We know this identitiy.
$$P(n)=\frac{1}{n}\sum_{k=0}^{n}\sigma _{1}(n-k)P(k)$$
where $P(n)$ is the partition function and $\sigma _{1}(n)$ is the divisor sum function. Can we pull partition function from this? I mean, i want to write partition function as $P(n)=f(\sigma _{1}(n))$ not $P(n)=f(P(n),\sigma _{1}(n))$.
It is a famous identity to express $\sigma(n)$ by $p(n)$ using pentagonal numbers, e.g., $$ σ(n) = p(n − 1) + 2p(n − 2) − 5p(n − 5) − 7p(n − 7) + 12p(n − 12) + 15p(n − 15) − 22p(n − 22) − 26p(n − 26)+· · · , $$ but conversely we only get $$ np(n) = σ(1)p(n − 1) + σ(2)p(n − 2) + σ(3)p(n − 3) · · ·+σ(n)p(0). $$ If we consider the smaller values of $p(k)$ as given, this would be a formula of $p(n)$ in terms $\sigma(n)$, e.g., $$ p(n)=\sigma(n)+\sigma(n-1)+2\sigma(n-2)+3\sigma(n-3)+5\sigma(n-4)+7\sigma(n-5)+\cdots $$