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