Closed-form of the sequence ${_2F_1}\left(\begin{array}c\tfrac12,-n\\\tfrac32\end{array}\middle|\,\frac{1}{2}\right)$ Is there a closed-form of the following sequence?
$$a_n={_2F_1}\left(\begin{array}c\tfrac12,-n\\\tfrac32\end{array}\middle|\,\frac{1}{2}\right),$$
where $_2F_1$ is the hypergeometric function and $n \in \mathbb{N}$. Maple could evaluate $a_n$ for arbitrary $n$. The exact values for $a_n$ from $n=0$ to $10$. 
$$1,\frac 56,{\frac {43}{60}},{\frac {177}{280}},{\frac {2867}{5040}},{
\frac {11531}{22176}},{\frac {92479}{192192}},{\frac {74069}{164736}},
{\frac {2371495}{5601024}},{\frac {9488411}{23648768}},{\frac {
126527543}{331082752}},\dots$$
It is interesting, that the first $7$ term of the numerator sequence matches with $\text{A126963}$ on OEIS, but after that it breaks.
 A: $$\tag{1}a_n:={_2F_1}\left(\begin{array}c\tfrac12\quad -n\\\tfrac32\end{array}\middle|\,\frac{1}{2}\right)$$
is defined as :
(with $(q)_j=q\,(q+1)\cdots(q+j-1)$ for $\,j>0\;$ the rising Pochhammer symbol ) :
\begin{align}
a_n&=\sum_{j=0}^{\infty}\frac{\left(\frac 12\right)_j\;(-n)_j}{\left(\frac 32\right)_j\;j!\,2^j}\\
&=\sum_{j=0}^{n}\frac{\frac 12\;(-n)_j}{\left(j+\frac 12\right)\;j!\,2^j}\\
&=\sum_{j=0}^{n}\frac{n!}{\left(2j+1\right)\;(n-j)!\;j!}\left(-\frac 12\right)^j\\
&=f_n\left(-\frac 12\right)\\
\end{align}
with $\;\displaystyle f_n(x):=\sum_{j=0}^{n}\binom{n}{j}\frac{x^j}{2j+1}$
Since $\;\displaystyle \sum_{j=0}^{n}\binom{n}{j}(t^2)^j=\left(1+t^2\right)^n\;$ and $\;\displaystyle\int t^{2j}\;dt=\frac{t^{2j+1}}{2j+1}\;$ we deduce that 
$$t\,f_n(t^2)=\int (1+t^2)^n\; dx$$
and (since $(i/\sqrt{2})^2=-1/2$) that 
$$\tag{2}a_n=-i\sqrt{2}\int_0^{i/\sqrt{2}}(1+x^2)^n\;dx$$
or simply that
$$\tag{3}a_n=\int_0^1\left(1-\frac {t^2}2\right)^n\;dt$$
Of course we could have found this directly using the appropriate integral for the hypergeometric series... Anyway such integral formulations may return clearer expressions for $\;\displaystyle\sum \frac {a_n}{n!}\;$ and so on.
A: 
$\qquad\qquad\qquad\qquad$ Hello, there! Cleo just asked me to post this:

Starting at $n=0$, we have $F(n)=\dfrac{a_{n+1}}{2^n~(2n+1)!!},$ where $a_{k>0}$ form the sequence described here.
