Collatz $2x + 1$ conjecture? 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 $5 \to 3 \to 2$.
 A: 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.
A: If $x$ is even, then $f(x)=2x+1$ is odd. So $$f(f(x))=\frac{(2x+1)+1}{2}= x+1,$$ which is also odd. So
$$ f(f(f(x))) = \frac{(x+1)+1}{2} =\frac{x}{2}+1 \leq x.$$
Thus no matter wheter $x$ is even or odd at the beginning, the number will always become smaller after a couple of iterations. Thus there is no number $x$ for which the series tends to infinity.
