The question is simple and direct: For a fixed $n$, how many (different) languages are accepted by a DFA of size $n$ (i.e. $n$ states)? I will formally state this:
Define a DFA as $(Q,\Sigma,\delta,q_0,F)$, where everything is as usual and $\delta:Q\times\Sigma\to Q$ is a (possibly partial) function. We need to establish this since sometimes only total functions are considered valid.
For every $n\geq 1$, define the (equivalence) relation $\sim_n$ on the set of all DFAs as: $\mathcal{A}\sim_n\mathcal{B}$ if $|\mathcal{A}|=|\mathcal{B}|=n$ and $L(\mathcal{A})=L(\mathcal{B})$.
The question is, then: for a given $n$, what is the index of $\sim_n$? That is, what is the size of the set $\{L(\mathcal{A})\mid\mathcal{A}\textrm{ is a DFA of size }n\}$?
Even when $\delta$ is a total function, it doesn't seem to be an easy count (for me, at least). The graph might not be connected, and the states in the connected component containing the initial state might all be accepting, so, for example, there are many graphs of size $n$ accepting $\Sigma^*$. Same with other trivial combinations for the empty language and other languages whose minimal DFA has fewer than $n$ states.
(A naïve) recursion doesn't seem to work either. If we take a DFA of size $k$ and add a new state, then, if we want to keep determinism and make the new graph connected (to try to avoid trivial cases), we have to remove a transition to connect the new state, but in that case we may lose the original language.
Any thoughts?
Note. I updated the question again, with a formal statement and without the previous distracting elements.