2

In December 2021, Robert Hills Nichols, Jr of Cumberland University claimed a proof of Collatz on arXiv. But I've not seen any reviews or comments on the validity of his claim, and it's certainly beyond me.

Is there any validity to his claim?

https://arxiv.org/abs/2112.07361

QED
  • 265
  • 1
  • 6
  • 2
    Took a look. I don't know about the rest, but the existence of an acyclic tree that covers the odds not divisible by $3$ doesn't directly imply to me that the Collatz conjecture is true, and I don't see where in the paper Nichols justifies this jump. – Rushabh Mehta Jan 30 '22 at 03:30
  • 3
    Beyond me as well, but a few observations give me doubts: (1) Terry Tao tried to tackle this and only got to "almost surely true" (https://terrytao.wordpress.com/2019/09/10/almost-all-collatz-orbits-attain-almost-bounded-values/) and (2) if he is confident he should submit to annals of mathematics or some other respected mathematical body -- not ArXiV (may be in the works) – Annika Jan 30 '22 at 03:34
  • 1
    You should be sceptical about the "almost proven" claim in wordpress as well. It is hard to predict the difficulty of a not yet solved problem, but my guess is that the Collatz conjecture is so difficult to be proven that Fermat's last theorem was child's play compared to it. – Peter Jan 30 '22 at 07:31
  • 1
    A preprint which made it to the General Mathematics [math.GM] class of arXiv rather than the supposed Number Theory [math.NT] subject area makes me suspect that the paper is flawed. However, I am not an expert on the Collatz Conjecture and cannot certainly comment on specific sections of Robert's "proof". That being said, I did not take a look at his preprint. – Jose Arnaldo Bebita Dris Jan 30 '22 at 09:37
  • 3
    @Bey It is common practice for mathematicians to post their work on arXiv, before submitting to journal. Indeed, In your link to the blog of Tao, it's clear that he also posted his preprint on arXiv. – Arctic Char Jan 30 '22 at 10:25
  • 1
    I gave my best shot at trying to understand Nichols' proof, but it is beyond me. Found it very hard to follow, not sure if I just don't understand it or the proof is making logical jumps. – Trevor Feb 08 '22 at 01:11
  • 1
    @RushabhMehta the existence of a single acyclic tree covering the odds not divisible by $3$ would indeed prove the conjecture since the multiples of $3$ are essentially the leaves of the tree and every even number leads to an odd. So such a tree would prove all the natural numbers to be connected. – it's a hire car baby Feb 18 '22 at 12:46

1 Answers1

9

Got a quick look and from what I see, there is a bunch of variable transformation and 2 main functions $p_n=p(2^{n+1}x+2^n-1)=x$ which gives the root $x$ of a number of the form $2^{n+1}x+2^n-1$ and some kind of 2-valuation $q_n=q(y)=\nu_2(y+1)+1$ used like this: $q(2^{n+1}x+2^n-1)=n+1$ which leads to the construction of two sequences: $v_n$ which are odd numbers and are the "bottom" of successive 1-cycles (the classical $a2^i-1$), and $u_n$ which are even and are the top of the successive 1-cycles (the classical $a3^i-1$)

e.g. from $v_0=7$, you get $u_1=26$, $v_1=13$, $u_2=20$, $v_2=5$,....

This re-arrangement gives another (Collatz) tree which has the same pitfalls as the classical (Collatz) tree: It fails to show the presence of all natural numbers in the tree (it uses the same argument as the classical tree), it fails to show there are no cycles (several trees), it fails to show there are no infinite sequences.

Collag3n
  • 2,556