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I've been thinking about this for a few days, but I haven't found a general solution yet. How many distinct simple, connected, undirected graphs are there of n labelled vertices? For example, there is one for n = 2 and there are four for n = 3. Thanks in advance!

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    Oh. Well, now I feel silly. Thank you very much for the answer.2012-05-27

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The number of all labelled simple graphs on $n$ vertices is $g_n=2^{\binom n 2}$ because you can decide for each edge whether to include it.

Now, let $G(x)=\sum_{n=0}^{\infty} g_n \dfrac {x^n}{n!}$, let $c_n$ be the number of connected labelled simple graphs on $n$ vertices and let $C(x)=\sum_{n=0}^{\infty} c_n \dfrac{x^n}{n!}$

Then, you have the relationship

$ G(x)=\exp (C(x)) $

which permits the calculation of the numbers $c_n$, but does not imply a simple formula.

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    When you say permits the calculation, how would you go about it?2014-12-07
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These numbers are given by the Tutte polynomial $T_n(x,y)$ of the complete graph at the point $(x,y)=(1,2)$. There are reasonably easy to compute formulae for this polynomial; e.g. this paper by Igor Pak gives the formula: $T_{n+1}(x,y)=\sum_{k=1}^n \binom{n-1}{k-1} (x+y+y^2+\cdots+y^{k-1})\ T_k(x,y)\ T_{n-k+1}(x,y).$

Here's some GAP code that implements this:

T:=[1];;  ComputeNextCoefficient:=function()   local n,f,k,q;   n:=Size(T)+1;   f:=0;   for k in [1..n-1] do     q:=1+Sum([1..k-1],i->2^i);     if(k=1) then       f:=f+q*T[n-1];       continue;     fi;     f:=f+Binomial(n-2,k-1)*q*T[k]*T[n-k];;   od;   T[n]:=f;; end;; 

Then we run it by something like:

while(Size(T)<600) do ComputeNextCoefficient(); od; 

It took 27 seconds to compute the numbers for $1,2,\ldots,600$.

gap> T[600];  
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    This has been awhile, but can this be used to compute the number of graphs on n vertices with k edges? I'm a little confused about what the recurrence is compute as the notation seems a little odd to me. Do the T_n terms represent individual coefficients?2014-12-16