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.
This seems oddly similar to an identity by the discoverer of the Catalan Numbers, Ming Antu, but I think the indices are shifted. They both seem to work. Per Pak's history in Stanley's Catalan Numbers, Ming Antu wrote:
$$ C_1=1, \ \ C_2=2, \ \ C_{n+1}= \sum_{k\ge 0} (-1)^k \binom{n+1-k}{k+1} C_{n-k} $$
That's slightly vague on the bounds, and unnecessarily excludes $C_1$ and $C_2$, so let's spiff up Ming Antu as: \begin{equation} C_0=1, \ \ C_{n+1}= \sum_{k=0}^{\lfloor n/2 \rfloor} (-1)^k \binom{n+1-k}{k+1} C_{n-k} \end{equation} I think it's easier to get as a matrix multiplication. \begin{equation} \begin{pmatrix} C_1 \\ C_2 \\ C_3 \\ C_4 \\ C_5 \\ C_6 \\ C_7 \end{pmatrix} \!\!=\!\! \begin{pmatrix} \binom{1}{1} & . & . & . & . & . & . \\ . & \ \ \binom{2}{1} & . & . & . & . & . \\ . & -\binom{2}{2} & \ \ \binom{3}{1} & . & . &. & . \\ . & . & -\binom{3}{2}& \ \ \binom{4}{1} & . &. & . \\ . & . & \ \ \binom{3}{3} & -\binom{4}{2} & \ \ \binom{5}{1} &. & . \\ . & . & . & \ \ \binom{4}{3} & -\binom{5}{2} & \ \ \binom{6}{1} & . \\ . & . & . & -\binom{4}{4} & \ \ \binom{5}{3} & -\binom{6}{2} & \ \ \binom{7}{1} \\ \end{pmatrix} \begin{pmatrix} C_0 \\ C_1 \\ C_2 \\ C_3 \\ C_4 \\ C_5 \\ C_6 \end{pmatrix} \end{equation} Let's call the matrix $P$; we have $P_{rc}=(-1)^{r-c} \binom{c}{r-c+1}$ with zero for the binomial coefficients outside the usual range.
The Parsnip identity in question is: $$\sum_{k=0}^n (-1)^k \binom{k+2}{n-k} C_k = 0\quad n\ge2$$
This generates binomial coefficients with top smaller than bottom, zero, which can probably be avoided but let that be Parsnip's problem. The last term is $(-1)^n \binom{n+2}{0} C_n= \pm C_n$ so we can view this as a recurrence similar to Ming Antu's. Let's sort terms and put them side by side as:
CatalanNumber = ParsnipIdentity = MingAntuRecurrence
\begin{align*} C_2 &= -\binom{2}{2} C_0 + \binom{3}{1} C_1 &&= \binom{2}{1} C_{1} \\ C_3 &= - \binom{3}{2}C_1 + \binom{4}{1}C_2 &&= - \binom{2}{2} C_1 + \binom{3}{1} C_2 \\ C_4 &= \binom{3}{3}C_1 - \binom{4}{2}C_2 - \binom{5}{1} C_3 &&= - \binom{3}{2} C_2+ \binom{4}{1} C_3 \\ C_5 &= \binom{4}{3}C_2 - \binom{5}{2} C_3 + \binom{6}{1} C_4 &&= \binom{3}{3} C_2 - \binom{4}{2} C_3 + \binom{5}{1} C_4 \end{align*}
Wow, it seems like there are two different ways to multiply that funky Pascal triangle matrix $P$ with the Catalans to get the Catalans. Ming Antu is:
$$[C_1, C_2, \ldots]^T = P [C_0, C_1, \ldots]^T$$
The Parsnip Identity is :
$$[?, ?, C_2, C_3, \ldots]^T = P [0, C_0, C_1, \ldots]^T$$
Interestingly, the actual result (transposed) is:
$$[0, C_0, C_1, \ldots] P^T = \left[\begin{matrix}0 & 2 & 2 & 5 & 14 & 42 & 132 & 429 & \ldots\end{matrix}\right]$$
so the two question marks don't give Catalans $C_{0}$ and $C_1$, which surprised me.
I suppose the obvious question is what happens if we prepend another zero? We get:
$$ [0, 0, C_0, C_1, \ldots] P^T =\left[\begin{matrix}0 & 0 & 3 & 1 & 5 & 14 & 42 & 132 & \ldots \end{matrix}\right] $$ which gets the Catalans right starting at $C_3$.
We seem to be doing a matrix multiplication one column at a time. Let's say matrix $C$ is $C_{rc}=C_{r-c}$, with zeros for $r<c$: $$C=\left[\begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\1 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\2 & 1 & 1 & 0 & 0 & 0 & 0 & 0\\5 & 2 & 1 & 1 & 0 & 0 & 0 & 0\\14 & 5 & 2 & 1 & 1 & 0 & 0 & 0\\42 & 14 & 5 & 2 & 1 & 1 & 0 & 0\\132 & 42 & 14 & 5 & 2 & 1 & 1 & 0\\429 & 132 & 42 & 14 & 5 & 2 & 1 & 1\end{matrix}\right] $$
Let's try 12x12. $PC=$ $$\left[\begin{array}{cccccccccccc}1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\2 & 2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\5 & 2 & 3 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\14 & 5 & 1 & 4 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\42 & 14 & 5 & -1 & 5 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\132 & 42 & 14 & 6 & -4 & 6 & 0 & 0 & 0 & 0 & 0 & 0\\429 & 132 & 42 & 14 & 9 & -8 & 7 & 0 & 0 & 0 & 0 & 0\\1430 & 429 & 132 & 42 & 13 & 15 & -13 & 8 & 0 & 0 & 0 & 0\\4862 & 1430 & 429 & 132 & 42 & 9 & 25 & -19 & 9 & 0 & 0 & 0\\16796 & 4862 & 1430 & 429 & 132 & 43 & -1 & 40 & -26 & 10 & 0 & 0\\58786 & 16796 & 4862 & 1430 & 429 & 132 & 48 & -21 & 61 & -34 & 11 & 0\\208012 & 58786 & 16796 & 4862 & 1430 & 429 & 131 & 63 & -56 & 89 & -43 & 12\end{array}\right] $$ Each subsequent column is shifted down by 1 and is correct one entry later.
The diagonal under the main diagonal, index $(n+2,n+1)$ (2,2,1,-1,-4,-8,-13, ...) is $2-\binom{n}{2} = (-n^2+n+4)/2 $.
It's a whole family of Catalan recurrences that only work after a while. Maybe there are some previous Catalan numbers $C_{-1}$ etc. that make it work.
If we look at Parsnip as $[C_0, C_1, C_2, C_3, \ldots]^T = P [C_{-1}, C_0, C_1, \ldots]^T$, we get $C_0 = C_{-1}, C_1=2 C_0$; the latter equation is definitely false, so I don't think there's any fixing this.
OK, I haven't proven anything here, but this is a fairly interesting conjecture; for $n\ge 0$:
$$ [\underbrace{?,?,?, \ldots, ?}_{2n \text{ question marks}}, C_{n+1}, C_{n+2}, \ldots]^T = P [ \underbrace{0, \ldots, 0}_{n \text{ zeros}}, C_0, C_1, \ldots]^T $$
where $P_{rc}=(-1)^{r-c} \binom{c}{r-c+1}$ and $C_m$ is the $m$th Catalan number.
