Q:

Collatz Conjecture (3n+1) variant

Let's consider the following variant of Collatz (3n+1) : 
if $n$ is odd then $n \to 3n+1$
if $n$ is even then you can choose : $n \to n/2$ or $n \to 3n+1$
With this definition, is it possible to construct a cycle other than the trivial one, i.e., $1\to 4 \to 2 \to 1$?
Best regards

A:

Yes! With the standard Collatz conjecture, every number must eventually end up at the cycle $4 \to 2 \to 1 \to 4 \cdots$ . This has been verified for all numbers up to $2^{60}$.
With your altered definition, you can start at $2$, apply $3n+1$ instead of $n/2$, and then continue like the standard Collatz again. 
$$2 \xrightarrow{3n+1} 7 \to  22 \to 11 \to \cdots ,$$
you'll eventually end at $2$ again, since this is one of the verified cases.

A:

$$7\to 22$$
$$22\to11$$
$$11\to34$$
$$34\to17$$
$$17\to52$$
$$52\to26\to13$$
$$13\to40$$
$$40\to20\to10\to5$$
$$5\to16$$
$$16\to8\to4\to2$$
$$2\to 3\cdot2+1=7$$

A:

$4\to13\to40\to20\to10\to5\to16\to8\to4$

