Dtjytyk

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 $log(S(n,n/2)) = 0.99n/2log(n/2) + + 0.82n - 0.62 log(n)$. 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!!

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

Edit

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

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

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.

For large $n$, and for $v=n/m $ constant,
we have the following known precise asymptotic approximation (my deduction here):

$$ S_{n,m} approx binom{n}{m} frac{(n-m)!}{sqrt{2 pi n (lambda -v+1)}}
frac{(2-lambda)^m}{lambda^{n-m}} tag1$$

where $lambda$ is the positive root of $ lambda = v (1-e^{-lambda}) $ (explicitly: $lambda=v+W(-v /e^{v})$ where $W()$ is the Lambert W function, first branch).

In our case, with $n=2m$, $v=2$, $lambda= 1.593624260040cdots$, so

$$ S_{2m,m}approx binom{2m}{m} frac{m!}{sqrt{4 pi m (lambda -1)}}frac{1}{(2lambda - lambda^2)^m} tag2$$

Plugging the Stirling approximation for the factorials we get

$$ log(S_{2m,m}) approx m log m + beta , m - frac12 log(m) + gamma + o(1) tag3$$
with $beta = -log( frac{2 lambda -lambda^2}{4})-1 =0.820760609$, and
$gamma = -frac12 log(2 pi ,(lambda-1)) = -0.65818417389$

This coincides with Claude Leibovici's answer.

Some computed values of $log(S_{2m,m})$ along with the approximation $(3)$:

 m    log(S)      approx     abs err

 10  29.408950    29.423980  0.015031

 20  74.166315    74.173807  0.007493

 40  177.879238  177.882979  0.003741

 60  292.198461  292.200954  0.002493

 80  413.371913  413.373782  0.001869

100  539.630815  539.632310  0.001495

120  669.937107  669.938352  0.001246

140  803.606351  803.607419  0.001068

160  940.152803  940.153737  0.000934

180 1079.213650 1079.214480  0.000830

200 1220.507505 1220.508252  0.000747