The Periodic Collatz Conjecture

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?

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.
• 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. – user334732 Oct 6 '18 at 18:52