Notice that it suffices to write $k$ as the product of some terms of the form $g(n)=\frac{[\frac{n}{2}]+1}{[\frac{n}{3}]+1}$. For this, we use strong induction. For $n=1, g(0)=1$. Now, suppose this holds for all $n<k$, let's show that it holds for $n=k$. First, notice that $g(2)=2$, $g(4)=\frac32$, $g(6k)=\frac{3k+1}{2k+1}$, and $g(6k+2)=\frac{3k+2}{2k+1}$. We're ready to finish, if $k=3t$, then $k=g(2)\cdot g(4) \cdot t$, if $k=3t+1$, then $k=(2t+1) \cdot g(6t)$ and if $k=3t+2$, then $k=(2t+1) \cdot g(6t+2)$. In all cases, we're done by the induction hypothesis.