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I know that it is possible to find the number of unlabelled graphs on $n$ vertices using Polya's theorem, but you get a horrible sum. This also tells you much more: it gives you the number of unlabelled graphs on $n$ vertices and $m$ edges.

If I don't want to know how many graphs there are with $m$ edges, just the total number, is there an easier way? I thought about using Burnside's lemma, but I don't think that this works because you use Polya's theorem on $S_{n}^{(2)}$, not $S_n$.

vukov
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    This is OEIS A000088; you might check the references, but I suspect that you won’t find anything very nice, since the formulas listed in that entry are all either ugly or asymptotic estimates. – Brian M. Scott Jul 14 '16 at 01:57
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    @vukov Is asymptotic results good enough? Most labeled graphs have a trivial automorphism group. In particular, the number of unlabeled graphs on $n$ vertices is $(1+o(1)) 2^{{n \choose 2}}/n!$ – D Poole Jul 14 '16 at 15:38
  • I was hoping for some sort of controllable sum. I guess it is probably possible to work with what you get out of Polya, but then it is a double sum over two or three products, with one of the sums over paritions. – vukov Jul 14 '16 at 20:39
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    This was also computed at the following MSE link using the Polya Enumeration Theorem to compute the generating function by the number of edges. For the case where we are not interested in classifying by the number of edges substitute all variables of the cycle index by two during the cycle index computation to get totals only. – Marko Riedel Aug 05 '17 at 20:56

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The first few values are given in OEIS sequence A000088. Since none of the references list a nice sum to calculate this number, I doubt that one is commonly known.

Mike Pierce
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