Define the recurrence sequence $\{s\}$ as
$$ s_{i+1} \longleftarrow \left\{\begin{array} {ll} 2s_i+1 & \text{for }\Omega(s_i)\text{ even} \\ \left\lfloor\frac{s_i}{2}\right\rfloor+1 & \text{for }\Omega(s_i) \text{ odd} \end{array}\right.. $$
Note $\Omega(n)$ is the number of non-distinct prime factors of $n$.
Example: for $s_0=17$, we get $\{s\}=\{17, 9, 19, 10, 21, 43, 22, 45, 23, 12, 7, 4, 9,\ldots\}$. As soon as a term repeats, we know it's cyclical and can stop.
I tested this with a few hundred small values, and can report that all $s_0>8$ I tried eventually hit that minimum of $4$ and cycled. The first troublesome value was $s_0=133$, which had over $10^4$ terms and a max value of over $10^{50}$, although it too eventually ran back down to $4$.
Question: Is it possible that for some value(s) of $s_0$, this sequence diverges to infinity? Or is this one of those problems that's anyone's guess?
I realize this is unlikely to get an actual solution, but what about heuristics that apply? Naively, I would think it should behave approximately like a random walk; if that were the case, that would guarantee that any given number (like $4$) must be reached after finite time, right?