How to prove a formula on sum of products of alternating sign Vandermonde convolution?

Question

The alternating sign Vandermonde convolution refers to the following sum \begin{align} V(r,s,N) = \sum_{k=0}^N (-1)^k \tbinom{r}{k}\tbinom{s}{N-k}. \end{align} Let $\Delta^{\pm} (m_0,m_1) := |m_0\pm m_1|$, how can I prove that \begin{align} \sum_{q=0}^n V(m_0,n-m_0,q) V(m_1,n-m_1,q) = \begin{cases} (-1)^{\frac{\Delta^-}{2}}\frac{\Delta^+!(2n-\Delta^+)!}{\frac{\Delta^+}{2}!(n-\frac{\Delta^+}{2})!n!}, & \Delta^-\text{ is even},\\ 0, & \Delta^-\text{ is odd}, \end{cases} \end{align} for $n\in \mathbb{Z}_{>0},\, m_{0,1}\in \mathbb{Z}_{\ge 0},\, m_{0,1}\le n$?

Background: I read this from Eq. (C5) in a research paper, where it is used without explanation.

My attempts: I did numerical check on this formula, then searched relevant questions on MSE. Thanks to alternating sign Vandermonde convolution, I learned that \begin{align} V(r,s,q) = \tbinom{s}{q} {}_2F_1(-r,-q;s-q+1;-1) \end{align} for $s\ge q$, but there are no hypergeometric closed formulas for the general case. Also I learned that \begin{align} V(r,s,q) = [x^q](1-x)^r(1+x)^s\,, \end{align} but for now I couldn't proceed further.

Update (refined statement):

From MSE 4131219, 4320991 and 4731417, V(m,n-m,q) is the Kravchuk polynomial \begin{align} \mathcal{K}_q(m;n,2) = \sum_{k=0}^q (-1)^k {m \choose k}{n-m \choose q-k}\,. \end{align} We are interested in \begin{align} S(n,m_0,m_1) = \sum_{q=0}^{n}\mathcal{K}_q(m_0;n,2)\mathcal{K}_q(m_1;n,2)\,. \end{align} Using the symmetry relations, \begin{align} \mathcal{K}_q(m;n,2) = \frac{n \choose q}{n \choose m}\mathcal{K}_m(q;n,2)\,, \end{align} the sum translates into \begin{align} S(n,m_0,m_1) = \tbinom{n}{m_0}^{-1}\tbinom{n}{m_1}^{-1}\sum_{q=0}^{n} \tbinom{n}{q}^2 \mathcal{K}_{m_0}(q;n,2)\mathcal{K}_{m_1}(q;n,2)\,. \end{align}

Answer 1

Introducing

$$V(r,s,N) = \sum_{k=0}^N (-1)^k {r\choose k} {s\choose N-k} = [z^N] (1+z)^s \sum_{k\ge 0} (-1)^k {r\choose k} z^k \\ = [z^N] (1+z)^s (1-z)^r$$

we seek a closed form of

$$S(n,m_0,m_1) = \sum_{q=0}^n V(m_0,n-m_0,q) V(m_1,n-m_1,q)$$

where $n\ge 1$ and $0\le m_{0,1}\le n.$ We obtain

$$\sum_{q=0}^n V(m_0,n-m_0,n-q) V(m_1,n-m_1,n-q) \\ = [w^n] (1+w)^{n-m_0} (1-w)^{m_0} [z^n] (1+z)^{n-m_1} (1-z)^{m_1} \sum_{q=0}^n w^q z^q.$$

Here we may extend the sum to infinity due to the coefficient extractors:

$$[w^n] (1+w)^{n-m_0} (1-w)^{m_0} [z^n] (1+z)^{n-m_1} (1-z)^{m_1} \frac{1}{1-wz}.$$

The contribution from $z$ is

$$-\frac{1}{w} \;\underset{z}{\mathrm{res}}\; \frac{1}{z^{n+1}} (1+z)^{n-m_1} (1-z)^{m_1} \frac{1}{z-1/w}.$$

Here the residue at infinity is zero (just barely) and residues sum to zero so we may evaluate using minus the residue at $z=1/w$, getting

$$[w^{n+1}] (1+w)^{n-m_0} (1-w)^{m_0} w^{n+1} \frac{(1+w)^{n-m_1}}{w^{n-m_1}} \frac{(w-1)^{m_1}}{w^{m_1}} \\ = (-1)^{m_1} [w^n] (1+w)^{2n-m_0-m_1} (1-w)^{m_0+m_1}.$$

This is

$$(-1)^{m_1} \;\underset{w}{\mathrm{res}}\; \frac{1}{w^{n+1}} (1+w)^{2n-m_0-m_1} (1-w)^{m_0+m_1}.$$

Note there are no poles at $\pm 1$ due to the stated boundary conditions. We evaluate using minus the residue at infinity (residues again sum to zero):

$$(-1)^{m_1} \;\underset{w}{\mathrm{res}}\; \frac{w^{n+1}}{w^2} (1+1/w)^{2n-m_0-m_1} (1-1/w)^{m_0+m_1} \\ = (-1)^{m_1} \;\underset{w}{\mathrm{res}}\; \frac{1}{w^{n+1}} (1+w)^{2n-m_0-m_1} (w-1)^{m_0+m_1} \\ = (-1)^{m_0} \;\underset{w}{\mathrm{res}}\; \frac{1}{w^{n+1}} (1+w)^{2n-m_0-m_1} (1-w)^{m_0+m_1}.$$

We thus have

$$S(n,m_0,m_1) = \frac{1}{2} ((-1)^{m_0}+(-1)^{m_1}) [w^n] (1+w)^{2n-m_0-m_1} (1-w)^{m_0+m_1}.$$

Here we clearly obtain a zero value when $\Delta^-$ is odd which is half the claim. Otherwise we have in terms of $\Delta^+$

$$(-1)^{m_0} \;\underset{w}{\mathrm{res}}\; \frac{1}{w^{n+1}} (1+w)^{2n-\Delta^+} (1-w)^{\Delta^+}.$$

Now put $w/(1+w)=v$ so that $w=v/(1-v)$ and $dw = 1/(1-v)^2 \; dv$ to get

$$(-1)^{m_0} \;\underset{v}{\mathrm{res}}\; \frac{1}{v^{n+1}} \frac{1}{(1-v)^{n-1-\Delta^+}} \frac{(1-2v)^{\Delta^+}}{(1-v)^{\Delta+}} \frac{1}{(1-v)^2} \\ = (-1)^{m_0} \;\underset{v}{\mathrm{res}}\; \frac{1}{v^{n+1}} \frac{1}{(1-v)^{n+1}} (1-2v)^{\Delta^+}.$$

Next put $v=(1-\sqrt{1-4u})/2$ so that $v(1-v)=u$ and $dv = \frac{1}{\sqrt{1-4u}}\; du$ to get

$$(-1)^{m_0} \;\underset{u}{\mathrm{res}}\; \frac{1}{u^{n+1}} \sqrt{1-4u}^{\Delta^+} \frac{1}{\sqrt{1-4u}} \\ = (-1)^{m_0} \;\underset{u}{\mathrm{res}}\; \frac{1}{u^{n+1}} \sqrt{1-4u}^{\Delta^+ -1} = (-1)^{m_0} \;\underset{u}{\mathrm{res}}\; \frac{1}{u^{n+1}} (1-4u)^{\Delta^+/2 -1/2} \\ = (-1)^{m_0} {\Delta^+/2 -1/2\choose n} (-1)^n 4^n.$$

This is an acceptable closed form but apparently OP ask for additional manipulation, which is (use $\Delta^+/2 \le n$ as well as the fact that now $\Delta^+$ is even)

$$(-1)^{m_0} {\Delta^+/2 -1/2\choose \Delta^+/2} (\Delta^+/2)! {-1/2\choose n-\Delta^+/2} (n-\Delta^+/2)! \frac{1}{n!} (-1)^n 4^n \\ = (-1)^{m_0+\Delta^+/2} {-1/2\choose \Delta^+/2} (\Delta^+/2)! \frac{(-1)^{n-\Delta^+/2}}{4^{n-\Delta^+/2}} {2n-\Delta^+\choose n-\Delta^+/2} (n-\Delta^+/2)! \frac{1}{n!} (-1)^n 4^n \\ = (-1)^{m_0} \frac{(-1)^{\Delta^+/2}}{4^{\Delta^+/2}} {\Delta^+\choose \Delta^+/2} (\Delta^+/2)! \frac{1}{4^{n-\Delta^+/2}} \frac{(2n-\Delta^+)!}{(n-\Delta^+/2)!} \frac{1}{n!} 4^n \\ = (-1)^{m_0+\Delta^+/2} \frac{\Delta^+!}{(\Delta^+/2)!} \frac{(2n-\Delta^+)!}{(n-\Delta^+/2)!} \frac{1}{n!}.$$

To conclude observe that $(-1)^{m_0+\Delta^+/2} = (-1)^{3m_0/2+m_1/2} = (-1)^{-m_0/2+m_1/2}$ to get

$$\bbox[5px,border:2px solid #00A000]{ (-1)^{\Delta^-/2} \frac{\Delta^+!}{(\Delta^+/2)!} \frac{(2n-\Delta^+)!}{(n-\Delta^+/2)!} \frac{1}{n!}.}$$