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 $n!$.

There can't be any multiplicative formula of the sort you describe for $4n\choose 2n$ in terms of $2n\choose n$, because there are prime factors of the former that aren't factors of the latter.  By Bertrand's Postulate there's a prime number between $k=2n$ and $2k$($=4n$); that prime will be a factor of $4n\choose 2n$, but can't be a factor of $2n\choose n$ or any factorial of the form $(2n)!$ or less.
