I know it might feel bad reading all the comments saying that this has been done before, but you should feel good. This video is what math is all about: Using your own logic and capabilities to teach yourself (and others) about a topic and perhaps discover something new. Bonus points because you are doing it to pursue your own interest and because you clearly enjoy it. Great video and keep at it!
I heartily agree! I'm coding up a fun problem right now, and I'm sure I won't contribute anything of any substance, but it's fun, I'm learning, and it also has cool graphs! 😁 Happy Holidays!
I actually had the same idea in the past when trying to find a way to optimize the search for Collatz conjecture counterexamples. I never noticed the 12-pattern, though, as I didn't actually think to check how many steps it took, only which numbers to rule out. Good eye. Unfortunately, whether this specific pattern continues doesn't prove the conjecture. If the collatz conjecture is false, then at least one of these "how many steps does it take to get to a lower number?" counters will be ∞ / undefined. And since you're analyzing each counter from least to greatest, whether the pattern continues says nothing about whether some of these numbers are ∞/undefined. Since, ∞ comes after all numbers, and undefined doesn't show up anywhere on the number line. Sorry.
It's so much more fun and satisfying to discover things for yourself. And you never forget something you learn that way. Sometimes it's a blessing to not know that a problem has already been solved.
This has been tried before (as practically all amateur approaches to collatz). It is not too difficult to generate a formula for all the different types of numbers that decrease below themselves in n steps as a function of n. What is difficult is showing that the set of all these numbers spans (or fills) the set of natural numbers. I can give such a formula for some simple cases inductively. Consider a number n=(4a-1)/3, where n is a natural number. We can see this implies a congruent 1 mod 3. This gives every number with 3 steps until it is smaller than itself. We can do the same with all number of steps by simply specifying the order of the sequence (div 2, do 3n+1) and then reverse it through modular multiplicative inverse. Again, the difficulty is in showing these sequences cover all the natural numbers.
Give me mod 4 and mod 8 and i will give you the rest only the mod 4(-1) and mod 8(+1) , don't always go down(one of the two) for all else it comes down to those two and i have the maths to prove that(prove those two eventually go to one and i will give the other 5/8 ths, splitting the prize 50/50). So if there is a loop it will be through some of those numbers.(you can't have a loop that don't contain them). The only going up is mod 4(-1) 3 goes to 5, 7 to 11, x to x+2n and that could make a runoff effect that always goes up in some distant range, but it 'should' get terminated(at 2^(some other n)).
For some interesting fun, consider expanding the domain of the problem into all integers: we quickly discover that there are other endpoints, and even loops: Some things eventually fall into the {-1, -2} loop, suggesting that negative numbers might just fall to -1... But then we then discover the {-5, -14, -7, -20, -10} loop that never falls to 1 It is trivial to see that no negative number ever 'flips' into the positive realm, and also vice versa Thus, the conjecture expands a little - there are more than two possibilities: A chain can go to 1 (and get into the {4, 2, 1} loop, or a chain could go off to infinity, or a chain could find itself caught in some other as yet unknown loop. Ultimately, a proof that every natural number eventually reaches a smaller natural number would, by induction, constitute a proof that all natural number chains go to 1, since no other loops of positive numbers exist within the 'base case' of the induction argument, so there are no other loops to fall into. If a loop exists somewhere in the higher numbers, then there would have to exist at least one natural number greater than 2 that never reaches a number smaller than itself. I love how tricky and weird this unsolved problem is while still being entire comprehensible and accessible. :)
at least one natural number >2 that never reaches a smaller number would prove either that a loop exists, or that a chain going off to infinity exists (in which case infinitely many such numbers exist). It's trivial to show that no loops of size two exist in the natural numbers, and a loop of size 3 (two downs, one up) seems to only exist as the familiar 4-2-1 loop. A 3-loop with two ups and one down is also trivially shown to be impossible, as it can be expressed as 3(3n+1)+1 = 2n -> 7n = -4, which clearly has no natural solutions.
This is really cool! This is one of the cases where yes someone has done it before, but that's almost everything in maths and that's not a good thing. I think this kind of amateur but well presented maths should be celebrated, even if someone did it in 2006. Treat maths like art!
@@peterboneg Oh wait no, I think the fact that everyone has done everything before in maths takes away some people's pride in rediscovering maths, and that I think is a bad thing
@@dirk-jantoot1029 I think it's less that and more that the collatz conjecture is 85 years old at this point and because Maths as a subject is so devoted to solutions, people get dismissive when you haven't engaged with 85 years of failed attempts already.
the piano part is new 😄. i have gone through this path as well. I've scripted and plotted graphs in python. but couldn't reach anywhere. Love to see similar minds. but to be honest, many professors and brilliant minds have worked on this. I'm certain that they must have considered these all. probably with in the first hour after hearing this problem. But then again i remember this quote all the time. " Sometimes it is the people no one imagines anything of who do the things that no one can imagine. ". So Keep working on it bro. I would really love it, if someone without phds solves it first. 😁
This quote: "Sometimes it is the people no one imagines anything of who do the things that no one can imagine." That quote is about me. Thanks for the quote! : ) MonsterPianoPlayer
whenever i find an integer sequence while playing with a math problem, i always scan it through the OEIS database! this one is code A102419, if you wanna go look at some of it. and don’t feel like your contributions to mathematics have to be solving impossibly difficult problems-i love math, and i know i won’t solve the Collatz conjecture. still, sometimes i play around with it and tons of other math problems too tough for me to solve just to scrape at how they work, like walking through a garden and thinking about the mycelial networks beneath and the xylem and phloem of the trees and the pollination symbiosis with the bees and letting it all accent the beauty of the flowers and world even though i’ll never be a true expert in ecology
The pattern repeats with powers of 2 works because the next N steps of the collatz conjecture are uniquely determined by the N least significant bits of a number, where the steps of the collatz conjecture are defined as: x/2 if x is even. and (3x+1)/2 if x is odd. Which makes sense because each of the above steps essentially removes the least significant bit of the number.
I did my own experiments using values up to 4.000.000. Cutting at every 12th- step, I got the following "piano"-patterns: [2, 3, 2, 3, 2, 3, 3, 2, 3, 2, 3, 3] [2, 3, 2, 3, 2, 3, 3, 2, 3, 2, 3, 3] [2, 3, 2, 3, 2, 3, 3, 2, 3, 2, 3, 3] [2, 3, 2, 3, 2, 3, 3, 2, 3, 2, 3, 3] [2, 3, 2, 3, 3, 2, 3, 2, 3, 2, 3, 3] [2, 3, 2, 3, 3, 2, 3, 2, 3, 2, 3, 3] [2, 3, 2, 3, 3, 2, 3, 2, 3, 2, 3, 3] [2, 3, 2, 3, 3, 2, 3, 2, 3, 2, 3, 3] [2, 3, 2, 3, 3, 2, 3, 2, 3, 3, 2, 5] [3, 2, 3, 3, 2, 3, 2, 3, 3, 12, 8, 21] (The vaules in the last two patterns are propably incorrect, since they still can change if we calculate more values) Note you get only 4 "piano"-patterns starting on the note f#. Then it changes to 4 "piano" -patterns starting on the note c#. Then it propably changes to a g# scale. So both times it went up 7 halfnotes (a perfect fith). Don't think this pattern continues forever thogh, as my mathematical instincts that the periodicity would be linked to log_2(3) being rational, which it is not. It'd be funny if the collatz-sereies generated a full cricle of fiths before the pattern breaks, though :D
The reason it looks like a piano is because in music the 3/2 ratio is also important. 19 is the difference (or sum of differences) over 12 steps. 2^19=524288 is close to 3^12=531441 but not exactly so it eventually diverges.
i understand that this is merely a coincidence that this pattern looks like a piano, but still, this is very very cool ! much cooler than many popular math coincidences. i am genuinely impressed with your ability to recognise a familiar pattern in such an unrelated context. this is the first sign of a great mathematician.
Interesting analysis. I also tried dealing with this and the Goldbach conjecture. My approach typically involves simple logic that translates the problem into another that is a corollary of the first. So in essence, you prove one, you prove another. Here are the rather simple ones I found myself after a few hours of thought (which I do believe others may have come up with too but I too found them without any inspiration): Collatz: Instead of proving every number goes to 1, you can prove every odd number goes to a number of type 4^k where k is a natural number. Proving one proves the other. Goldbach: Instead of proving that every even composite number is a sum of two primes, you can prove that every number greater than 1 is equidistant from at least one set of 2 primes. For the latter, the primes themselves have trivial solution with a distance of 0. For example, 3 is equidistant from 3 and 3 at a distant of 0. For 4, it is (3, 5) at distance of 1. For 10, it is (7, 13) at distance of 3.
There is a simple formula for the calculation of distances. Basically, only those numbers are interesting which increase at least after the first sequence step. One can write these numbers in the form 4x+3. For example, if we start with the number 135, the sequence results: 135 (2) 203 (2) 305 (2) 458 (2) 229 (2) 344 (2) 172 (2) 86 We multiply the denominators and get: 2 * 2 * 2 * 2 * 2 * 2 * 2 = 128 We subtract 128 from 135 and get 7 with the sequence: 7 (2) 11 (2) 17 (2) 26 (2) 13 (2) 20 (2) 10 (2) 5 2 * 2 * 2 * 2 * 2 * 2 * 2 = 128 The general formula: div(u) = div(u-2^n) = div(u+2^n) = 2^n with div(u-2^n) > 0 If we take u, we get an infinite number of distances with powers of 2: div(u) = div(u+i*2^n) In the interval [u,u+2^n] there can be further equal div(u), so that we finally get a sequence pattern with distances of 2^n, that repeats infinitly. Since there are infinitely many different div(u), this approach is not suitable to prove the Collatz conjecture.
This was a fun watch and keep up trying to do things like this, even if people already discovered this path, it just proves that it is correct and that even without outside knowledge it can be done.
It makes sense that 2^n would show up in multiple places when looking at this function. Because an entailment of the Collatz conjecture is that, if you the function for long enough, it will always hit a power of 2 in a finite amount of steps.
Interesting! Despite every sequence ending at 1, or it shooting off to infinity (which itself would need a proof if such a case would be found) there could also be the possibility of having a loop, like 1-4-2, but with a lot more and huge numbers.
We know that if such a loop exists, it's billions of numbers long, and we also know that all the numbers are more than 2^62 (since we know all values up to 2^62 follow the conjecture)
for it to shoot off to infinity is impossible though, all series dont really have a limit to the amount of steps so the probability of them reaching a power of 2 is >0 which means that there can't exist a series that reaches infinity, for that to happen a series needs to never reach a power of 2. which is impossible in itself due to probability. If something is possible and is given enough time it will happen.
@@GodbornNoven That's not a proof, what if the rule was 3n + 2 instead... would it still have a >0 probability to land on a power of 2? How can you know a given starting number has a >0 probability of hitting a power of 2, just because powers of 2 exist, doesn't mean they will be landed on
@@veno2195 A cycle (if it were to exist) would actually be a lot longer. I found it to have a length of at least 2^68 + 68 numbers. Didn't get around publishing it yet though... I'm not in any math circles & I'm not confident enough in my math & writing skills. Maybe me making this comment will push me to make my findings public sooner rather than later (already found this back in Nov 2021)
As a next step, you can consider 'what does it mean that the pattern repeats every 32', and make the parallel that this means 'the last five bits, if the number is written in binary'. Then you might think how that related to the underlying 3k+1, k/2 mechanism, and think not of steps, but how many divisions by 2 you encounter. The patterns then become even more apparent...
I feel like the Collatz Conjecture would be fun for people, even those who don't like math, if it could be gamified in a fun way. Might be something I look into in the future.
I like the 2-adic extensions of the Collatz function. Basically since even numbers always become smaller, consider a function defined from positive odd integers to positive odd integers: f(x)=(3x+1)/2^a where a is the largest power of 2 that divide 3x+1. This can be restated as 3x+1 divided by the even part of 3x+1. Division by the even part of a number is known as the 2-adic norm ||•||_2. So, f(x) is easily extended as f(x)=(3x+1)||3x+1||_2 which can be applied to all 2-adic rationals.
That's exactly the approach I took when I tried to look for a way to make progress on this. If you only focus on odd numbers, you can write them in the form of 4n+k, with k = 1 or 3. 4k+1 numbers are trivial to handle, but 4k+3 cannot be handled in the general sense, you always will see numbers which do not conform to a any rule.
It breaks earlier, if n grows. Seems like every count of steps is possible. 1.999.975 ->86, 1.132.159->87, 1.001.375->88, 100.099.631->23, 100.000.415->24 Every count except 2. All even numbers one step and all odd numbers at least 3 steps, because (3n+1)/2 can't be lower than n. Every count except 2 and 4, because 3(((3n+1)/2)+1)/2 is always greater than n and (3n+1)/2/2 is already lower than n, so a count of 4 makes no sense.
I think if you replace the 2s and 3s with 1s and 0s it’s something like the base 2 decimal expansion if log(2)/log(3). Might not be exactly that, but there’s some fundamental relationship like that.
Yes. And there are numbers that take 138 steps (138 mod 31 = 14). If the pattern was true we would expect 137 steps which would be 53 steps up (x3) and 84 steps down (/2) in any order. But 3^53 in slightly more than 2^84 (by about 0.2%) so these steps are close but not enough to get below starting number.
Great video, this is exactly how I started my thought process as well and then translated it into algebra. So, the idea behind this is basically look at different modulo with base 2 until a smaller number is found. So obviously 0mod2 will decrease in 1 step. But then the question is what happens with 1mod2, well it depends and it depends because if you try to do the collatz on a generic odd number this is what happens: 2n + 1 (odd) -> 6n +4(even) -> 3n + 2 and 3n +2 can be either odd or even and depends if n is odd or even so what you can do is split up 1mod2 into 1mod4 and 3mod4. Lets look at 1mod4 and represent an arbitrary number of the set as 4n + 1 which is of course odd so 4n + 1 -> 12n +4 (even) -> 6n +2 (even) -> 3n +1 so it will be less than the original number (3n + 1 < 4n + 1)(except when n = 0 which explains the loop 1 -> 4 -> 2 -> 1). Which is fairly interesting (although unimportant) because it makes it simple to calculate what the first number less than the original will be. for example, 401 = 4(100) + 1 so the first smaller number will be 301. Anyways, so we know that even numbers go down and numbers in the set 1mod4 go down. The question is what happens with 3mod4? 4n +3 (odd) -> 12n +10 (even) ->6n +5 (odd) ->18n +16 (even) -> 9n +8 (depends on the parity of n). So we get another depends, and basically you will keep finding some patterns that will always go down and some will depend if n is odd or even. I wrote a program that calculates these patterns in a low amount of time by splitting any "depends", but with a space complexity that my computer can't handle past when the modulo is 2^32, as the number of modulos that need to be split grows rather quickly.(It grows less than linear, but still when dealing with numbers this large a normal computer can't handle it, and likely it doesn't matter since there almost certainly isn't a point where every modulo case is covered and Im satisfied calculating the first million+ patterns (11mod32 is considered separate from 23mod32)). But anyways, that is the reason why you see patterns repeating in the form 2^n, and if you were curious why it skipped some powers of 2, it will skip whenever there isn't a number of the form 3^n that goes in between. For example, there aren't any patterns mod8 because there isn't a power of 3 that goes between 4 and 8.
I did went onto the same road but not that far but i realised that if a number goes to infinity then it cannot periodically goes up->down because each number carries only a finite amount of an+b information.
Working with binary alot I the first thing I noticed was the 2^n trunk of the tree. I'm glad someone put in the work to take a look thanks man. It was definitely gave some good insight to why this may be the case. Maybe there's a good way to redefine the problem using boolean logic
Well, finally here is the clue… you can calculate the number of times that a trajectory has to go up respect de total steps and this percentage has to be over than 38,7%. Then you can study a specific subsequence of the trajectory mod(4) and realize that this is a Markovian chain for a specific probability space. Using the ergodic theorem for Markovian chains, you can show that with probability one the collatz trajectory goes to one. It doesn’t prove the conjecture, but it tells you that the probability to find one divergent or cyclic trajectory is cero. So I’m not going to try to find one.
If it eliminated the possibility of divergent or cyclic trajectories, then logically that only leaves convergent ones. If a collatz trajectory converged on some number, then that number would be self-looping, as it were, since your theory is assumed to have eliminated all cyclic trajectories (which I took to mean chains of two or more numbers forming a loop). If a number as an input gives the self-same output it'll come from one or other of the formulas. Using 0.5x would mean that 0.5x=x, making x=0, but now we have to reach 0 first to get stuck in the loop, and we can't reach it from 0.5x, as that requires us to already be at 0 in order to reach 0, so we'd have to reach it using 3x+1, so now we have 3x+1=0, so 3x=-1, so x=-(1/3), which isn't a valid input as the theory is confined to the integer space. So maybe our self-looping number actually uses 3x+1, so that 3x+1=x, which means that 2x+1=0, and 2x=-1, so x=-(1/2), which also isn't a valid input, and even if we decide to have extra fun and extend the theory to encompass rational numbers rather than integers we'd have the thorny issue of how to define if non-integers are even or odd for the purposes of deciding which formula was to be used on them. In other words the tl:dr is that if divergence and cyclic trajectories (other than 4-2-1) were definitely ruled out then that in itself would be enough to rule out anything but ending in a 4-2-1 loop in all cases.
I really like this. you came to a similar conclusion as myself through a different method I've been using which in itself seems new. I might have some additional insight into your proof. I'm in agreeance that it is true; so that kinda makes it impossible to prove but I think I have come a bit further in my conclusion of it being infinite beyond a reasonable doubt.
4:52 the numbers in column E represent every number k s.t. there exists a sequence S of length k consisting of operations n -> 3n + 1 and n -> n/2, s.t. n/2 < S(n) < n. if you were to use n -> 3n instead, this results in the same list.
AFAIK, the Collatz Conjecture has only three possible outcomes: 1. Every positive whole number reduces to a 4-2-1 loop Or... 2. Some numeral can reduce to an entirely different loop Or... 3. Some numeral spirals out to infinity. Seems you just eliminated #3 as a possibility Good show.
You know what's wild?! I had the same idea and starting noticing patterns with the numbers of steps just like you did Then I wrote a program to print out every number it calculates on the way to 1 When you do that, you'll notice some wild patterns!!
This explains why The Guardian of this realm seems to favour The Collatz' Conjecture, aswell as the timeless design of pianos. They are, from The Guardian's perspective, one and the same. I salute your discovery, brother!
I think it's really simple. 3n+1 can only apply once then the number is even. /2 can be applied as many times as needed to get to uneven. The probability of 3n+1 against /2 can easily be calculated and it turns out that /2 just outperforms 3n+1 so that you always end up with 1. I guess that 5n+1 would always lead to infinity but I have not tried it yet.
In short, you'll have to prove there's no odd number you can keep multiplying by 3 adding 1 and it'll never hit an even number, since even numbers will immediately get smaller to a number we've already proved and there's no other way to get smaller other than to hit an even number by definition, which also proves there's no loop other than {4, 2, 1} since a loop must hit an even number to get smaller and then bigger to back where it was, or vice-versa
If x is odd, 3^n is odd, (3^n)x is odd, (3^n-1)/2 is odd half of the time, odd + odd = even, therefore unless my math is flawed it must hit an even number eventually
if you start with 1 it will never get to a number lower than itself, thus the y value on that point should be infinity. if your conjecture is provable then it means that the next number that won't lower than itslef can't follow a simple pattern
if you generalize to 3x+k problem (which is just the 3x+1 problem but with rational numbers), you can tabulate all possible cycles of low lengths, and you can arrange the odd numbers of the cycle with some analogous cycles of the same length in the same problem to form a sort of discrete simplex, with layers depending on the spacing between the odd numbers
sometimes, a "miracle" happens: 3x+k problem has a cycle where all numbers are divisible by a factor of k, so a 3x+(k/d) problem gets a mysterious cycle from above!! The smallest miracle cycle is 3x+11 having {1, 14, 7, 32, 16, 8, 4, 2}, which is from 3x+55 having {5, 70, 35, 160, 80, 40, 20, 10}. however, 3x+55 also has {7, 76, 38, 19, 112, 56, 28, 14}; analogous in 3x+55, but can't appear in whole numbers of 3x+11! as these cycle have only two odd numbers each, the odd numbers in 3x+55 are arranged in a line (1D simplex): 5 7 11 19 35 (the eleven comes from a degenerate cycle that repeats itself halfway through.)
another funny one is the fact that 3x+61 has a cycle of length 107 starting at 235, but is a miracle from 3x+37313979917667420061!!!! it is divisible by a HUGE factor!
Thanks for this. I like what you have done here. Am I correct in stating that, an actual Proof would need to demonstrate the inner logic as to why ""any number reverts back to 4,2,1"" ??
I once worked on it too but I didn't tried to find out if it goes to infinity or not but rather if there is another cycle like 4-2-1and I actually found requirements a cycle has to have but I don't know how to find a way to proof If the requirements are filled another time.
The "piano" regularity is easy to explain: one needs two pair numbers to go lower than one odd. Once in a while, the rest of this difference interferes.
the pattern exists and it gives you some idea of how does stuff work, but what if it actually broke somewhere? we would prob need some other proof for that but once we prove 1 thing, the rest could get much easier kinda funny to see how such a simple problem became so stupidly impossible so quickly
I tried doing this a few months ago and started looking for a pattern. It seemed like every other odd number had one rule, and every other number of the set that remained had another rule, and if that continued I was going to try to figure out the pattern. I also figured I was probably treading old ground, so as soon as it got more complicated on the next step I stopped working on it.
I don't know if this has been discussed elsewhere, but when I wrote various numbers going through the conjecture rules in Google Sheets (first number in the first column, next step in the next column, and so on) and then looked at each column for each step, I noticed that they seemed to follow regular patterns. So maybe if it can be proved that each step has a corresponding cycle, or that one of them doesn't, or whatever, it might help advance the conjecture? I don't have a lot of hope for it, since this is just some dingus messing with spreadsheets on a whim instead of professional math-ing, but I didn't see it discussed in any of the materials I found.
1:06 thats not rephrasing it thats one implication (though not equivalent). The sequence could also not shoot to infinity by ending in another loop not including a power of 2.
This reminds me of someone who put out a failed proof where they used modular arithmetic with divisor 6 and induction. For five of the equivalence classes, the induction step was trivial. For one of the equivalence classes, no proof was in sight. One might think that the proof is 83% complete, but in reality, it's more like the proof is 0% complete.
is there a subreddit or forum for people attempting the collatz conjecture? might as well have the attempts easily accessible so people can build off them, cause when i tried it i started from scratch
The reason why 2^a appears is that if you have a number n which takes k steps to go below itself, with m steps multiplying by 3, then n+b2^(k-m) will likely take k steps (more likely if k is small), for every integer b>0. But it necessarily takes k or less steps. This could be sufficient to explain what you saw, as you only look at small enough such k. As for the piano pattern, it is linked to this : given an integer m>0, what is the least integer n such that 3^m < 2^n ? Here is a list for (m,n) : (1,2) (2,4) (3,5) (4,7) ... Add up n+m and you find your list. I let you see why it is not a surprising fact. That's why it repeats itself in such a way. I fear it may not repeat indefinitely because this inequality wont necessarily be sufficient at some point, for large k. Even if it holds (but it probably doesn't), I don't see how it could provide a proof of Collatz conjecture. The way you look at patterns in a list reminds me of when I first began studying math. I don't think this will lead anywhere, but if you enjoy it then do it. It's like looking for something in the sand with your hands, thinking maybe there is some diamond. But mathematicians have x rays glasses ^^. What you did is essentially getting some clues that probably dividing by 2 more (in proportion) than multipliying by 3 is what we should expect to go below the number. But the piano pattern is lovely ;)
You know, something interesting about the collatz problem is that if you reduce the problem to even -> n/2 and odd -> (n3+1)/2, you can entirely predict a numbers path for a number of steps m by looking at it mod 2^m
I know it hasn’t been proven but my best bet is that it will ALWAYS go to 1. After all is said and done, you really only have even numbers which give you two probabilities: 50% reduce by 1/2, 50% reduce by 3/4. Either way, no matter what you do, the number which results is going to be smaller than the even number you started with, and you can’t have an infinitely descending sequence of natural numbers without it terminating at 1.
I used the same approach as you. I even had proven the collatz as long as something was true for all numbers less than 100 but it failed for only one of them.
I don't see it stated this way, but the conjecture is equivalent to saying every series eventually produces a power of 2. THe powers of2 that you observed in the differences is intriguing.
Unfortunately, I think there is an error with the final implication that the pattern being finite proves the collatz conjecture. Even if the pattern held, it wouldn't prove the collatz conjecture. Assuming the 2,3 pattern holds, does not necessarily imply that every number reduces in size. It just means that if a number reduces in size, then it must do so according to the 2,3 pattern. There still might exist a number that blows up to infinity and doesn't apply to the pattern even if all other numbers reduce in size. Basically, you still need to prove that the pattern applies to all natural numbers and not just a specific subset of natural numbers. Still though, the 2,3 pattern is very interesting
I found a flaw in your logic: You try to derive a formula from a bunch of data and use the formula to prove the conjecture is true. But if the conjecture is false, there is no formula. So you assume the conjecture is true to find a formula which proves the conjecture is true.
I think he's rather using the formula to make the nature of the Collatz function clearer, which may make the conjecture easier to prove. That is why he called it a step towards proving it, not a proof. There's also the option of proving the formula does hold for all numbers (without assuming the conjecture is true) but someone mentioned that it doesn't work for 138 so I don't know about that.
@@vikraal6974 Exactly, this is what makes this problem very difficult to solve. No matter how many numbers you check, its not enough to prove its true.
Es un algoritmo sobre los números. Sólo hay que demostrar que en algún momento baja, para todos los números, pues los ciclos son bien conocidos. Haciéndolo para todos los números se demuestra el teorema
So basically what you were trying to show is that this function is strictly decreasing, and therefore theres no number which will diverge to infinity. Even if you somehow managed to prove this (which no one has yet), this is still not a complete proof, because there is still the possibility that the number forms a closed loop, and you also have to prove that too
Recently some dude put a statement that if number has a step where it become less than it was on the start, than whole sequence can be reduced to 1. But there were no solid and sound proof from him, but there were still kinda ok attempt
Probably it's along these lines: We've tested all numbers up to X and they call go to 1. If we can show that X+1 eventually goes to X or smaller, then it falls in the realm of numbers we know go to 1, so we can add this new number. This means that if we can prove for *any* number that it will always end up on a number less than the original number, then we have shown that the conjecture is true.
In simpler terms. If all numbers decrease then all numbers reduce to 1. It's a way of looking at the problem but doesn't really break a ton of ground towards actually solving it
The relation to music theory may be seemingly unexpected at first but think about that music scales and human pitch perception is exponential in nature and "dividing by 2" is basically the same as going down an octave.
What? I thought the collatz conjecture fell under non-archemedian spectral theory or something? so...there's like wayyy more to it than just thid right?
this is already known if you can speak french, see the video of EL JJ, he talk about modular arithmetic and about cycle of length 2^n but he also say “we can’t eliminate every number with this”
your pattern breaks - your pattern is pseudo-primes - the break is the order of "mensnem" primes 2^q- 1 'prime' with q prime of the form 2^p-1 with p prime.
@@makessenseright sorry no, it is big big number - 3^^^3 @bout :(, 2^X - 1 => 3*[2^X] - 3 + 1 == 3*[2^X] - 2 ==> 3*[2^(X-1)] -1 =>=>> 3^X -1 So 3^X becomes the longer point.
Here is a rigorous proof that for any number N that Collatz-Desends in SO odd steps and in SE even step such that SE/SO > ln3/ln2 then any number A (mod 2^SE) that is congruent to N must also Collatz-Desend in at most SE even and SO odd steps and in the same order: I claim that the procedure (ie. number and order of Collatz steps) must be same for N and A for the first SE + SO steps. Further, let A = N + l*2^SE w.l.g. at the M-th step with SEM even and SOM odd steps I claim the C(A, M) = C(N, M) + 3^SOM*l*2^(SE - SEM), where C(n,m) is the output of the Collatz procedure at the m-th step starting with the number n. I will prove this via induction on the step number. Base case: A = N + l*2^SE since the second term (l*2^SE) is even, it cannot effect the parody of the number. So, the first step in the Collatz procedure for A must be the same as the first step for N. If N is even, then the first step is even and the theorem is trivially true since dividing by two always decrease a natural number. If N is odd, then C(A, 1) = 3*(N + l*2^SE) +1 = 3*N+1+3*l*2^SE = C(N,1) + 3*l*2^SE. Since the first step is odd, SOM = 1 and SEM = 0; therefore, this matches the assumption for step 1. Inductive Step: Consider the M-th step in the procedure. By the inductive assumption, to this point the procedures have been the same. Let SEM be the number of even steps at the M-th step and let SOM be the number of odd steps at the M-th step. By the inductive assumption, C(A, M) = C(N, M) + 3^SOM*l*2^(SE - SEM). We know that SE is grater than SEM, since SE is the total number of even steps and SEM is only the number of even steps at step M. Therefore, SE - SEM > 0 which implies that 3^SOM*l*2^(SE - SEM) is even. Therefore, C(N, M) + 3^SOM*l*2^(SE - SEM) must have the same parody as C(N, M). Therefore, the M+1-th step must be the same for A and N. If C(N, M) is even, then C(A, M+1) = C(N, M)/2 + 3^SOM*l*2^(SE - SEM-1) = C(N, M+1) + 3^SOM*l*2^(SE - (SEM+1)). Since there would be SEM + 1 even steps and SOM odd steps at step M+1, this matches the assumption for step M+1. If C(N,M) is odd, then C(A, M+1) = 3*C(N,M) + 1 + 3^(SOM+1)*l*2^(SE - SEM) = C(N, M+1) + 3^(SOM+1)*l*2^(SE - SEM). Once again that matches the formula in the assumption for the M+1 step. Therefore, C(A, SE+SO) = C(N, SE+SO) + 3^SO*l*2^(SE - SE) = C(N, SE+SO) + 3^SO*l. Since N Collatz-Desends in SE + SO steps, C(N, SE+SO) < N. By the assumption in the theorem, SE/SO > ln3/ln2 which implies 2^SE > 3^SO. Therefore, 3^SO*l < l*2^SE. Adding those inequalities C(N, SE+SO) + 3^SO*l < N + l*2^SE. Finally, using the equalities above for each side C(A, SE+SO) < A. So A Collatz-Desends in at most SE + SO steps.
The assumption SE/SO > ln3/ln2 can be remove if you instead consider the number of steps to reach 1. If you take sequence of the Collatz procedure for the number N, and replace the odd function f(n) = 3*n+1 with g(n) = 3*n. The outcome must be less then or equal the Collatz output. Since all of the functions are monotonic, if there is at least one odd step evaluating g(n) < f(n) so at step m latter this must be smaller. For any addition odd steps, they only continue to decrease the output. Therefore, 3^SO/2^SE < C(N, SO + SE). Once N converges to 1, 3^SO/2^SE < 1 which implies SE/SO > ln3/ln2. Therefore, following the same proof you would find that A (mod 2^SE) congruent to N Collatz-Desends in number of steps that it takes N to reach 1.
What if proving 3kn+3^(k-1)+...3^0 always reaches a power of 2... sure that assumes division doesn't occur in even cases but that might not be so important. The conjecture always feels like inevitable power of 2 is the solution.
An infinite sequence of step sizes isn't enough. Not every infinite sequence of sets covers all natural numbers. Take for example the primes, prime squares, prime cubes and so on. Those are infinitely many infinite sequences with no overlap, and yet the numbers 6 and 10 are not contained in them.
Quote: »If this formula is correct, then it means, that every amount of steps will always be a finite number; which means the conjecture is true.« But you haven't proved whether your formula is correct or not. You have only found that it is true for all numbers below a certain limit (i.e., for all numbers you tested). It is possible that this pattern does not hold above this limit. In that case, your formula is wrong, and you can't conclude anything from it. The first exception to this pattern could occur for the very first number you didn't test, or it could occur for a number that requires millions of trillions of decimal places when you write it down. Mathematicians must accept that, unfortunately, identifying a pattern in a finite range of numbers proves nothing. That's what makes mathematics so difficult: you have to prove it for ALL numbers, even numbers that are so insanely large that no one can ever test them.
The Collatz Conjecture is not really a true equation or a true algorithm because there is an "if" within it... it is a "function" though, but that is NOT nessisarily an equation. Equations may have an "if" outside the equation, like: x+1=1 if and only if x=0. In that case the "if" is not within the equation but rather an external "requirement" of the equation. What this Collatz Conjecture function is saying (in pseudo code): x={if odd(x)=true then: x=x*3+1 else: x=x/2} So, it is closer to a system of equations rather than a pure equation or pure algorithm. Look at the Wikipedia page he shows, see the "if"s? The equation has two "elements" to it... that's the giveaway. So trying to resolve this for infinity is pointless, because that would say that ALL systems of equations resolve for ALL inputs, no matter how MANY other elements are within it... that is: an infinite number of equations, with an infinite number of elements, with an infinite number of results. We can make conclusions that there are certain functions that provide predictable results for an infinite series inputs, but once the factors are "multiples" of infinity (like what I stated above which is infinity^3) then you can no longer make any proof or assumption about any single infinity - like a function that has infinite elements within it. So it's pointless. Like trying to prove: 1=1 | if over an infinite number of evaluations. ...well, duh! We assume that 1=1 period... forever in the past, now, and forever in the future. So if you want to waste your time trying to resolve the Collatz Conjecture, may I suggest you first prove: not(1=1)=true | if "ever" over an infinite number of evaluations. Hmmm, the Madelbrot set is jumping out at me on this. I wonder if "order out of chaos" could come out of the Collatz Conjecture if plotted by color vs. resolution rate.
@@makessenseright Thanks! I am glad someone was pickin' up what I was puttin' down [smile]. By your family name I assume you might be of Russian heritage? My family name was shortend from Romanowsky to Romsky when my grandfather imigrated from Poland in 1939 - for obvious reasons. Before that, the name was changed from Romanov (they changed the 'v' to a 'w' and added 'sky' to make it sound more Polish) when my family fled Russia as the Czar was being executed. So far, here in the USA, nobody has tried to hunt down my family for 83 years... that's a record [laugh]. Ok, I am not too proud of my great, great, grandfather, but I might be an heir to the throne of Russia... so I have that going for me [laugh]. Anyway... we seem to have a great affinity for math. Great video!
The idea of calculating the steps for the first 9999 numbers and then checking if our current number ever goes bellow 9999 is not a new idea. Im computer programming its used all the time. It even has a name: dynamic programing. Im sure that when researchers checked the first 2^60 numbers on a super computer they used this technique.
Hi Igor! I too am thinking about this problem. Intuitively, it kinda makes sense. For example, if we use x+1, instead of 3x+1, it is very clear as to why it will always end in 2-1-2-1 cycle. I am trying to see it as, we have to make any number reach the magical line of 2-4-8-16-32.. This is the last road, our goal. With x+1, you push even numbers down, which halves their distance from the magical line, and add 1 to odd number, which will also reduce the distance to the bigger number on the magical line. So ultimately, we will always reach our goal. No issues What happens with 3x+1 is that. It increases the distance sometimes. Not consistent like x+1. Which is why it seems that there could be some unfortunate number which keeps getting their distance increased over time. I can't see the possibility of any loop. But if we can figure out a way to make sure that this distance will always eventually be following a reducing trend, I think that could be another step to proving this
theres already proof of both reducing trend and increasing trend increasing trend: all odd numbers increase by a factor of ~3/2 [(3x+1)/2 is always an integer and since 1/2 is negligible for big x) and the proportion of all odds that go down to 1 reaches 1/1 (proof is complex, basically says that first apply 3x+1 to the odd number, then every 2^nth number, the result of 3x+1 is divisible by 2^n(obvious for integers, not so obvious for collatz stuffz))
there could be a loop if the number also unfortunately decreases into a number that increases and decreases pingponging at least ~1B times (proven minimum is greater than that) until it reaches the first number
Even if this pattern was true (it isn't, it breaks when we get 138 instead of 137), I don't see how it could take us closer to the proof on the conjecture. Anyway I always welcome new ideas for other approaches to the conjecture, so keep trying and good luck.
I know it might feel bad reading all the comments saying that this has been done before, but you should feel good. This video is what math is all about: Using your own logic and capabilities to teach yourself (and others) about a topic and perhaps discover something new. Bonus points because you are doing it to pursue your own interest and because you clearly enjoy it. Great video and keep at it!
My rediscovery of the Pythagorean Triples was like that. I didn't know that was what they were called, so it wasn't something I could easily find.
I heartily agree! I'm coding up a fun problem right now, and I'm sure I won't contribute anything of any substance, but it's fun, I'm learning, and it also has cool graphs! 😁 Happy Holidays!
I actually had the same idea in the past when trying to find a way to optimize the search for Collatz conjecture counterexamples. I never noticed the 12-pattern, though, as I didn't actually think to check how many steps it took, only which numbers to rule out. Good eye.
Unfortunately, whether this specific pattern continues doesn't prove the conjecture. If the collatz conjecture is false, then at least one of these "how many steps does it take to get to a lower number?" counters will be ∞ / undefined. And since you're analyzing each counter from least to greatest, whether the pattern continues says nothing about whether some of these numbers are ∞/undefined. Since, ∞ comes after all numbers, and undefined doesn't show up anywhere on the number line. Sorry.
It's so much more fun and satisfying to discover things for yourself. And you never forget something you learn that way. Sometimes it's a blessing to not know that a problem has already been solved.
I figured out the difference between square numbers on my own and felt really cool I had discovered something like that
This has been tried before (as practically all amateur approaches to collatz). It is not too difficult to generate a formula for all the different types of numbers that decrease below themselves in n steps as a function of n. What is difficult is showing that the set of all these numbers spans (or fills) the set of natural numbers. I can give such a formula for some simple cases inductively. Consider a number n=(4a-1)/3, where n is a natural number. We can see this implies a congruent 1 mod 3. This gives every number with 3 steps until it is smaller than itself. We can do the same with all number of steps by simply specifying the order of the sequence (div 2, do 3n+1) and then reverse it through modular multiplicative inverse. Again, the difficulty is in showing these sequences cover all the natural numbers.
Give me mod 4 and mod 8 and i will give you the rest only the mod 4(-1) and mod 8(+1) , don't always go down(one of the two) for all else it comes down to those two and i have the maths to prove that(prove those two eventually go to one and i will give the other 5/8 ths, splitting the prize 50/50).
So if there is a loop it will be through some of those numbers.(you can't have a loop that don't contain them). The only going up is mod 4(-1) 3 goes to 5, 7 to 11, x to x+2n and that could make a runoff effect that always goes up in some distant range, but it 'should' get terminated(at 2^(some other n)).
For some interesting fun, consider expanding the domain of the problem into all integers: we quickly discover that there are other endpoints, and even loops:
Some things eventually fall into the {-1, -2} loop, suggesting that negative numbers might just fall to -1...
But then we then discover the {-5, -14, -7, -20, -10} loop that never falls to 1
It is trivial to see that no negative number ever 'flips' into the positive realm, and also vice versa
Thus, the conjecture expands a little - there are more than two possibilities: A chain can go to 1 (and get into the {4, 2, 1} loop, or a chain could go off to infinity, or a chain could find itself caught in some other as yet unknown loop.
Ultimately, a proof that every natural number eventually reaches a smaller natural number would, by induction, constitute a proof that all natural number chains go to 1, since no other loops of positive numbers exist within the 'base case' of the induction argument, so there are no other loops to fall into. If a loop exists somewhere in the higher numbers, then there would have to exist at least one natural number greater than 2 that never reaches a number smaller than itself.
I love how tricky and weird this unsolved problem is while still being entire comprehensible and accessible. :)
at least one natural number >2 that never reaches a smaller number would prove either that a loop exists, or that a chain going off to infinity exists (in which case infinitely many such numbers exist). It's trivial to show that no loops of size two exist in the natural numbers, and a loop of size 3 (two downs, one up) seems to only exist as the familiar 4-2-1 loop. A 3-loop with two ups and one down is also trivially shown to be impossible, as it can be expressed as 3(3n+1)+1 = 2n -> 7n = -4, which clearly has no natural solutions.
This is really cool! This is one of the cases where yes someone has done it before, but that's almost everything in maths and that's not a good thing. I think this kind of amateur but well presented maths should be celebrated, even if someone did it in 2006. Treat maths like art!
Do you mean 'and that's not a *bad* thing'?
@@peterboneg Yes hahaha
@@peterboneg Oh wait no, I think the fact that everyone has done everything before in maths takes away some people's pride in rediscovering maths, and that I think is a bad thing
@@evanev7 We have 8 billion people. You're just not likely to try something no one has ever tried.
@@dirk-jantoot1029 I think it's less that and more that the collatz conjecture is 85 years old at this point and because Maths as a subject is so devoted to solutions, people get dismissive when you haven't engaged with 85 years of failed attempts already.
the piano part is new 😄. i have gone through this path as well. I've scripted and plotted graphs in python. but couldn't reach anywhere. Love to see similar minds.
but to be honest, many professors and brilliant minds have worked on this. I'm certain that they must have considered these all. probably with in the first hour after hearing this problem.
But then again i remember this quote all the time. " Sometimes it is the people no one imagines anything of who do the things that no one can imagine. ". So Keep working on it bro. I would really love it, if someone without phds solves it first. 😁
Great problems in mathematics are solved by young mathematicians who are too naive to know that they are impossible.
This quote: "Sometimes it is the people no one imagines anything of who do the things that no one can imagine."
That quote is about me. Thanks for the quote! : )
MonsterPianoPlayer
whenever i find an integer sequence while playing with a math problem, i always scan it through the OEIS database! this one is code A102419, if you wanna go look at some of it. and don’t feel like your contributions to mathematics have to be solving impossibly difficult problems-i love math, and i know i won’t solve the Collatz conjecture. still, sometimes i play around with it and tons of other math problems too tough for me to solve just to scrape at how they work, like walking through a garden and thinking about the mycelial networks beneath and the xylem and phloem of the trees and the pollination symbiosis with the bees and letting it all accent the beauty of the flowers and world even though i’ll never be a true expert in ecology
also the piano key layout is TREMENDOUS it made my brain tickle that’s a really neat find
Beautifully said. My thoughts as well
The pattern repeats with powers of 2 works because the next N steps of the collatz conjecture are uniquely determined by the N least significant bits of a number, where the steps of the collatz conjecture are defined as:
x/2 if x is even.
and (3x+1)/2 if x is odd.
Which makes sense because each of the above steps essentially removes the least significant bit of the number.
I did my own experiments using values up to 4.000.000. Cutting at every 12th- step, I got the following "piano"-patterns:
[2, 3, 2, 3, 2, 3, 3, 2, 3, 2, 3, 3]
[2, 3, 2, 3, 2, 3, 3, 2, 3, 2, 3, 3]
[2, 3, 2, 3, 2, 3, 3, 2, 3, 2, 3, 3]
[2, 3, 2, 3, 2, 3, 3, 2, 3, 2, 3, 3]
[2, 3, 2, 3, 3, 2, 3, 2, 3, 2, 3, 3]
[2, 3, 2, 3, 3, 2, 3, 2, 3, 2, 3, 3]
[2, 3, 2, 3, 3, 2, 3, 2, 3, 2, 3, 3]
[2, 3, 2, 3, 3, 2, 3, 2, 3, 2, 3, 3]
[2, 3, 2, 3, 3, 2, 3, 2, 3, 3, 2, 5]
[3, 2, 3, 3, 2, 3, 2, 3, 3, 12, 8, 21]
(The vaules in the last two patterns are propably incorrect, since they still can change if we calculate more values)
Note you get only 4 "piano"-patterns starting on the note f#. Then it changes to 4 "piano" -patterns starting on the note c#. Then it propably changes to a g# scale. So both times it went up 7 halfnotes (a perfect fith). Don't think this pattern continues forever thogh, as my mathematical instincts that the periodicity would be linked to log_2(3) being rational, which it is not. It'd be funny if the collatz-sereies generated a full cricle of fiths before the pattern breaks, though :D
The reason it looks like a piano is because in music the 3/2 ratio is also important. 19 is the difference (or sum of differences) over 12 steps. 2^19=524288 is close to 3^12=531441 but not exactly so it eventually diverges.
Thank you, that really seems to explain it!
i understand that this is merely a coincidence that this pattern looks like a piano, but still, this is very very cool ! much cooler than many popular math coincidences. i am genuinely impressed with your ability to recognise a familiar pattern in such an unrelated context. this is the first sign of a great mathematician.
Interesting analysis. I also tried dealing with this and the Goldbach conjecture. My approach typically involves simple logic that translates the problem into another that is a corollary of the first. So in essence, you prove one, you prove another. Here are the rather simple ones I found myself after a few hours of thought (which I do believe others may have come up with too but I too found them without any inspiration):
Collatz: Instead of proving every number goes to 1, you can prove every odd number goes to a number of type 4^k where k is a natural number. Proving one proves the other.
Goldbach: Instead of proving that every even composite number is a sum of two primes, you can prove that every number greater than 1 is equidistant from at least one set of 2 primes. For the latter, the primes themselves have trivial solution with a distance of 0. For example, 3 is equidistant from 3 and 3 at a distant of 0. For 4, it is (3, 5) at distance of 1. For 10, it is (7, 13) at distance of 3.
There is a simple formula for the calculation of distances. Basically, only those numbers are interesting which increase at least after the first sequence step. One can write these numbers in the form 4x+3.
For example, if we start with the number 135, the sequence results:
135 (2) 203 (2) 305 (2) 458 (2) 229 (2) 344 (2) 172 (2) 86
We multiply the denominators and get:
2 * 2 * 2 * 2 * 2 * 2 * 2 = 128
We subtract 128 from 135 and get 7 with the sequence:
7 (2) 11 (2) 17 (2) 26 (2) 13 (2) 20 (2) 10 (2) 5
2 * 2 * 2 * 2 * 2 * 2 * 2 = 128
The general formula:
div(u) = div(u-2^n) = div(u+2^n) = 2^n
with div(u-2^n) > 0
If we take u, we get an infinite number of distances with powers of 2:
div(u) = div(u+i*2^n)
In the interval [u,u+2^n] there can be further equal div(u), so that we finally get a sequence pattern with distances of 2^n, that repeats infinitly.
Since there are infinitely many different div(u), this approach is not suitable to prove the Collatz conjecture.
Thank you for your comment, this really seems to work!
This was a fun watch and keep up trying to do things like this, even if people already discovered this path, it just proves that it is correct and that even without outside knowledge it can be done.
Thank you!
Fun fact: that repeating pattern of 12 is also the pattern of the lengths of months in a year (30 or shorter (Feb.) -> 2, 31 -> 3)
i never thought of connecting the days in the months with the piano layout, the fact that they match perfectly is so cool !!!
It makes sense that 2^n would show up in multiple places when looking at this function.
Because an entailment of the Collatz conjecture is that, if you the function for long enough, it will always hit a power of 2 in a finite amount of steps.
Nice job. There's no better feeling than pulling at a thread to see where it goes.
Interesting! Despite every sequence ending at 1, or it shooting off to infinity (which itself would need a proof if such a case would be found) there could also be the possibility of having a loop, like 1-4-2, but with a lot more and huge numbers.
We know that if such a loop exists, it's billions of numbers long, and we also know that all the numbers are more than 2^62 (since we know all values up to 2^62 follow the conjecture)
for it to shoot off to infinity is impossible though, all series dont really have a limit to the amount of steps so the probability of them reaching a power of 2 is >0 which means that there can't exist a series that reaches infinity, for that to happen a series needs to never reach a power of 2. which is impossible in itself due to probability. If something is possible and is given enough time it will happen.
@@GodbornNoven That's not a proof, what if the rule was 3n + 2 instead... would it still have a >0 probability to land on a power of 2? How can you know a given starting number has a >0 probability of hitting a power of 2, just because powers of 2 exist, doesn't mean they will be landed on
@@GodbornNoven By the same argument the sequence 4,6,8,10,... has probability 1 of eventually hitting a prime.
@@veno2195 A cycle (if it were to exist) would actually be a lot longer. I found it to have a length of at least 2^68 + 68 numbers. Didn't get around publishing it yet though... I'm not in any math circles & I'm not confident enough in my math & writing skills. Maybe me making this comment will push me to make my findings public sooner rather than later (already found this back in Nov 2021)
As a next step, you can consider 'what does it mean that the pattern repeats every 32', and make the parallel that this means 'the last five bits, if the number is written in binary'. Then you might think how that related to the underlying 3k+1, k/2 mechanism, and think not of steps, but how many divisions by 2 you encounter. The patterns then become even more apparent...
Do people have the patterns already understood? Where are people actually stuck, then?
I feel like the Collatz Conjecture would be fun for people, even those who don't like math, if it could be gamified in a fun way. Might be something I look into in the future.
I like the 2-adic extensions of the Collatz function. Basically since even numbers always become smaller, consider a function defined from positive odd integers to positive odd integers: f(x)=(3x+1)/2^a where a is the largest power of 2 that divide 3x+1. This can be restated as 3x+1 divided by the even part of 3x+1. Division by the even part of a number is known as the 2-adic norm ||•||_2. So, f(x) is easily extended as f(x)=(3x+1)||3x+1||_2 which can be applied to all 2-adic rationals.
I never noticed how the process relates to p-asic numbers. Thank you for that, it's prerty cool!
That's exactly the approach I took when I tried to look for a way to make progress on this. If you only focus on odd numbers, you can write them in the form of 4n+k, with k = 1 or 3. 4k+1 numbers are trivial to handle, but 4k+3 cannot be handled in the general sense, you always will see numbers which do not conform to a any rule.
The piano patten isn't cyclical. It breaks at 138. Still I really enjoyed the video.
It breaks earlier, if n grows. Seems like every count of steps is possible. 1.999.975 ->86, 1.132.159->87, 1.001.375->88, 100.099.631->23, 100.000.415->24
Every count except 2. All even numbers one step and all odd numbers at least 3 steps, because (3n+1)/2 can't be lower than n.
Every count except 2 and 4, because 3(((3n+1)/2)+1)/2 is always greater than n and (3n+1)/2/2 is already lower than n, so a count of 4 makes no sense.
I think if you replace the 2s and 3s with 1s and 0s it’s something like the base 2 decimal expansion if log(2)/log(3). Might not be exactly that, but there’s some fundamental relationship like that.
@@threinhardt66 1132159 takes 112, or did I calculate it wrong? And 100099631 takes 73 for me...
Missing a 6 on 345
Another 115
"piano pattern" stops working when we calculate count of steps for 57115. (200steps mod 31 = 14 unfortunetly )
Yes. And there are numbers that take 138 steps (138 mod 31 = 14). If the pattern was true we would expect 137 steps which would be 53 steps up (x3) and 84 steps down (/2) in any order. But 3^53 in slightly more than 2^84 (by about 0.2%) so these steps are close but not enough to get below starting number.
Great video, this is exactly how I started my thought process as well and then translated it into algebra. So, the idea behind this is basically look at different modulo with base 2 until a smaller number is found. So obviously 0mod2 will decrease in 1 step. But then the question is what happens with 1mod2, well it depends and it depends because if you try to do the collatz on a generic odd number this is what happens: 2n + 1 (odd) -> 6n +4(even) -> 3n + 2 and 3n +2 can be either odd or even and depends if n is odd or even so what you can do is split up 1mod2 into 1mod4 and 3mod4. Lets look at 1mod4 and represent an arbitrary number of the set as 4n + 1 which is of course odd so 4n + 1 -> 12n +4 (even) -> 6n +2 (even) -> 3n +1 so it will be less than the original number (3n + 1 < 4n + 1)(except when n = 0 which explains the loop 1 -> 4 -> 2 -> 1). Which is fairly interesting (although unimportant) because it makes it simple to calculate what the first number less than the original will be. for example, 401 = 4(100) + 1 so the first smaller number will be 301. Anyways, so we know that even numbers go down and numbers in the set 1mod4 go down. The question is what happens with 3mod4? 4n +3 (odd) -> 12n +10 (even) ->6n +5 (odd) ->18n +16 (even) -> 9n +8 (depends on the parity of n). So we get another depends, and basically you will keep finding some patterns that will always go down and some will depend if n is odd or even. I wrote a program that calculates these patterns in a low amount of time by splitting any "depends", but with a space complexity that my computer can't handle past when the modulo is 2^32, as the number of modulos that need to be split grows rather quickly.(It grows less than linear, but still when dealing with numbers this large a normal computer can't handle it, and likely it doesn't matter since there almost certainly isn't a point where every modulo case is covered and Im satisfied calculating the first million+ patterns (11mod32 is considered separate from 23mod32)). But anyways, that is the reason why you see patterns repeating in the form 2^n, and if you were curious why it skipped some powers of 2, it will skip whenever there isn't a number of the form 3^n that goes in between. For example, there aren't any patterns mod8 because there isn't a power of 3 that goes between 4 and 8.
I did went onto the same road but not that far but i realised that if a number goes to infinity then it cannot periodically goes up->down because each number carries only a finite amount of an+b information.
that can easily show that any number that can be written as 2^k n + (k-1) will decrease eventually, but that a good corollary
That was an awesome video, the piano comparison is really interesting
Working with binary alot I the first thing I noticed was the 2^n trunk of the tree. I'm glad someone put in the work to take a look thanks man. It was definitely gave some good insight to why this may be the case. Maybe there's a good way to redefine the problem using boolean logic
Excellent work. The search for patterns is all the beauty of mathematics.
Well, finally here is the clue… you can calculate the number of times that a trajectory has to go up respect de total steps and this percentage has to be over than 38,7%. Then you can study a specific subsequence of the trajectory mod(4) and realize that this is a Markovian chain for a specific probability space. Using the ergodic theorem for Markovian chains, you can show that with probability one the collatz trajectory goes to one. It doesn’t prove the conjecture, but it tells you that the probability to find one divergent or cyclic trajectory is cero. So I’m not going to try to find one.
If it eliminated the possibility of divergent or cyclic trajectories, then logically that only leaves convergent ones. If a collatz trajectory converged on some number, then that number would be self-looping, as it were, since your theory is assumed to have eliminated all cyclic trajectories (which I took to mean chains of two or more numbers forming a loop). If a number as an input gives the self-same output it'll come from one or other of the formulas. Using 0.5x would mean that 0.5x=x, making x=0, but now we have to reach 0 first to get stuck in the loop, and we can't reach it from 0.5x, as that requires us to already be at 0 in order to reach 0, so we'd have to reach it using 3x+1, so now we have 3x+1=0, so 3x=-1, so x=-(1/3), which isn't a valid input as the theory is confined to the integer space. So maybe our self-looping number actually uses 3x+1, so that 3x+1=x, which means that 2x+1=0, and 2x=-1, so x=-(1/2), which also isn't a valid input, and even if we decide to have extra fun and extend the theory to encompass rational numbers rather than integers we'd have the thorny issue of how to define if non-integers are even or odd for the purposes of deciding which formula was to be used on them.
In other words the tl:dr is that if divergence and cyclic trajectories (other than 4-2-1) were definitely ruled out then that in itself would be enough to rule out anything but ending in a 4-2-1 loop in all cases.
remember: there is no better discovery in math than rediscovery, for rediscovery must mean you took steps in the right direction
I really like this. you came to a similar conclusion as myself through a different method I've been using which in itself seems new. I might have some additional insight into your proof. I'm in agreeance that it is true; so that kinda makes it impossible to prove but I think I have come a bit further in my conclusion of it being infinite beyond a reasonable doubt.
4:52 the numbers in column E represent every number k s.t. there exists a sequence S of length k consisting of operations n -> 3n + 1 and n -> n/2, s.t. n/2 < S(n) < n.
if you were to use n -> 3n instead, this results in the same list.
AFAIK, the Collatz Conjecture has only three possible outcomes:
1. Every positive whole number reduces to a 4-2-1 loop
Or...
2. Some numeral can reduce to an entirely different loop
Or...
3. Some numeral spirals out to infinity.
Seems you just eliminated #3 as a possibility
Good show.
You know what's wild?!
I had the same idea and starting noticing patterns with the numbers of steps just like you did
Then I wrote a program to print out every number it calculates on the way to 1
When you do that, you'll notice some wild patterns!!
Would you share? :)
Interesting. It may be useful to look at this for the 3n-1 alternative problem (since there is the 5, 14, 7, 20, 10 loop).
*thinking*
negative numbers?
oh that's just like with positive numbers, except +1 therefore -1 for negative numbers
This explains why The Guardian of this realm seems to favour The Collatz' Conjecture, aswell as the timeless design of pianos. They are, from The Guardian's perspective, one and the same. I salute your discovery, brother!
That 2323233 sequence is awesome. From there you can find the closed form for a function that covers that sequence
115?
I literally got goosebumps when I saw piano. Math has a beauty!!!!
I think it's really simple. 3n+1 can only apply once then the number is even. /2 can be applied as many times as needed to get to uneven. The probability of 3n+1 against /2 can easily be calculated and it turns out that /2 just outperforms 3n+1 so that you always end up with 1. I guess that 5n+1 would always lead to infinity but I have not tried it yet.
In short, you'll have to prove there's no odd number you can keep multiplying by 3 adding 1 and it'll never hit an even number, since even numbers will immediately get smaller to a number we've already proved and there's no other way to get smaller other than to hit an even number by definition, which also proves there's no loop other than {4, 2, 1} since a loop must hit an even number to get smaller and then bigger to back where it was, or vice-versa
So, prove that there's an x for which (3^n)x + ((3^n)-1)/2 is odd for every n
If x is odd, 3^n is odd, (3^n)x is odd, (3^n-1)/2 is odd half of the time, odd + odd = even, therefore unless my math is flawed it must hit an even number eventually
if you start with 1 it will never get to a number lower than itself, thus the y value on that point should be infinity. if your conjecture is provable then it means that the next number that won't lower than itslef can't follow a simple pattern
You prove 1 separately 1->4->2->1. Proven that 1 returns to 1. Now you can start the pattern for 2+.
if you generalize to 3x+k problem (which is just the 3x+1 problem but with rational numbers), you can tabulate all possible cycles of low lengths, and you can arrange the odd numbers of the cycle with some analogous cycles of the same length in the same problem to form a sort of discrete simplex, with layers depending on the spacing between the odd numbers
sometimes, a "miracle" happens: 3x+k problem has a cycle where all numbers are divisible by a factor of k, so a 3x+(k/d) problem gets a mysterious cycle from above!!
The smallest miracle cycle is 3x+11 having {1, 14, 7, 32, 16, 8, 4, 2}, which is from 3x+55 having {5, 70, 35, 160, 80, 40, 20, 10}. however, 3x+55 also has {7, 76, 38, 19, 112, 56, 28, 14}; analogous in 3x+55, but can't appear in whole numbers of 3x+11!
as these cycle have only two odd numbers each, the odd numbers in 3x+55 are arranged in a line (1D simplex):
5 7 11 19 35
(the eleven comes from a degenerate cycle that repeats itself halfway through.)
the miracles, for obvious reasons , can't be arranged in a simplex; only the original undivided ones can
another funny one is the fact that 3x+61 has a cycle of length 107 starting at 235, but is a miracle from 3x+37313979917667420061!!!! it is divisible by a HUGE factor!
cool result man. i love finding this kind of pattern. 🎉
Very cool that you were able to get that far!
Thanks for this. I like what you have done here. Am I correct in stating that, an actual Proof would need to demonstrate the inner logic as to why ""any number reverts back to 4,2,1"" ??
I love this insight! It's a unique way of seeing how the numbers are working together to build this complicated, but simple pattern.
I once worked on it too but I didn't tried to find out if it goes to infinity or not but rather if there is another cycle like 4-2-1and I actually found requirements a cycle has to have but I don't know how to find a way to proof If the requirements are filled another time.
The "piano" regularity is easy to explain: one needs two pair numbers to go lower than one odd. Once in a while, the rest of this difference interferes.
the pattern exists and it gives you some idea of how does stuff work, but what if it actually broke somewhere? we would prob need some other proof for that
but once we prove 1 thing, the rest could get much easier
kinda funny to see how such a simple problem became so stupidly impossible so quickly
I tried doing this a few months ago and started looking for a pattern. It seemed like every other odd number had one rule, and every other number of the set that remained had another rule, and if that continued I was going to try to figure out the pattern. I also figured I was probably treading old ground, so as soon as it got more complicated on the next step I stopped working on it.
I don't know if this has been discussed elsewhere, but when I wrote various numbers going through the conjecture rules in Google Sheets (first number in the first column, next step in the next column, and so on) and then looked at each column for each step, I noticed that they seemed to follow regular patterns. So maybe if it can be proved that each step has a corresponding cycle, or that one of them doesn't, or whatever, it might help advance the conjecture?
I don't have a lot of hope for it, since this is just some dingus messing with spreadsheets on a whim instead of professional math-ing, but I didn't see it discussed in any of the materials I found.
It does not only mean that it does not increase without bounds, it also means that it does not go into a loop.
1:06 thats not rephrasing it thats one implication (though not equivalent). The sequence could also not shoot to infinity by ending in another loop not including a power of 2.
Great video. You explained it better than anyone else I've come across. You have a gift.
The correlation with piano keyboard is an interesting find! It seems that you only need to prove that it's true.
This reminds me of someone who put out a failed proof where they used modular arithmetic with divisor 6 and induction. For five of the equivalence classes, the induction step was trivial. For one of the equivalence classes, no proof was in sight. One might think that the proof is 83% complete, but in reality, it's more like the proof is 0% complete.
is there a subreddit or forum for people attempting the collatz conjecture? might as well have the attempts easily accessible so people can build off them, cause when i tried it i started from scratch
The reason why 2^a appears is that if you have a number n which takes k steps to go below itself, with m steps multiplying by 3, then n+b2^(k-m) will likely take k steps (more likely if k is small), for every integer b>0. But it necessarily takes k or less steps. This could be sufficient to explain what you saw, as you only look at small enough such k.
As for the piano pattern, it is linked to this : given an integer m>0, what is the least integer n such that 3^m < 2^n ?
Here is a list for (m,n) :
(1,2)
(2,4)
(3,5)
(4,7)
...
Add up n+m and you find your list. I let you see why it is not a surprising fact.
That's why it repeats itself in such a way. I fear it may not repeat indefinitely because this inequality wont necessarily be sufficient at some point, for large k. Even if it holds (but it probably doesn't), I don't see how it could provide a proof of Collatz conjecture.
The way you look at patterns in a list reminds me of when I first began studying math. I don't think this will lead anywhere, but if you enjoy it then do it. It's like looking for something in the sand with your hands, thinking maybe there is some diamond. But mathematicians have x rays glasses ^^.
What you did is essentially getting some clues that probably dividing by 2 more (in proportion) than multipliying by 3 is what we should expect to go below the number. But the piano pattern is lovely ;)
Thank you! Yes, I am a software engineer, not a mathematician at all. And yes, I do enjoy playing in sand :)
Шикарное видео!! Спасибо!! Продолжай!!
I’ve been working on this problem algebraicly so much sometimes I forget what the actual conjecture is
You know, something interesting about the collatz problem is that if you reduce the problem to even -> n/2 and odd -> (n3+1)/2, you can entirely predict a numbers path for a number of steps m by looking at it mod 2^m
I know it hasn’t been proven but my best bet is that it will ALWAYS go to 1. After all is said and done, you really only have even numbers which give you two probabilities: 50% reduce by 1/2, 50% reduce by 3/4. Either way, no matter what you do, the number which results is going to be smaller than the even number you started with, and you can’t have an infinitely descending sequence of natural numbers without it terminating at 1.
I used the same approach as you. I even had proven the collatz as long as something was true for all numbers less than 100 but it failed for only one of them.
Awesome video, nice work!
I don't see it stated this way, but the conjecture is equivalent to saying every series eventually produces a power of 2. THe powers of2 that you observed in the differences is intriguing.
Unfortunately, I think there is an error with the final implication that the pattern being finite proves the collatz conjecture. Even if the pattern held, it wouldn't prove the collatz conjecture.
Assuming the 2,3 pattern holds, does not necessarily imply that every number reduces in size. It just means that if a number reduces in size, then it must do so according to the 2,3 pattern.
There still might exist a number that blows up to infinity and doesn't apply to the pattern even if all other numbers reduce in size.
Basically, you still need to prove that the pattern applies to all natural numbers and not just a specific subset of natural numbers.
Still though, the 2,3 pattern is very interesting
I found a flaw in your logic:
You try to derive a formula from a bunch of data and use the formula to prove the conjecture is true. But if the conjecture is false, there is no formula. So you assume the conjecture is true to find a formula which proves the conjecture is true.
Assume the conjecture is true and patterns hold. Use the patterns to derive a formula.
Use formula to prove the conjecture.
Kind of circular.
I think he's rather using the formula to make the nature of the Collatz function clearer, which may make the conjecture easier to prove. That is why he called it a step towards proving it, not a proof. There's also the option of proving the formula does hold for all numbers (without assuming the conjecture is true) but someone mentioned that it doesn't work for 138 so I don't know about that.
@@vikraal6974 Exactly, this is what makes this problem very difficult to solve. No matter how many numbers you check, its not enough to prove its true.
He admitted that already. Watch the entire video
Es un algoritmo sobre los números. Sólo hay que demostrar que en algún momento baja, para todos los números, pues los ciclos son bien conocidos.
Haciéndolo para todos los números se demuestra el teorema
So basically what you were trying to show is that this function is strictly decreasing, and therefore theres no number which will diverge to infinity. Even if you somehow managed to prove this (which no one has yet), this is still not a complete proof, because there is still the possibility that the number forms a closed loop, and you also have to prove that too
Recently some dude put a statement that if number has a step where it become less than it was on the start, than whole sequence can be reduced to 1. But there were no solid and sound proof from him, but there were still kinda ok attempt
Probably it's along these lines: We've tested all numbers up to X and they call go to 1. If we can show that X+1 eventually goes to X or smaller, then it falls in the realm of numbers we know go to 1, so we can add this new number. This means that if we can prove for *any* number that it will always end up on a number less than the original number, then we have shown that the conjecture is true.
In simpler terms. If all numbers decrease then all numbers reduce to 1. It's a way of looking at the problem but doesn't really break a ton of ground towards actually solving it
The relation to music theory may be seemingly unexpected at first but think about that music scales and human pitch perception is exponential in nature and "dividing by 2" is basically the same as going down an octave.
What? I thought the collatz conjecture fell under non-archemedian spectral theory or something? so...there's like wayyy more to it than just thid right?
this is already known if you can speak french, see the video of EL JJ, he talk about modular arithmetic and about cycle of length 2^n but he also say “we can’t eliminate every number with this”
Thank you! After discovering the conjecture a bit deeper I learned that as well :)
It doesn't stop at one. It reaches an equilibrium, hypothetically.
Of course, if a series doesn't end at 1, does not imply that it increases without bound.
Good show for a first video.
Wow, well done! I teally think you made a significant step forward!
NICE JOB DUDE 👍
It's a logistics problem, so it will find a solution within logic and we got logic that's how we construct abstract thinking...
very very interesting
Interesting. Congrats
your pattern breaks - your pattern is pseudo-primes - the break is the order of "mensnem" primes 2^q- 1 'prime' with q prime of the form 2^p-1 with p prime.
Could you tell the sample number where the pattern breaks?
@@makessenseright sorry no, it is big big number - 3^^^3 @bout :(, 2^X - 1 => 3*[2^X] - 3 + 1 == 3*[2^X] - 2 ==> 3*[2^(X-1)] -1 =>=>> 3^X -1 So 3^X becomes the longer point.
Great video
Here is a rigorous proof that for any number N that Collatz-Desends in SO odd steps and in SE even step such that SE/SO > ln3/ln2 then any number A (mod 2^SE) that is congruent to N must also Collatz-Desend in at most SE even and SO odd steps and in the same order:
I claim that the procedure (ie. number and order of Collatz steps) must be same for N and A for the first SE + SO steps. Further, let A = N + l*2^SE w.l.g. at the M-th step with SEM even and SOM odd steps I claim the C(A, M) = C(N, M) + 3^SOM*l*2^(SE - SEM), where C(n,m) is the output of the Collatz procedure at the m-th step starting with the number n. I will prove this via induction on the step number.
Base case: A = N + l*2^SE since the second term (l*2^SE) is even, it cannot effect the parody of the number. So, the first step in the Collatz procedure for A must be the same as the first step for N. If N is even, then the first step is even and the theorem is trivially true since dividing by two always decrease a natural number. If N is odd, then C(A, 1) = 3*(N + l*2^SE) +1 = 3*N+1+3*l*2^SE = C(N,1) + 3*l*2^SE. Since the first step is odd, SOM = 1 and SEM = 0; therefore, this matches the assumption for step 1.
Inductive Step: Consider the M-th step in the procedure. By the inductive assumption, to this point the procedures have been the same. Let SEM be the number of even steps at the M-th step and let SOM be the number of odd steps at the M-th step. By the inductive assumption, C(A, M) = C(N, M) + 3^SOM*l*2^(SE - SEM). We know that SE is grater than SEM, since SE is the total number of even steps and SEM is only the number of even steps at step M. Therefore, SE - SEM > 0 which implies that 3^SOM*l*2^(SE - SEM) is even. Therefore, C(N, M) + 3^SOM*l*2^(SE - SEM) must have the same parody as C(N, M). Therefore, the M+1-th step must be the same for A and N. If C(N, M) is even, then C(A, M+1) = C(N, M)/2 + 3^SOM*l*2^(SE - SEM-1) = C(N, M+1) + 3^SOM*l*2^(SE - (SEM+1)). Since there would be SEM + 1 even steps and SOM odd steps at step M+1, this matches the assumption for step M+1. If C(N,M) is odd, then C(A, M+1) = 3*C(N,M) + 1 + 3^(SOM+1)*l*2^(SE - SEM) = C(N, M+1) + 3^(SOM+1)*l*2^(SE - SEM). Once again that matches the formula in the assumption for the M+1 step.
Therefore, C(A, SE+SO) = C(N, SE+SO) + 3^SO*l*2^(SE - SE) = C(N, SE+SO) + 3^SO*l. Since N Collatz-Desends in SE + SO steps, C(N, SE+SO) < N. By the assumption in the theorem, SE/SO > ln3/ln2 which implies 2^SE > 3^SO. Therefore, 3^SO*l < l*2^SE. Adding those inequalities C(N, SE+SO) + 3^SO*l < N + l*2^SE. Finally, using the equalities above for each side C(A, SE+SO) < A. So A Collatz-Desends in at most SE + SO steps.
The assumption SE/SO > ln3/ln2 can be remove if you instead consider the number of steps to reach 1. If you take sequence of the Collatz procedure for the number N, and replace the odd function f(n) = 3*n+1 with g(n) = 3*n. The outcome must be less then or equal the Collatz output. Since all of the functions are monotonic, if there is at least one odd step evaluating g(n) < f(n) so at step m latter this must be smaller. For any addition odd steps, they only continue to decrease the output. Therefore, 3^SO/2^SE < C(N, SO + SE). Once N converges to 1, 3^SO/2^SE < 1 which implies SE/SO > ln3/ln2. Therefore, following the same proof you would find that A (mod 2^SE) congruent to N Collatz-Desends in number of steps that it takes N to reach 1.
Remember, proving it doesn't go to infinity is not enough. You have to prove there is no loop other Than 4 2 1 4 2 1 also.
1:03 Couldn't it instead get stuck in some other loop rather than diverging to infinity?
Yes, good point!
What if proving 3kn+3^(k-1)+...3^0 always reaches a power of 2... sure that assumes division doesn't occur in even cases but that might not be so important. The conjecture always feels like inevitable power of 2 is the solution.
the terminating sequence could be other than 4->2->1
Amazing ✨
what is the piano keyboard doing there!?
1:00 no, that is not true ("In other words that should go off to infinite"). It can have another loop somewhere else
An infinite sequence of step sizes isn't enough. Not every infinite sequence of sets covers all natural numbers. Take for example the primes, prime squares, prime cubes and so on. Those are infinitely many infinite sequences with no overlap, and yet the numbers 6 and 10 are not contained in them.
Quote: »If this formula is correct, then it means, that every amount of steps will always be a finite number; which means the conjecture is true.« But you haven't proved whether your formula is correct or not. You have only found that it is true for all numbers below a certain limit (i.e., for all numbers you tested). It is possible that this pattern does not hold above this limit. In that case, your formula is wrong, and you can't conclude anything from it. The first exception to this pattern could occur for the very first number you didn't test, or it could occur for a number that requires millions of trillions of decimal places when you write it down. Mathematicians must accept that, unfortunately, identifying a pattern in a finite range of numbers proves nothing. That's what makes mathematics so difficult: you have to prove it for ALL numbers, even numbers that are so insanely large that no one can ever test them.
The Collatz Conjecture is not really a true equation or a true algorithm because there is an "if" within it... it is a "function" though, but that is NOT nessisarily an equation. Equations may have an "if" outside the equation, like: x+1=1 if and only if x=0.
In that case the "if" is not within the equation but rather an external "requirement" of the equation.
What this Collatz Conjecture function is saying (in pseudo code):
x={if odd(x)=true then: x=x*3+1 else: x=x/2}
So, it is closer to a system of equations rather than a pure equation or pure algorithm. Look at the Wikipedia page he shows, see the "if"s? The equation has two "elements" to it... that's the giveaway.
So trying to resolve this for infinity is pointless, because that would say that ALL systems of equations resolve for ALL inputs, no matter how MANY other elements are within it... that is: an infinite number of equations, with an infinite number of elements, with an infinite number of results.
We can make conclusions that there are certain functions that provide predictable results for an infinite series inputs, but once the factors are "multiples" of infinity (like what I stated above which is infinity^3) then you can no longer make any proof or assumption about any single infinity - like a function that has infinite elements within it. So it's pointless. Like trying to prove:
1=1 | if over an infinite number of evaluations.
...well, duh! We assume that 1=1 period... forever in the past, now, and forever in the future. So if you want to waste your time trying to resolve the Collatz Conjecture, may I suggest you first prove:
not(1=1)=true | if "ever" over an infinite number of evaluations.
Hmmm, the Madelbrot set is jumping out at me on this. I wonder if "order out of chaos" could come out of the Collatz Conjecture if plotted by color vs. resolution rate.
I think this is one of the most beautiful comments received so far. Thank you so much.
@@makessenseright Thanks! I am glad someone was pickin' up what I was puttin' down [smile]. By your family name I assume you might be of Russian heritage? My family name was shortend from Romanowsky to Romsky when my grandfather imigrated from Poland in 1939 - for obvious reasons. Before that, the name was changed from Romanov (they changed the 'v' to a 'w' and added 'sky' to make it sound more Polish) when my family fled Russia as the Czar was being executed. So far, here in the USA, nobody has tried to hunt down my family for 83 years... that's a record [laugh]. Ok, I am not too proud of my great, great, grandfather, but I might be an heir to the throne of Russia... so I have that going for me [laugh]. Anyway... we seem to have a great affinity for math. Great video!
2:53 This looks very similar to the logistic map
The idea of calculating the steps for the first 9999 numbers and then checking if our current number ever goes bellow 9999 is not a new idea. Im computer programming its used all the time. It even has a name: dynamic programing. Im sure that when researchers checked the first 2^60 numbers on a super computer they used this technique.
Mathematicians also generalized it to a principal called “strong induction”
Hi Igor! I too am thinking about this problem. Intuitively, it kinda makes sense. For example, if we use x+1, instead of 3x+1, it is very clear as to why it will always end in 2-1-2-1 cycle. I am trying to see it as, we have to make any number reach the magical line of 2-4-8-16-32.. This is the last road, our goal. With x+1, you push even numbers down, which halves their distance from the magical line, and add 1 to odd number, which will also reduce the distance to the bigger number on the magical line. So ultimately, we will always reach our goal. No issues
What happens with 3x+1 is that. It increases the distance sometimes. Not consistent like x+1. Which is why it seems that there could be some unfortunate number which keeps getting their distance increased over time. I can't see the possibility of any loop.
But if we can figure out a way to make sure that this distance will always eventually be following a reducing trend, I think that could be another step to proving this
theres already proof of both reducing trend and increasing trend
increasing trend: all odd numbers increase by a factor of ~3/2 [(3x+1)/2 is always an integer and since 1/2 is negligible for big x) and the proportion of all odds that go down to 1 reaches 1/1 (proof is complex, basically says that first apply 3x+1 to the odd number, then every 2^nth number, the result of 3x+1 is divisible by 2^n(obvious for integers, not so obvious for collatz stuffz))
there could be a loop if the number also unfortunately decreases into a number that increases and decreases pingponging at least ~1B times (proven minimum is greater than that) until it reaches the first number
@Creatotron how is this a contradiction exactly? + this chain is 1month old.
We can reform the conjucture from getting to 1 TO getting a power of 2.
Even if this pattern was true (it isn't, it breaks when we get 138 instead of 137), I don't see how it could take us closer to the proof on the conjecture.
Anyway I always welcome new ideas for other approaches to the conjecture, so keep trying and good luck.
What do you mean by "I calculated the steps of amounts"? I don't understand how you got that table.
It's just a set (no duplicate values) of random glide values for numbers that has been sorted and reassigned indices.
I was hoping to hear some music from the piano interpretation :)
That's a great idea actually... I might think about it for the next video :)