Interested by this question, I wondered how easily we could "solve" for $n$ the equation $${2n\brack n}=k$$ where the left hand side is the unsigned Stirling number of the first kind.
I really do not know if this problem has been already adressed.
My idea was to take advantage of the splendid approximation proposed by Vaclav Kotesovec in $2011$ $$\color{blue}{{2n\brack n}\sim\frac 1 {\sqrt {2\pi}}\left(\frac{2n}{e(1-z) z}\right)^n \sqrt{\frac{1-z}{n (2 z-1)}}}\qquad \text{where}\qquad \color{blue}{z=1+\frac{1}{2 W_{-1}\left(-\frac{1}{2 \sqrt{e}}\right)}}$$
To make the problem better conditioned, it was reset as : find the zero of function $$f(n)=\log\left({2n\brack n}\right)-\log(k)$$ In a first step, I performed a quick and dirty regression work (using $n=10,20,\cdots,990,1000$) to get (using whole numbers to look nicer) $$\log\left({2n\brack n}\right) \approx \frac{3756}{1075} n^{1424/1267}-\frac{30042}{1003}$$ which hides a logarithmic term somewhere. For this range, $R^2 > 0.999999$.
This gives an easy and simple first estimate $n_0$ of the solution.
Now, naming $g(n)$ Vaclav Kotesovec's approximation and trying to find the zero of function $$h(n)=\log(g(n))-\log(k)$$ the first iterate of Newton method is just $$n_1=n_0-\frac{\log(g(n_0))-\log(k)}{\log \left(\frac{2 n_0}{z(1-z)}\right)-\frac{1}{2 n_0} }$$ and this seems to be a quite good approximation.
For example,
- using $k=10^{500}$, we have $n_0=177.906$ and $n_1=178.535$ while $${356\brack 178}\approx 1.838 \times 10^{498}\qquad \text{and}\qquad {358\brack 179}\approx 3.213 \times 10^{501}$$
- using $k=10^{1500}$, we have $n_0=465.715$ and $n_1=465.557$ while $${930\brack 465}\approx 9.144 \times 10^{1497}\qquad \text{and}\qquad {932\brack 466}\approx 4.176 \times 10^{1501}$$
- using $k=10^{2500}$, we have $n_0=731.440$ and $n_1=731.151$ while $${1462\brack 731}\approx 2.627 \times 10^{2499}\qquad \text{and}\qquad {1464\brack 732}\approx 1.886 \times 10^{2503}$$
- using $k=10^{5000}$, we have $n_0=1352.13$ and $n_1=1355.57$ while $${2710\brack 1355}\approx 4.597 \times 10^{4997}\qquad \text{and}\qquad {2712\brack 1356}\approx 6.117 \times 10^{5001}$$
Just out of curiosity, I did the same for ${3n\brack n}$ and ${4n\brack n}$ for which Vaclav Kotesovec also provided extremely good approximations.
Do you know if similar things have be done earlier ?