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Are there any non-trivial lower bounds on the number of isomorphism classes for a graph with $N$ vertices?

For example there are at least $N(N-1)/2$ isomorphism classes (counting one for the number of possible edges in our graph) but as $N$ increases, there will clearly be a lot more.

rwolst
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You really need to say what you mean by non-trivial. What do you want to use the bound for? But, gripes aside, $2^{\binom{n}2}/n!$ is a lower bound and asymptotically tight.

Chris Godsil
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  • I meant $N(N-1)/2$ as the trivial lower bound. Do you have a reference for this? – rwolst Oct 28 '14 at 19:04
  • The number of graphs in the isomorphism class of $G$ is $n!/|\mathrm{Aut}(G)|$, so it's at most $n!$. So the number of isomorphism classes is at least $2^n/n!$. I do not think a reference is necessary. – Chris Godsil Oct 28 '14 at 20:20
  • $2^{{n \choose 2}}/n!$. Yes that is enough. – rwolst Oct 28 '14 at 22:05