Solution:

We show that to obtain $f(n)$, one writes $n$ in base 2 and then reads it in base 3. For example, $12 = 1100_2$, so $f(12) = 1100_3 = 36$. Let $g(n)$ be defined in this way. Then certainly $g(1) = 1$. Now $2n+1$ has the same binary expansion as $2n$ except for a final 1, so $g(2n+1) = g(2n) + 1$. Similarly, $2n$ has the same binary expansion as $n$ with the addition of a final zero. Hence $g(2n) = 3g(n)$. So $g$ is the same as $f$. Hence the set of all $m$ such that $m = f(n)$ for some $n$ is the set of all $m$ which can be written in base 3 without a digit 2.