How does $\tbinom{4n}{2n}$ relate to $\tbinom{2n}{n}$? I got this question in my mind when I was working on a solution to factorial recurrence and came up with this recurrence relation: $$(2n)!=\binom{2n}{n}(n!)^2$$ which made me wonder: is there also a recurrence relation for $\tbinom{4n}{2n}$ in terms of $\tbinom{2n}{n}$? Please use no factorials greater than $(2n)!$, preferably not greater than $n!$.

Here is an estimate that gives a good approximation of $\binom{4n}{2n}$ in terms of $\binom{2n}{n}$.
Using the identity
$$
(2n-1)!!=\frac{(2n)!}{2^nn!}\tag{1}
$$
it is straightforward to show that
$$
\frac{\binom{4n}{2n}}{\binom{2n}{n}}=\frac{(4n-1)!!}{(2n-1)!!^2}\tag{2}
$$
Notice that
$$
\begin{align}
\frac{(2n-1)!!}{2^nn!}
&=\frac{2n-1}{2n}\frac{2n-3}{2n-2}\frac{2n-5}{2n-4}\cdots\frac12\\
&=\frac{n-\frac12}{n}\frac{n-\frac32}{n-1}\frac{n-\frac52}{n-2}\cdots\frac{\frac12}{\;1}\\
&=\frac1{\sqrt\pi}\frac{\Gamma(n+\frac12)}{\Gamma(n+1)}\tag{3}
\end{align}
$$
By Gautschi's Inequality, we have
$$
\frac1{\sqrt{n+1}}\le\frac{\Gamma(n+\frac12)}{\Gamma(n+1)}\le\frac1{\sqrt{n}}\tag{4}
$$
Thus, $(3)$ and $(4)$ yield
$$
\frac1{\sqrt{\pi(n+1)}}\le\frac{(2n-1)!!}{2^nn!}\le\frac1{\sqrt{\pi n}}\tag{5}
$$
and
$$
\frac1{\sqrt{\pi(2n+1)}}\le\frac{(4n-1)!!}{2^{2n}(2n)!}\le\frac1{\sqrt{\pi 2n}}\tag{6}
$$
Dividing $(6)$ by $(5)$ gives
$$
\sqrt{\frac{n}{2n+1}}\le\frac{(4n-1)!!}{(2n-1)!!^2}4^{-n}\le\sqrt{\frac{n+1}{2n}}\tag{7}
$$
Combine $(2)$ and $(7)$ to get
$$
4^n\sqrt{\frac{n}{2n+1}}\le\frac{\binom{4n}{2n}}{\binom{2n}{n}}\le4^n\sqrt{\frac{n+1}{2n}}\tag{8}
$$
or asymptotically
$$
\binom{4n}{2n}\sim\frac{4^n}{\sqrt2}\binom{2n}{n}\tag{9}
$$
