A possible way to prove non-cyclicity of eventual counterexamples of the Collatz conjecture?

I've been recreatively working on the Collatz conjecture for a few months now, and I think I may have found something that could potentially prove at least half of the conjecture, which is the non-existence of non-trivial cycles. $\textbf{If you want to tl;dr}$, just check the framed equations. The first one is my conjecture, and the second one is a corollary which shows that if the conjecture is correct with all the conditions and everything, it would contradict the existence of non-trivial cyclic patterns. $\textbf{This is supposed to lead to a proof by contradiction}$ and so far, it seems to work. Otherwise, you could what I've done to get to this conjecture idea (because I'm narrating it chronologically so you can sort of get my process). I've not seen any peer-reviewed proof of their inexistence, so I guess it's still an open problem by itself. The fact is, I really think this conjecture is manageable, I just think I don't have the level required to tackle this kind of thing. Anyway, first things first, I didn't use the usual $$a_0\in\mathbb N,~a_{n+1}=\left\{\begin{array}{cc}(3a_n+1)/2&a_n~\rm odd\\a_n/2&\rm otherwise\end{array}\right.$$ but a more dynamical subsequence, which I randomly called $(e_n)$, defined with $$e_0=\frac{a_0}{2^{\nu_2(a_0)}},~e_{n+1}=\frac{3e_n+1}{2^{\nu_2(3e_n+1)}}$$ where $\nu_2$ is the 2-adic valuation. This basically chops off all the even numbers and basically keeps the core dynamics of the sequences. First off, I had to prove by induction that $$\begin{array}{ccccc} e_{n+1}&=&3^n\left(3e_0+1+\sum\limits_{k=1}^n\frac1{3^k}\prod\limits_{\ell=0}^{k-1}2^{\nu_2(3e_\ell+1)}\right)\prod\limits_{k=0}^n\frac1{2^{\nu_2(3e_k+1)}}&n\ge1&(1)\\ &=&3^n\left(3e_0+\left(\sum\limits_{k=0}^n\frac1{3^k}\prod\limits_{\ell=k}^n\frac1{2^{\nu_2(3e_\ell+1)}}\right)\prod\limits_{k=0}^n{2^{\nu_2(3e_k+1)}}\right)\prod\limits_{k=0}^n\frac1{2^{\nu_2(3e_k+1)}}&n\ge0&(2) \end{array}$$ However, $\nu_2(3e_k+1)$ has a very chaotic behaviour for $k\in\mathbb N$, so I had to bound it in some way or another. First obvious bound is that $\nu_2(3e_k+1)\ge1$, since from how the sequence is defined, $3e_k+1$ is even. Hence, I deduced that $$e_{n+1}\prod_{k=0}^n2^{\nu_2(3e_k+1)}\le3^{n+1}e_0+\frac{3^n}{2^{n+1}}\left(\sum_{k=0}^n\left(\frac23\right)^k\right)\prod_{k=0}^n2^{\nu_2(3e_k+1)}$$ Since $\sum\limits_{k=0}^n\left(\frac23\right)^k<3$ for all $n\in\mathbb N$, I found out that $$e_{n+1}\prod_{k=0}^n2^{\nu_2(3e_k+1)}<3^{n+1}e_0+\frac{3^{n+1}}{2^{n+1}}\prod_{k=0}^n2^{\nu_2(3e_k+1)}\\ \iff\frac1{e_0}\left(e_{n+1}-\left(\frac32\right)^{n+1}\right)\prod_{k=0}^n2^{\nu_2(3e_k+1)}<3^{n+1}$$ Now, I need to use a bit of a trick here. I'll assume $e_0$ to be minimal. In fact, for all $(e_n)$ which doesn't get to the trivial sequence, it can be shown that there are infinitely many $k\in\mathbb N$ such that for all $n\ge k$, $e_k\le e_n$, so this trick can describe literally any counterexample of the Collatz conjecture. Therefore, we get $$\prod_{k=0}^n2^{\nu_2(3e_k+1)}<\frac{3^{n+1}}{1-\frac1{e_0}\left(\frac32\right)^{n+1}}$$ if and only if $n+1 < \log_{3/2}e_0$. Since we know that for all $e_0\le87\times2^{60}$, $(e_n)$ is not a counterexample, we have $$\prod_{k=0}^n2^{\nu_2(3e_k+1)}<\frac{3^{n+1}}{1-\frac1{87\times2^{60}}\left(\frac32\right)^{n+1}}$$ for all $n+1 < \log_{3/2}(87\times2^{60})\approx113.58\ldots$ Hence, we got that $$\sum_{k=0}^n\nu_2(3e_k+1)<(n+1)\log_23-\log_2\left(1-\frac1{87\times2^{60}}\left(\frac32\right)^{113}\right)$$ for $n\le112$. So, to sum it up, we just bounded $\sum\limits_{k=0}^n\nu_2(3e_k+1)$ is bounded from above by $(n+1)\log_23+c$ for some constant $c$. Yet, we can also derive that for all $n\le107$, $$\sum_{k=0}^n\nu_2(3e_k+1)<(n+1)\log_23$$ (NB : The $107$ is here because $\left\lfloor(n+1)\log_23\right\rfloor=\left\lfloor(n+1)\log_23-\log_2\left(1-\frac1{87\times2^{60}}\left(\frac32\right)^{n+1}\right)\right\rfloor$ for all natural $n\le107$). Anyway, basically, here is my conjecture :

If $(e_n)$ doesn't converge to 1 and that for all $n\in\mathbb N$ we have $e_0\le e_n$, then for all $n\in\mathbb N$, $$\begin{array}{|c|}\hline\sum\limits_{k=0}^n\nu_2(3e_k+1)<(n+1)\log_23\\\hline\end{array}$$ I even have some numerical evidence supporting it. With a little algorithm which basically computes, for any $e_0$, the sum $\sum\limits_{k=0}^n\nu_2(3e_k+1)$ and checks whether or not it's below $(n+1)\log_23$ for as long as for all $k\le n$, we have $e_0\le e_k$. Checked all odd $e_0$ from $3$ to $29\;322\;479$ and it worked, so I'm pretty confident with that ! Now, how is this even related to the non-existence of cyclic sequences ? Well, if we assume this conjecture and using formula $(2)$, we'd have for minimal $e_0$ and $n\ge1$ $$\begin{array}{|c|}\hline e_{n+1}\ge 3^{n+1}\left(e_0+1/3+2/9\right)\frac1{3^{n+1}}=e_0+5/9>e_0\\\hline\end{array}$$ But this means that we could only reach $e_0$ once, which is a contradiction to cyclicity if it works for all minimal $e_0$. So basically, if my upper bound turns out to be correct for all minimal $e_0$ and $n\ge0$ (or $n\ge1$ to be cautious but anyway), this would essentially imply that there is no non-trivial cycle ! I'm putting this here so people could eventually work out a way to prove it. Obviously tried by myself, but I figured out I might not be good enough for this !

$$\frac{3e_0+1}{2^{\nu_2(3e_0+1)}}=e_1$$ can be rewritten as $$(3+\frac{1}{e_0})=2^{\nu_2(3e_0+1)}\frac{e_1}{e_0}$$ Now you have

$(3+\frac{1}{e_0})=2^{\nu_2(3e_0+1)}\frac{e_1}{e_0}$

$(3+\frac{1}{e_1})=2^{\nu_2(3e_1+1)}\frac{e_2}{e_1}$

...

$(3+\frac{1}{e_n})=2^{\nu_2(3e_n+1)}\frac{e_{n+1}}{e_n}$

You multiply every LHS/RHS to get

$(3+\frac{1}{e_0})(3+\frac{1}{e_1})...(3+\frac{1}{e_n})=\frac{e_{n+1}}{e_0}\prod_{k=0}^n2^{\nu_2(3e_k+1)}$

From here you get

$$(3+\frac{1}{e_{max}})^{n+1}\leq\frac{e_{n+1}}{e_0}\prod_{k=0}^n2^{\nu_2(3e_k+1)}\leq (3+\frac{1}{e_{min}})^{n+1}$$

But it means that in a cylce where $e_{n+1}=e_0$ you have

$\prod_{k=0}^n2^{\nu_2(3e_k+1)}\gt 3^{n+1}$ or $\begin{array}{|c|}\hline\sum\limits_{k=0}^n\nu_2(3e_k+1)>(n+1)\log_23\\\hline\end{array}$

Unless I messed something in translating to your notations, it does not match what you get.

I am not sure if this helps: in our working paper we studied cycles in Collatz sequences for $3n+1$ and the generalized form $kn+1$. We empirically found out that cycles only occur, if the condition $\alpha=\lfloor n*log_2k\rfloor+1$ is met. That is close to the considerations above. The variable $\alpha$ is the number of divisions that is performed to get from the first odd number $v_1$ to the odd number $v_{n+1}$ that forms the cycle. The variable $n$ is the length of the cycle.

Example for $v_1=13$, $k=5$ and $n=3$:

• $v_{n+1} = 5^3 * 13 * (1 + \frac{1}{5 * 13}) * (1 + \frac{1}{5 * 33}) * (1 + \frac{1}{5 * 83}) * 2^{-7}$ = 13
• $\alpha = \lfloor 3*log_25\rfloor+1$ = 7

Our hypothesis holds for all known cycles. Maybe this information is usefull for your further analysis.

This is a matrix of n as a function of k = odd steps. The even numbers descend to an odd number (divided by 2) and the odd numbers jump to the left column (3n +1). It will be useful for visualizing because the cycle 1,4,2,1 is the only possible cycle. Otherwise,

$f\left ( n \right )= n$ , and this is only possible when n = 1, taking the odd n. This implies that the function takes a value from the form

$f_{0}^{k}\left ( n \right )= \frac{n\times 2^{x}}{2^{x}}$

and the even numbers of the form $n\times 2^{x}$ are the numbers where n comes from and therefore the function does not go through those numbers again. In the matrix, they are always above n and the function always moves downwards for even and to the left-up for the odd ones, looking for its corresponding pair number. It is easy to see that the function from n always leaves behind the numbers that would give rise to another cycle other than 4, 2.1. Mathematically, for now, I do not know how to express it, it’s as if the function had to do the reverse cycle for this to happen. P.D.: I do not know about your proof(correct or incorrect?) but I think it could be useful to prove the inexistence of another cycle than 1,2,4,1.