I am in my own Automaton (finite-state deterministic automata) research, so i have four sets of automata. 2 states automata, 3 states, 4 states and 5 states.
Input alphabet $\{0,1\}$
so... the amount of automata for each set of states is:
$$ 2^n n^{2n}$$
where: 2= number of state-transition function(depend on intput alphabet)
n= number of states.
for example for 3 states:
$$ 2^3 \cdot 3^{2\cdot3} = 5832 $$
by some method(computational) i get two type of automaton, then i get the next results:
$$\begin{array}{r|rrr} \text{states} & \text{amount of combinations*} & \text{amount of Type $B$} & \text{amount of Type $A$**} \\ \hline 1 & 2 & 0 & 2\\ 2 & 64 & 38 & 26\\ 3 & 5\;832 & 4\;778 & 1\;054\\ 4 & 1\;048\;576 & 991\;508 & 57\;068\\ 5 & 312\;500\;000 & 308\;737\;626 & 3\;762 374\\ 6 & 139\;314\;069\;504 & ? & ? \end{array} $$
*Note that: amount type B + amount type A = amount of combinations per state,
** with type B and A, i mean: Isomorphic Automaton and Not Isomorphic Automaton
Isomorphic and Not Isomorphic sample:
if We have 5 automata that they are Isomorphic among them, 1 of them go to the subset of Not Isomorphic, and the another 4 to the subset of Isomorphic.
My question is if, can you help me to get the formula/way to calculate the amount of Type A or Type B, depending by the states number?
and i am looking for the 6 states results
must be a way to know
i am pretty sure that we could get it with just a modification of this:
$$ 2^n n^{2n}$$ I have tried with Newton's method(numeric method, matrices) but didnt work(better say, I didnt get it).
Not exactly an answer, but maybe this observation may help you. For amount of type A you can write
$$26=2^5-2^2-2^1$$ $$1054=2^{11}-2^9-2^8-2^7-2^6-2^5-2^1$$ $$57068=2^{16}-2^{13}-2^8-2^4-2^2$$ $$3762374=2^{22}-2^{18}-2^{17}-2^{15}-2^{12}-2^{10}-2^9-2^8-2^5-2^4-2^3-2^1$$
and I suspect that for the type B you can find something similar.