Pentagonal number recurrence relations for p(n)

We revisit Euler's partition function recurrence, which asserts, for integers $n\ge 1$, that$p\left(n\right)=p(n-1)+p(n-2)-p(n-5)-p(n-7)+\dots =\sum _{k\in Z\setminus \left\{0\right\}}{(-1)}^{k+1}p(n-\omega (k\left)\right),$ where $\omega \left(m\right):=(3m^2+m)/2$ is the mth 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$p\left(n\right)=\frac{1}{g_{\nu }(n,0)}({\alpha }_{\nu }\cdot {\sigma }_{2\nu -1}(n)+{\mathrm{Tr}}_{2\nu }(n)+\sum _{k\in Z\setminus \left\{0\right\}}{(-1)}^{k+1}{g}_{\nu }(n,k)\cdot p(n-\omega (k\left)\right)),$ where ${\sigma }_{2\nu -1}\left(n\right)$ is a divisor function, ${\mathrm{Tr}}_{2\nu }\left(n\right)$ is the nth weight 2ν 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${\mathrm{Tr}}_{12}\left(n\right)=-\frac{33108590592}{691}\cdot \tau \left(n\right).$