Solutions for $n$? Use Stirling approximations if needed

Question

$$(2n)! = a^{2n}$$ where $a \in \mathbb R$, and $n \in \mathbb N$.

This is relevant because of a research question I'd asked and received an answer to by Sotiris here

Answer 1

If you look at this old question of mine, @robjohn's approximation gives

$$\large n\sim \frac a 2 \, e^{1+W(t)}-\frac 14 \qquad \text{where} \qquad t=-\frac{\log (2 \pi a)}{2 e a}$$ $W(t)$ being the principal branch of Lambert function.

Using $a=1234.56789$, it gives as a real $n=\color{red}{1675.4611}4575$.

Answer 2

If we use Stirling as you suggest we have $$(2n)!=a^{2n}\\\frac {(2n)^{2n}}{e^{2n}}\sqrt {4\pi n}=a^{2n}\\ (4\pi n)^{1/(4n)}=\frac {ae}{2n}$$ Let us define $x=\frac 1n$. This becomes $$(\frac {4\pi}x)^{(x/4)}=\frac 12aex\\ x=\frac 2{ae}(\frac {4\pi}x)^{(x/4)}$$ Some might find a way to use the Lambert W function, but I see this as a fixed point iteration. Choose a starting value $x_0$, plug it into the right, compute $x_1$, plug that in, and iterate to convergence. I just started with $x_0=0.1$ and it converged quickly for $a=3,10,50,1000,1000000$ Starting with $x_0=\frac 1a$ may be even better.