Modified HRR partition function to detect primes without factorisation

Question

The generating function for integer partitions (see section "generating function") can be defined as: $$ \sum_{n=0}^{\infty}P(n)\,x^n := \prod_{k=1}^{\infty}\left(\frac{1}{1-x^k}\right)=(1+x+x^2+\dots)(1+x^2+x^4+\dots)(1+x^3+x^6+\dots)\dots$$ for $x\in\mathbb{C}$ inside the unit circle. There is the much celebrated Hardy–Ramanujan–Rademacher asymptotic expansion for $P(n)$, which allows for a direct numerical calculation of an approximation of $P(n)$ (the number of different ways to express $n\in\mathbb{N}$ as a sum of positive intergers without regarding order of terms) without recurring to factorisations or the like.

I thought about a modification of this to construct the function $$\sum_{n=0}^{\infty}P'(n)\,x^n = \require{cancel} (\cancel 1+\cancel{x}+x^2+\dots)(\cancel1+x^2+x^4+\dots)(\cancel1+x^3+x^6+\dots)\dots$$ where all $1$'s in each factor and the only one $x^1$ occurring in the first factor are omitted.

The special interest of $P'(x)$ would be that it would for exactly all primes $p_n$ evaluate to 0: $P'(p_n)=0$ and $P(nm)>0$ for $n,m>1$. So the hope would be to be able to somehow express $P'(n)$ in a closed from in terms of $P(n)$ or maybe recursively in terms of the $P(i) \forall i\le n$.

I have tried eg. to remove the $1$'s by eg. trying to use something like $$\frac{q+q^2+q^3+\dots}{1+q+q^2+q^3+\dots} = \frac{\frac{q}{1-q}}{\frac{1}{1-q}}=q$$ But that would mean that for each factor $\frac{1}{1-x^k}$ where you subtract 1 you get $P^{l+1}(n)\rightarrow P^{l}(n+k)$ understanding $P^{0}=P$ and $P^{\infty}=P'$, and this looks suspiciously like some convergence trouble.

So my question is, if this ansatz for $P'(n)$ makes sense and if and how $P'(n)$ could be expressed in "closed form".

Answer 1

The generating function for integer partitions (see section "generating function") can be defined as: $$\sum_{n=0}^{\infty}P(n)\,x^n := \prod_{k=1}^{\infty}\left(\frac{1}{1-x^k}\right)=(1+x+x^2+\dots)(1+x^2+x^4+\dots)(1+x^3+x^6+\dots)\dots$$ for $x\in\mathbb{C}$ inside the unit circle.

No. A generating function is a formal series: its radius of convergence is emphatically not part of the definition.

I thought about a modification of this to construct the function $$\sum_{n=0}^{\infty}P'(n)\,x^n = \require{cancel} (\cancel 1+\cancel{x}+x^2+\dots)(\cancel1+x^2+x^4+\dots)(\cancel1+x^3+x^6+\dots)\dots$$ where all $1$'s in each factor and the only one $x^1$ occurring in the first factor are omitted.

The special interest of $P'(x)$ would be that it would for exactly all primes $p_n$ evaluate to 0: $P'(p_n)=0$ and $P(nm)>0$ for $n,m>1$.

Again, no. You have $$\sum_{n=0}^{\infty}P'(n)\,x^n := \left(\frac{1}{1-x} - 1 - x\right) \prod_{k=2}^{\infty}\left(\frac{1}{1-x^k} - 1\right) = \frac{x^2}{1-x} \prod_{k=2}^{\infty}\frac{x^k}{1-x^k}$$

If we consider just the first $m$ terms of the RHS we have $$\frac{x^2}{1-x} \prod_{k=2}^m\frac{x^k}{1-x^k} = x^{m!+1} \prod_{k=1}^m\frac{1}{1-x^k}$$

As $m$ tends to infinity, clearly the index of the first non-zero coefficient also tends to infinity, so that $P'(n) = 0$ for all finite $n$.