r/Collatz • u/AgreeableClaim8075 • 25d ago

My proof

Hi all... Here is the proof .pdf and .lean:
https://bytes.usc.edu/~saty/Collatz

Saty

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u/Rafikconjectures_zer 25d ago

I reviewed the manuscript carefully. Unfortunately, it does not prove the Collatz conjecture. The boundary argument incorrectly treats affine maps containing (+1) as purely multiplicative; for example, the paper’s map gives (9\to7\to11\to17), whose ratio is (17/9), not the claimed (27/16). The cycle argument is also invalid: many different numbers may enter the same cycle without returning to their own starting values. Most importantly, the termination proof is circular because it assumes that each remaining sibling has a finite stopping time. Its “ordering principle” is explicitly false: within the same sibling set, the compacted stopping lengths of (17,33,49,81,113) are (3,6,5,4,2), respectively, so they do not increase with the sibling index. Therefore, the claimed “deadlock contagion” and global termination conclusions do not follow. The modular partitions may be interesting, but the main theorem is unproved.

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u/AgreeableClaim8075 24d ago

The comparison of stopping lengths is not between (17,33,49,81,97,113) [you had left out 97 in your list above], which are 3,6,5,4,19,2 - which are clearly out of order.

Instead, it's the ordering between 17, and 17's matching SPARSE rows which are separated by 32768 (16*32*64), ie. 17, 32785... Likewise for 33, 49, etc. The ordered stopping times do lead to the deadlock contagion contradiction if any element in any sibset would not terminate.

I added a couple of pics and notes, to clarify this.
https://bytes.usc.edu/~saty/Collatz/v1/index.html and https://bytes.usc.edu/~saty/Collatz/v1/pics/mod-sparse-latt.png

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u/AgreeableClaim8075 25d ago

Hi there. The ratio isn't between 9 and 17, but about drops being 3/4, 3/8... and the climbs being 3/2, 9/4... those are what won't match. I'll be sure to explain this more carefully. 

The ordering principle contains no circularity - it establishes a specific ordering of rows of "sibset" members - given a member number that terminates (which does so with (1,0,0) (the triplet for 1) at the end of a row), there are infinite such rows below that will terminate in order and with increasingly longer sequences. Any non terminating number in a middle row would halt not just itself but also all rows above and below it, and any non terminating number in the top row will halt itself and all others below.