The classical Euler's gamma function $\Gamma(z)$ can be defined by \begin{equation} \Gamma(z)=\lim_{n\to\infty}\frac{n!n^z}{\prod_{k=0}^n(z+k)}, \quad z\in\mathbb{C}\setminus\{0,-1,-2,\dotsc\}. \end{equation} The extended binomial coefficient $\binom{z}{w}$ for $z,w\in\mathbb{C}$ is defined by \begin{equation} \binom{z}{w}= \begin{cases} \dfrac{\Gamma(z+1)}{\Gamma(w+1)\Gamma(z-w+1)}, & z\not\in\mathbb{N}_-,\quad w,z-w\not\in\mathbb{N}_-;\\ 0, & z\not\in\mathbb{N}_-,\quad w\in\mathbb{N}_- \text{ or } z-w\in\mathbb{N}_-;\\ \dfrac{\langle z\rangle_w}{w!},& z\in\mathbb{N}_-, \quad w\in\mathbb{N}_0;\\ \dfrac{\langle z\rangle_{z-w}}{(z-w)!}, & z,w\in\mathbb{N}_-, \quad z-w\in\mathbb{N}_0;\\ 0, & z,w\in\mathbb{N}_-, \quad z-w\in\mathbb{N}_-;\\ \infty, & z\in\mathbb{N}_-, \quad w\not\in\mathbb{Z}, \end{cases} \end{equation} where \begin{align} \mathbb{Z}&=\{0,\pm1,\pm2,\dotsc\}, & \mathbb{N}&=\{1,2,\dotsc\},\\ \mathbb{N}_0&=\{0,1,2,\dotsc\}, & \mathbb{N}_-&=\{-1,-2,\dotsc\} \end{align} and \begin{align} \langle\alpha\rangle_n&=\prod_{k=0}^{n-1}(\alpha-k)\\ &= \begin{cases} \alpha(\alpha-1)\dotsm(\alpha-n+1), & n\ge1\\ 1, & n=0 \end{cases} \end{align} is called the falling factorial.
My question is: what are the general formulas of the finite sums $$ \sum_{k=0}^{n}(\pm1)^{k} \binom{n}{k}\binom{k/2}{m} $$ for $n\ge0$ and any suitable number $m$?
- When $m\ne0$ and $n\ge0$, it is known that the identity $$ \sum_{k=0}^{n}\binom{n}{k}\binom{k/2}{m}=\frac{n}{m}\binom{n-m-1}{m-1}2^{n-2m} $$ is valid.
- When $2m\ge n+1\ge1$, it is known that the identity \begin{align} \sum_{k=0}^{n}(-1)^{k} \binom{n}{k}\binom{k/2}{m} &=(-1)^m\biggl[\binom{2m-n-1}{m-1}-\binom{2m-n-1}{m}\biggr]2^{n-2m}\\ &=(-1)^m\frac{n}{m}\binom{2m-n-1}{m-1}2^{n-2m} \end{align} is valid.
These two answers are slightly modifications in form of Items (3.163) and (3.164) on pages 91--92 in the monograph:
R. Sprugnoli, Riordan Array Proofs of Identities in Gould’s Book, University of Florence, Italy, 2006. (Has this monograph been formally published somewhere?)
- The identity \begin{equation} \sum_{k=0}^{r}(-1)^k\binom{r}{k}\binom{k/2}{q} =\frac{(-1)^q}{2^{2q-r}}\frac{r}{2q-r}\binom{2q-r}{q}, \quad 0\le r\le q \end{equation} at Finite summation with binomial coefficients, $\sum (-1)^k\binom{r}{k} \binom{k/2}{q}$ has a more strict restriction $0\le r\le q$ than the above restriction $2m\ge n+1\ge1$.
Finally and essentially speaking, my question is: if without the restriction $2m\ge n+1\ge1$, what is the general formula of the sum $\sum_{k=0}^{n}(-1)^{k} \binom{n}{k}\binom{k/2}{m}$ for $m\in\mathbb{N}$ and $n\ge0$? even for $m\in\mathbb{C}$ and $n\ge0$?