In the classical problem of bracketing $n$ numbers, I know the response is $C_{n-1}$. I find this $$C_{n-1}=\sum_{i=1}^{\left\lfloor\frac{n}{2}\right\rfloor}(-1)^{i+1}\binom{n-i}{i}C_{n-1-i}$$ but I can't prove that. Can someone help me? Regards.

Let $$C(x) = \sum_{n=1}^\infty C_{n-1} x^n = \frac{1-\sqrt{1-4x}}{2}.$$ Then the identity in question follows easily from $C(x(1-x)) = x$.

ear-vertexof a triangulation $T$ if no diagonal of $T$ contains $v$. Then, every triangulation has at least one ear-vertex (why?), and the ear-vertices of a given triangulation form a lacunar subset of the set of all vertices (where "lacunar" means that it contains no two adjacent vertices). For every nonnegative integer $i$, there are precisely $\binom{n-i}{i}$ lacunar subsets of the set of all vertices of $N$, and if $L$ is any such subset, then there exist precisely ... $\endgroup$