# Summing nearly successive falling factorials

Define the functions $Q_b(a) = \prod_{k=1}^{2b} \big( a +1 - \frac{k}{2}\big)= (a+\frac{1}{2})a(a-\frac{1}{2}) \dotsb (a+1-b)$, and consider the sum $$\sum_{a=0}^{r-2} Q_b(a+m)\,.$$ This sum is polynomial in $r$ and $m$, and I am interested in the value $r=\frac{1}{2}$ of this polynomial, which results in a polynomial in $m$ of degree $2b$, let's call it $P(m)$. Doodling around in Wolfram Mathematica, it seems that $$P(m) = -\frac{(2b-1)!!}{2^{3b+1}}\quad m = 0,\dotsc, b\,.$$ This is true for $b \leq 50$ at least, but how to prove it in general?

Some more information: when considering the polynomial $P(m) + \frac{(2b-1)!!}{2^{3b+1}}$, these seem to be the only integral zeroes, all single except for the middle one, if it exists. All other zeroes seem real and between $0$ and $b$ as well (checked for $b\leq 10$), and the polynomial is symmetric around $\frac{b}{2}$. Rewriting $Q_b(a) = \frac{1}{2^{2b}}(2a+1)^{\underline{2b}}$ (using falling factorials), the problem is related to this question, but it is not as simple: the sum is done with steps of two, and there is no easy guess for a general form of the polynomials.

• $Q_b(a) = (a+\frac{1}{2})a(a-\frac{1}{2}) \dotsb (a+1-b)$ I'm sorry? It's not at all obvious what the dots are in this case. At least to me. – orlp Oct 10 '17 at 15:23
• @orlp I clarified the definition, hope this helps. – Reinier Kramer Oct 11 '17 at 9:20