# Sum of Product of Central Binomial Coefficients

The following constants have appeared in my research and I was wondering if they have a simpler expression. I have computed some values and I don't see any obvious combinatorial answer.

$$c_n:= n! \sum_{i+j+k+l=n} \dbinom{2i}{i}\dbinom{2j}{j}\dbinom{2k}{k}\dbinom{2l}{l}$$ Here $i,j,k,l$ are nonnegative integers. At the moment I don't think I need to know anything in particular about these constants, but it would be nice to know if they had a nice expression.

• It's not a constant, it's a sequence. – Robert Israel Mar 19 '18 at 4:16

Using generating functions, it's not hard to show that $c_n = 4^n (n+1)!$.
• Thanks! Looking back, there must have been an error in my program. I had written something which correctly computed the first three values of $c_n$ but $c_4$ for example had very large prime factors. I'm now getting your numbers, and I was able to show it with a generating function! – Dylan Yott Mar 19 '18 at 16:56
Following Robert Israel's suggestion and using the generating function here: $$\frac{1}{\sqrt{1-4x}} = \sum_{k=0}^{\infty} \dbinom{2k}{k}x^k.$$ We're looking for the $n^{th}$ coefficient of the fourth power of this generating function. So $$\left( \frac{1}{\sqrt{1-4x}} \right)^4 = \frac{1}{(1-4x)^2}.$$ Recall that $$\frac{1}{1-x} = \sum_{k=0}^\infty x^k$$ so that when we differentiate $$\frac{1}{(1-x)^2} = \sum_{k=0}^{\infty} (k+1)x^{k}$$ and substitute in $4x$ $$\frac{1}{(1-4x)^2} = \sum_{k=0}^{\infty} (k+1)4^k x^{k}.$$ We see that the $n^{th}$ term is $(n+1)4^n$, so that when we multiply by $n!$ as in the definition of $c_n$ we get $$c_n=4^n(n+1)!.$$ Thanks again Robert!