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.

  • $\begingroup$ How this is different from mathoverflow.net/questions/103716/… ? $\endgroup$ Commented Dec 3, 2014 at 21:11
  • $\begingroup$ @Max Alekseyev It was just about the influence of symmetric groups I'am still unsure. Moreover, generally speaking I am not sure what to answer. By now I can see a reply from Jan-Christoph Schlage-Puchta to study. Any ideas from you? $\endgroup$ Commented Dec 4, 2014 at 18:45

1 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.