The total number of simple undirected graphs of order $n$ is
$\sum\limits_{i = 0}^{\frac{n(n-1)}{2}}{\binom{\frac{n(n-1)}{2}}{i}} = 2^{\frac{n(n-1)}{2}}$.
What is the number of simple undirected graphs (including isomorphisms) of order $n$, maximum degree $k$, and diameter at most $d$? Is there a similar, concise representation? Are there close bounds?
The maximum degree states, that every node can be connected to at most $k$ other nodes.
The maximum diameter states, that the length of the longest shortest path between two nodes must be at most $d$ (i.e., each nodes eccentricity is at most $d$).
Since it is not yet known whether, e.g., a graph with $n=3250$, $d=2$, $k=57$ exists, I do not expect an exact solution. However an approximation or bounds would be helpful.
Edit: Maybe a slightly different point of view helps: Assuming there exists at least one graph of order $n$, maximum degree $k$, and diameter at most $d$. How many more exist if we increase the maximum degree by one? A lower bound suffices.
