While exploring the Baxter sequences from my earlier MO post, I obtained a rather curious identity (not listed on OEIS either). I usually try to employ the Wilf-Zeilberger (WZ) algorithm to justify such claims. To my surprise, WZ offers two different recurrences for each side of this identity.
So, I would like to ask:
QUESTION. Is there a conceptual or combinatorial reason for the below equality?
$$\frac1n\sum_{k=0}^{n-1}\binom{n+1}k\binom{n+1}{k+1}\binom{n+1}{k+2}
=\frac2{n+2}\sum_{k=0}^{n-1}\binom{n+1}k\binom{n-1}k\binom{n+2}{k+2}.$$
Remark 1. Of course, one gets an alternative formulation for the Baxer sequences themselves:
$$\sum_{k=0}^{n-1}\frac{\binom{n+1}k\binom{n+1}{k+1}\binom{n+1}{k+2}}{\binom{n+1}1\binom{n+1}2}
=2\sum_{k=0}^{n-1}\frac{\binom{n+1}k\binom{n-1}k\binom{n+2}{k+2}}{\binom{n+1}1\binom{n+2}2}.$$
Remark 2. Yet, here is a restatement to help with combinatorial argument:
$$\sum_{k=0}^{n-1}\binom{n+1}k\binom{n+1}{k+1}\binom{n+1}{k+2}
=2\sum_{k=0}^{n-1}\binom{n+1}k\binom{n}k\binom{n+1}{k+2}.$$
Best Answer
Just playing around with it: The RHS multiplied by $n$ is the same as
$$2 \sum_{k=0}^{n-1} \binom{n+1}{k} \binom{n}{k} \binom{n+1}{k+2}.$$
Subtracting this from $n$ times the LHS gives
$$\sum_{k=0}^{n-1} \binom{n+1}{k} \binom{n+1}{k+2} \left( \binom{n}{k+1} - \binom{n}{k} \right).$$
Now you check that replacing $k$ with $n-1-k$ changes the sign of the summand, so the sum is zero.