**Could you please recommend a reference to or supply a proof of the following identity \eqref{combin-ID-Maclaurin}, or \eqref{first-equiv-form}, or \eqref{combin-ID-Mac-Equiv}, or \eqref{combin-ID-Mac-Reform} for $m\ge2$?**


For non-negative integers $k,n\ge0$, I guess that
\begin{equation}\label{combin-ID-Maclaurin}\tag{Q1}
\sum\_{\ell=0}^{n}\binom{2n+1}{2\ell+1} \binom{n-\ell}{n-k}
=2^{2k}\frac{2n+1}{2k+1}\binom{n+k}{2k}.
\end{equation}



> 
> The identity \eqref{combin-ID-Maclaurin} is equivalent to
> \begin{equation}\label{first-equiv-form}\tag{Q2}
> \sum\_{\ell=0}^{k} \binom{2n+1}{2k-2\ell+1} \binom{\ell+n-k}{\ell}
> =2^{2k}\frac{2n+1}{2k+1}\binom{n+k}{2k}, \quad k,n\ge0
> \end{equation}
> and is equivalent to
> \begin{equation}\label{combin-ID-Mac-Equiv}\tag{Q3}
> \sum\_{\ell=0}^{n}\binom{2n+1}{2\ell+1} \binom{n-\ell}{m}
> =2^{2(n-m)}\frac{2n+1}{2(n-m)+1}\binom{2n-m}{m}, \quad n\ge0,\quad n\ge m\in\mathbb{Z}. 
> \end{equation}
> The identity \eqref{combin-ID-Mac-Equiv} can be further reformulated as
> \begin{equation}\label{combin-ID-Mac-Reform}\tag{Q4}
> \sum\_{\ell=0}^{n}\binom{2n+1}{2\ell} \binom{\ell}{m}
> =2^{2(n-m)}\frac{2n+1}{2(n-m)+1}\binom{2n-m}{m},\quad n\ge0,\quad n\ge m\in\mathbb{Z}.
> \end{equation}
> 
> 
> 



> 
> 
> > 
> > When $m<0$, it is trivial that both sides of \eqref{combin-ID-Mac-Reform} are equal to $0$.
> > 
> > 
> > 
> 
> 
> 



> 
> 
> > 
> > When $m=0$, the identity \eqref{combin-ID-Mac-Equiv} or \eqref{combin-ID-Mac-Reform} becomes
> > \begin{equation}\label{m=0-(1.93)Sprug}\tag{Q5}
> > \sum\_{\ell=0}^{n}\binom{2n+1}{2\ell}
> > =\sum\_{\ell=0}^{n}\binom{2n+1}{2\ell+1}
> > =2^{2n}, \quad n\ge0.
> > \end{equation}
> > This is just the identity (1.93) on page 38 in the monograph [1] below.
> > 
> > 
> > 
> 
> 
> 



> 
> 
> > 
> > When $m=1$, the identity \eqref{combin-ID-Mac-Equiv} or \eqref{combin-ID-Mac-Reform} reduces to
> > \begin{equation}\tag{Q6}
> > \sum\_{\ell=0}^{n}(n-\ell)\binom{2n+1}{2\ell+1}
> > =\sum\_{\ell=0}^{n}\binom{2n+1}{2\ell}\ell
> > =2^{2(n-1)}(2n+1), \quad n\ge1.
> > \end{equation}
> > This follows from combining \eqref{m=0-(1.93)Sprug} with the identity
> > \begin{equation}\tag{Q7}
> > \sum\_{\ell=1}^{n}\binom{2n+1}{2\ell+1}\ell=(2n-1)4^{n-1}, \quad n\ge1,
> > \end{equation}
> > which can be found in (1.100) on page 40 of the monograph [1] below.
> > 
> > 
> > 
> 
> 
> 


**Could you please recommend a reference to or provide a proof of the above identity \eqref{combin-ID-Maclaurin}, or \eqref{first-equiv-form}, or \eqref{combin-ID-Mac-Equiv}, or \eqref{combin-ID-Mac-Reform} for $m\ge2$?**


References


1. R. Sprugnoli, *Riordan Array Proofs of Identities in Gould’s Book*, University of Florence, Italy, 2006.
2. F. Qi, *Diagonal recurrence relations for the Stirling numbers of the first kind*, Contrib. Discrete Math. **11** (2016), no. 1, 22--30; available online at <https://doi.org/10.11575/cdm.v11i1.62389>.
3. F. Qi, *Diagonal recurrence relations, inequalities, and monotonicity related to the Stirling numbers of the second kind*, Math. Inequal. Appl. **19** (2016), no. 1, 313--323; available online at <https://doi.org/10.7153/mia-19-23>.
4. F. Qi and B.-N. Guo, *A diagonal recurrence relation for the Stirling numbers of the first kind*, Appl. Anal. Discrete Math. **12** (2018), no. 1, 153--165; available online at <https://doi.org/10.2298/AADM170405004Q>.



Ask for a reference or a proof of a combinatorial identity $\sum_{k=0}^n\binom{2n+1}{2k}\binom {k}{m} =2^{2(n-m)}\frac{2n+1}{2(n-m)+1}\binom{2n-m}{m}$
We will prove the identity **(A8)** that qifeng618 alluded to. The method is called the Wilf-Zeilberger methodology. To this end, define the two functions (suppressing $x$)
$$F(n,k):=\frac{\binom{2x+1}{2k+1}\binom{x-k}{n-k}(2n+1)}{(2x+1)\binom{x+n}{2n}4^n}
\qquad \text{and} \qquad
G(n,k):=-\frac{F(n,k)\,2k\,(2k+1)}{2(x+n+1)(n+1-k)}.$$
Next, verify (preferably using a math software) that
$$F(n+1,k)-F(n,k)=G(n,k+1)-G(n,k).$$
Now, sum both sides of this equation over all integers $k$. The right-hand side then vanishes, showing that $f(n+1)-f(n)=0$ where $f(n):=\sum\_kF(n,k)$. Since $f(0)=1$, it follows that $f(n)=1$ for all $n$. That is,
$$\sum\_{k=0}^n\frac{\binom{2x+1}{2k+1}\binom{x-k}{n-k}(2n+1)}{(2x+1)\binom{x+n}{2n}4^n}=1 \qquad \text{or} \qquad
\sum\_{k=0}^n\binom{2x+1}{2k+1}\binom{x-k}{n-k}=\frac{2x+1}{2n+1}\binom{x+n}{2n}4^n,$$
as desired. $\,\,\,\,\square$


