Is there a counting known for $3$-regular (or any $n$-regular) pseudo graph of a given order $|V|$? An asymptotic result for connected (or disconnected) graphs is good enough. It would be even better to know an asymptotic counting with $2$-labeling, i.e. pseudo graph a given order $|V|$ labelled with two labels (say 0,1).
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$\begingroup$ could you edit to specify what is meant by “counting with $2$-labelling”? $\endgroup$– Zach HunterCommented Jan 19, 2022 at 19:04
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1$\begingroup$ Have you tried calculating a few values and then consulting the Online Encyclopedia of Integer Sequences? $\endgroup$– Gerry MyersonCommented Jan 20, 2022 at 6:37
2 Answers
There is an article by Catherine Greenhill (UNSW Sydney) and Brendan McKay (ANU Canberra), Asymptotic enumeration of sparse multigraphs with given degrees
Theorem 1.1 yields a more general results, but Corollary 1.2 should be want you wanted. If I'm not mistaken, using their notation, you should input $y_1=x_2=x_3=1$, and you obtain that the number of $k$ regular multigraphs is, for $k=o(n^{1/2})$, $$\frac{(kn)!}{(kn/2)!2^{kn/2}(k!)^n}\exp(-Q(k,n)+O(k^2/n)),$$ where $$Q(k,n)= \frac{-1}{4}(k-1)(k+1)+\frac{k^3}{12n}.$$
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1$\begingroup$ This is for labelled graphs. Probably the number of isomorphism classes is get by dividing by $n!$ but I'd have to check that. $\endgroup$ Commented Jan 25, 2022 at 12:52
I think that this is essentially due to
Read, R. C., The enumeration of locally restricted graphs. I, J. Lond. Math. Soc. 34, 417-436 (1959). ZBL0089.18301.
see https://oeis.org/A129427 Nowadays, combinatorial species provide a nice framework. For $r$ vertices, you want $$ \langle E_r(X E_3(Y)), E_{3r/2}(E_2(Y))\rangle_Y, $$ where $E_r$ is the species of sets of $r$ elements (with cycle index series, or Frobenius character, $h_r$). Think of $X$ as the species of vertices and $Y$ as of the species of half-edges. Then $X E_3(Y)$ are the $3$ half-edges incident to an edge, and $E_r(X E_3(Y))$ are sets of $r$ such things. On the other hand, $E_{3r/2}(E_2(Y))$ is the species of partitions of the label set of half-edges into pairs. Taking the scalar product with respect to $Y$ means that we want to superimpose these two structures, and then forget about the labels.
The good thing about the formula is that you get labelled and unlabelled structures in one go. The bad thing is that it is terrible for computations.