Background: Lately, I have fallen down the rabbit hole of partition numbers. Specifically the partition function, $p(n)$. It's well known that no closed-form expression (with only finitely many elementary operations within a single equation) for the partition function is known. There are, however, a few explicit, but not "closed-form", expressions, such as Rademacher's formula. This has got me curious, so I've been playing around with the idea, and have come across a method that I haven't seen documented anywhere so far.
Recursive formulae: The first observation is noting Euler's pentagonal number formula $$p(n)=p(n-1)+p(n-2)-p(n-5)-p(n-7)+\cdots.$$ If we only consider the first two terms, then we get the Fibonacci sequence $$F_n=F_{n-1}+F_{n-2},$$ which does have a closed-form. Similarly, any recursive formula of the form $$f(n)=a_1f(n-i_1)+\cdots+a_mf(n-i_m)$$ can explicitly be solved similarly for as $$f(n)=\sum_{k=1}^m\frac{r_k^{n + d - 1}}{\prod_{j=0,j\neq k}^m(r_k-r_k)},$$ where $r_k$ is the $k^{\text{th}}$ root of the polynomial $$x^d-\sum_{k=1}^m a_k x^{d-i_k},$$ and $d=\max(i_1,\cdots,i_m)$.
Applying this to the partition function: What this means is that (assuming it exists) we can do this for limiting terms of Euler's formula for the partition numbers. Namely, construct the polynomial $$Q_N(x)=x^{\mathcal{P}(N)}\left(1-\sum_{k=1}^N(-1)^k\left(x^{-\mathcal{P}(-k)}+x^{-\mathcal{P}(k)}\right)\right),$$ where $\mathcal{P}(k)=\frac{k(3k+1)}{2}$ is the generalized pentagonal number function. Then, the partition function is given by $$p(n)=\lim_{N\to\infty}\sum_{k=1}^{\mathcal{P}(N)}\frac{r_k^{n + \mathcal{P}(N) - 1}}{\prod_{j=1, j\neq k}^{\mathcal{P}(N)}(r_k-r_j)},$$ such that $r_k$ is the $k^{\text{th}}$ root of $Q_N(x)$. Numerically, solving for the roots of $Q_N(x)$ is quite doable, and I have done so and verified that this does indeed (at least computationally) converge to $p(n)$! In fact, for any given $N$, we seem to have $\{p(n)\}$ correct for $0\leq n\leq \mathcal{P}(N)$, where $\{\cdot\}$ is the rounding function.
Power series: However, we can go further with this method, as this is not so difficult to convert into a power series for $p(n)$. Letting $\gamma_k = \prod_{j=1, j\neq k}^{\mathcal{P}(N)}\frac{r_k^{\mathcal{P}(N) - 1}}{(r_k-r_j)},$ since $p(n)=\lim_{N\to\infty}\sum_{k=1}^{\mathcal{P}(N)}\gamma_k r_k^n,$ $$ \begin{align*} \gamma_k r_k^n &= \gamma_k e^{n\ln r_k} = \gamma_k \sum_{j=0}^\infty\frac{(n \ln r_k)^j}{j!},\\ \implies p(n) &=\lim_{N\to\infty}\sum_{k=1}^{\mathcal{P}(N)} \gamma_k \sum_{j=0}^\infty\frac{(n \ln r_k)^j}{j!},\\ &= \lim_{N\to\infty} \sum_{j=0}^\infty \frac{n^j}{j!}\sum_{k=1}^{\mathcal{P}(N)} \gamma_k (\ln r_k)^j,\\ \implies c_j &= \frac{1}{j!}\lim_{N\to\infty}\sum_{k=1}^{\mathcal{P}(N)} \gamma_k (\ln r_k)^j, \end{align*} $$ where $c_j$ is the $j^\text{th}$ coefficient in the power series for $p(n)$ centered at $n=0$. I.e; $$p(n)=\sum_{j=0}^\infty c_j n^j.$$ This method too, seems to numerically converge.
Euler function: One final observation (one which my question is centered around), is the pentagonal number theorem $$\phi(x)=\sum_{k=1}^\infty(1-x^k)=1+\sum_{k=1}^\infty(-1)^k\left(x^{\mathcal{P}(-k)}+x^{\mathcal{P}(k)}\right),$$ which bares a striking resemblence to the formula for $Q_N(x)$ I gave above. I'm a little unsure of how the limits work here, but I think that we can use the Euler function to find $$\lim_{N\to\infty}\frac{Q_N(x)}{x^{\mathcal{P}(N)}}=2-\phi\left(\frac{1}{x}\right).$$ So my questions are,
Is the above limit correct? If so, what can be known about the roots of $2-\phi\left(\frac{1}{x}\right)$, and can this be used to find an explicit formula for the partition function?
Any insight would be greatly appreciated (and sorry for the long read!).