TITLE: An identity involving Catalan numbers and binomial coefficients.
QUESTION [7 upvotes]: 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.

REPLY [5 votes]: 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.