Total number of parts in the all partitions of $n$

Question

Let's denote $N_k(n)$ as the number of partitions $n$ into at most $k$ parts. Prove that the total number of parts in the all partitions of $n$ is equal to: $$\sum_{a=1}^n \sum_{b=1}^{\lfloor n/a \rfloor} \sum_{k=0}^{n-ab} N_a(k)N_{b-1}(n-ab-k)$$

Sincerely speaking, that task appears difficult for me. What I was able to notice is a fact that expression $$ \sum_{k=0}^{n-ab} N_a(k)N_{b-1}(n-ab-k)$$ can be some kind of the convolution $ \sum_{k=0}^{m} N_a(k)N_{b-1}(m-k)$. But I am not sure, because of the lower indices $a$ and $b-1$.

Could you give any hints how to start with this task?

Answer 1

Let $f(a,b)=\sum_{k=0}^{n-ab}N_a(k)N_{b-1}(n-ab-k)$.

Claim: $f(a,b)$ is the number of Ferrers diagrams of partitions of $n$ in which there is a dot in row $a$, column $b$ with no dot immediately below it. In other words, it counts the partitions $\lambda\vdash\lambda_1,\dots,\lambda_r$ of $n$ such that $r\ge a$, $\lambda_a\ge b$, and either $a=r$, or $\lambda_{a+1}<b$. These diagrams all have an $a\times b$ block in the upper lefthand corner. Suppose that there are $k$ dots to the right of that block (or in other words that $\sum_{i=1}^a\lambda_i-ab=k$); those dots are the Ferrers diagram of a partition of $k$ into at most $a$ parts, so there are $N_a(k)$ possibilities for them. The $n-ab-k$ dots below the $a\times b$ block lie in at most $b-1$ columns, so there are $N_{b-1}(n-ab-k)$ possibilities for them. $\dashv$

In other words, there are $f(a,b)$ partitions of $n$ in which the dot at the bottom of column $b$ is in row $a$. Taking conjugates, we see that there are $f(a,b)$ partitions $\lambda\vdash\lambda_1,\dots,\lambda_r$ of $n$ such that $\lambda_b=a$: $f(a,b)$ counts the $b$-th parts that are of size $a$. And $\sum_{a=1}^n\sum_{b=1}^{\lfloor n/a\rfloor}f(a,b)$ just sums $f(a,b)$ over all ordered pairs $\langle a,b\rangle$ of positive integers such that $ab\le n$, i.e., over all ordered pairs $\langle a,b\rangle$ such that a partition of $n$ can have a $b$-th part of size $a$, so it counts the parts of all partitions of $n$.

(I could have avoided taking conjugates by reversing the rôles of $a$ and $b$, but this was the way I actually saw it.)