Can I map $\Gamma(z+\frac{1}{2})$ to $f(z)\Gamma(z)$?

Question

I am looking for a similar property $\Gamma(z+1)=z\Gamma(z)$ but with $\Gamma(z+\frac{1}{2})$. I suspect it has something to do with "duplication formula", maybe.

Of course $f(z) = \frac{\Gamma(z+1/2)}{\Gamma(z)}$ works and is the only function that works. It does not appear that there is any simpler form for $f(z)$. — Greg Martin, Jun 19 '22 at 19:48

score 0 · Answer 1 · answered Jun 20 '22 at 05:35

For $z>1$, you could write $$f(z)=\sqrt z \,\sum_{n=0}^\infty \frac{a_n}{b_n} z^{-n}$$ where the $a_n$ and $b_n$ form respectively sequences $A143503$ and $A061549$ in $OEIS$.

But, using the Gaussian hypergeometric function, there is an identity (have a look here) $$\color{blue}{f(z)=\sqrt{\left(z-\frac{1}{2}\right) \, _2F_1\left(-\frac{1}{2},-\frac{1}{2};z-\frac{1}{2};1\right)}}$$

Can I map $\Gamma(z+\frac{1}{2})$ to $f(z)\Gamma(z)$?

1 Answers1