Consider a function $C(n, d)$ which counts the number of well formed, i.e, balanced, parenthetical 'words' of length $n$ and maximal nested depth $\leq d$.
That is, $(())$ has $n = 4, d = 2$. $()((())())$ has $n = 10, d = 3$ and so on.
$C(n, d)$ counts all well formed strings above (i.e, $($ and $)$ are balanced), of length exactly $n$ and depth at most $d$ (that is, smaller depths are allowed).
There should be a recursive expression for $C(n, d)$.
The base cases are easy enough:
$C(n, d) = 0$ if $n$ is odd regardless.
$C(n, d) = 0$ if $d = 0$ but $n> 0$
$C(n, d) = 1$ if $d = n= 0$ (the empty string)
Now for the recursive step.
I've been thinking about this but there are a lot of overlap which makes things difficult.
For instance, I first guessed $\sum_{i}\sum_j C(n- i, d-j)C(i, j)$ That is, you mixing all strings of length $i$ with strings of length $n-i$ should give you strings of length $n$. I think the depth in general will remain $\leq d$ without problems. But the problem with this approach is that I'm pretty sure there will be plenty of double/extra counting of the same strings more than once.
I have a few other candidates but again I think they largely double count in general.
I was also thinking if I could express it solely in terms of $C(n-2, d-1)$ as ${n-2 \choose 2}C(n-2, d-2)$ (placing $(, )$ at any positions in an $n-2$ string) but then this misses out on potential $d$ depth strings. But doing ${n-2 \choose 2}C(n-2, d)$ might inculcate depth of $d$.
