Let $\Sigma$ be an alphabet of size $2$, and consider minimal DFAs whose size is bounded by at most $m$. Let $f(m)$ denote the number of different such minimal DFAs.
Can we find a closed-form formula for $f(m)$?
Considering that for $|\Sigma|=2$ the transition function of a DFA of size at most $m$ is a graph. Since the nodes degree is bounded by $2$, for each node there are $m^2$ possibilities of pairs of arcs (as suggested in the comments). In this graph there are at most $m$ possible choices of initial state and at most $2^m$ possible choices of final states sets. Thus, the maximum number of DFAs of size at most $m$ is $f(m) \leq m^{2m}\cdot m\cdot2^m = 2^m\cdot m^{2m+1}$.
We can generalize to an arbitrary alphabet $\Sigma$: the bound becomes $f(m) \le 2^m\cdot m^{|\Sigma|m+1}$.
But we bounded here arbitrary DFAs and I'm interested in bounding the number of minimal DFAs. Thus, it looks like this bound could be tighter... Does someone have a better estimate?
I would appreciate if possible, some papers related to this problem or a proof/counter-example.