A distributive identity for products of partition functions

Question

An $r$-composition of a non-negative integer $s$ is an expression $s=s_1+s_2+\cdots+s_r$ where the $s_i$ are also non-negative integers. Define $k(r,s):=\sum \pi(s_1)\pi(s_2) \cdots \pi(s_r)$ where the sum runs over all $r$-compositions of $s$ and $\pi$ is the partition function. I came across the following identity for which I was unable to find a proof. Any help would be greatly appreciated.

For positive integers $a,b,c,d,e$ it seems $$k(a(b+c),d)=\sum_{e=0}^d k(ab,e)\sum k(a,f_1)k(a,f_2)\cdots k(a,f_c)$$ where the sum runs over all $c$-compositions of $d-e = f_1+f_2 + \cdots + f_c$. For example, when $d=1$, $$k(a(b+c),1) = ab+ac = k(ab,1) \cdot k(a,0)+k(ab,0)\cdot c \cdot k(a,1)$$ and when $d=2$, $$k(ab,0) \cdot \left(c \cdot k(a,2) + {c \choose 2} k(a,1)^2 \right)+k(ab,1) \cdot c \cdot k(a,1) + k(ab,2) = k(a(b+c),2) $$

since $k(r,2)=(r^2+3r)/2$.

Answer 1

Your identity has nothing to do with the partition function. In fact it is true for any function $\pi$ defined on the non-negative integers. Let's define the generating function $F(x)=\sum_{i\geq 0}\pi(i)x^i$. Then $k(r,s)=\sum \pi(s_1)\pi(s_2) \cdots \pi(s_r)$ is simply the coefficient of $x^s$ in the formal power series $F(x)^r$. To abbreviate this we can use the notation $k(r,s)=[x^s]F(x)^r$. Now, returning to your identity, the right hand side is simply $$\sum_{e=0}^d \left([x^e]F(x)^{ab}\right)\left([x^{d-e}](F(x)^a)^c\right)$$ which simplifies to $[x^{d}]F(x)^{ab+ac}$ which is the left hand side.