematics you're not allowed to say "These numbers will do exactly this" without offering proof, so simply saying "As long as the even number divides into a multiple of 2 it will divide all the way down to 1" doesn't solve the problem. I'm pretty much sure that after reaching 5x2^60 mathematicians are pretyt certain that it works with any number but till now there is still no proof written down. That's what this video is about.
Kind regards,
Meta Custom Computers
+Meta Custom Computers I know that I was just ansering the previous comment
The point is here's no obvious relation between the starting number and the number of steps required to get to 1. That's be problem.
Someone please correct me if Im wrong but I see a very clear/definitive solution to this. First, let's say you start with an even number. You will divide by 2 until you either reach 1 as your first odd number and you will get the 4, 2, 1 cycle. Or, you will reach an odd number before 1. then you multiply it by 3 and add 1. this will always result in a even number again because an odd number x 3 is always odd, then once you add 1 it will be even. Basically what this means is that it is impossible to have 3n+1 twice in a row but is not impossible to have n/2 twice in a row. This means that as our base number grows larger due to a repeating series of 3n+1, n 2, 3n+1, n/2.... the statistical probability of getting more n/2 twice (or more) in a row grows, at some point becoming nearly impossible to avoid. When this happens it will more than counteract the 3n+1 and will continually happen until the series converges to 1 because the probability of dividing will always be higher than multiplying.
Note that 3 (the factor by which you multiply) is larger than 2 (the factor by which you divide). Therefore, it is possible to divide by 2 more times than you multiply by 3 and still end up with a number larger than the number that you started with.
Cool
+supergreatsuper no if you divide a number by 2 twice, then you are dividing it by 4 which will shrink number (compared to operation of multiplying it by 3)
If you divide by 2 three times and only multiply by 3 two times, you end up with a number larger than the number you started with. Therefore, "it is possible to divide by 2 more times than you multiply by 3 and still end up with a number larger than the number that you started with" as claimed.
Oh true
Seeing how few people actually appreciate this problem makes me sad =-\
I think I understand the problem..
The bgm in this video .. that's the primary problem
I have a truly beautiful proof for this, but alas the RUclips comment box is too narrow to contain it.
Ok Fermat.
+mum pert the truth hurts lol, u bursting his bubble of vocabulary
A person who prefers typing clean words instead of cuss words
Iunderstoodthatreference.jpg
Fermat's last theorem reference?
The problem is that there is no exception, regardless of the number, and that there is no real pattern correlated to any of the other starting numbers. For people who were confused.
*****
They're not a pattern because no matter what number you start with, there is no matching patterns with any other starting number, ever.
But the system always ends with 16,8,4,2,1, and then, 4,2,1 repeating. Mathematically isn't that the only outcome, when a number reaches such close amounts to 0 it has to eventually falls into a pattern. Doesn't that mean that a number, no matter how big or small, can be divided by 2 if even, and multiplied by 3 then have 1 added to that if odd? Why isn't that a solution?
There is clear pattern dude, I'm going to prove that soon)
For example, I found that in average ratio between odd numbers and even numbers in those sequences is 1 to 2, thats why /2 prevails over *3.
But a lot of work is still needed, won't disclose anything until figure that out)
+Neon Lights Also, the reason for that is obvious, because 3*n+1 ALWAYS leads to even number as a result, but /2 could result in even number, in fact, half of even numbers could be divided more than once by 2.
That is why we have this pattern
3 times an odd number is always an odd. Adding 1 makes the number even which means it will be divided by two. If the last digit in the number is a 2, 6, or 0 the number will be odd repeating the cycle until 1. Dividing a whole number by 2 will always be a positive integer and since 1 is odd 3(1)+1=4 which then causes the cycle to repeat
"Dividing a whole number by 2 will always be a positive integer."
Are you sure?
@@lillilacac maybe he thinks in binary
@@lillilacac 🤣
-4/2
You only have to check all possible combinations of 1 , 2 , 4 , 8 , 16 , 32 ,64etc. to get the shortest possible count or , all other numbers are poluted , this is clean halving time i.e. 64^32 / 2*191 = 1 , 64^32 each time divided by 2 takes you 191 steps to get 64^32 to 1 OR from 64 - 32 - 16 - 8 - 4 - 2 - 1 = 1 , six subtractions from 64 to 1 , reminds me of bits and bytes All combinations from these number to the power of (mix them) will result in 'low' steps to get to the one , tho
Engineers build rockets, skyscrapers and computers... meanwhile mathematicians try to find a solution to the 3x +1 problem. For fucks sakes.
Mathematics has had unexpected applications before... which is why we pay them.
Oh fuck off. I'm going into engineering, but pure mathematics great too. It's kinda like an exploratory mission into the system that we live in [5]. And engineers couldn't do what the do without the underlying math behind it.
Have you ever heard of the Fibonacci Sequence? 1 1 2 3 5 8 13 21 34... you get it by adding the previous two numbers in the sequence 1+1=2, 1+2=3, 2+3=5 and so on. Just looking at it, it might seem to be just folly but it appears everywhere, from snail shells to plant growth to the shape of the galaxy.
Maybe the solution to this problem will become part of the foundation of the next paradigm shift in physics and in 100 years every engineer will have to study it in university. Or maybe the proof techniques invented to solve this problem will be applied to other "impossible" problems. Or maybe this problem is truly impossible to prove which will have powerful implications on the field of mathematics. Or maybe the solution of this problem is completely inconsequential and unimportant
The fact of the matter is that we don't know. Engineers - for all their importance to society - work firmly within the bounds of human knowledge, and thusly for all the work they do it is immediately obvious how it is beneficial. Mathematicians (and scientists as well) work outside the bounds of human knowledge and don't have this privilege of knowing beforehand what is "important".
Well bud, we'd love to hear how you'd prove it wouldn't we?
+Nukestarmaster Well put
Wait.. how is this a problem?
3n+ 1 (n being uneven) always results in an even number you'll eventually hit a factor of 2 resulting in more divisions than multiplications.
But 3>2, so it's possible to divide more times than you multiply and still end up with a larger number than you started with. How do you know there are more "enough" divisions?
supergreatsuper 3*5+1 = 16 look how many times you can divide vs multiply
You'll always reach a chain of divisions..
Depending on your definition of "chain", I agree that you will always reach a chain of divisions. However, how do you know that these chains will be long enough, and that there will be enough of them?
your example is trivial, as you purposefully chose a power of two.
But then again, you are claiming to understand the solution to a problem that has never been cracked by the most brilliant minds for the past century or so; so clearly I fell to a troll. Stupid me.
+wiwh you fell for a troll?
most comments prove that math is being taught the wrong way...
Guys come in, the problem is to prove it, the fact that every number we have plugged goes to 1 doesn't mean it will happen with all numbers.
0.000000000000000004248134.
dont think that will go to 1 anytime soon...
Delacroid NewGen if you start with one you go to 0 not to one I am the smartest of them all!
If you start with 1 you already reached 1 with 0 steps and then the repetition starts (1 4 2 1 4 2 1 4 2 1 etc.)
Delacroid NewGen dont do maths lmfao
Actually that is 100% logical:
Every odd number when tripled gives an odd number, add 1, it's even, you divide it by 2, if its even, keep dividing, if it is odd, get it even again, you will keep doing so until you either progressively reach 4, or until you reach a power of 2, it may near an infinity of tries, but at the end , it works out, the difference in number of tries is related to how far away is that number from the nearest power of 2
I'm amused by the number of people who think they have this solved in the youtube comments...
+supergreatsuper You have been FANTASTIC at responding to comments. Your sensible and articulate responses have been spot on. Much appreciated!
The only reason I can think for this is because every time you use an odd number, it will go to an even number. As an even number is every other number, the probability of you getting a chain of them is 1/4, 1/8 etc...
Due to this, it is unavoidable that you will get a chain of numbers that will continuously divide by 2. As the 2nd or 3rd division is less effective, you are in effect multiplying it by 1/4. Which, if you were to multiply by 3 and add 1, you would get 3/4 + 1, which, unless the number is 4, this produces a smaller number. Due to probability, every number will eventually reach a point where it will hit a chain of divisions, causing the number to decrease by a factor of 2. This is what results in all numbers reaching the chain.
+Tipping Point Math Thanks for the support!
+Callum Chiverton I can't really tell if you're claiming you have a proof or not so I'll go ahead and assume you are just in case. I agree that any number will eventually be divided by 2 twice in a row. How do you know this will be enough to counteract the gains due to the multiplications by 3?
supergreatsuper See, that's the thing. There is almost no way to guarantee that this can happen, the only thing that can make it "certain" is the probability of you getting a longer chain with the more times you cycle through the calculation. You could even go on long enough and, based on probability, you could get 10000 divisions in a row. There is just no way to prove every single number with just probability, as this applies to everything in life
2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
1 2 3 4 5 6 7 8
1 2 3 4
1 2
Using this as a base example, we can see that as you go up in numbers, that every 2 numbers you can divide by 2 (obviously). Out of these, numbers it produces, you have another 1/2 chance that it will divide. This means you get a 1/4 chance that you can counteract the gains of the multiplication by 3.
If we were to take this probability into account and say that it would happen EVERY 2nd or 4th or 8th division.
Keeping n as the same number throughout, but applying the logic
n/2
3n/2 + 1
9n/2 + 1/2
9n/4 + 1/4
27n/4 +5/4
27n/8 +5/8
81n/8 +13/16
81n/16+13/32
81n/32+13/64
81n/64+13/128
This could go on for nearly forever. However, it proves that with time, the probability of the division occurring more than once can counteract the multiplication by 3.
I wouldn't be surprised if I messed up here somehow, I rushed this, but for me this is proof, even though it isn't hard mathematical proof, which can only come from having a calculation with n that then results in the amount of steps needed to get it to 1.
For me, however, this is proof because you can go to infinity, with each number you use having a chance to divide more than once. This means that you can theoretically go to an almost unlimited number with almost unlimited steps, however, the more steps you perform the higher the chance there is for you to get chains of divisions, so a near unlimited number of steps in my eyes will eventually bring a number of divisions that will result in 1
Suppose just a few numbers were to form a loop or a chain that did not include 1 (that is, most numbers reach 1, but a few (still infinitely many) don't). Would this affect the overall probability? The next question is: How strong must the definition of "few" be to not affect the overall probability? Could it be almost all numbers?
an odd number multiplied by three is always an odd number. an odd number + 1 is always an even number. therefore, following this formula, if you have an odd number the next number is guaranteed to be an even number. HOWEVER, when you have an even number there is no guarantee. you can have an odd number, or an even number. when you get an even number from an even number, the number you get is typically lower than the odd number you got there from. over time, this formula harbors a disproportionate amount of even numbers, and therefore, a disproportionate amount of division. this leads to the unavoidable cycle of 4 > 2 > 1. am i missing anything here?
tl;dr you're always more likely to get an even number, so more division
well more divisions isnt the odd thing causing the repeating pattern like with 28
14 7 22 11 34 17 52 26 13 40
the numbers are going from high to low there
whats odd is that eventually one of those odd numbers when multipled by 3 +1 is a power of 2 so therefore you keep dividing by 2 until you get one 256 128 64 32 16 8 4 2 1 then the pattern starts
thats why 8192 had so few steps because its a power of 2 and goes straight into the pattern
How does continuously adding 1 make any odd number eventually get to a power of 2? Don't forget that you are doing lots of operations between the adding 1.
+supergreatsuper because the additional one forces an even number. Take any odd number and apply this. It will come out even. Then division. If it's a multiple of 4 it's gonna come out even. Then divide again. It doesn't necessarily have to become a power of 2 to get to smaller numbers, but at that point there are more multiples of two. 2,4,8,16,32,64,128. Get in this range and the multiplication is more likely to hit a power of two than in higher numbers on your next multiplication. Once you hit it, it's inevitable it will reach 1.
Do you know you will eventually get in the range where you are guaranteed to eventually hit a power of 2?
it's more about the power of 2, infinity makes the probability tend to 0% but will make it happen eventually
*I thought its when your girlfriend seems mad at you and you ask her if everything is ok and she says yes and is still clearly mad at you specifically...*
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Of course it will always end at 1. .... it could NEVER end any other way !
It's the "divide in half" part of the process that will always prove that.
Well of course you're right! I mean it's not like as if much more talented and experienced mathematicians worked on this problem! What I'm trying to say is - the answer is not so easy!
I don't see why it couldn't end any other way (unless you're using the word "end" literally, as in the sequence stops. Which technically it does not stop at 1 either - it just goes back to 4, 2, 1, ...).
On the contrary, it is the geniuses that look for the simple answer. In fact, mathematicians in general look for the elegant solutions (e.g. proofs without words).
This is explained in the video. The other possibilities are that it has some long cycle or that it approaches infinity.
***** I agree with you, but you should really calm down on the run-on sentences. They make your paragraph really difficult to follow.
i have another one: if number is odd, add 1, if it's even, divide by 2. you always end with a sequence of 2,1,2,1. can i have my cookie now?
The difference is that your problem is easy to prove, while the Collatz Conjecture is hard to prove. Why don't you give your conjecture a go?
supergreatsuper
I will try but I didn't see math for many years :)
No cookie you forgot about negative numbers it will end at 0
-1
+Rejanth Thevalingam it will not go to 0 it will get lower
I know the answer just don't wanna tell
lol
0?!
0 is not positive ;-;
42 , we all know ...
I can prove it too but the proof is too long to fit into this comment.
Not a mathematician by any stretch, but this seems very intuitive.
Half of numbers are odd, half are even. So 50/50 chance of performing each operation.
However the 3n+1 always yields an even number which will be halved so the step should really be described as (3n+1)/2.
The n/2 can yield an even or odd number with equal likelihood so that single calculation can be considered a complete step.
So the equally likely options are multiply by 1.5 or by .5 (you cN ignore the +.5 because that just rounds the number up to the next number which may be odd or even and just invokes the above method anyway.
So, to simplify if I have a 50/50 chance of multiplying by 1.5 or by .5, the expected impact is to multiply by .75 over each set of 2 goes, so the odds are stacked towards a downward direction.
Sure you can find numbers that will have a 'hot streak' of being multiplied by 1.5 several times im a row, but eventually the methods weighting towards 1 will kick in.
Well, if it ever reaches a power of two, it gets caught in the cycle of 4-2-1, as it keeps halving until it reaches one. Multiplying an odd whole number by three and adding one will always produce an even number, which will let you get loser to small numbers. As you get smaller numbers, you get a higher probability of the 3x+1 function producing a power of two. But, to get to the 4-2-1 cycle, you HAVE to reach a power of two. There is no other method of reaching one. For example, to get one, you know there are only two formulas that can produce each number: 3x+1 and x/2. For one of these two produce one, the previous number would either have to be two or zero. It can't be zero, as that would require starting at zero or a negative number, so we know you have to reach two first. If we do the same to two, we must either get four or one third. Again, we know the sequence will never produce any number other than a positive integer, so it can't be one third. We must first reach four. To reach four, we must either have eight or one. We know that to get one, we must first get four. So, getting one will not give us a solid method of reaching the cycle. We must first get eight. Since 8-1, or seven, is not divisible by three, we know the only way to reach eight is by getting twice its value, which is 16. To get one, you HAVE to reach 16. Now, this is where the methodology begins to split. Both methods (3x+1 and x/2) will produce a positive integer for the answer. Having the first expression equal 16 gives us five as our previous number, while the second expression gives us 32. Five gives us ten, ten gives us 20 and 3, 3 gives us six, and twenty gives us forty. If we continue this on forever, we will eventually include every single whole number in our pattern. We've already got one through six in our pattern, plus many other greater numbers. It always loops back.
"If we continue this on forever, we will eventually include every single whole number in our pattern." How do you know?
supergreatsuper Because the field of numbers keeps expanding before going to much higher numbers.
I agree that it will include arbitrarily large numbers. However, I do not see why it must contain every number.
supergreatsuper The rate of expansion for the set is too rapid for it to not include every number. The rate just keeps getting faster as you come across specific numbers that work for both 3n+1 and n/2, like 16.
It's the infinite monkeys, writing on infinite typewriters quip. You may not understand, or comprehend, that every number will arise, it might sound unlikely every number will pop up even once, but in an infinite sequence, every number will show up an infinite number of times.
I see no problem with it. It's just a fact and something inevitable that will happen when you apply these specific rules.
Why must it be a problem?
It is common usage of the word "problem" to denote (from the second definition from Google) "an inquiry starting from given conditions to investigate or demonstrate a fact, result, or law."
huh... you learn something new everyday. Thanks.
but it's really mind bugging that the same words have different meanings. When someone tells me they have a problem I assume that it is something that is making their life harder in some way.
E-gre-gioous
1. Outstandingly bad;
2. Remarkably good;
The same word not only has different meanings, it literally has polar opposite meanings.
According to Wiktionary, the "good" connotation is outdated (see en.wiktionary.org/wiki/egregious)
However, there are such words and they are called Janus words (see en.wikipedia.org/wiki/Auto-antonym)
(Yeah I just looked this all up)
my exact thoughts
This series always occurs because the pattern can cause up to four divisions backwards but only one multiplication forwards at a time this leads to statistically more divisions backwards in the series which leads to the enevitable approach of the 8-4-2-1 at this point the pattern because infinitely repeating simply because the numbers cannot escape the pattern
I think ^
You made the mistake of calling it inevitable without any proof of that. Also, statistically speaking, airplanes tend to land on airports. Somehow, 2 of them still managed to land inside the twin towers.
+kek u wot m9 HUEHUEHUEHUEHUEHUEHUEHUEHUEHUEHUEHUE I would agree with you but in terms of airplanes there are a lot more variables to consider that could cause statistic variation, but in this the only variable is what we are in fact measuring, it would be like saying an airplane might not land because it in fact is not an airplane
godoforanges
well, my analogy is not be the best, i'll grant you that.
That's not a proof
this video explained it better than veritasium under 4 minutes
okay, stupid here, why is this a problem? watched the video 3 times and dont understand where the problem situation lies. I was taught that every even number can be divided by two and if you add two odd numbers you get a even number. So it makes sense they get in this 4-2-1 circle in the end. There is no problem. That is just proofing the rule.
The problem is in showing that every starting number eventually iterates to 1. As noted in the video, the number 27 wanders a lot before eventually reaching 1.
Tipping Point Math lol and now? of course because of the law of math. I don't get why men wonder about the own laws they created.
Tipping Point Math and even if it would go on in another way or only would proof the law of simple mathematics. That's wrong with humanity, they want to see a problem where no problem is on the columns they created
Supposedly by your rules, there is no problem, because you're only looking at the conditions. The actual problem is proving that there are no exceptions to the conditions. If you're going to assume that there are no exceptions to all rules, then that's a pretty naive and ignorant way to live. It's like seeing one of those erroneous headlines nowadays and immediately believing it.
Konohura
well, there are exceptions in patterns but not in rules. If you dont follow a rule in math its plain wrong. If its a exception in a pattern, this exception is created by a rule. Comparing this to believing in headlines is not an appropiate analogy.
Once a result becomes a power of 2, the pattern is unavoidable. That's the pattern. The proof required is for any given positive number m, what its probability of avoiding being a power of 2. Any even number that is not a power of 2 is the product of 2 and an odd number and it will thereby avoid the pattern for at least one more round as the result will be odd and the function 3n+1 will be applied. Another way to look at the issue is for any given odd number greater than 1, what is the probability that 3n+1 is not a power of 2?
+Bryon Lape After looking at the comments, I do want to be clear. I am not stating the probabilities are the proof, but the start of the idea. I've put many different numbers into the calculator. I am amazed how many reach a result of 5 and then fall into 16->8->4->2->1.
+Bryon Lape
The probability that 3n+1 is a power of 2 is 0. More precisely, the density (natural density, but replace the set of natural numbers with the set of positive odd integers) of the set of odd numbers for which 3n+1 is a power of 2 is 0. This is not hard to prove, and I encourage you to try and prove this.
+supergreatsuper First I want to be clear that Bryon's assertion does not solve the conjecture as the assumption does not prove the recursion will not spiral out of control (get rid of or try substituting another number for 3 and you can easily find examples that will).
But, I want to address 3n+1's ability to evaluate to a power of 2. If n is a sum of sequential powers of 4 starting from 4^0 then 3n+1 will evaluate to a power of 4 and, by default, a power of 4 is a power of 2.
This is easy though a bit long to prove. First remember the following rule of summation:
now replace i=0, c=1 and r=4 we get the sum of powers of 4 from 4^0 to 4^j
simplifying we get
multiply top and bottom by -1 and flipping around we get
now, lets substitute this for n
and put it into
we get
The 3s cancel
-1+1=0 so, when n=sum of powers of 4 we get:
Which is a power of 4 which is also a power of 2. So, it is not impossible that 3n+1 can not evaluate to a power of 2.
Try it out with sums of powers of 4 (the powers of 4 are 1, 4, 16, 64, 256, . . . )
I'll start the list of sums so there is no confusion: 1, 5, 21, 85, 341, 1365, 5461, 21845, . . .
Q.E.D. :)
+Robert Dostal Darn thing go rid off all the formulas....
Will it take html? (hard to write out sigmas any other way) [tried, nope]
From Google docs? (they don't make great but will do)
k=ijcrk =c(ri-rj+1)/(1-r)
nope, that doesn't work either.
+Robert Dostal
Did you write that using LaTeX? I can read LaTeX code, so it would be just find if you wrote out the source code. If not, what did you use? There is a chance I will be able to read that as well.
At any rate, I said that the probability that it becomes a power of 2 is 0, not that it is impossible (actually, because a uniform probability distribution is impossible on the set of natural numbers, I used natural density instead of probability). What this means is if you "pick a random odd natural number," then the probability that it goes to a power of 2 in one step tends to 0 as the numbers get arbitrarily large.
I interpreted Bryon's claim is "all numbers eventually reach a power of 2," which would clearly solve the conjecture.
The solution is simple. 3n+1 always equals an even number. If n is odd, it WILL be followed by an even result. There cannot be 2 odd numbers following one after the other.
More than 2/3 of all n values will be even. Odd numbers increase the value to 3/1(plus 1). Even numbers decrease the value to 1/4 because you will always end up halving twice. half of a half is a quarter.
1/4 of 3 = 3/4
25% of 300% = 75%
The math is forcing the value of n towards 1. When it hits an odd value, it jumps up to an even value and starts going down a different sequence. The math is designed to search for the correct sequence to reach 1.
The only way that this sequence will not reach {4,2,1,...} is if every even result is an even number that is made up of two separate odd numbers, or if the sequence repeats a multi-digit n value (odd or even) more than once.
This would be a great way to stump a class of third graders. Bonus points: tell them that if any of them can find a number that doesn't hit 4-2-1, you'll order the whole class pizza. xD And then if any of them come up with a cool insight about the problem, order pizza anyway.
It would take the kids forever and they would still come out with nothing
@@therandomwizard188 incorrect. The solution is solved. 😀
does this problem even matter?
does it help us solve, or prevent us from solving anything of importance?
just askin...
the point of math research is not to solve real life problems, but is to solve math problems
Did you watch the part of the video he said the different types of maths at the start
He pointed out that this is purely just to create problems.
Mathematicians pride themselves in being useless
If we don't solve it trump will become president
isn t this the type of problem you solve with mathematical induction?
Many people have tried to use induction, but it has not worked. Indeed, this problem seems infuriating because it resists every approach. Of course many newcomers to the problem may try induction or modular arithmetic, but all of the low-hanging fruit was picked long ago.
Tipping Point Math Hi, could you tell me what is the name of the music you used in the video?
Sheepy Thanks.
Monado Boy How did I find a fellow Xenoblade fan here of all places
yes but u will also need to conduct a proof
this video doesn't really explain why its a problem. The actual problem that mathematicians are trying to solve is to repeat the process indefinitely without invoking the Collatz Conjecture. It's an impossible problem because it was designed to take advantage of how we define even and odd. If anything, the Collatz Conjecture is a problem of language not mathematics.
not true, even and odd are well defined
How can so many people not see the problem? Did they not listen at all to the video?
Katie Inman they don’t understand why people find the problem so hard. It’s more that they don’t understand the question than the problem.
Since you claim to be a mathematical genius, why hasn't this conjecture been solved yet?
if I start with an even no., the next no. can either be an odd no. or an even no. if the no. I'd even, the process continues and if the even no is a power of 2, the end will have a series of 4,2,1.
and if initially the no. was odd, multiplying 3 and adding 1 makes it even again putting it back into the loop.
so no matter which no. you start with you will eventually reach a no. that is a power of 2.
unless the no. is a power of 2, in which case the answer will lead to the series(4,2,1) the no. will eventually become a power of 2 since only then will it stop. the stopping point is the series.
Wait, how do you get that every number will eventually become a power of 2? I agree any odd number will become even in one step.
+Christian McDaniel 10 is not a power of 2...
exactly
I don't see why this is a 'problem', as you don't fail at doing it.
You're just following a set of instructions.
According to Google (and personal experience), a problem is "an inquiry starting from given conditions to investigate or demonstrate a fact, result, or law." Surely you'll agree the Collatz Conjecture fits that description?
When adding, you follow the instructions of putting them together. So what's the difference here?
the problem is, why does every number end with 4, 2, 1?
there doesn't seem to be any pattern in which it should, yet it does every single time with any number you choose to start with, as if there should be a pattern in which it should end this way, so why does it??
that's the problem
No the problem is "does every number end at 1?". It is fairly obvious why numbers that do end at 1 do so, namely the pattern eventually ends on some power of two, which then reduces all the way to 1. Every number that has been tried has ended at one, but that is not a proof.
Ah, well if the problem is "Does every number end at one?", then I'll accept that as a valid problem.
There is no limit to the number of numbers you can try this with, so we will never know. That sounds like a genuine problem. Thanks for the input :)
In both results, integer/2 or integerX3+1, more possible outcomes that are even exist than that are odd. Because an even result requires a halving, eventually the result is going to degrade to 1. It's even number survivor bias and the result is a continual down counting.
Interesting observation!
an odd number when multiplied by 3 and added to 1 will be even as well.
I just came to the same conclusion.
My thoughts exactly. I don't see how it is a "problem". There's no solution other than if what you said is considered the solution.
Jordan Boyer The solution would be a mathematical proof, which is a demonstrable and repeatable verification using mathematical language.
The point is to reach a power of 2 which will then become 1. This means that only powers of 2 that are of the form 3x+1 where x is odd will work. That would be 4, 16, 64, 256, 1024, and so on. These are the powers of 4. So numbers that reach a power of 4 will work. The numbers that do this are 1, after which you've already finished, 5, 21, 85, 341, etc. This is a sequence with a common difference of powers of 4. These numbers are only reachable by dividing by 2, from 2, 10, 42, 170, 682, etc. Thus, only numbers that reach 4, 7, 20, 31, 84, 127, 340, 511, 1364, 2047, etc. will work, plus those from earlier lists: 2, 4, 5, 7, 10, 16, 20, 21, 31, 42, 64, 84, 85, 127, 170, 256, 340, 341, 511, 682, 1024, 1364, 2047, etc. Keep increasing the size of this list.
If you expected me to prove it here, I'm sorry.
There's no problem, just a waste of time
agree
I have a simple and elegant proof for this problem, but I can't fit it in this comment section
wtf fermat
check out my reply to Damon Harris Brennan. I also have an answer, and it's long.
Sammy I
I fucking hate you for doing this but i also know Its a Joke ... And Its funny , now I dont know what to do
Schau dir das da nochmal an Hannah
If came to look for a solution in the comments, I am sorry to disappoint you.
Yeah he just said this problem has puzzled mathematicians since the 1930s but I was thinking one of you shitheads might have cracked it
HeatyFrog
Number 15
Burger king foot lettuce
Will I be the first one to break the conjecture let's find out !
@@PandaFan2443 don't waste your time!
This is a shitty problem. Its more like "You dont have a formular." Everyone with brain knows how to solve it but it is not like a formular..
If it hits y=2^x than its shrinks to 4 2 1
And because infinity its 100% possible.. -> Solved :I
The reason it always ends back up at the sequence 4, 2, and 1 are multiple factors.
First, I will explain why the numbers will always end at 4, and then 2. If you notice, no matter what number you start with, you always end up with far more evens than there are odds. This is because of the algorithms that have been chosen. Every other number divided by 2 will give you another even or an odd. And if it’s even, you divide again, (remember that most numbers in these sequences are even), so therefore you will always go down, bit by bit, and all even numbers can be divided by the numbers in the sequence at the end: 4, and 2. And of course 2/2 will always give you the 1 in the sequence.
Now to further explain: when you get an odd, you must multiple that number by 3, (an odd number), and then add 1 to the whole. This will always leave you with an even number. (Again, the even numbers will always bring you back down eventually, and back to 4, 2, and then 1). If you take any odd number, (remember that the first number doesn’t matter, but the last, so therefore keep in mind there’s only actually 10 whole numbers you have to worry about), like 3 for instance, or even 23, or 43, you then multiply by 3. So, 3x3= 9, and now you add the 1 to get you to the even number. 9+1=10. Or, 23x3= 69+1= 70. Or, 43x3= 129+1= 130. Let’s try 7. 7x3= 21+1= 22. Or, 17x3= 51+1=52. Starting to see a pattern here? They always end up at the same number (referring to the last number on the right), and then always end up with an even.
And since you’re only multiplying every once in a while to get , and you’re dividing evens most of the time throughout the sequence, you will always end the sequence with 4, 2, and then 1. It is simple really. I figured this out by viewing the sequence a couple times and within maybe 30 minutes?
Also, a lot of times (if not every time), you will send back up with a number that does already end the sequence back down to 4, 2, 1. As you can see in the case of 11 and its sequence, and then 29 and its sequence which eventually leads to 11 and it's own sequence which leads to the infamous 4, 2, 1.
And if you look at all the sequences shown in this video, you can see a pattern, not with the sequences themselves persay (even though there somewhat is), but with the percentage of how many times the odds are compared to all the numbers in the sequence. If you count all the numbers including the 4, 2, 1, and either exclude or include the number you start at, odd shows up in a range of 25% - 37% of the time. Usually around 33% of the time.
And if you were to count all the odds and compare to all theevents, with or without counting the starting number and the 4, 2, 1, you'll end up with the odds only being 25% - 37% of theevents, again with 33% being the usual percentage.
Also, also, if you take the starting number and find how many times the designated number goes into it, a strange and certain pattern exists again. (Odds only).
For example, let’s look at the starting number 11. So, this is an odd number which will mean you have to multiply by 3 and add 1. If you find out how many times 3 goes into 11, (3.67 times), and you take those 3.67 times and divide it by the starting number 11, (3.67/11), you will get 33%. Basically giving an obvious glimpse at the approximate percentage the odds will show in the sequence.
Let’s try 29. 3 goes into 29, 9.67 times. 9.67/29= 33%.
Let’s do 8,199. 8,199/3= 2,733. This divided by 8,199= 33%.
I’m basically pointing out that this should be obvious with every odd number divided by 3, and then taking the answer and dividing the original odd number to get 33%. This problem is simple yet not, because the “patterns” and answers are right in front of our eyes, just hidden in plain sight.
And to add to this, if you do the evens only. The starting number is even, so we’ll do 10. You must divide by 2, which will give you the amount of times 2 goes into the starting number. 2 goes into 10, 5 times. And then 5/10 is 50%. Basically this is alluding to the fact that after every time you divide, every other time you will end up at an even or an odd. 50% of the time you will end up at and even. And like I said, after you do the 3n+1, 100% of the time you will end up at an even. So approximately 33% (from previous paragraph) of the numbers in any sequence from this conjecture will be an odd number (this isn’t a rule of thumb, but a better understanding), and the rest will be evens.
And since the majority of the numbers will end on an even, and the even you have to divide compared to only multiplying about 33% of the time, you will always end back up at 4, then 2, then 1. Like I said, 4 and 2 go into all even numbers (although 4 obviously don’t go into 2), and 2/2 will always equal 1. This is why all numbers you start with (all sequences) will end up back at 4, 2, and 1.
Oh, and it don’t matter what numbers you do these to. The starting numbers or any even number in the sequence or just period. You will always get that 50% chance that you’ll end up with an even or odd. 20/2=10 (Even). 10/2=5 (Odd). 12/2=6 (Even). 14/2=7 (Odd). 16/2=8 (Even). Etc, etc. 50%.
Same with doing that with odds. It doesn’t matter what number; starting or in the sequence or just period. You’ll always have an answer that’s even (after adding 1), and in the sequence with these equations of sorts, you’ll have about 33% of the sequence be odd and the rest be even. Again, this isn’t a set rule, but more of an answer that just gives you a rounded reasoning for the problem.
I don’t have too high of a vocabulary to better and fully explain this. Hopefully people can understand what I’m referring to and how I got the answers I did. I didn’t show the math too much, but if you do it yourself you’ll see. I can’t exactly show a picture of the math notes in my journal.
cool, but this is not rigorous enough to constitute a proof...
Unfortunately, while what you said constitutes ideas for a potential proof, it does not solve the problem at all. The real question is whether there exists any loop patern other than 4 -> 2 -> 1. Note that it does not have to be of size 3, but at least size 2 (although size 2 is impossible). There needs to only exist one to provide a counter-example to Collatz conjecture.
Why bother with a RUclips comment. If you’re right and have a proof, there is fame and fortune waiting for you.
@@codycast well I don’t know how to actually properly “proof” conjectures. But I would like to work with someone who does. I know for a fact I’m right, so I can really help someone who actually knows how to proof math equations and conjectures.
I looked up 1 video on how to do so and I immediately just went to sleep bc of how complicated it was 😅
@@alexandrezeddam7817 which one does not exist. In this conjecture the other digits do not matter whatsoever. The only thing that matters is the last digit (since this is based on odd or even). Odd and even numbers are actually just 1 digit, and will always end up to another ONE digit. The other numbers/digits do not matter at all.
That’s the whole point of my argument on this matter and why I postulate no one or no computer will ever find a different ending sequence.
I'm not sure what the big question is here. All it seems this problem is likely asking is how likely it is for this equation to get several even numbers in a row? I mean, any even number is divided by two. and anytime you get an odd number the 3n+1 ensures you get an even number 100% of the time. So then you just need several even numbers in a row to ensure that the division overtakes the multiplication right? I mean, as unscientific as it might sound, but shouldn't the law of large numbers kinda answer that? I mean there is literally an infinite amount of numbers, so either the law of large numbers ensures that you would eventually reach the pattern, like it should, or we accept that we just can't ever test it because it is literally infinite.
As a side note, I could be wrong here, but a pattern almost identical to this should technically also exist for 5n+3 right? It would likely exponentially increase the amount of times it would take to reach the pattern, but it should still work.
Edit: Actually I guess a simpler way to look at it would be, instead of trying to get several even numbers in a row, all you are really doing is trying to hit a power of 2. That's also part of the idea behind the 3n+1 and 5n+3. As long as even numbers are divided by 2 then you can create a repeating pattern with odd numbers. Basically find an equation for odd numbers, that when applied to the number one ends in a power of 2. You also need to make sure that both numbers in the equation are odd numbers(Such as the 3 and 1, or 5 and 3 in my example), so that way all odd numbers will end up even. And then once again in the end, the law of large numbers should ensure you eventually hit a power of 2.
Anyone else have thoughts on this, or did I miss something and look stupid? Or did I look stupid anyway lol
THe problem is that when N is odd, we will always get an even number afterwards by using 3n+1. Hence if we start the problem with an odd number, that will be the only odd number in the sequence. If we start by an even number, n might be an odd number as the 2nd number of the sequence but never odd again. Therefore, no matter how big the number is, everything will always keep getting divided by 2 as there will no longer be any odd number ever in the sequences. That is my theory at least :)
Kevin Nguyen Mmm, are you saying that no number divided by 2 can be odd again? What about 42,18, and 6, just to name a few. Sorry, but I think I just got confused on what you were saying.
That's what I was thinking.
huh. That is true. I only tested out a few and they didnt turn out to be odd for like a 2nd time within a sequence. Then again, I did think about it for like a minute :) it seems that I were wrong haha. Thank you man.
Get The Chillz
There is no positive number that doesn't reach 1, because there are infinite positive numbers and once the number hits a power of 2, it is able to hit 1. Also, the chance for higher numbers to reach a power of 2 declines by half each number in the 2 squared chain.
so the challenge is pick a number that will not end in 4... 2.... and 1?
+Zara auto that will only work as a counter example if one actually exists which is what the computers are currently trying to do. If no numbers exist that do not end in 4, 2, 1 then your challenge becomes impossible.
+Oliver Foggin why WOULD it ever NOT end 4 2 1? You perform a simple conditional check to decide which function to use, if it's even, you half, if it's odd, you increase the size of the number but make it even. You then feed that number back into the algortihm and repeat until it reaches the end cyclical pattern.
The ONLY variable in this algorithm is the starting number. If the starting number is not a power of two, and thus can't be collapsed in minimal steps, you perform arbitrary steps on it, until, eventually, it WILL hit a power of two number, and collapse down.
There is no problem or challenge to be found here, but it's hard to see a pattern because the algorithm is arbitrary. 3n+1 could literally ANY other equation that ensures an even number... such as n+1. Eventually this pattern will also always end cyclically (although as 2, 1... 2, 1 in this case.).
This thing is stupid and I don't even know why it is here nor why people can't wrap their heads around it. Eventually you hit a power of two, so the n/2 will collapse to 1, which will be the first odd number in sequence once n becomes a power of two, at which point you once again start from whatever results from the OTHER equation when n is 1, and then collapse back down. As long as you turn the odd n into any even number, and don't have ANY other variables in this process, this will ALWAYS result in an eventual repeating pattern. Can I get my nobel prize in mathematics now please?
WilliumBobCole that's what the entire video is about. What you have done is claimed to have solved the problem... that many many maths geniuses have not been able to solve with multiple super computers.
Yeah... right.
+WilliumBobCole But you're making the assumption that you *do* eventually hit a power of 2, which is essentially what you are trying to prove in the first place. Based on intuition, this may be true, although being the genius you are, you have excluded this portion of the proof, due to it being too simple. So no, you don't get a nobel prize quite yet, unless you include the proof of reaching a power of 2, so we inferior humans can understand.
*****
what I did was explain how and why the end sequence is the same. The video never explained what the problem WAS, so to me it seemed like they were saying that maybe there is a number somewhere that breaks the pattern...
but I explained that no, that will never be the case, and simplified the algorithm since the 3n is completely unnecessary and arbitrary. No problem was presented here FOR maths geniuses or super computers TO solve, just an infinitely large pool of numbers to run through this simple algorithm for no reason.
If there WAS a number that broke the pattern of reaching a cyclical loop, then there would be the problem of working out WHY, and are there MORE like it, but no such number CAN exist because we can prove that logically it will always become a power of two after a varying amount of steps...
so I don't understand your point, and again, the point for this video's existence... unless you are suggesting the problem is why some numbers require more steps than others before 'completion'?...
Try going reverse, from 1->2->4 and proceed using assumptions whether the number after 4 will be odd or even on each step
It will be 1-2-4-2-1
2:25 why people invented a problem? thanks for your problem, Collatz.
Suppose you're trying to solve a problem, but you only have so many proof techniques. It is completely possible that no combination of the proof techniques you have already will solve the problem; you might need to come up with hundreds of new techniques to solve it, which is too hard to do if you're only working on this problem. So one way of coming up with new proof techniques is making new problems that may be easier than the original problem. They might only require one or two new techniques to solve, which is more feasible. Eventually, you might get all of the techniques you need to solve the original problem.
Along the way, you might have created some problems that you still can't solve. These can also be used as milestones: as you keep doing more math, every once in a while you will solve one of those unsolved problems, which can sort of "keep track" of how far mathematics has come.
u put a random number and it took 420 times to get it to 0
I mean 1
shut up
Math suggests we smoke weed every dayy
+Gabe Cook 100% agree
Smoking qeed ebery dayy suggesrs a typo.
that number starting from 27 is really over 9000
ITS OVER 9000!!!
Once you reach a power of two, then the process is trivial as you keep halving to 1. Since 3n+1 will always produce an even number from an odd number you keep reducing the number of odd factors in a number. What’s left to prove is that indefinitely repeating f(y) = (y-1)/3 covers the whole natural numbers for y in 2**b, b in (1,2,3,...)
if x is odd then x * 3 is odd, this + 1 is even, divide by two and continue, this means that the number will always get smaller, leading to the cycle
F4T4L 405 holy shit lol I think you just solved it
F4T4L 405 That's not a valid explanation, you don't have prove that after dividing any even number by two, you don't just immediatly get an odd number again, then divide by 2, get an odd number again, etc. If a number would go like that it would continuesly become bigger, seeing as 3n+1increases the number by more than n/2 can decrease it.
Thats not a proof :v
But the number doesn't get smaller always, for example, here I ran a script which uses that logic, and with starting N 99 it becomes 298 at next step.
$ python solver.py 99
N is 99
N is 298
N is 149
N is 448
N is 224
N is 112
N is 56
N is 28
N is 14
N is 7
N is 22
N is 11
N is 34
N is 17
N is 52
N is 26
N is 13
N is 40
N is 20
N is 10
N is 5
N is 16
N is 8
N is 4
N is 2
N has reached 1
Yes...10 billion...that's a big cycle that we can't really solve the question by hand.
As someone said in comments, it won't work for 5n+1, so that makes it even more challenging.
What about, say 7n+1? Since 7 is another Mersenne Prime like 3, if this holds good for 7n+1, I guess it wouldn't remain as hard as it seems.
For the modified problems with 5n+1, 7n+1, 9n+1, etc, the conjecture is that there are at most a finite number of cycles but "most" starting points run away to infinity.
Tipping Point Math No, I mean if 7n+1 also reaches to 1 in the cycle.
The problem here is that after about 5 steps, it becomes too extensive for humans. Though if all numbers of form 2^x-1 follow the exact same pattern for mn+1, ie. leading to 1, then perhaps it can have a simpler way to solve.
no it wont work with 5n+1. but if you look closly. 1 --> 2 ---> 4 thats byte logic. if we use the same concept, it works.
0n + 1 = 1
1n + 1 = 1, 2
3n + 1 = 1, 2, 4
6n + 2 = 1, 2, 4, 8
12n + 4 = 1, 2, 4, 8, 16
(still missing the formula to explain how to get 0n and 1n, but they work)
well you didnt fail math
lol
My programming book calls this the Syracuse Sequence.
+ikirstenHD This name was given because someone gave a talk in Syracuse about this problem and many people there started working on it.
While in elementary with a calculator:
What do pimps say every morning?
“30,401,134”
*Turns the calculator upside down
Our reaction: 🤣😂🤣😂🤣😂🤣
Now I wonder, what does this problem look like in binary?
The even rule would simply shift everything to the right.
The odd rule would be multiplying by 11 (3 in binary) and adding 1.
Now what?
This is an interesting approach. Think of it as follows:
- If there are any zeros on the right, eliminate them (this is the division by 2 for even numbers)
- with no zeros on the right, add a zero on the right then add this to the unshifted number (this is multiplying by 3). Now add 1. This will produce at least one zero on the right. Remove all of those zeros and repeat this process.
I think mathematicians should try different approaches like this to help them find a pattern.
all you need to do now is show that the numbers always tend to get smaller over time. try using a string of 1's as s starting number.
I wrote a program to do this in java, printing the number in each iteration in binary and decimal. I noticed that the 3n + 1 step tends to slowly turn numbers into alternating 1s and 0s. Multiplying that by 3 turns it into a string of 1s, and adding 1 turns it all into a power of two, which then divides down to 1. I think the next question is, how does 3n + 1 achieve this?
+Tipping Point Math I just realized your graph look wrong. The powers of two should be incrementing with a slope of one, but your 8 value appears to be less than your 4 value.
When I saw that super computer my first thought was! This is one way to slow things down, next he tells me it's a conspiracy to slow things down LOL
Why not reverse the problem ? Considering the main tree that is made by the powers of 2 & the branches with their predecessors ... The question would be demonstrated if we provide a necessary condition to reach all integers comprized between 2^n & 2^(n+
1) .
They tried that as well, still no luck...
Reversing the problem is not logic ? Why ?
Instead of checking each number 1,2,3,4,... etc .. & apply the rule : divide by 2 when even or multiply by 3 & add 1 if odd , we can reverse the problem & consider the main tree of the powers of 2 which obviously leads to the final loop 4,2,1 for you always divide by 2 in this case . Then for each power of 2 , namely 2,4,8,16,32,... we check their predecessors . For example 16 has two predecessors : 32 & 5 , we have multiplied by two & substracted 1 & divided by three to reach there , each predecessor has a maximum of two predecessors & we keep on going like this , reaching number after number . The theorem would be demonstrated if we provide a necessary condition to fill all gaps between two consecutive powers of 2 . Is this clear enough ?
Take the normal way : 5 is odd so it gives 3x5 +1 =16 ; 32 is even so it gives 32/2=16 . 5 & 32 are the two predecessors ( or previous values ) leading to 16 . Taken the reversed way : 16x2=32 , (16-1)/3=5 . We keep on going like this for each new value we get ; for ex. 5x2=10 ; (5-1)/3=(4/3) in this case the result is not an integer so we reject it . Any number "n" has one or two previous values depending on the cases . The idea with this method is to see if we can find a necessary condition to fill all the gaps between two given values , in this case 2^n & 2^(n+1) . Of course these are only some personal ideas to deal with the problem , i didn't pretend i had solved it . The advantage with reversing the problem is that instead of getting a single image for each value you can have two sometime ...
Bikko induction and proof by contradiction is valid
It’s simple. If we multiply an odd number by 3 and add 1, we will always get an even number. If we divide by 2, we have a chance of either getting an even or an odd number. This means that overall it is more likely that we will get an even number as an outcome or performing one of the two operations. Because we are decreasing in number when we divide by two, and we are dividing by 2 more than multiplying by 3, we will eventually reach a lower number.
We only need to look at the ratios of even to odd numbers in the 1’s place that can come as a result of specific 1’s place values to see this.
0 has a ratio of 1:1 (will either have a 0 or 5
1 has a ratio of 1:0 (4)
2 has a ratio of 1:1 (6 or 1)
3 has a ratio of 1:0 (0)
4 has a ratio of 1:1 (2 or 7)
5 has a a ratio of 1:0 (6)
6 has a ratio of 1:1 (8 or 3)
7 has a ratio of 1:0 (2)
8 has a ratio of 1:1 (4 or 9)
9 has a ratio of 1:0 (8)
That’s an overall ratio of 10:5. So a rough estimate could say that for ever 15 we multiply by, we divide by 20, or for every 3 we multiply by, we divide by 4. This, of course, is not exactly true if we have some of the crazy numbers, the ratios will be a bit different just by the nature of probability, but overall there should always be more even results, unless there is some weird starting number I am missing. If that is the case, tell me, please.
Also I realize this probably is not entirely a valid proof, but maybe it helps someone to get an exact proof with the right mathematical terms, or maybe it’s just worthless, idk.
I got 99 problems but this ain't it...
No seriously... Ahhhh homework!
:p
it would be perfect if you changed the last word to one instead of it.
Mathematics is all about patterrrrns.
Nope.
@@patricksalhany8787 why not? and if not then what?
Mathematics is about logic, reasonning, patterns,... not only about patterns.
OOOOOHHH so that's the problem
@@patricksalhany8787 yes which is why Mathematics has a major flaw we can't see, EPR Paradox tells us clearly so
0:09 There is one that looks like Aperture Science logo, the 7 1.
8 1 would look more like it.
god damn
Multiplication is repeated addition therefore 3x = x + x + x +1 the size of x can really be ignored because you are creating 3 instances of x which can be considered as 3 then adding 1 to create an even number. An even number will eventually reduce to multiples of 2. Divide 2 by 2 and get 1 and we have the pattern: 4 from 3 + 1, 2 from reducing multiples of 2, and 1 which is the result of 2/2. 4 - 2 - 1.