Pentagonal number recurrence relations for $p(n)$

Abstract

We revisit Euler's partition function recurrence, which asserts, for integers $n\ge 1,$ that $$p(n)=p(n-1)+p(n-2)-p(n-5)-p(n-7)+\cdots =\sum _{k\in \mathbb{Z}\setminus \{0\}}(-1{)}^{k+1}p(n-\ddot{I}‰(k)),$$ where $\ddot{I}‰(m):=(3m^2+m)/2$ is the $m$th pentagonal number. We prove that this classical result is the $\hat{I}½=0$ case of an infinite family of ``pentagonal number'' recurrences. For each $\hat{I}½\ge 0,$ we prove for positive $n$ that $$p(n)=\frac{1}{g_{\hat{I}}½(n,0)}\left(\hat{I}{\pm }_{\hat{I}}½\cdot \ddot{I}{ƒ}_{2\hat{I}½-1}(n)+{\mathrm{Tr}}_{2\hat{I}½}(n)+\sum _{k\in \mathbb{Z}\setminus \{0\}}(-1{)}^{k+1}{g}_{\hat{I}}½(n,k)\cdot p(n-\ddot{I}‰(k))\right),$$ where $\ddot{I}{ƒ}_{2\hat{I}½-1}(n)$ is a divisor function, ${\mathrm{Tr}}_{2\hat{I}½}(n)$ is the $n$th weight $2\hat{I}½$ Hecke trace of values of special twisted quadratic Dirichlet series, and each $g_{\hat{I}}½(n,k)$ is a polynomial in $n$ and $k.$ The $\hat{I}½=6$ case can be viewed as a partition theoretic formula for Ramanujan's tau-function, as we have $${\mathrm{Tr}}_{12}(n)=-\frac{33108590592}{691}\cdot \ddot{I}„(n).$$

A few minor typos corrected. This paper will appear in Advances in Mathematics