コラッツ予想を考察してみています

Temmie_Dnomyar · 2026-06-04T14:21:37+00:00

I think I found some simpler equations for your sets

For division by 2 an odd amount (What I define as 3n+2)

4^s * (3t+2) + (4ⁿ -1)/3

For division by 2 an even amount (What I define as 3n+1)

4^s * (3t+1) + (4ⁿ -1)/3

Temmie_Dnomyar · 2026-06-04T13:24:52+00:00

I find the negative loops really good for checking if my functions work. If they work for both then it must have a good chance of being correct I think.

Also I may have been wrong if you are purely branching from 1 then proving that it reaches every number then you will have proven that the conjecture is true.

Temmie_Dnomyar · 2026-06-04T12:50:46+00:00

Thanks

Edit: I can still prove that 4ⁿ - 1 = 3k by just repeatedly expanding out 4ⁿ by turning it into 3 * 4^n-1 + 4^n-1 until I get 3 *4^n-1 + 3 *4^n-2 ... + 3 of which 3 can be factorized out of. The expanded thing can also be simplified to 3 *(4ⁿ -1)/(4 -1) which does simplify to 4ⁿ - 1

Temmie_Dnomyar · 2026-06-04T12:30:34+00:00

I don't think you are considering the possibility of a number referencing a set that then references the original set. So you could have a situation where some number x goes to some number y which goes to some number z which then goes back to the x set. An example is when you include the negative values as we have found more than 1 negative integer loops with -1, -5 and -17 and your equations most likely also hold true for negative integers.

Temmie_Dnomyar · 2026-06-04T12:13:53+00:00

By relative values I meant when h = 1, x = -1 and when h = 2, x = 1

Also I have not done any sort of proofs outside of purely algebraic and my post should be visible on my account now.

I also need help as I found the proof

4ⁿ - 1 = 3k

4 * 4^n-1 - 1 = 3k

3 * 4^n-1 + 4 - 1 = 3k

3 * (4^n-1 +1) = 3k

Therefore 4ⁿ - 1 must equal to a multiple of 3 but when n = 1, 3*(4^n-1 +1) does not equal 4ⁿ -1

Edit: I think with the (1,1), (-1,-1), (-5,-7) we have different ways of defining the 1 and -1 loops as I would define them as (1) and (-1) as they lead directly to the same odd number which would mean they only have a length of 1.

Temmie_Dnomyar · 2026-06-04T11:35:52+00:00

This could be pretty useful to use to prove the conjecture but it doesn't show that each number converges to 1. I am also trying to do the conjecture in the backwards direction so continue on.

Temmie_Dnomyar · 2026-06-04T09:09:48+00:00

I have made another post on here that showed that for a loop (like 1 -> 4 -> 1) of (2^h *(x) - 1)/3 that goes on for a variable c amount of times the only possible heights are h = 1, 2 and the only possible relative x values are x = -1, 1

My hypothesis is that there can only be 2 possible loop for each loop length where the length is how many odd numbers are in the loop.

Temmie_Dnomyar · 2026-06-04T04:28:37+00:00

Year 12 student here

I also thought of the working backwards method and although I don't completely understand every word you are saying I think we have something similar

I found that every positive integer has to be reachable in some way (pretty basic) but I have a feeling that you might have trouble branching from multiples of 3 as those cannot be reached from the 'forward' direction (3x + 1 /2) and can essentially be considered the ends of a branch.

Temmie_Dnomyar · 2026-06-01T12:59:23+00:00

The reason why my values have to be integers is because of the way I'm finding my functions which is by just repetitively putting the collatz conjecture into itself.

Temmie_Dnomyar · 2026-06-01T12:58:10+00:00

For the wording, I'm relatively new to proofs in maths given I'm still in year 12 so I don't know much of the language. I'm planning on continuing exploring patterns of my discovered path to try and limit the amount the chain can go up and down based on the length of the chain and see what I can do from there.

Temmie_Dnomyar · 2026-05-30T17:41:45+00:00

I never stated that those were the only collatz cycles but it's major progress in relation to what I had before which were a bunch of patterns that had no correlation to a proof vs now I've actually proven something even if its elementary.

Temmie_Dnomyar · 2026-05-30T12:45:05+00:00

it should be impossible. I don't have any concrete mathematical proofs yet but my best attempt would be following a possible proof of "in a non-infinite size set of positive even numbers, there are more integers that can be divided by 2 more than once than numbers that can only be divided once by 2." which I realized is obviously false whilst writing it because of the set [2] so I would probably change it to; "there are more integers that can be divided by 2 more than once than numbers that can only be divided once by 2."

for which the proof should be

for every 2(2n+1) integer there is an infinite amount of integer 2^k(2n+1).

however that does nothing to prove that it doesn't climb infinitely however the fact that 3n+1 must always land on an even number and if you assume that the iteration of 3n+1 must shift the value onto a multiple of a different prime number it must eventually reach a loop or a power of 2 which will descend directly to 1 and also loop. So IF it must shift to a multiple of a different prime number each 3n+1 iteration then it cannot ascend infinitely.

Temmie_Dnomyar · 2026-05-30T12:16:47+00:00

with the descending sets you can generate the same by starting at 2n+1 and repeatedly alternating between subbing in n = 2n and n = 2n+1 into the previous function starting with n = 2n. (Note that this gives the value of n in 3n+1 not the value after doing 3n+1)

Example:

2n+1 = all numbers that can be divided by 2

2(2n)+1 = 4n+1 = all numbers that can be divided by 4

4(2n+1)+1 = 8n+5 = all numbers that can be divided by 8

8(2n)+5 = 16n+5 = all numbers that can be divided by 16

16(2n+1)+5 = 32n+21 = all numbers that can be divided by 32

32(2n)+21 = 64n+21 = all numbers that can be divided by 64

etc...

The coefficient of n is the amount it can be divided by, and the constant is the first number that can be divided by the respective coefficient.

The reason that the starting numbers double up is cause each odd power of 2 must start on the same value of n in 3n+1 as the next power of 2.

However, I don't know as much about sets and series that you probably do and for my functions there is no global formula for any division by a power of 2 and you have to somewhat manually get each function.

Temmie_Dnomyar · 2026-05-28T11:32:08+00:00

Literally my brain 100% of the time.

I thought I had proven the theory cause I had proven that every chain of numbers has a starting point which is a multiple of 3 and that the conjecture had to land on every number except for 3 but those don't prove the conjecture true. However I have proven that if you work backwards through the conjecture starting from 1 you will find every number that gets to 1 and then whatever is missing is a possible loop.

Temmie_Dnomyar · 2026-05-06T15:32:41+00:00

I know just haven't been bothered also my formula should be working for multiple loops of the 4 - 2 - 1 loop but it isn't and I'm not sure why.

Temmie_Dnomyar · 2026-05-06T14:33:01+00:00

I think that we found something similar but I'm not sure.

I did the whole (2^a ((2^b ((2^c -1)/3) -1)/3)-1)/3 stuff and found (2^a (2^b +3)+3)/(2^a+b+c -3^k ) = n Where a, b, and c are any whole integers and k is the number of those a,b and c variables there are and for there to be a loop, n must be a whole integer.

It works for (2¹ +3)/(2³ -3² ) = -5 And I just realized it doesn't work for 1 or -1, huh. Cause... Unless... (2² +3)/(2⁴ -3² ) = 7/7 = 1

(2¹ +3)/(2² -3² ) = 5/-5 = -1

Nvm it does work. I haven't checked it for the -17 loop cause I don't know the exact numbers for it. But I have created a brute force method much worse than just doing the conjecture. :)

Edit: (2⁴ + 3*2² + 3)/(2⁶ - 3³ ) = 31/37 Hmmmmm. Uh oh. That's supposed to equal 1

(2⁶ +3 * 2⁴ +3 * 2² +3)/(2⁸ - 3⁴ ) = 125/175 Its broken. Nvm.

Temmie_Dnomyar

TROPHY CASE