For $n=0,1,2,\ldots$, let us define the polynomial $$S_n(x):=\sum_{k=0}^n\binom{x/2}k\binom{(x-1)/2}k\binom{-(x+1)/2}{n-k}\binom{-(x+2)/2}{n-k}.$$ Such polynomials occur in some series for $1/\pi$ discoveried by me in 2011, see Conjecture 4 of my paper List of conjectural series for powers of $\pi$ and other constants.
In 2011, I found the following two novel identities for the polynomials $S_n(x)$: $$S_n(x)=\binom{-1/2}n\sum_{k=0}^n(-1)^k\binom{x}k^2\binom{-1-x}{n-k}\tag{1}$$ and $$\begin{aligned}&\sum_{k=0}^n\binom nk(-1)^kS_k(x) \\=&\sum_{k=0}^n\binom{x/2}k\binom{-(x+1)/2}k\binom{(x-1)/2}{n-k}\binom{-(x+2)/2}{n-k}.\end{aligned}\tag{2}$$
Note that $(1)$ implies the symmetric identity $$\sum_{k=0}^n(-1)^k\binom xk^2\binom{-1-x}{n-k}=\sum_{k=0}^n(-1)^k\binom{-1-x}k^2\binom x{n-k},$$ which was proved without computer in my paper Supercongruences involving dual sequences, Finite Fields Appl. 46(2017), 179-216.
By the Zeilberger algorithm, if $u_n$ is the left-hand side or the right-hand side of $(1)$ then we have the recurrence relation: \begin{align}4(n+2)^3u_{n+2}=&2(2n+3)(2n^2+6n+x^2+x+5)u_{n+1} \\&-(n+1)(2n+1)(2n+3)u_n.\end{align} Similarly, if $v_n$ denotes the left-hand side or the right-hand side of $(2)$, then we have the recurrence \begin{align}4(n+2)^3v_{n+2}=&2(2n+3)(2n^2+6n-x^2-x+5)v_{n+1} \\&-(n+1)(2n-2x+1)(2n+2x+3)v_n.\end{align} So both $(1)$ and $(2)$ have proofs via a computer.
Question. How to provide human proofs of the identities $(1)$ and $(2)$?
Your comments are welcome!
