Let:
- $x_i > 1$ be any odd integer that is not divisible by $3$
- $\nu_2(x)$ be the 2-adic valuation of $x$
- $x_{i-1} = \begin{cases} \text{undefined}, && \text{if }x_i\equiv 0 \pmod 3\\ \frac{4x_i - 1}{3}, && \text{if }x_i\equiv 1 \pmod 3\\ \frac{2x_i - 1}{3}, && \text{if }x_i\equiv 2 \pmod 3\\ \end{cases}$
When defined, $x_{i-1}$ is an odd integer such that $x_i$ results from applying the collatz operation for odd integers since:
$$x_i = \frac{3x_{i-1} + 1}{2^{\nu_2(3x_{i-1}+1)}}$$
Note: Details on the collatz operation for odd integers are found in this Wikipedia article.
Since each $x_{i-1}$ is either divisible by $3$ or it has a predecessor $x_{i-2}$, for any $x_i$, either there exists an integer $c$ where $x_{i-c} \equiv 0 \pmod 3$ or there is an infinite sequence where no such $c$ exists.
It seems to me that it should be straight forward to show that $c$ exists for any such $x_i$. When I try to prove this, I am finding that the argument is trickier than I expected.
Is there a straight forward argument to show that there always exists some integer $c$ so that the sequence is never infinite. Or is this problem more challenging than it would appear on first glance?
I checked the Wikipedia article on the Collatz conjecture in reverse but did not find any information related to my question.
Update: Added $x_i > 1$ which was implied since for $x_i=1$, $\dfrac{4(1)-1}{3}=1$ and $x_i=-1$, $\dfrac{2(-1)-1}{3}=-1$.
Update 2: Restated the title to replace "Collatz Operation in reverse" with the "following operation" which is then defined in the question.
Gottfried Helms has pointed out that this statement was not accurate.