Combinatorics- Stirling numbers of second kind

Question

I am looking for an approximation of $S(n,n/2)$ where $S$ denotes the second stirling numbers. The wikipedia article gives quite a few approximations but they do not fit my problem. I now tried to Plot the stirling numbers $S(n,n/2)$ in mathematica for $n$ up to about 10000 and fitted them. This way I received $\log(S(n,n/2)) = 0.99n/2\log(n/2) + + 0.82n - 0.62 \log(n)$. This model fits pretty good however I have no idea how to justify it or how to approximate the value of $S(n,n/2)$ in another way. Thanks for any tips!!

Answer 1

Wikipedia gives lower and upper bounds $L(n,k)$ and $U(n,k)$ for $\left\{n \atop k\right\}$. For $k = \frac n2$, dropping some less-significant factors in $L(n,k)$, we get $$\left(\frac{n}{2}\right)^{n/2} < \left\{n \atop n/2\right\} < \binom{n}{n/2} \left(\frac{n}{2}\right)^{n/2}$$ so $$\frac n2 \log_2 \frac n2 < \log_2 \left\{n \atop n/2\right\} < \frac n2 \log_2 \frac n2 + n.$$

This confirms at least the leading term of your estimate, though we need better bounds if we want to improve it and get the coefficient of $n$ right: these bounds are genuinely off by a factor of $2^n \cdot \operatorname{poly}(n)$, which is why I take the log base $2$ above.

Answer 2

Looking at sequence $A007820$ in $OEIS$, there is an asymptotics by Vaclav Kotesovec (2011) which write $$\left\{{2p \atop p}\right\}\sim\frac{2^{2 p-\frac{1}{2}} e^{-p} \left(\frac{p}{(2-z) z}\right)^p}{\sqrt{\pi } \sqrt{p (z-1)}}\qquad \text{where}\qquad z=2+W\left(-\frac{2}{e^2}\right)\approx 1.59362$$ Replacing $z$ by its numerical value, this gives $$\log\left(\left\{{2p \atop p}\right\}\right)\approx p \log (p)+0.820761 p-0.5 \log (p)-0.658184$$

Edit

Repeating the curve fit for $100 \leq p\leq 5000$, what I obtained is $$\log\left(\left\{{2p \atop p}\right\}\right)=1. p \log (p)+0.820757 p-0.49903 \log (p)-0.663801$$