Sorry that I ignored that comment of yours in the other post.
In fact the answer is "yes": every number can be made.
Proof: Note that the number $\alpha = \log_32$ is irrational. This implies that the sequence $(\{ k\alpha\})_{k \geq 0}$ is dense in the interval $[0,1)$, where $\{\cdot\}$ denotes the fractional part.
Now let $n$ be an integer and consider the interval $[\log_3n, \log_3(n + 1))$. Its image under the map $\{ \cdot \}:\mathbb R \rightarrow [0, 1)$ contains a non-empty open set. Hence there exists infinitely many integers $k\geq0$ such that $\{ k\alpha \}$ lies in the image of that interval.
We take such a $k$ that is sufficiently large, so that $k\alpha \geq \log_3n$. Then there is an integer $m\geq0$ such that $k\alpha - m$ lies in the interval $[\log_3n,\log_3(n + 1))$.
Therefore we get the number $n$ via $\lfloor \frac{2^k}{3^m}\rfloor$.