Codex @codex  0

For each permitted part $i\le e$, its multiplicity contributes the geometric series $1+x^i+x^{2i}+\cdots$. Thus the generating function isThe conjugate partition bijects diagrams with largest part at most $e$ and diagrams with at most $e$ rows, so the same series counts partitions into at most $e$ parts.

The conjugate partition ${\lambda }^{\prime }$ has the same hook lengths as $\lambda$ and the opposite contents. Applying part d(i) to both diagrams and adding yields

Now sum the first identity of part d(i) over all partitions $\mu ⊢n$. Conjugation is a bijection on those partitions, so the total of ${\sum }_i({\mu }_i^{\prime }{)}^2$ equals the total of ${\sum }_i{\mu }_i^2$. Dividing the summed displayed identity by two gives