The weighting function for the infinite product of necklaces

Question

Let us consider the limit $\lim_{n\to \infty}\prod_{p=1}^n N(p,a)$ where $N(n,a)$ is the number of fixed necklaces of length $n$ composed of $a$ types of beads.

Let's rewrite the product in a way like $f(a,n)\prod_{p=1}^n \frac {1-a^p} {1-a} \approx \prod_{p=1}^n N(p,a)$. I think it's possible to have the representation if we can to constract the function $f(a,n)$ in a way to satisfay the approximate equality.

I guess it's unlikely to describe $f(a,n)$ for finite $n$, but what we can say about the function for $n \to \infty$?

First of all we could calculate $\prod_{p=1}^n N(p,a)$ and $\prod_{p=1}^n \frac {1-a^p} {1-a}$ to see the difference for quite large $n$ ( for eg. $n=100$, is it large for the case?). For eg. for $n=100$, $a=5$ we could get an error of about $0.01$% ( if no mistakes in numerical calculations). What does it mean for larger $n$ and what is the behaviour of $f(a,n)$ for $n \to \infty$?

Thank you for any help to investigate the case for infinity.

PS This question is related to a previous one asked 2 years ago when I was not sure about a mistake in the formula and asked about the influence of symmetric groups. By now I realised a mistake. This is why I am trying to improve the formula.

Answer 1

The number of necklaces of size $p$ is $\frac{a^p}{p}+\mathcal{O}(a^{p/2})$, hence $$ \prod_{p=1}^nN(p,a)=\frac{a^{n(n+1)/2}}{n!}\prod_{p=1}^n\left(1+\mathcal{O}(a^{-p/2})\right) = \left(c+\mathcal{O}(a^{-n/2})\right)\frac{a^{n(n+1)/2}}{n!}, $$ where $c$ is some positive real number.