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