# counting simple, connected graphs

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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What is a "simply connected" graph? A tree? – Phira May 27 '12 at 9:04
I count three for $n=3$. – Chris Eagle May 27 '12 at 9:12
@Phira: why is my definition relevant? It's the OP we need to hear from. – Chris Eagle May 27 '12 at 9:14
Ach, sorry! I meant "simple, connected," not "simply connected," which I don't think is a term in graph theory. That must be the source of confusion. – Chris May 27 '12 at 9:17
This is OEIS sequence A001187. When looking for this sort of sequence, it's a good idea to first search OEIS, both by text search and by determining the first few terms and searching with them. – joriki May 27 '12 at 9:20

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];
<integer 123...752 (54096 digits)>

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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? – gct Dec 16 '14 at 4:40

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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The function G(x) (above) is undefined for x != 0 as g_n grows faster than n!. To see this, take the log_2 of both 2^(n choose 2) and n! leaving (n choose 2) = n(n-1)/2 and SUM log_2(k) for k from 1 to n. It is also unclear how you get from the definitions of G(x) and C(x) to the relationship G(x) = exp(C(x)) as there is no clear use of simplicity or connectedness shown. Is there a piece missing? – user59414 Jan 23 '13 at 20:56
@Just, it's a formal power series, it doesn't have to converge. A common technique in the use of generating functions. – Gerry Myerson Jan 24 '13 at 4:51
When you say permits the calculation, how would you go about it? – Theo Belaire Dec 7 '14 at 18:57