Given any integer $n$, how many ways can it be partitioned in which the number $1$ is not allowed? For instance, if $n=6$, then the partitions obeying the aforementioned rule are $6+0$, $4+2$, $3+3$, and $2+2+2$.
I'm trying to find a nice formula for this. I believe (correct me if I'm wrong) the generating function would be $(1+x^2 +x^3 + x^4 + \dots)^{\lfloor \frac{n}{2} \rfloor}$. The $x$ term vanishes because $1$ is forbidden. The $\lfloor \frac{n}{2} \rfloor$ comes from that fact that the longest string $n$ can be partitioned is from a string of $2's$ or a string of $2's$ and one $3$ (verify this).
Are there any better counting methods? Is there an easy way to find these coefficients?