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 !