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: $$8 \rightarrow 25 \rightarrow 76 \rightarrow 38 \rightarrow 19 \rightarrow 58 \rightarrow 29 \rightarrow 88 \rightarrow 44 \rightarrow 22 \rightarrow 11 \rightarrow 34 \rightarrow 17 \rightarrow 52 \rightarrow 26 \rightarrow 13 \rightarrow 40 \rightarrow 20 \rightarrow 10 \rightarrow 5 \rightarrow 16 \rightarrow 8$$
And another one:
$$16 \rightarrow 49 \rightarrow 148 \rightarrow 74 \rightarrow 37 \rightarrow 112 \rightarrow 56 \rightarrow 28 \rightarrow 14 \rightarrow 7 \rightarrow 22 \rightarrow 11 \rightarrow 34 \rightarrow 17 \rightarrow 52 \rightarrow 26 \rightarrow 13 \rightarrow 40 \rightarrow 20 \rightarrow 10 \rightarrow 5 \rightarrow 16$$
Both avoid the 4, 2, 1 cycle.
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: $4\to13\to40\to20\to10\to5\to16\to8\to4$
