As explained in Isaac Solomon's answer just adding $1$ would make the problem simple. Multiplying by $2$ (or another even number) and adding $1$ would not work, so there one is at $3$.

One could ask the same question for $5$ instead of $3$, or there are other variants, but note the following:

If you assume a probabilistic model then in the multiplying by $3$ case you are in the following situation: in "half" the cases (even $n$) you divide by $2$, in "half" the cases (odd $n$) you take $3n+1$, but then you are guaranteed to divide by $2$ so effectively the situation is:

- "half" the time multiply by $\frac{1}{2}$
- "half" the time multiply by (roughly) $\frac{3}{2}$ (as the combination of $3n+1$ and the then forced $\frac{(3n+1)}{2}$).

The product of these two $\frac{1}{2}$ and $\frac{3}{2}$ is $\frac{3}{4}$ and this is less than 1, so in the long run you'd expect a decrease.

So the heuristic suggests a decrease in the long run. (This argument is rough but one could make it a bit more precise. However, I think to get a rough idea this might do.)

If you do the same with $5$ (or something still larger) instead of $3$, you'd have $
\frac{1}{2}$ and $\frac{5}{2}$ instead. So you'd get $\frac{5}{4}$ which is greater than $1$. So in the long run you'd expect an increase.

Indeed, one can also study this problem; so $5$ instead of $3$, but then the situation is different in that one will (it is believed) have some starting sequence that go to infinity *and* several small loops. So the question remains interesting, but it somehow changes as then one will have these values that escape to infinity.

More generally, there are numerous Collatz-like problems that are considered. But for some it can even be shown that they are formally undecidable.