TITLE: Identity with Catalan numbers
QUESTION [17 upvotes]: How would you prove the following identity
$$\sum_{1\ \leq\ j\ <\ j'\ \leq\ n}\
\prod_{k\ \neq\  j,\,j'}^{n}
{\left(\, j + j'\,\right)^{2} \over \left(\, j - k\,\right)\left(\, j' - k\,\right)} =C_{n - 2}
$$
where $C_{k}$ is the $k$-th Catalan number defined as 
$$
C_k\equiv{\left(\, 2k\,\right)! \over k!\left(\, k + 1\,\right)!}$$

REPLY [2 votes]: I am leaving this proofless for now; one day I hope to get the time.
Let $\mathbb{N}$ be the set $\left\{  0,1,2,\ldots\right\}  $. If
$n\in\mathbb{N}$, then I shall use the notation $\left[  n\right]  $ for the
set $\left\{  1,2,\ldots,n\right\}  $.
Let $\mathbb{K}$ be a commutative ring. Fix $k\in\mathbb{N}$. Consider the
polynomial ring $\mathbb{K}\left[  y_{1},y_{2},\ldots,y_{k}\right]  $ in $k$
indeterminates $y_{1},y_{2},\ldots,y_{k}$ over $\mathbb{K}$. A polynomial
$P\in\mathbb{K}\left[  y_{1},y_{2},\ldots,y_{k}\right]  $ is said to be
symmetric if any permutation of the indeterminates $y_{1},y_{2},\ldots
,y_{k}$ leaves it unchanged (that is, $P\left(  x_{\sigma\left(  1\right)
},x_{\sigma\left(  2\right)  },\ldots,x_{\sigma\left(  k\right)  }\right)  =P$
for each permutation $\sigma$ of $\left[  k\right]  $).

Definition 1. Let $P\in\mathbb{K}\left[  y_{1},y_{2},\ldots,y_{k}\right]
$ be a symmetric polynomial. Let $S$ be a $k$-element set. Let $\left(
x_{h}\right)  _{h\in S}$ be a family of $k$ elements in some commutative
  $\mathbb{K}$-algebra $\mathbb{L}$. Then, we define the value $P\left(  \left(
x_{h}\right)  _{h\in S}\right)  \in\mathbb{L}$ as follows: Choose any list
  $\left(  h_{1},h_{2},\ldots,h_{k}\right)  $ containing each element of $S$
  exactly once, and set $P\left(  \left(  x_{h}\right)  _{h\in S}\right)
=P\left(  x_{h_{1}},x_{h_{2}},\ldots,x_{h_{k}}\right)  $. This does not depend
  on the list $\left(  h_{1},h_{2},\ldots,h_{k}\right)  $, since $P$ is
  symmetric (and since any two such lists $\left(  h_{1},h_{2},\ldots
,h_{k}\right)  $ are permutations of each other). Roughly speaking, $P\left(
\left(  x_{h}\right)  _{h\in S}\right)  $ means the result of substituting the
  $k$ values $x_{h}$ with $h\in S$ into $P$, in any order.
Theorem 2. Let $n\in\mathbb{N}$ and $k\in\left\{  0,1,\ldots,n\right\}  $.
  Let $x_{1},x_{2},\ldots,x_{n}$ be $n$ elements of $\mathbb{K}$. Assume that
  for any two distinct elements $i$ and $j$ of $\left[  n\right]  $, the element
  $x_{i}-x_{j}$ of $\mathbb{K}$ is invertible. Let $P\in\mathbb{K}\left[
y_{1},y_{2},\ldots,y_{k}\right]  $ be a symmetric polynomial such that $\deg
P<k\left(  n-k\right)  $. Then,
  \begin{equation}
\sum_{\substack{S\subseteq\left[  n\right]  ;\\\left\vert S\right\vert
=k}}P\left(  \left(  x_{h}\right)  _{h\in S}\right)  \prod_{\substack{i\in
\left[  n\right]  \setminus S;\\j\in S}}\dfrac{1}{x_{j}-x_{i}}=0.
\end{equation}

For the next theorem, we need more notations.

Definition 3. In the following, the word "$k$-part partition" will mean an
  integer partition
  with at most $k$ parts. If $\lambda$ is a $k$-part partition, then we let
  $s_{\lambda}$ be the Schur
  polynomial in $\mathbb{K}
\left[  y_{1},y_{2},\ldots,y_{k}\right]  $ corresponding to $\lambda$. It is
  well-known that the family of the $s_{\lambda}$ (where $\lambda$ ranges over
  all $k$-part partitions) is a basis of the $\mathbb{K}$-module of symmetric
  polynomials in $\mathbb{K}\left[  y_{1},y_{2},\ldots,y_{k}\right]  $. This
  basis is called the Schur basis. If $\mu$ is any $k$-part partition, and if
  $P$ is a symmetric polynomial in $\mathbb{K}\left[  y_{1},y_{2},\ldots
,y_{k}\right]  $, then $\left[  s_{\mu}\right]  P$ shall mean the coefficient
  of $s_{\mu}$ in the expansion of $P$ in this Schur basis.

Now, we can state the following extension of Theorem 2:

Theorem 4. Let $P\in\mathbb{K}\left[  y_{1},y_{2},\ldots,y_{k}\right]  $
  be a symmetric polynomial such that $\deg P\leq k\left(  n-k\right)  $. Let
  $\square$ denote the rectangular partition $\left(  \underbrace{n-k,n-k,\ldots
,n-k}_{k\text{ entries}}\right)  $. Then,
  \begin{equation}
\sum_{\substack{S\subseteq\left[  n\right]  ;\\\left\vert S\right\vert
=k}}P\left(  \left(  x_{h}\right)  _{h\in S}\right)  \prod_{\substack{i\in
\left[  n\right]  \setminus S;\\j\in S}}\dfrac{1}{x_{j}-x_{i}}=\left[
s_{\square}\right]  P.
\end{equation}

[EDIT: Theorem 4 is essentially Theorem 2 in Ömer Egecioglu, On Böttcher's mysterious identity, Australasian Journal of Combinatorics 43, 2009, pp. 307--316. I am saying "essentially" because Egecioglu only states it for symmetric polynomials $P$ that are homogeneous of degree $k\left(  n-k\right)$, but the proof applies in the general case.
Also, Theorem 4 is essentially Theorem 1 in Dang Tuan Hiep, Identities involving (doubly) symmetric polynomials and integrals over Grassmannians, arXiv:1607.04850v3. This time, I'm saying "essentially" because Hiep never mentions Schur polynomials and, instead of $\left[ s_{\square}\right]  P$, uses the coefficient of $y_1^{n-1} y_2^{n-2} \cdots y_k^{n-k}$ in the polynomial $P \cdot \prod_{i<j} \left(y_i-y_j\right)$ (modulo typos in his paper); but it follows easily from the alternant definition of Schur polynomials that this latter coefficient is precisely $\left[ s_{\square}\right]  P$.]
Theorem 4 has the following consequence:

Corollary 5. The sum
  \begin{equation}
\sum_{\substack{S\subseteq\left[  n\right]  ;\\\left\vert S\right\vert
=k}}\prod_{\substack{i\in\left[  n\right]  \setminus S;\\j\in S}}\dfrac
{\sum_{h\in S}x_{h}}{x_{j}-x_{i}}
\end{equation}
  is the number of standard Young
  tableaux of
  rectangular shape $\left(  \underbrace{n-k,n-k,\ldots,n-k}_{k\text{ entries}
}\right)  $.

My proof of Theorem 4 uses Vandermonde determinants
and the alternant definition of Schur functions (see John R. Stembridge, A
Concise Proof of the Littlewood-Richardson
Rule
for all you need to know). Corollary 5 is obtained from Theorem 4 by setting $P = \left(y_1 + y_2 + \cdots + y_k\right)^{k\left(n-k\right)}$, because of the well-known fact that every $m \in \mathbb{N}$ satisfies
\begin{align}
& \left(y_1 + y_2 + \cdots + y_k\right)^m \\
&= \sum_{\lambda \text{ is a } k\text{-part partition of } m} \left(\text{number of standard tableaux of shape } \lambda\right) s_{\lambda} .
\end{align}
If you set $k=2$, $\mathbb{K}=\mathbb{Q}$ and $x_{i}=i$ in Corollary 5, you
recover the original question, since the number of standard Young tableaux of
rectangular shape $\left(  n-2,n-2\right)  $ is $C_{n-2}$ (indeed, these
tableaux are in bijection with Dyck paths with $n-2$ upsteps and $n-2$ downsteps).