I have been looking for the function that generates the partitions of $n$ into $k$ parts of prime numbers (let's call it $Pi_k(n)$). For example: $Pi_3(9)=2$, since $9=5+2+2$ and $9=3+3+3$.
I know that if $P(n)$ denotes the partitions of $n$ then: $$\sum_{n=0}^{\infty} P(n)z^n=\prod_{k=1}^{\infty}\frac{1}{1-x^k}$$$$\sum_{n=0}^{\infty} P(n)x^n=\prod_{k=1}^{\infty}\frac{1}{1-x^k}$$ where can you get that $$\sum_{n=0}^{\infty} P_k(n)z^n=\frac{x^k}{(1-x)(1-x^2)\cdots (1-x^k)}$$$$\sum_{n=0}^{\infty} P_k(n)x^n=\frac{x^k}{(1-x)(1-x^2)\cdots (1-x^k)}$$ where $P_k(n)$ are the partitions of $n$ into $k$ parts.
Similarly we can obtain that if $Pi(n)$ denotes the partitions of n$n$ into prime parts, then $$\sum_{n=0}^{\infty} Pi(n)z^n=\prod_{\textit{p}\; \text{prime}} \frac{1}{1-x^p}$$$$\sum_{n=0}^{\infty} Pi(n)x^n=\prod_{\textit{p}\; \text{prime}} \frac{1}{1-x^p}$$ With this, I thought it would be "logical" to have something similar for $Pi_k(n)$, that is: $$""\sum_{n=0}^{\infty} Pi_k(n)z^n=\frac{x^k}{(1-x^2)\cdots (1-x^{p_k})}""$$$$""\sum_{n=0}^{\infty} Pi_k(n)x^n=\frac{x^k}{(1-x^2)\cdots (1-x^{p_k})}""$$ But doing the product in Mathematica does not give me the number of partitions of $n$ into $k$ raw parts. I have also tried to make modifications but I still can't get to what I want.
Any suggestions on how to get to the desired generating function?
