Let $N$ be an even integer. Assume a class of functions $g(b_1,...,b_N)$ with $b_i\neq b_j$ for all $i,j$. These functions are defined recursively by the symmetric function $g(b_i,b_j)=g(b_j,b_i)$ with the formula \begin{align}\tag{*}\label{1} g(b_1,...,b_N)=\sum_{l=1}^{\frac{N-2}{2}}\frac{g(b_1,...,b_{2l})g(b_{2l+1},...,b_N)-g(b_2,...,b_{2l+1})g(b_{2l+2},...,b_N,b_1)}{(b_1-b_{2l+1})(b_2-b_N)}. \end{align} It is knwon that any function has a cyclic property $g(b_1,...,b_N)=g(b_2,...,b_N,b_1)$ and a reflection property $g(b_1,...,b_N)=g(b_N,b_{N-1},...,b_1)$.
The recursive formula \eqref{1} has a assigned variable $b_1$. Now the goal is to proof that the $g(b_1,...,b_N)$ is independent of the choice of the assigned variable, and to find a formula for $g(b_1,...,b_N)$ directly depending on $g(b_i,b_j)$.
The recursive formula shows that the function splits in noncrossing partitions on a circle (like the combinatorial problem of the Catalan numbers), where the independence of the choice of the assigned variable is due to the denominator.