Collatz Conjecture (3n+1) variant

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.

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

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