# Counting Ribbon graphs

Let $G$ be a ribbon graph (sometimes called fat graph) with $v$ vertices and $e$ edges. Furthermore each vertex is of degree $d$.

Q) What is the number of $G$ with the above properties? I mean does there exist a closed formula for the number of such $G$ in terms of $v$, $e$, and $d$.

I don't know any reference or result in this area. I searched but I could not find anything of this type. Any reference, link, suggestion will be helpful.

• Could you clarify your question, are you looking for some kind of finite formula relating this number to standard functions like polynomials and factorials and such? It sounds like you know what the definition of the number is but perhaps that's what you're after. – Ryan Budney Nov 24 '14 at 5:52
• @Ryan You are right. I want a closed formula in terms of $v$, $e$ and $d$. I don't know anything about the number. I am editing the question. Thanks for the suggestion. – Cusp Nov 24 '14 at 6:19
• @Cusp: The data is redundant since $e = vd/2$. If you want isomorphism classes, this seems impossible (in the moral, not the technical sense) for the same reason that it is impossible to write down a closed formula for the number of isomorphism classes of degree $d$ graphs with $v$ vertices. – Andy Putman Nov 24 '14 at 6:34
• @AndyPutman I know the data is redundant. I just kept it there to emphasise that every vertex has same order. But fat graph have an ordering at each vertex, shouldn't it reduce the complexity of the fat graph. – Cusp Nov 25 '14 at 4:36

This is a difficult problem. As usual, it is better to count a certain weighted number, and one should take the genus into account. There are some partial results for special dessins, Here is a citation from the review MR2428514:

A paper of I. P. Goulden and D. M. Jackson [Adv. Math. 219 (2008), no. 3, 932–951; MR2442057] gives a recursion for the total mass of dessins with ramification invariants $(3^{2m},2^{3m},\lambda)$ with $\lambda$ varying over all partitions with $m+2−2g$ parts. In the case of genus $g=0$, this recursion can be solved giving a total mass of $$\frac{F(m,0)}{6m}=\frac{4m(3m−2)!!}{m(m+2)!(m−2)!!}.$$