Consider a H-graph. It has $6$ vertices and $5$ edges (H-graph).
Now let's enumerate vertices of graph with $\mathbb{Z_{+}}$ numbers , such as sum of all numbers will give us $n$.
So we want to compute number of such different (up to isomorphism) graphs (let's denote this number as $a_n$). The last sentence means, that if we select $v_{1} = 6$ and $v_{i} =0$, then this equivalent to $v_{3} = 6$ $v_{i} = 0$ (where $v_{1}, v_{3}, v_{4},v_{6}$ are corner vertices and $v_{2}, v_{5}$ middle ones).
My attempt was :
Let's compute it using Polya's enumeration theorem. First of all we need to determine group, which map this graph into itself. That's was the first problem. I think this is the group generates by $(i,j)(k,l)$ and $(i,j)(k,l)(m,n)$ (we may swap two left corner vertices and two right, or we may swap left part by right part). It give us $45$ and $15$ motions.
Let's write a cycled index : $I_G = \frac{1}{60}\left(45z_{1}^2z_2^2 + 15z_2^3 \right)$. Now let's say that $z_{i}$ correspond to $x^{0} + x^{i} + x^{2i} + \dots $ - it means that we make infinity task. It will be helpful, because then we can make generating function and find our $a_n$.
Hence we have $\frac{1}{60}\left(45\frac{1}{(1-x)^2 (1-x^2)^2} + 15\frac{1}{(1-x^2)^3} \right)$. Wolfram said that $a_n$ is so. Unfortunatly this is incorrect for $n = 2$. Where is my problem? Maybe I've found a wrong group.
I get cycle index polynomial $$\frac{1}{8}\left(z_1^6+2 z_1^4 z_2+z_1^2 z_2^2+2 z_2^3+2 z_2 z_4\right),$$ from which substituting $z_i = 1/(1-x^i)$ yields generating function $$\sum_{n=0}^\infty a_n x^n = \frac{1-x+2 x^2}{(1-x)^6 (1+x)^3 (1+x^2)}.$$ WolframAlpha then claims that $a_n$ is $$\frac{1320 + 1873n + 1070n^2 + 300n^3 + 40n^4 + 2n^5 + 15(-1)^n(32 + 17n + 2n^2) + 120\cos\left(\frac{n\pi}{2}\right)}{1920}$$ The first several values match what I computed via enumeration: $${1,2,7,14,31,54,97,154,246,364,538}$$