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, $${_2F_1}\left(\tfrac{1}{2},-n;\tfrac{3}{2};\tfrac{1}{2}\right) = \frac{(2n)!!}{(2n+1)!!}\, \sum_{k=0}^n \binom{-\tfrac{1}{2}}{k}\big(-\tfrac{1}{2}\big)^k$$
