Number of minimal DFAs of size at most $m$? 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.
According to Ishigami Y., Tani S. (1993) The VC-dimension of finite automata with $n$ states,
http://link.springer.com/chapter/10.1007/3-540-57370-4_58
,
the VC-dimension of the concept class of $n$-state DFAs over an alphabet of size $k$ is
$$ d=d(n,k) := (k-1+o(1))n\log_2 n.$$
It follows that there are at least $2^d$ distinct $n$-state automata on a $k$-letter alphabet. The upper bound on the number such automata follows from a simple counting argument (given in the paper), and is at most $2^d$.