I stumbled upon the following identity $$\sum_{k=0}^n(-1)^kC_k\binom{k+2}{n-k}=0\qquad n\ge2$$ where $C_n$ is the $n$th Catalan number. Any suggestions on how to prove it are welcome!
This came up as a special case of a generating function for labeled binary trees. Actually I can directly prove the identity is zero by showing that certain trees don't exist, but I expect that seeing a direct proof will help me find nice closed formulae for other coefficients of the generating function.
My answer uses the same methods as that of tc2718's answer to a very similar problem.
Let $a_n$ be the quantity in question, and let $F(x)=\sum_{n\ge 0}a_nx^n$. We will evaluate $F$ by switching the order of summation, applying the binomial theorem to the inner sum, rearranging a bit, and then recognizing the expression as the Catalan generating function evaluated at $-x-x^2$. $$ \begin{align} F(x) &=\sum_{n\ge 0}\sum_{k=0}^n (-1)^k C_k \binom{k+2}{n-k}x^n \\&=\sum_{k=0}^\infty (-1)^k C_k \sum_{n= k}^\infty\binom{k+2}{n-k}x^n \\&=\sum_{k=0}^\infty (-1)^k C_k \; x^k(1+x)^{k+2} \\&=(1+x)^2\sum_{k=0}^\infty C_k \; (-x-x^2)^k \\&=(1+x)^2 \frac{1-\sqrt{1-4(-x-x^2)}}{2(-x-x^2)} \\&=(1+x)^2 \frac{1-\sqrt{(1+2x)^2}}{2(-x-x^2)} \end{align} $$ Here, we must choose the branch of $\;\sqrt{\cdot}\;$ where $\sqrt{(1+2x)^2}=+(1+2x)$, since the other choice causes a pole at $x=0$, and our definition of $F(x)$ makes it clear there is no pole there. Finally, \begin{align} F(x)=(1+x)^2 \cdot \frac{1-(1+2x)}{-2x(1+x)}=1+x \end{align}
This closed form expression for $F(x)=\sum_{n\ge 0}a_nx^n$ shows that $a_0=a_1=1$, while $a_n=0$ for $n\ge 2$.
This answer uses Umbral calculus.
Given the formal variable $\,c,\,$ define the formal power series linear operator
$$ L\!\!\left[\sum_{n=0}^\infty a_nc^n\right] := \sum_{n=0}^\infty a_nC_n. $$
Define the Catalan number generating function
$$ C(x) := L\left[\frac1{1-c\,x}\right] = \sum_{n=0}^\infty C_n x^n = \frac{1-\sqrt{1-4x}}{2x}. $$
Define the generating function
$$ F(x) := \sum_{n=0}^\infty (-1)^kC_k\binom{k+2}{n-k}x^n. $$
Define the generating function
$$ f(x,y) := \sum_{n=0}^\infty \left(\sum_{k=0}^n (-1)^k{k+2\choose n-k}y^k\right)x^n. $$
Now, by previous definitions
$$ F(x) = L[f(x,c)]. $$
Verify that
$$ f(x,y) = \frac{(1 + x)^2}{ 1 + y (x + x^2) }. $$
Now get
$$ F(x) = (1+x)^2 C(-(x+x^2)) = 1+x $$
from
$$ C(-(x\!+\!x^2)) \!=\! \frac{1\!-\!\sqrt{1\!+\!4(x\!+\!x^2)}} {-2(x\!+\!x^2)} \!=\! \frac{1-(1\!+\!2x)}{-2(x\!+\!x^2)} \!=\! \frac1{1\!+\!x}. $$
Notice, my answer to MSE question 4317353 also uses Umbral calculus to prove another Catalan number recursion identity. Also, the accepted answer to this question is very similar to mine except it does not use the Umbral calculus linear operator.
