Find the exact Pochammer notation for $\binom{-2a}{n}$ and its binomial expression

Question

Find the exact Pochammer notation for $\binom{-2a}{n}$ and its binomial expression, whch I have tried to do by the following $$\binom{-2a}{n}=\frac{(-2a)!}{n!}=\frac{(-1)^n(2)^na(a+\frac{1}{2})(\cdots)(a+\frac{n}{2}-\frac{1}{2})}{n!} \\ (-1)^n(2^n)\frac{\binom{a+\frac{n}{2}-\frac{1}{2}}{a+\frac{1}{2}}}{n!}=(-1)^n(2)^n\frac{\left(\frac{\Gamma(a+\frac{n}{2}+\frac{1}{2})}{\Gamma(a+\frac{1}{2})}\right)}{n!}$$ However, when I reintroduce the value $2^n$ into the gamma, I get the following $$(-1)^n\frac{\left(\frac{\Gamma(2a+n+1)}{\Gamma(2a+1)}\right)}{n!}$$ But when I expand the rising factorial, I get the following $\Gamma$ result $(-1)^n\frac{\Gamma(2a+n)}{\Gamma(2a)}$ however its obvious to me the fractions $+\frac{1}{2}$ is screwing this approac, for example I know $x\Gamma(x)=\Gamma(x+1), $ and $\Gamma(x)=(x-1)!$, how to properly understand the role of fractions from Pochammer $\to$ Binomial $\to$ $\Gamma$

Answer 1

On the one hand we have \begin{align*} \color{blue}{\binom{-2a}{n}}&=\frac{(-2a)(-2a-1)\cdots(-2a-n+1)}{n!}\\ &=(-1)^n\frac{(2a)(2a+1)\cdots(2a+n-1)}{n!}\tag{1}\\ &\,\,\color{blue}{=\frac{(-1)^n}{n!}\,(2a)_n=\frac{(-1)^n}{n!}\,\frac{\Gamma(2a+n)}{\Gamma(2a)}}\tag{2} \end{align*}

Comment:

• In (1) we factor out $(-1)^n$.

• In (2) we use the Pochhammer symbol and the formula $(x)_n=\frac{\Gamma(x+n)}{\Gamma(n)}$.

On the other hand we obtain from (1) for even $n=2m$ \begin{align*} \color{blue}{\binom{-2a}{n}} &=(-1)^n2^n\frac{a\left(a+\frac{1}{2}\right)\left(a+1\right)\cdots \left(a+\frac{n-1}{2}\right)}{n!}\\ &=(-1)^n2^n\frac{a\left(a+\frac{1}{2}\right)\left(a+1\right)\cdots \left(a+m-\frac{1}{2}\right)}{n!}\\ &=(-1)^n2^n\frac{(a)_m\left(a+\frac{1}{2}\right)_{m}}{n!}\tag{3}\\ &=\frac{(-1)^n2^n}{n!}\,\frac{\Gamma(a+m)\Gamma\left(a+m+\frac{1}{2}\right)}{\Gamma(a)\Gamma\left(a+\frac{1}{2}\right)}\\ &=\frac{(-1)^n2^n}{n!}\,\frac{2^{1-2(a+m)}\sqrt{\pi}\,\Gamma(2a+2m)}{2^{1-2a}\sqrt{\pi}\,\Gamma(2a)}\tag{4}\\ &=\frac{(-1)^n2^n}{n!}\,\frac{2^{-2m}\Gamma(2a+2m)}{\Gamma(2a)}\\ &\,\,\color{blue}{=\frac{(-1)^n}{n!}\frac{\Gamma(2a+n)}{\Gamma(2a)}}\\ \end{align*} in accordance with (2). The calculation with $n=2m+1$ odd can be done similarly.

Comment:

• In (3) we observe consecutive factors differ by $\frac{1}{2}$. We can write it as product of two Pochhammer symbols.

• In (4) we use the Legendre duplication formula.