On a "coincidence" of two sequences involving $a_n = {_2F_1}\left(\tfrac{1}{2},-n;\tfrac{3}{2};\tfrac{1}{2}\right)$

Question

This was inspired by this post. Define,

$$a_n = {_2F_1}\left(\tfrac{1}{2},-n;\tfrac{3}{2};\tfrac{1}{2}\right)$$

$$b_n = \sum_{k=0}^n \binom{-\tfrac{1}{2}}{k}\big(-\tfrac{1}{2}\big)^k$$

where $_2F_1$ is the hypergeometric function and binomial $\binom n k$. The first few numerators $N$ are,

$$N_1(n) = \color{brown}{1, \,5, \,43, \,177, \,2867, \,11531, 92479}, \,74069, 2371495,\dots$$

$$N_2(n) = \color{brown}{1, \,5, \,43, \,177, \,2867, \,11531, 92479}, \,370345, 11857475,\dots$$

respectively, and where the latter is sequence A126963. As the OP from the other post pointed out, the first seven numerators match.

I thought it was a peculiar coincidence so, after some tinkering, found a possible relationship.

Q: How do we show that, $$(2n+1)!!\, a_n = (2n)!!\, b_n$$ or equivalently, $${_2F_1}\left(\tfrac{1}{2},-n;\tfrac{3}{2};\tfrac{1}{2}\right) = \frac{(2n)!!}{(2n+1)!!}\, \sum_{k=0}^n \binom{-\tfrac{1}{2}}{k}\big(-\tfrac{1}{2}\big)^k$$

Answer 1

Converting factorials in the binomial term to $\Gamma$-functions we have $$ b_n = \sum_{k=0}^n \binom{-1/2}{k}(-1/2)^k=\surd \pi \sum_{k=0}^n \frac{1}{\Gamma(1/2-k)} \frac{(-1/2)^k}{k!} $$ and we can flip the direction of summation in the finite sum $$ = \surd \pi \sum_{k=0}^n \frac{1}{\Gamma(1/2-n+k)} \frac{(-1/2)^{n-k}}{(n-k)!} $$ $$ = (-1/2)^n \surd \pi \sum_{k=0}^n \frac{1}{\Gamma(1/2-n+k)} \frac{(-2)^{k}}{(n-k)!} $$ Convert factors to Pochhammer symbols $(a)_n\equiv \Gamma(a+n)/\Gamma(a)$ $$ = \frac{(-1/2)^n}{\Gamma(1/2-n)} \surd \pi \sum_{k=0}^n \frac{1}{(1/2-n)_k} \frac{(-2)^{k}}{(n-k)!} $$ $$ = \frac{(-1/2)^n}{\Gamma(1/2-n)} \surd \pi \sum_{k=0}^n \frac{1}{(1/2-n)_k} \frac{(-2)^{k}}{(1)_{n-k}} $$ use Slater's conversion for the negative indices in the Pochhammer Symbol (L. J. Slater, Generalized hypergeometric functions, Cambridge University Press 1966 ISBN 978-0521090612) $$ = \frac{(-1/2)^n}{\Gamma(1/2-n)} \surd \pi \sum_{k=0}^n \frac{1}{(1/2-n)_k} \frac{(-2)^{k}(1-1-n)_k}{(-1)^k(1)_n} $$ $$ = \frac{(-1/2)^n}{\Gamma(1/2-n)} \surd \pi \sum_{k=0}^n \frac{1}{(1/2-n)_k} \frac{2^{k}(-n)_k}{(1)_n} $$ $$ = \frac{(-1/2)^n}{\Gamma(1/2-n)n!} \surd \pi \sum_{k=0}^n \frac{(-n)_k}{(1/2-n)_k} 2^{k} $$ Multiply by $k!$ to match the definition of the Gaussian Hypergeometric Function $$ = \frac{(-1/2)^n}{\Gamma(1/2-n)n!} \surd \pi \sum_{k=0}^n \frac{(-n)_k(1)_k}{(1/2-n)_k} \frac{2^{k}}{k!} $$ $$ = \frac{(-1/2)^n}{\Gamma(1/2-n)n!} \surd \pi {}_2F_1(1,-n; 1/2-n;2) $$ Look into wikipedia for the well-known $\Gamma(1/2-n)$, $$ = \frac{(-1/2)^n(2n-1)!!}{n!(-2)^n} {}_2F_1(1,-n; 1/2-n;2) $$ $$ = \frac{(2n-1)!!}{n!4^n} {}_2F_1(1,-n; 1/2-n;2) $$ Use Chaundy's first formula in chapter 6 of https://doi.org/10.1093/qmath/os-14.1.55 with $a=1$, $1-c=1/2$, $c=1/2$, $p=2$, % $$ % = (-1/2)^n \binom{-1/2}{n} {}_2F_1(1,-n; 1/2-n;2) % $$ $$ = \frac{(2n-1)!!}{n!4^n} \frac{(3/2)_n}{(1/2)_n} {}_2F_1(1,-n; 3/2;-1) $$ Use the first of the two Pfaff transformations in the wikipedia article at $a=1$, $b=-n$, $c=3/2$, $z=-1$ $$ = \frac{(2n-1)!!}{n!4^n} \frac{(3/2)_n}{(1/2)_n} 2^n {}_2F_1(-n,1/2;3/2;1/2) $$ replace Pochhammer's symbols by definition $$ = \frac{(2n-1)!!}{n!4^n} \frac{\Gamma(3/2+n)\Gamma(1/2)}{\Gamma(3/2) \Gamma(1/2+n)} 2^n {}_2F_1(-n,1/2;3/2;1/2) $$ with the recurrence equation $z\Gamma(z)=\Gamma(z+1)$ $$ = \frac{(2n-1)!!}{n!2^n} \frac{(1/2+n)}{1/2} {}_2F_1(-n,1/2;3/2;1/2) $$ $$ = \frac{(2n-1)!!}{(2n)!!} (1+2n) {}_2F_1(-n,1/2;3/2;1/2) $$ $$ = \frac{(2n+1)!!}{(2n)!!} {}_2F_1(-n,1/2;3/2;1/2) $$ which was to be shown.