Hockey-stick identity with terms of alternating sign $\sum_{k=0}^{m} (-)^k \binom{k+m}{k}$ I'm currently looking at Hermite polynomials, looking at sums of the form $\sum_k \frac1{(\gamma_i-\gamma_k)^{n}}$ for $n\in\mathbb{Z}^+$, and an identity that popped up is of the form
\begin{equation}
\sum_{k=0}^{m} (-)^k \binom{k+m}{k},
\end{equation}
which reminds me a lot about hockeystick.
So, I found a closed form of this in terms of exotic functions:
\begin{equation}
\sum_{k=0}^m (-)^k \binom{k+m}{k} = \frac1{2^{m+1}} \left[1- (-)^m \frac{(2m)!}{m!(m-1)!}B\left(\frac12,m,-2m\right)\right] + (-)^{m} \binom{2m}{m},
\end{equation}
but it's not very insightful how the result depends on $m$, notably because of the Beta function on the RHS.
Does anybody know how to improve on this answer? Either with another derivation or with a nice representation of the Beta function (e.g. with another binomial coefficient, or some elementary function)?
 A: Here is some additional information which might be useful. Calculating the first few terms of the sequence of numbers
\begin{align*}
\color{blue}{a_m=\sum_{k=0}^{m}(-1)^k\binom{k+m}{k}\qquad\qquad m\geq 0}\tag{1}
\end{align*}
gives
\begin{align*}
\left(a_m\right)_{m\geq 0}=(1,     -1,      4,    -13,     46,   -166,    610,  -2\,269,\ldots)\tag{2}
\end{align*}
OEIS A026641:
The sequence $a_m, m\geq 0$ is not stored in OEIS, but we find the sequence A026641 of absolute values
\begin{align*}
\left((-1)^ma_m\right)_{m\geq 0}=(1,     1,      4,    13,     46,   166,    610,  2\,269,\ldots)
\end{align*}
giving the number $(-1)^ma_m$ of nodes of even outdegree (including leaves) in all ordered trees with $m$ edges. We can derive from the stated information the generating function
\begin{align*}
\color{blue}{\sum_{m=0}^{\infty}a_mz^m}&=\sum_{m=0}^\infty \sum_{k=0}^{m}(-1)^k\binom{k+m}{k}z^m\\
&\,\,\color{blue}{=\frac{1}{\sqrt{1+4z}}\,\frac{2}{3-\sqrt{1+4z}}\qquad\qquad |z|<\frac{1}{4}}
\end{align*}
We can  also derive some nice formulas of (1) for instance from one of the formulas from Richard Choulet we get
\begin{align*}
\color{blue}{a_m=(-1)^m\sum_{k=0}^m(-1)^k\binom{2m-k}{m}\qquad\qquad m\geq 0}
\end{align*}
OEIS A188289:
Another closely related sequence can be found when we skip the summand $k=0$ in (1) and consider
\begin{align*}
a_m-1=\sum_{\color{blue}{k=1}}^{m}(-1)^k\binom{k+m}{k}\qquad\qquad m\geq 0
\end{align*}
We get
\begin{align*}
\left(a_m-1\right)_{m\geq 0}=(0,     -2,      3,    -14,     45,   -167,    609,  -2\,270,\ldots)
\end{align*}
and find again the sequence of absolute values of $a_m-1, m\geq 0$ in OEIS as A188289. We derive for instance from the first stated formula
\begin{align*}
(-1)^m\left(a_m-1\right)=\binom{2m}{m}-(-1)^m-\sum_{k=0}^{m-1}\binom{2k}{m-1}
\end{align*}
from which the nice identity
\begin{align*}
\color{blue}{a_m=(-1)^m\left(\binom{2m}{m}-\sum_{k=0}^{m-1}\binom{2k}{m-1}\right)\qquad\qquad m\geq 0}
\end{align*}
follows. Note in the last line the sum is empty if $m=0$ and all terms $\binom{2k}{m-1}$ with $k<\left\lfloor\frac{m-1}{2}\right\rfloor$ vanish.
A: What follows is more of a comment. We can find the OGF using the residue operator. We have
$$S_m = \sum_{k=0}^m (-1)^k {k+m\choose k}$$
and we use an Iverson bracket
$$\sum_{k\ge 0} (-1)^k {k+m\choose k} [[k\le m]]
= \sum_{k\ge 0} (-1)^k {k+m\choose k} [z^m] \frac{z^k}{1-z}
\\ = [z^m] \frac{1}{1-z} \frac{1}{(1+z)^{m+1}}
= \; \underset{z}{\mathrm{res}} \;
\frac{1}{z^{m+1}} \frac{1}{1-z} \frac{1}{(1+z)^{m+1}}.$$
Now we put $z=(-1+\sqrt{1+4w})/2$ so that $z(1+z)=w$ and $dz = 
1/\sqrt{1+4w} \; dw$ to get
$$\; \underset{w}{\mathrm{res}} \;
\frac{1}{w^{m+1}} \frac{2}{3-\sqrt{1+4w}}
\frac{1}{\sqrt{1+4w}}.$$
