The 3x+1 Problem: Status and Recent Work Part 1
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- Опубликовано: 14 апр 2013
- Friday, April 12th
Marc Chamberland , Grinnell College
"The 3x+1 Problem: Status and Recent Work"
Time: 4:00 PM
Location: Hill 705
Abstract: The 3x+1 Problem is a long-standing conjecture. Let T be a map from the positive integers into itself, where T(x)=x/2 if x is even and T(x) = (3x+1)/2 if x is odd. The conjecture asks whether, under iteration of the map T, any positive integer eventually reaches the value one. This talk gives a survey of the various approaches and results, intersecting areas such as number theory, dynamical systems, and functional equations.
When math kooks can't get published they can try the next best publishing journal: RUclips comments section. Forgetting their math, the logic behind what most people in the comment say is so stupid. I'm no mathematician but the amount trying solve this thing on RUclips is amazing. I'm sure one day some kid will publish a real solution to a real hard problem on a place like RUclips just for kicks but not today boys.
This is an amazing comment.🐢 Perfect.
8:25
"It is a proof [if you forget about infinity]."
-Dorron Zeilberger everybody!
How useful are null results if they can be obtained?
If a number, n, is found to fall to the ground then are there any facts that can be stated about all numbers < n?
3n+1 reminds me of the Mersenne Primes, a function that bumps the system over by one unit and gets an amazing result.
You've got another subscriber
they should all have a finite stopping time. Maybe we just cannot compute that much yet.
So. Fundamentally, is the question and proof, "why does ever decreasing numbers occur, and not ever increasing numbers??"
28:30 love the generalization. But where on earth does the second line come from =x+1/4... ?
It just a rearrangement using the identities:
sin^2(x)+cos^2(x) = 1
cos^2(x) = (1/2)(cos(2x)+1)
Great - I managed to show it. Thank you.
thanks for posting :)
+Quadrazar i have a complete solution for the collatz solution that shows only one trivial cycle and no infinity mail mkontra@ hotmail.com i need help to publish
You will eventually reduce to the Collatz sequence (4,2,1) if you also use the following formula > (If EVEN n/2 if ODD 2n +2)
Which is easily proven!
Since we know 2n+2 is even, divide by 2 to get to n+1, which is even if n is odd, to get to (n+1)/2, making your map equivalent to
if even, n/2, and if odd, (n+1)/2
which is strictly decreasing for all n>1 because of the 1/2 factor. QED.
It's the 1 that created all that followed no matter which point in time you are and it is the 1 that all that followed will return.
Who's eating potato chips goddamn it 😂
isn't it sufficient to say that if collatz is untrue then there must be a sequence such that a_{n+1}=(3a_n+1)/2 is never odd for all n? I don't think i'm about to beat a load of professional mathematician. I don't know how to disprove the existence of such a sequence
err... always odd
It may not be odd, it maybe even and then drop down to odd. For ex:- 102 drops to 7, so there may be some number ...
Yes, and that sequence is a(n)=2^n. What you've just discovered is that the Collatz algorithm terminates if it reaches a power of two.
@@firstnamegklsodascb4277 And I have a fornula that proves that every integer eventualy gets to a power of two. My method also tells us how many steps are needed to resch 1 without iterating. Do you know any serious mathemTician that would review my work ? I am a trained Physicist. Not a mathematician. How could I publish my results If I do not know the proper jargon? Can I just submit a paer in the mathematical arxiv without being refereed? Would that be valid to make it certain thst nobody "steals" my idea ? I could I do ?
@@jrsleao good luck found you is a hope I have for you. I just roll back around once in a while and check my last thoughts.
Check this v1xr4 2105.0003. The proof consists of two parts. The first, shows that the Collatz conjecture is the equivalent of the statement that every positive integer can be presented as a certain equation. In the second part, we prove that for every initial positive integer, this equation can be found. To achieve this, we propose a procedure that can be iterated, and we prove that by doing this we arrive at this equation. We also prove that any initial positive integer can be presented in an infinite number of ways in the form of needed equation. Each such form represents the loop occurring when number 1 is reached. The analysis is conducted using binary representation of numbers.
At 7:00, talks about strongest partial result. Is he saying that it has been proven for most numbers that there is a finite stopping time? He doesn't say so in so many words. Any clarification of his point appreciated.
I think what is being pointed at is that for example we can say all numbers of specific classes are forced to end for example 4^c is reachable from (4^c-1)/3 always etc. so you can prove a certain amount about which numbers we know have to hit 1 eventually. 2^x*(4^n-1)/3 will also hit 1 as these numbers will go down to (4^n-1)/3
its kinda dumb to work on this ive tried to work it doesnt work out. its always a loop. 4 2 1
I can get nice prove and get knew science according to collatz conjucter it will be amazing
So, it is obvious that if any odd number is multiplied by 3, the result will also be odd, e.g., 3 x 3 = 9 and 3 x 9 = 27, etc. This is simply because the last digit in the multiplication which is odd, multiplied by an odd number, structurally cannot be but odd. It follows then, logically that the addition of 1 will make it even, i.e., it changes the last digit (or the only digit if a one digit number) to an even number. Then the division by 2 will always be possible because the number being divided, as per the above, is always going to be even. The problem then arises when dividing an even number which can result in two odd numbers of equal value such as in 10/2 = 5. If then there is a particular structure in the application of this process, it should be predictable.
Note an example of a structure…….in each column the black numbers are increments of 10 as would be expected. The red numbers in the columns are separated each by 30. The columns of black numbers each of which is the product of halving the even numbers are separated each by 5. The red numbers in rows are separated by 6. The black numbers in rows, the divided product of the numbers above are separated by 1. I believe that given this structure, it should be understood that the progression would have to always come back down to 1 and loop.
1 2 3 4 5 6 7 8 9 10
4 1 10 2 16 3 22 4 28 5
11 12 13 14 15 16 17 18 19 20
34 6 40 7 46 8 52 9 58 10
21 22 23 24 25 26 27 28 29 30
64 11 70 12 76 13 82 14 88 15
31 32 33 34 35 36 37 38 39 40
94 16 100 17 106 18 112 19 118 20
41 42 43 44 45 46 47 48 49 50
124 21 130 22 136 23 142 24 148 25
51 52 53 54 55 56 57 58 59 60
154 26 160 27 166 28 17 29 178 30
31 32 33 34 35 36 37 38 39 40
94 16 100 17 106 18 112 19 118 20
41 42 43 44 45 46 47 48 49 50
124 21 130 22 136 23 142 24 148 25
Repeating 9s is the closest we have to breaking this. AS multiplying LARGE #s by 3 +1, ends up netting more than dividing by 3. 999999s are uniform and go to X plus 9s plus 8 at the end, then divide by 2, 9998 divided by 2 is 4999. 4999X 3 plus 14,998. 7,499 x3+1. is 22498. 22498. Etc. It starts breaking down when the 9s start flowing backwards. But with enough 9s would it loop? I keep having it fluctuate. As long as the 9s keep coming back, its actually kinda cool. But yeah. Eventually it seems like, even with a rediculously LARGE number, it would loop back to 1.
9s looping is kinda the best example, it goes above and below its self, but it ALWAYS increases when the 9 is the last number, and a 9 is infront. It slowly works its way back down. From right to left. So even with 9s, yes after like billions of spaces and commas etc, it would take a VERY VERY large number of moves, but i am pretty certain it would loop back around.
And its because odd by odd is odd, +1 is even. Divide by 2 is now even or odd. If even keep dividing. If odd, make it even. If even divide, if odd, make it even. etc.
Probably not. So you would need to math.
@@eon2330 thanks for responding. very cool stuff. Though i have no illusions about being able to solve this, it is fund to play with.
Yes. They iterate to 1.
Have you checked for googolplex+1?
if 1x2 - 2x2 - 4x-2 - 8x2 - 16x2 - 32x2 - is a infinite sequense that will always lead to 1 if divided - (32/2 - 16/2 - 8/2) then 3x+1 will do nothing to it - infinite is a really big number, so that sequence will always bring it to 1, unless infinite means something else than all the numbers.. 3x+1 is just a red herring that cant win against infinity
I wrote a Python program for this. If anyone wants it, here it is:
n=int(input(""))
while n!=1:
if n%2==0:
n=n/2
else:
n=3*n+1
print(n)
' by the even and by the odd"
if you can make sense it starts light, joy,feel,emote,make,sense,space,time,dark before light again
Someone eating crisps at a lecture??
Bloody good show
Let's see what AI comes up with...
Just map every Odd number to an Even number and visa versa which shows for all numbers except infinity that it will resolve to one. then the problem is about when the operation produces a smaller number. Just saying! Is it the number of operations it takes to get to one that is realy of interest?
+Andrew Meldrum To map all odd numbers to the even numbers just add 1 so if the result of dividing by 2 is odd it has an equivilent map number by adding one. That the operation 3x+1 is also mapped to the even numbers with 3(n+1)+1. This should reduce the problem into working with only one type of number within the same infinite set. Or I am just blowing wind!
+Andrew Meldrum thats a rather difficult approach still. It still takes an infinate amount of time to calculate the infinate number of series you would create. For example the next number that follows the same path that 27 takes is 576460752303423515. This is because it takes 59 steps to reduce that number bellow itself therefore it must be 2^59 higher than the first number in the series. Congratz you just eliminated the need to check (1/(2^59)) of all numbers.
+Wemdiculous by doing this you can create a geometric series 1/2+1/4+1/8 +....of all numbers. But theres always the possibility that one of those numbers will be reduced to itself instead of a number that is lower than itself :(
+Wemdiculous On further consideration but not complete. Could the problem be restated as: prove for all odd integers that 3n+1 will eventually reach a value of 2 to some integer power. Take the set of all n such that 3n+1 is a member of the set 2 to the power m. Manipulating this we find n=(2^m-1)/3 and for all even values of m have an integer result. (trivial binary number pattern as proof) Thus if all 3n+1 all sequences reach one of the values of this set the result of division by 2 ends in 1. --- consider a wall 3 bricks High with 1 brick put next to it. we can take one brick off the 3 brick post and put it on the extra brick and now have a 'wall' of 2by2 bricks, (case of n=1 in 3n+1) Now as n is always odd n=1+2r where r=0,1,2 and 3n=3+3.2.r ... Visually we are adding multipples of 6 bricks to our wall that is 3high and we can by comutativeity of multiplication turn the wall into a wall of bricks only 2 high. We can see that for the 'vanishing' bricks on division by 2 that for every other odd number, i.e multiples of 2 on 6 bricks or for all even values or r and divides by the number of twos in r. After the 'vanishing' we are left with either 1 brick or a 'foundation' of an odd number of bricks to rebuild our 3 brick high wall plus 1 brick. So we only need to look at potential growth numbers i.e odd values of r.
You can look at 3n+1 and the next odd number of n as n+2 where n=1,3,5,7 ... and interestingly you get 3(n+2)+1 as the result of 3n+1 on the next odd number up in the sequence. Looking at the difference between latter and the former i.e. 3(n+2)+1 -(3n+1)=6! Thus it can be said that the result for each successive odd number of n will be +6 of the previous odd number in the sequence. 3n+1 with n=1 is 4 and 3n+1 with n=3 is10 and so on. 2(2+3+3+3....) but 3 is 2+1 thus when the 'left over' 1's will increment by 1 to eventually reach a power of 2 such that the n of 3n+1 will be a terminating number for the theory roughly speaking. Hence there will always be an odd number that can terminate the sequence.
I don't understand what mathematicians don't understand about this problem. Maybe someone can help me with this. The way I see this problem is that it is set up intentionally to give you more divisions than multiplication. The rules are, you have to divide even numbers, and odd numbers you have to put in this formula x*3+1 or odd*3+1, which will ALWAYS give you an even number back. So this problem makes you either divide even number if you have it, or if you got the odd number you have to convert it to even using the formula odd*3+1=even and then divide it again. So it works like this. I randomly choose even or odd, regardless of what I choose I put it in this formula, and the formula will give me either odd or even, also random. But this is only the first step that creates this feeling of randomness, after that, I either get even by chance and divide it, or I don't get even and will have to convert the odd number I've got into the even number by using this formula and then divide it again. Of course sooner or later by doing all these divisions I will get down to 1 where I would start infinitely converting this 1 into an even number and then divide it back to 1 all over again. Not sure I understand what mathematicians don't understand about this problem. Can someone please help me and explain the deeper meaning that mathematicians are searching for?
The question is whether you'll always run up against a number of the form 2^n, or if it's possible to get back to your starting number without hitting 1
you have nothing to understand bro, you just prove the conjecture, congrats !
start with -1 try it
I have a complete solution for the problem no other cycle except 142 and no infinity run need help to publish
There is no prize man if you know how its done and youre 100% just make a youtube video called collatz conjecture proof
did you really solve it?
RIP all the fools who have been nerd sniped by this problem
I can give you n biger than 2^100000-1 and i can difind steps to 1
Numbers belonging to the group 4k+5 are all reduced by the step (3n+1)/4 and apparently you cant create a loop other than 4,2,1 with less than 245000 steps. Why wasnt this considered solved in 1984? Was it big brother? Srsly if it was i now need to know pls tell me.
Same question here.
4k+1 [= 4k+5] will almost always reach a lower number after 3 steps. (Exception k=0 -> n=1: 1,4,2,1)
So if you *start there* you will always fail because C^3(n) = m < n (and m should have been tested before n).
(16k+3 could be skipped too. 6 steps to reach a smaller number.)
I hope they used this when they ran these long tests.
------------------
I know that it is impossible that there are loops that are shorter than 34 steps.
How did you find the information about this:
>you cant create a loop other than 4,2,1 with less than 245000 steps
It's actually pretty easy no n-circuits exist, for any fixed n.
(However I don't claim a general method, at all.)
An element of an n-cycle must satisfy one of finitely many linear equations, which are easily solved, and well, you can just check if they spit out a natural number or not... Yeah... :/ It's not much...
I cannot see any way an algorithm that says Use only non-zero whole numbers. Divide that number by the smallest even number if it produces a whole number. If not multiply by the smallest odd number that will change the number, then add 1 to make it even. Repeat. Would yield anything other than 1, 4, 2, 1. At any stage. Like... its literally designed not to fail.
Maybe someone some day will find a number like, 987654329876543298765432987654329999999999999999999999999999 That just so happens to abuse the fact of the 9s doing that odd even thing over and over but also working some magic on the front numbers. To go infinitely.
Good luck everyone.
The
think I have solve this problem
first I realize that if there is only 1 cycle=> there is only 1 root number ( the number where it recycle in this case it is 1)
to proof that
let G(n) define for (3n+1)/2
G(N)/(2^(k1))=S (so that S is an odd number)
G(S)/(2^(k2))= N ( so that N is an odd number)
=> (3n+1)/2^(k1)=S
(3S+1)/2^(k2)=N
replace S by (3N+1)/2^(k1)
=>(3(3N+1)/2^k1+1)/2^K2=N
=> N =(3+2^k1)/(2^(k1+k2)-9)
since N is define as a positive integer
=> (3+2^k1) is divisible for 2^(k1+k2)-9)
also ( 3+2^k1) is greater or equal to (2^(K1+K2)-9)
let us take the difference
3+2^2^k1-2^(K1+K2)+9 >0
12+ 2^(k1)*(1-2^(k2)) >0
=> max of k2= 3, at that time, K1 can be 0,1
let us calculate N
K2=3, K1=0
N= -4 not right because N has to be bigger and N belong to Z)
K2=3, K1=1
N= 5/7 same reason
For K2=2, K1 could be 1,2,0
K2=2, K1= 3
N= 11/23 not accepted
K2= 2, K1=2
N=1 accepted
k2=2, k1=1
N= -5 not accepted
K2=2, k1=0
N=-4/5 Not accepted
for K2=1, K1 could be 3,2,1,0
for K2=1
K1=3 => N= 1/5 not accepted
K1=2=> N= 7/23 not accepted
K1=1 => N=5/7 not accepted
K1=0=> N=-4 not accepted
one might ask a question why negative number is not valid and the reason for a/b is not valid for b≠0, b≠1 is obvious
(3n+1)/2 is alway positive and divided by 2 is also positive
the collate conjecture is true
Yes, but this is only a solution for a two-way cycle - to prove this conjecture fully, you must prove it for all possible cycles
Lol
LOL and prove for no divergent numbers
You should call it by its proper name, the Collatz conjecture.
That's exactly what I thought. It's just the Collatz conjecture and the Collatz conjecture has been solved.
@@williejohnson5172 No it hasn't.
@@Kalumbatsch : Yes is has . Google
Proving the Sum of All Integers Equal 1
timestamp 36:45
Google
A Margin proof of the Collatz Conjecture
@@williejohnson5172 Sorry, that's nonsense.
@@Kalumbatsch : Sorry I disturbed your ovine grazing. Carry on
1:49 for the person making eating sounds: common man, this is the reason I quit this video.
It is not necessary to show that every sequence of numbers ends at 1. You "only" have to show that every initial number becomes smaller at some point through a sequence.
I would write all initial numbers in the form 2^m + n (e.g. 35 = 2^5 + 3). Also, you only have to consider the evolution of n. There are also only certain modulo values to check.
Suppose we assume m = 7 and check all numbers 2^7 + n.
Uncritical are values for which holds: n modulo 4 = 1.
These values become smaller than the initial value after three runs.
More exact consideration earns only:
n modulo 128 = 27
n modulo 128 = 31
n modulo 128 = 39
n modulo 128 = 47
n modulo 128 = 63
n modulo 128 = 71
n modulo 128 = 79
n modulo 128 = 91
n modulo 128 = 95
n modulo 128 = 103
n modulo 128 = 111
n modulo 128 = 123
n modulo 128 = 127
At 2^m = 128, 13 critical modulus values are found,.
Number of critical modulus values
2^m = 4: 3
2^m = 8: 2
2^m = 16: 3
2^m = 32: 4
2^m = 64: 8
2^m = 128: 13
2^m = 256: 19
Maybe these findings will help one or the other ;-)
Mathematicians hate this!
So I think the way this works is
Any odd is always turned into a even and an even has about a 1/2 chance of being a even number so on every number mow fast it goes down would be like 3/4 a step till you get to 1
That was easy
The problem is if it loops. There is a miniscule probability that after following these iterations you will get back to the number that you started with. The probability might be zero it might not be. And there are infinate numbers so if the probability isnt zero it must eventually happen
the chance is less than 0% because there is no prime number so there is no decimal and it steadily goes down.because it is more than likely that a odd number will be a even number than a even number be a odd number.
the equation goes down and there is no decimal so it is doomed to repeat.
If this argument is correct then why does replacing 3n+1 with 3n-1, or with 3n+5, lead to multiple loops?
solved
no
start with -1 try it . this is the solution
light is = to dark+1. Then it evolved to 3x=1 right now it should be 5x=1 because it should have evolved with space time make sense of 5 sense in space and
time
I think I found the solution.
I would consider this a waste of time.
I have got a legit proof. We have to prove that it is true for all odd numbers, since even numbers at some time will turn odd.
Let the number be n = 2^a1 + 2^a2 + 2^a3 ........................... 2^an +1 which does not stop at 1, for a1 > a2 > a3 > a4 ........ a(n-1) > an. Since it is odd, ( assuming 2^an is not equal to 2 ), we get that 3n+1 is divisible by 4. I means we can divide it by 2 2 times, and the resulting number is smaller than n, which has to end at 1, since it isn't the smallest contradictory number meaning that n is going to end at 1. If 2^an =2, then 3n+1 is of form 4k + 2 and hence, by dividing it by 2, we get numbers of either form 4k +1 and 4k+3.We again see, for it to be not possible, the number chain has to be going infinitely times to 4k+3. It means that for any contradictory number n, (3n + 1)/2 +1 = 4z. But, if n itself is of the form 4k+3, then some number of the form either 4x+1 or 4x+3 must be multiplied by 3 and added 1 to it to get n itself, meaning that number d would itself be contradictory number smaller than n, a contradiction. Hence, all number must be able to go to 1.
Rusty Fox is this a meme?
How is 3n+1 for odd n divisible by 4? If we replace n with 2k+1 (for k=0,1,2) so that n=2k+1 is odd, then 3n+1=6k+4. 6k+4 is only divisible by 4 if k is even.
Math should be more creative rather then creating more problems that can't be solved with less and less people in the world to help solve then. it just like the which one came first the chicken or the egg you just don't see reality correctly. you'll never evolve this way you'll only destroy everything you can see unless you learn to see it correctly. you need to make sense
start with -1 try it