It is well known that the sum of squares of all binomial coefficients ${n\choose k}$ with a fixed $n$ is ${2n\choose n}$. That is, $\sum_{k=0}^n {n\choose k}^2 = {2n\choose n}$.
But do we know what the value of the sum of squares of multinomial coefficients is? In particular, I am interested in $\sum_{a,b,c,d} {n\choose a,b,c,d}^2$, where the sum is taken over all 4-tuples ${a,b,c,d}$ of nonnegative integer whose sum is $n$.
Thanks.
Yes, indeed, it is unlike the case $\sum_{a+b=n}\binom{n}{a,b}^2=\binom{2n}n$ which is annihilated by the (first order) operator $\mathcal{L}=(n + 1)N-2(2n+1)$.
If we denote $$f(n):=\sum_{a+b+c+d=n}\binom{n}{a,b,c,d}^2$$ then the sequence satisfies the three-term recurrence $$n^3f(n) = 2(2n-1)(5n^2-5n+2)f(n-1) - 64(n-1)^3f(n-2).$$ For $f(n)$ to have a closed form, the recursive relation should be of first order, which it is not. Further, the operator $\mathcal{L}$, $$\mathcal{L}:=n^3N^2-2(2n-1)(5n^2-5n+2)N+64(n-1)^3$$ is irreducible. Here $Ng(n-2)=g(n-1)$ is the forward shift operation.
By the way, even the following sum does not achieve a closed form (apart from variations): $$\sum_{a+b+c=n}\binom{n}{a,b,c}^2=\sum_{k=0}^n\binom{n}k^2\binom{2k}k= \sum_{m=0}^n\binom{n}m\sum_{k=0}^m\binom{m}k^3.$$
