Pentagonal number theorem

Remarks

The exponents $k(3k-1)/2$ for $k = 0, \pm 1, \pm 2, \ldots$ are the generalized pentagonal numbers $0, 1, 2, 5, 7, 12, 15, 22, 26, \ldots$. Let $E(m)$ and $O(m)$ be the numbers of partitions of $m$ into an even and an odd number of distinct parts. The coefficient of $x^m$ on the left is $E(m) - O(m)$. Franklin's bijection pairs these two kinds of partitions, and the only unpaired partitions occur when $m$ is a generalized pentagonal number, with sign $(-1)^k$. Example: for $m = 3$, the distinct-part partitions are $3$ (one part) and $2 + 1$ (two parts), so the coefficient is $1 - 1 = 0$; indeed $3$ is not a generalized pentagonal number. The generating function for partitions is $\prod_n (1 - x^n)^{-1} = \sum_m p(m) x^m$. Multiplying the theorem by it gives the recurrence $$p(m) = p(m-1) + p(m-2) - p(m-5) - p(m-7) + p(m-12) + p(m-15) - \cdots,$$ with $p(0) = 1$ and $p(j) = 0$ for $j < 0$.

Going deeper

For a student

Euler found the identity, and it gives a fast recurrence for the partition function $p(m)$. It is a special case of the Jacobi triple product identity. The left side is, up to the factor $x^{1/24}$, the Dedekind eta function, so the theorem is also a statement about a modular form.