Pentagonal number recurrence relations for $p(n)$

We revisit Euler’s partition function recurrence, which asserts, for integers $n\ge 1,$ that $$p(n)=p(n-1)+p(n-2)-p(n-5)-p(n-7)+\cdots =\sum _{k\in \mathbb{Z}\setminus \{0\}}(-1{)}^{k+1}p(n-\omega (k)),$$ where $\omega (m):=(3m^2+m)\mathrm{/}2$ 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 $\nu \ge 0,$ we prove for positive $n$ that $$p(n)=\frac{1}{g_{\nu }(n,0)}({\alpha }_{\nu }\cdot {\sigma }_{2\nu -1}(n)+{\mathrm{T}\mathrm{r}}_{2\nu }(n)+\sum _{k\in \mathbb{Z}\setminus \{0\}}(-1{)}^{k+1}{g}_{\nu }(n,k)\cdot p(n-\omega (k))),$$ where ${\sigma }_{2\nu -1}(n)$ is a divisor function, ${\mathrm{T}\mathrm{r}}_{2\nu }(n)$ 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 $k\mathrm{.}$ The $\nu =6$ case can be viewed as a partition theoretic formula for Ramanujan’s tau-function, as we have $${\mathrm{T}\mathrm{r}}_{12}(n)=-\frac{33108590592}{691}\cdot \tau (n)\mathrm{.}$$