Is there a closed form formula for counting 2-regular labelled graphs?

Question

Do we have a closed form formula for counting undirected 2-regular labelled graphs ? The sequence for there enumeration is given here.

Answer 1

Initially the question said "$2$-regular graphs". It was changed to labelled graphs after I had written this answer. So, the first part of the answer addresses unlabelled graphs, and the second part (after the EDIT) explains why I think it would be hard to extend this to labelled graphs.

A finite $2$-regular graph is a disjoint union of cycles, so the number of $2$-regular graphs on n vertices is the number of partitions of n into parts of size$\geq 3$.

As far as I can determine, there isn't a closed form known for this. At least, I haven't found one by searching the Web. One can give a generating function of course. If $Q(n)$ is the number of partitions of $n$ into parts of size $\geq3$ then $Q(n)$ is the coefficient of $x^n$ in $$\prod_{k=3}^\infty\frac1{1-x^k}=(1-x)(1-x^2)\prod_{k=1}^\infty\frac1{1-x^k}$$

The last infinite product is the generating function for the ordinary partition function, so if $p(n)$ is the number of partitions of $n$, then $$Q(n)=p(n)-p(n-1)-p(n-2)+p(n-3)$$

We can also see this by inclusion and exclusion. To get the partitions into parts $\geq3$ we must exclude the partitions with a $1$ and the partitions with a $2$, but the partitions with both a $1$ and a $2$ must be added back in.

A recursive formula for the partition formula is known of course. (See the last section these notes.)

EDIT

Offhand, the case of labelled graphs sounds harder. Take the $n=8$ case. There are three $2$-regular graphs: $$C_8\\ C_4\cup C_4\\ C_5\cup C_3$$ where $C_k$ is a $k$-cycle. There are $\frac{7!}2$ labelled graphs corresponding to the first case. For the second case, there $\binom84$ ways to choose the labels of the first cycle, and then $(3)^2$ ways to label the vertices, but we must divide by $2$, to take account of isomorphism of the two cycles, so $\frac 12\binom84(3)^2$ labellings. In the third case, we have $\binom83\cdot12$ labellings.

This approach would require listing all the partitions of parts of size $\geq3$.

Answer 2

We can construct a recurrence for these numbers. Using combinatorial classes we have the following class $\mathcal{Q}$ of sets of undirected cycles of length at least three:

$$\def\textsc#1{\dosc#1\csod} \def\dosc#1#2\csod{{\rm #1{\small #2}}} \mathcal{Q} = \textsc{SET}( \textsc{DHD}_{=3}(\mathcal{Z}) + \textsc{DHD}_{=4}(\mathcal{Z}) + \textsc{DHD}_{=5}(\mathcal{Z}) + \cdots).$$

This gives the EGF (the dihedral group $D_n$ has order $2n$)

$$Q(z) = \sum_{n\ge 0} Q_n \frac{z^n}{n!} = \exp\left( \frac{1}{2} \frac{z^3}{3} + \frac{1}{2} \frac{z^4}{4} + \frac{1}{2} \frac{z^5}{5} + \cdots\right) \\ = \exp\left(- \frac{z}{2} - \frac{z^2}{4} + \frac{1}{2} \log\frac{1}{1-z}\right) = \frac{\exp(-z/2-z^2/4)}{\sqrt{1-z}}.$$

Differentiating we find

$$Q'(z) = - \frac{1}{2} (1+z) \frac{\exp(-z/2-z^2/4)}{\sqrt{1-z}} + \frac{1}{2} \frac{1}{1-z} \frac{\exp(-z/2-z^2/4)}{\sqrt{1-z}}.$$

Extracting the coefficient on $[z^{n-1}]$ for $n\ge 2$ on both sides yields

$$\frac{Q_n}{(n-1)!} = - \frac{1}{2} \frac{Q_{n-1}}{(n-1)!} - \frac{1}{2} \frac{Q_{n-2}}{(n-2)!} + \frac{1}{2} \sum_{m=0}^{n-1} [z^m] \frac{\exp(-z/2-z^2/4)}{\sqrt{1-z}} \\ = - \frac{1}{2} \frac{Q_{n-1}}{(n-1)!} - \frac{1}{2} \frac{Q_{n-2}}{(n-2)!} + \frac{1}{2} \sum_{m=0}^{n-1} \frac{Q_m}{m!} = \frac{1}{2} \sum_{m=0}^{n-3} \frac{Q_m}{m!}.$$

This gives the recurrence

$$\bbox[5px,border:2px solid #00A000]{ Q_n = \frac{(n-1)!}{2} \sum_{m=0}^{n-3} \frac{Q_m}{m!}.}$$

with base cases $Q_0=1$ and $Q_1=0.$ To simplify this further we introduce for $n\ge 3$

$$Q_{n-1} = \frac{(n-2)!}{2} \sum_{m=0}^{n-4} \frac{Q_m}{m!}$$

and get

$$Q_n = \frac{(n-1)!}{2} \left(\frac{2}{(n-2)!} Q_{n-1} + \frac{Q_{n-3}}{(n-3)!}\right)$$

which yields

$$\bbox[5px,border:2px solid #00A000]{ Q_n = (n-1) Q_{n-1} + \frac{1}{2} (n-1)(n-2) Q_{n-3}}$$

for $n\ge 3$ with base cases $Q_0=1, Q_1=0$ and $Q_2=0.$ This will produce the sequence

$$1, 0, 0, 1, 3, 12, 70, 465, 3507, 30016, 286884, \ldots$$

which points us to OEIS A001205 where we find the EGF and the recurrence, so the above is sound.

Addendum. If we ask for a closed form we get e.g.

$$\begin{align*} & n! [z^n] \frac{\exp(-z/4(z+2))}{\sqrt{1-z}} = n! [z^n] \sum_{k=0}^n \frac{1}{k!} \frac{(-1)^{k}}{4^{k}} z^{k} (z+2)^{k} \frac{1}{\sqrt{1-z}} \\ & = n! \sum_{k=0}^n \frac{1}{k!} \frac{(-1)^k}{4^k} [z^{n-k}] (z+2)^k \frac{1}{\sqrt{1-z}} \\ & = n! \sum_{k=0}^n \frac{1}{k!} \frac{(-1)^k}{4^k} \sum_{j=0}^{n-k} {k\choose n-k-j} 2^{2k+j-n} \frac{1}{4^j} {2j\choose j}. \end{align*}$$

We obtain

$$\frac{n!}{2^n} \sum_{k=0}^n \frac{(-1)^k}{k!} \sum_{j=0}^{n-k} \frac{1}{2^j} {k\choose n-k-j} {2j\choose j}.$$

This is one of several possibilities.

Remark. Wilf in generatingfunctionology page 180 gives a proof for the asymptotic

$$Q(n) \sim n! \exp(-3/4) {n-1/2\choose n} = \exp(-3/4) (n-1/2)^{\underline{n}}.$$

This can be further simplified to

$$Q(n) \sim \frac{n! \exp(-3/4)}{\sqrt{n\pi}}.$$