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It is known that $F(x)$ is the generating function of the counting sequence of connected simple graphs with N vertices is given by:

$F(x) = x + x^2 + 2x^3 + 6x^4 + 21x^5 + 112x^6 + 853x^7...$

where the total number of connected graphs of $n$ vertices is obtained by the coefficient corresponding to $x^n$.

The question here is: How to find the series (or formula) to get the counting of connected simple graphs with $n$ vertices of maximum degree up to $D$? (generalizing with at least $1$ vertex with maximum degree equal to $D$)

For example, for $n = 5$ with $D=3$, we have $8$ connected graphs with at least $1$ vertex with maximum degree equal to $3$, and for the same $n$ with for $D=4$, we have $11$ connected graphs with at least $1$ vertex with maximum degree equal to $4$, as shown by the distribution of degrees in the figure below: enter image description here

Follows a table of the count of the first groups of simple graphs according to the number of vertices and the maximum degree: enter image description here My question in brief is to determine the number of connected graphs with max. degree $D$ for a given number of Nodes $n$?

OBS: Do not get confused! the question does not include Disconnected Graphs, Digraphs, Labeled Graphs, number of cubic or quartic graphs!

Anyone know a solution to this intriguing problem? help me!

Note:

$F(x)$ is the generating function of the counting sequence of connected structures, then the corresponding generating function $G(x)$ of the counting sequence of all structures is given by:

$$G(x) = exp \sum_{k>=1} F(x^k)/k = \sum_{i=0} b_i x^k$$

Applying a variant of the Mobius inversion to equation above, it is possible to express $F(x)$ in terms of $G(x)$:

$$F(x) = exp \sum_{k>=1} [\mu(x)/k]log G(x^k) = \sum_{i=0} a_i x^k$$

Where $\mu$ stands for the Mobius function.

Let $a_n$ be the number of connected graphs on $n$ vertices and $b_n$ the number of all graphs on N vertices. Then:

$F(x) = x + x^2 + 2x^3 + 6x^4 + 21x^5 + 112x^6 + 853x^7 + ...$

and

$G(x) = 1 + x + 2x^2 + 4x^3 + 11x^4 + 34x^5 + 156x^6 + ...$

I saw this on "Counting Disconnected Structures: Chemical Trees, Fullerenes, I-graphs" on the link: https://hrcak.srce.hr/file/4232 but I did not understand how I can use it in favor of my problem.

An observation:

If $T(n)$ represents the total of connected graphs of $n$ nodes and $D_p(n)$ represents the total of connected sub-graphs group with max. degree $P$, and knowing that $P_{max} = (n-1)$, we can write: $$ T(n) = D_1(n) + D_2(n) + D_3(n) + ... + D_{n-1}(n)$$

For $n>2$, we know that $D_1(n) = 0$ and $D_2(n) = 2$, so we have to: $$T(n) = 2 + D_3(n) + ... + D_{n-1}(n)$$

Note also that $D_{n-1}$ can be obtained by the coefficient of $x^{n-1}$ of $G(x)$!

But how to calculate $D_p(n)$?

OBS: Another way to attack the problem I posted on the link: About counting the number of graphs by the maximum degree $D$.

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    $\begingroup$ It seems worth noting that you are counting isomorphism classes of graphs, not graphs in total. This usually makes counting harder (think, e.g., counting partitions vs ball-and-bin problems). $\endgroup$
    – user44191
    Commented Feb 22, 2020 at 13:34
  • $\begingroup$ There is unlikely to be a formula as such, although as noted by others it can be computed from the values for not necessarily connected graphs. (See A263293 in OEIS for values up to 10 nodes - these values were obtained by brute force enumeration of all graphs). Better methods exist, since it isn't necessary to enumerate every graph - only every degree sequence, but to my knowledge this has not been done. This technique, might give values up to about 15 nodes, but it is still exponential. $\endgroup$
    – Andrew
    Commented Feb 22, 2020 at 18:31
  • $\begingroup$ I agree with Andrew that a usable exact formula is unlikely except in special cases. An asymptotic formula for fixed $D$ would be not so difficult and perhaps it is published somewhere. $\endgroup$ Commented Feb 23, 2020 at 0:21
  • $\begingroup$ I think a general formula here does exist, but its complexity is similar to that of Jovovic's formula. $\endgroup$ Commented Sep 16, 2020 at 19:25

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