Approximation of Stirling numbers of the second kind ${2n \brace n}$

Question

I want an approximation of ${2n \brace n}$ as $n\to\infty$, also ${\cdot\brace\cdot }$ is the Stirling numbers of the second kind.

Right now, I know an evaluation \begin{equation} {2n \brace n}=O\left(n^n\binom{2n}{n}\right) \end{equation} holds up, but I don't know an accurate evaluation or approximation.

Could you help me finding a good approximation? I would appreciate it if you do!

Answer 1

OEIS A007820 seems to give an asymptotic formula from Vaclav Kotesovec of $$\left(\dfrac{4n}{e z(2-z)}\right)^n\Big/\sqrt{2\pi n(z-1)}$$

where $z = 1.59362426...$ is a root of the equation $\exp(z)(2-z)=2$

Answer 2

In this paper from year $2016$ is given the more general approximation $$\mathcal{S}_n^{(m)}=\frac 1{2 \sqrt{\pi }}\,\frac{n! \,\left(e^R-1\right)^m }{R^n\, m!\, \sqrt{m H R}}\,\,\left(1+O\left(\frac 1 {m^{1/3} } \right)\right)$$ where

$$R=k+W\left(-k\,e^{-k}\right) \qquad \text{and} \qquad H=\frac{k \left(1+W\left(-k\,e^{-k} \right)\right)}{2 \left(k+W\left(-k\,e^{-k} \right)\right)}\qquad \text{and} \qquad k=\frac n m$$

Using it for the specific case $$\mathcal{S}_{2 n}^{(n)}\sim \frac{(2 n)! \left(-W\left(-\frac{2}{e^2}\right) \left(2+W\left(-\frac{2}{e^2}\right)\right)\right)^{-n}}{2 \sqrt{\pi n}\, n! \sqrt{ 1+ W\left(-\frac{2}{e^2}\right)} } \tag1$$ while the asymptotic formula from Vaclav Kotesovec write $$\mathcal{S}_{2 n}^{(n)}\sim \frac{2^{2 n-\frac{1}{2}} n^{n-\frac{1}{2}} \left(-e W\left(-\frac{2}{e^2}\right) \left(W\left(-\frac{2}{e^2}\right)+2\right)\right)^{-n}}{\sqrt \pi \sqrt{1+ W\left(-\frac{2}{e^2}\right)}}\tag 2$$

Trying to improve the approximation, a purely empirical approach gives $$\mathcal{S}_{2 n}^{(n)}\sim \frac{(2 n)! \left(-W\left(-\frac{2}{e^2}\right) \left(2+W\left(-\frac{2}{e^2}\right)\right)\right)^{-n}}{2 \sqrt{\pi n}\, n! \sqrt{ 1+ W\left(-\frac{2}{e^2}\right)} } \left(1-\frac{1}{\pi \sqrt{\pi } \left(2+W\left(-\frac{2}{e^2}\right)\right) n}\right)\tag3$$

Computing their logarithms, it seems that $(1)$ is slightly better than $(2)$.

$$\left( \begin{array}{ccccc} n & \text{from }(2) & \text{from }(1) & \text{from }(3) & \text{exact} \\ 5 & 10.6881 & 10.6798 & 10.6570 & 10.6578 \\ 10 & 29.4240 & 29.4198 & 29.4085 & 29.4089 \\ 15 & 50.9200 & 50.9172 & 50.9096 & 50.9100 \\ 20 & 74.1738 & 74.1717 & 74.1661 & 74.1663 \\ 25 & 98.7233 & 98.7216 & 98.7171 & 98.7173 \\ 30 & 124.300 & 124.299 & 124.295 & 124.295 \\ 35 & 150.728 & 150.727 & 150.724 & 150.724 \\ 40 & 177.883 & 177.882 & 177.879 & 177.879 \\ 45 & 205.673 & 205.672 & 205.669 & 205.669 \\ 50 & 234.025 & 234.024 & 234.022 & 234.022 \end{array} \right)$$

In fact $$\frac{(1)}{(2)}=\frac{e^n \Gamma \left(n+\frac{1}{2}\right)}{\sqrt{2 \pi }\,n^n}=1-\frac{1}{24 n}+\frac{1}{1152 n^2}+O\left(\frac{1}{n^3}\right)$$