TITLE: Prove $\sum_k{2k\choose k}^2{2\left(n-k\right)\choose n-k}^2=\sum_k(-1)^k16^k{n-k \choose k}{2\left(n-k\right)\choose n-k}^3$
QUESTION [6 upvotes]: So I've come across this one :
$$\forall n\in\mathbb{N},\sum\limits_{k=0}^{n}\left[{2k\choose k}{2\left(n-k\right)\choose n-k}\right]^2=\sum\limits_{k=0}^{\left\lfloor{\frac{n}{2}}\right\rfloor}\left(-1\right)^k16^k{n-k \choose k}{2\left(n-k\right)\choose n-k}^3$$
It looks nice, but since I am not really familiar with combinatorics, I have no clue as how to prove it.
What would you suggest ?

REPLY [3 votes]: Not  an answer  but too  long for  a comment.  The Maple  sumtools
package features an implementation of Zeilberger's algorithm, exported
through  the  sumrecursion command.  This  will  produce the  same
recurrence for the LHS and the RHS. Then we just need to check the two
initial values. This is the Maple transcript.

> with(sumtools);
[Hypersum, Sumtohyper, extended_gosper, gosper, hyperrecursion, hypersum,

    hyperterm, simpcomb, sumrecursion, sumtohyper]

> sumrecursion(binomial(2*k,k)^2*binomial(2*n-2*k,n-k)^2, k,S(n));
              3                             2                             3
   256 (n - 1)  S(-2 + n) - 8 (2 n - 1) (2 n  - 2 n + 1) S(n - 1) + S(n) n

> sumrecursion((-1)^k*16^k*binomial(n-k,k)*binomial(2*n-2*k,n-k)^3, k, S(n));
              3                             2                             3
   256 (n - 1)  S(-2 + n) - 8 (2 n - 1) (2 n  - 2 n + 1) S(n - 1) + S(n) n

> A := n -> add(binomial(2*k,k)^2*binomial(2*n-2*k,n-k)^2, k=0..n);
                                 2                           2
   A := n -> add(binomial(2 k, k)  binomial(2 n - 2 k, n - k) , k = 0 .. n)

> seq(A(n), n=1..5);
                          8, 88, 1088, 14296, 195008

> B := n -> add((-1)^k*16^k*binomial(n-k,k)*binomial(2*n-2*k,n-k)^3, k=0..n);
B := n ->

            k   k                                              3
    add((-1)  16  binomial(n - k, k) binomial(2 n - 2 k, n - k) , k = 0 .. n)

> seq(B(n), n=1..5);
                          8, 88, 1088, 14296, 195008

The common recurrence is
$$256\, \left( n-1 \right) ^{3}S \left( n-2 \right) -8\, \left( 2\,n-1
 \right)  \left( 2\,{n}^{2}-2\,n+1 \right) S \left( n-1 \right) +S
 \left( n \right) {n}^{3} = 0.$$
This  is a  classic example  from the  foreword to  the book  A=B by
Petkovsek,  Wilf, and  Zeilberger,  and identified  as  such at  OEIS
A036917.