Number of bits in a stirling number

Question

I know that the number of bit required for $N!$ is $\Theta(n\log n)$ by the stirling approximation. But what is the number of bits required for a stirling number of the first kind of the form $\left[\begin{smallmatrix} n \\ k \end{smallmatrix}\right]$?

I know that we can upper bound the stirling numbers since, $\sum_{k=0}^{n}\left[\begin{smallmatrix} n \\ k \end{smallmatrix}\right] = n!$

Answer 1

It's a good question. I don't know of a nice satisfying answer for the full two-parameter space. Wikipedia has good introductory discussion. You'd probably have to dive into papers to see what cases are well known if you're not interested in one of the simple ones they list.