@EmilJeřábek I did come up with a formula for the size, but the problem is that the formula is a sum, indexed by varying values for $n_1$, because for every two unary symbols I add to the degree profile a binary and a nullary symbol is taken out. Each summand of this is then a multiple of the Catalan number for the number of binary symbols. I call this "a problem", because the sums are not any easier to compute than the recurrences, and there is very little room for simplification when computing the quotients for the probabilities.

The trees are only labelled with the function symbols. Two trees are the same if they have the same shape and the same labelling with function symbols. "calculate to floating point precision" is not applicable here, as we ran out of exponents, e.g. the one number I mentioned I computed was roughly 10^8000.