While exploring the Baxter sequences from [my earlier MO post](https://mathoverflow.net/questions/412324/closed-formula-for-1-baxter-sequences), 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:
$$\sum\_{k=0}^{n-1}\binom{n+1}k\binom{n+1}{k+1}\binom{n+1}{k+2}
=2\sum\_{k=0}^{n-1}\binom{n+1}k\binom{n}k\binom{n+1}{k+2}.$$



A rather curious identity on sums over triple binomial terms
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.


