TITLE: Proving that a recursive sequence will eventually reach an even number.
QUESTION [8 upvotes]: Take an arbitrary odd natural number $x$, take the recursive sequence given by
$$a_0 = x, a_n= \left \lfloor{ \frac{3 a_{n-1}}{2}} \right \rfloor$$
where $\left \lfloor{ .} \right \rfloor : N \rightarrow N$ is the floor function.
I would like to prove that for any $x$ chosen this sequence eventually has an even number in it, I thought this would be doable but I am stuck. How could one prove this?

REPLY [4 votes]: The iterates of $2n+1$ are
$$3^{k+1}\left\lfloor\frac{2n+1}{2^{k+1}}\right\rfloor+1.$$
This is an even number when $k$ equals the number of trailing zeroes in the binary representation of $n$.
For instance, $2017=11111100001_b$ gives $3^5\cdot111111_b+1=15310$ after $5$ iterations.