Generating function of partitions of $n$ in $k$ prime parts.

Question

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: $$\sum_{n=0}^{\infty} P(n)x^n=\prod_{k=1}^{\infty}\frac{1}{1-x^k}$$ where can you get that $$\sum_{n=0}^{\infty} P_k(n)x^n=\frac{x^k}{(1-x)(1-x^2)\cdots (1-x^k)}$$ where $P_k(n)$ are the partitions of $n$ into $k$ parts.

Similarly we can obtain that if $Pi(n)$ denotes the partitions of $n$ into prime parts, then $$\sum_{n=0}^{\infty} Pi(n)x^n=\prod_{\textit{p}\; \text{prime}} \frac{1}{1-x^p}$$ With this, I thought it would be "logical" to have something similar for $Pi_k(n)$, that is: $$""\sum_{n=0}^{\infty} Pi_k(n)x^n=\frac{x^k}{(1-x^2)\cdots (1-x^{p_k})}""$$ 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?

Answer 1

You can derive a bivariate generating function, where the $x^n y^k$ coefficient is the desired number of partitions: \begin{align} \sum_{n=0}^\infty \sum_{k=0}^\infty Pi_k(n) x^n y^k &= (1 + x^2 y + (x^2 y)^2 + (x^2 y)^3 + \dots) (1 + x^3 y + (x^3 y)^2 + (x^3 y)^3 + \dots) (1 + x^5 y + (x^5 y)^2 + (x^5 y)^3 + \dots) \cdots \\ &= \prod_{p\ \text{prime}} \left(1 + x^p y + (x^p y)^2 + (x^p y)^3 + \dots\right) \\ &= \prod_{p\ \text{prime}} \sum_{m=0}^\infty (x^p y)^m \\ &= \prod_{p\ \text{prime}} \frac{1}{1-x^p y} \end{align}