Let $ f: (\mathbb{N}, \mathbb{N}) \to \mathbb{N} $ be defined recursively as follows:
$ f(0, 0) = 1 $
$ f(0, n) = 0 \quad $ for $ n > 0 $
$ f(s, n) = \sum_{k=0}^{n} \binom{n}{k} (n-k)^{n-k} f(s-1,k) \quad $ for $ s > 0 $
Using this definition, it is impractical to calculate $ f(s, n) $ for large $ s $ and $ n $, e.g. $ f(300, 10^6) $, even with a computer. Is there a closed-form approximation for $ f(s, n) $ or $ log \, f(s, n) $ that is easy to compute for large values?
For reference, here are some values of $ f(s, n) $:
s| 0 1 2 3 4 5
n +-------------------------------------------------
0 | 1 1 1 1 1 1
1 | 0 1 2 3 4 5
2 | 0 4 10 18 28 40
3 | 0 27 78 159 276 435
4 | 0 256 824 1848 3496 5960
5 | 0 3125 10970 26595 54020 98345
6 | 0 46656 176112 456048 984384 1896480
7 | 0 823543 3309110 9073911 20655796 41828255
Background:
For all $ n $, $ f(1, n) = n^n $. $ f (2, n) $ is A063170 on OEIS and $ f(3, n) $ is A089901. (The higher columns don't seem to be on OEIS.) The diagonal is A239761.
$ f(s, n) $ can equivalently be defined as a combinatorial sum. Let the alphabet $ A_s $ be any set of size $ s $ . $ (A_s)^n $ is then the set of words over this alphabet with length $ n $ . Let $ count_w(a) $ be the number of occurrences of a character $ a $ in a word $ w $. Then we have:
$$ f(s, n) = \sum_{w \in (A_s)^n} \prod_{a \in A_s} count_w(a)^{count_w(a)} $$
$ 0^0 = 1 $ in this whole question.