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Consider the function

$$f(n)=\begin{cases}n/2&\mbox{if }n\mbox{ is even}\\3n+1&\mbox{otherwise}\end{cases}.$$

Starting from any positive integer $x_0$, we can iterate the sequence $x_1=f(x_0)$, $x_2=f(x_1)$, and so on, with $x_n=f^n(x_0)$. The Collatz conjecture is a famous unsolved problem that claims that the sequence eventually reaches $1$ for any starting value $x_0$.

We can separate the steps of the sequence into "even" and "odd" moves, based on whether $x_n$ is even or odd. This leads us to the following, much easier conjecture:

There is no divergent Collatz sequence such that $x_n\bmod 2$ is periodic.

(This does not rule out the possibility of a nontrivial cycle; here I am talking only about sequences that go to infinity.) Can we show this?

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Any sequence of parities corresponds to specific composition of the two possible linear transformations, hence to a transformation of the form $n\mapsto \frac{3^an+b}{2^c}$ with non-negative integers $a,b,c$. For this to produce an integer, it must be the case that $2^c\mid 3^an+b$, i.e., $n$ must be in one specific residue class modulo $2^c$. As we can consider several periods instead of one, it follows that $n$ must be in a specific residue class modulo $2^{kc}$ for every $k\in\Bbb N$. This means that there can be only one such $n$. But then $\frac{3^an+b}{2^c}$ must be the same number and hence the sequence does not diverge.

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  • $\begingroup$ Does this result imply that, supposing there were some cycle or sequence diverging to infinity, the limit point of the iterated function $f(x)=3x+2^{\nu_2(x)}$ would either be irrational over $\Bbb Z_2$ or diverge by $\lvert\cdot\rvert_2$? If that question even makes sense? Because I'm pretty sure we know that limit to be $0$ for all integer inputs in which case we would have a contradiction. $\endgroup$ – user334732 Oct 6 '18 at 18:52

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