Do we know of any Collatz theorem involving similar functions. For example what do we know about iterations of:
$$ f(x) = \begin{cases} \dfrac{x + 1}{2} \text{, if } x \text{ is odd}. \\ 2x + 1, \text{ if } x \text{ is even}.\end{cases} $$
Since this one doesn't involve $3$ maybe it is easier to solve? Why isn't this one the lowest hanging fruit rather than the $3x +1$ problem?
Examples:
$$ f(1) = 1 \\ f(2) = 5 \to 3 \to 2 \to \dots \\ f(4) = 9 \to 5 \dots \\ f(6) = 13 \to 7 \to 4 \to \dots $$
So one of the finite number of conjectured cycles would be $5 \to 3 \to 2$.
The Collatz conjecture is interesting because it is hard. If you use the $-1$ version, every even $x$ goes to $2x+1$, which goes back to $x$, so we can just say every pair $(2k, 4k+1)$ is a cycle including $(0,1)$ and be done.
Even for the $+1$ version, every doubling is followed by a divide by $2$, which might be followed by another divide by $2$, so numbers can't grow much beyond twice the start. There is a cycle $(2,5,3)$ which everything except $1$ seems to fall into. We can prove this by considering what happens to numbers of the form $4k$ and $4k+2$. A number of the form $4k$ goes to $8k+1$, which then goes to $4k+1$. A number of the form $4k+1$ goes to $2k+1$ which then goes to $k+1$, so any large number gets divided by $2$ in a few steps. We can expand this to show every number except $1$ goes to the $(2,5,3)$ cycle.
The multiplication by $3$ in the Collatz conjecture is just right to balance the up steps and the down steps if you consider the chances of odd numbers in a row. If you increase the multiplier most numbers will go off to infinity. If you decrease it, all numbers fall into a small cycle and we can prove that. Chaos lives on the boundary between two or more well behaved regions.