TITLE: Why the number of unlabeled graphs on $n$ vertices is not exactly $\frac{2^{n \choose 2}}{n!}$?
QUESTION [7 upvotes]: OEIS A000088 sequence lists the number of unlabeled graphs on $n$ vertices.
Very naively, I would have expected it to be exactly (well, it can't be, at the least because it's not an integer):
$$\frac{2^{n \choose 2}}{n!}$$
thinking about the possible ${n \choose 2}$ edges present/not present, and then dividing by all the possible permutations of the vertices.
Can someone explain simply why, even if tending asymptotically to that value, there are more than that?

REPLY [9 votes]: Let's write out the argument for $2^{\binom n2}/n!$ carefully in order to see why it doesn't work.
If we fix a vertex set $\{v_1, v_2, \dots, v_n\}$, there are $\binom n2$ possible edges $v_i v_j$ between those vertices: $\{v_1 v_2, v_1 v_3, \dots, v_{n-1} v_n\}$. There are $2^{\binom n2}$ subsets of this set, and every one of these subsets specifies a graph.
However, if we list out all $2^{\binom n2}$ graphs we get in this way, an unlabeled graph (that is, an isomorphism class of graphs) can appear on this list multiple times. If you a graph from this list, and permute the labels on the vertices, you will get an isomorphic graph with a different edge set - another entry on this list! There are $n!$ ways to permute the vertices, so - you'd think - there are $n!$ entries on this list isomorphic to any given $n$-vertex graph.
If this were true, we could divide by $n!$ to correct for the overcount, and get the true number of graphs. Unfortunately, it is not true.
Consider a particular example: the complete graph $K_n$. There are still $n!$ ways to permute the labels on the vertices, but when you do that, you don't get a different entry on the list - you get the same entry! So $K_n$ only appears on the list once, not $n!$ times.
Or take the $n$-vertex path $P_n$. For $n\ge2$, this appears on the list $\frac12 n!$ times, because if your permutation reverses the vertices of the path, you get a graph with the same set of edges.
In general, a graph with automorphism group $H$ only appears on the list $\frac{n!}{|H|}$ times. So dividing by $n!$ gives an underestimate: we've assumed all graphs appear $n!$ times when in fact some appear less often.
(We could also see that $2^{\binom n2}/n!$ can't possibly be a correct answer because it's not an integer for $n\ge 3$, but that doesn't tell us why it's wrong.)