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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.

user153012
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2 Answers2

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$$\tag{1}a_n:={_2F_1}\left(\begin{array}c\tfrac12\quad -n\\\tfrac32\end{array}\middle|\,\frac{1}{2}\right)$$ 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.

Raymond Manzoni
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  • I've got very similiar expression using an Euler's integral transormation. After this two answers as I see there is no hope to get a closed-form of $a_n$ without using integral or summation. Thank you for your answer. By the way the indefinite version of your result gives $t$ times a hypergeom function at $t^2/2$ instead of $1/2$. As we can see here. +1. – user153012 Oct 18 '14 at 14:05
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    @user153012: As explained by Lucian in the comments we may rewrite it as an incomplete beta function (with the substitution $t:=\sqrt{2}\sin(x)$) getting : $$a_n=\sqrt{2}\int_0^{\pi/4}\cos(x)^{2n+1};dx=\frac 1{\sqrt{2}}B\left(\frac 12,\frac 12, n+1\right)$$ as you may try in Wolfram (which doesn't seem to help...) – Raymond Manzoni Oct 18 '14 at 15:21
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$\qquad\qquad\qquad\qquad$ 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.

Lucian
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  • Sadly using your $a_n$ your answer just shows a connection between ${_2F_1}(1/2,1/2+n;3/2;-1)$ and my $a_n$. Thank you anyway. I guess the @Cleo thing was just a joke because of this answer style. +1. – user153012 Oct 18 '14 at 09:18
  • @user153012: So basically you've been trying to find a general formula for integrals of the form $\displaystyle\int_0^\tfrac\pi4\cos^{2n+1}(x)~dx$ ? – Lucian Oct 18 '14 at 10:01
  • The problem appeared while I tried to get a closed-form of the series in this answer. – user153012 Oct 18 '14 at 10:07
  • @user153012: As I already explained, a closed form for the definite integral in your other post only exists if we are willing to accept the existence of incomplete Struve functions. – Lucian Oct 18 '14 at 10:25
  • Yes I've read that, but if we have a cloed-form of this hypergeom function, then at least we'll have a series for that integral without double summation. – user153012 Oct 18 '14 at 10:29
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    @user153012: You can use $\displaystyle\int_0^\tfrac\pi4\sin^{2n+1}(x)~dx = \dfrac12~B\bigg(\dfrac12~,~n+1~,~\dfrac12\bigg)$ right after the second row of Harry's answer. – Lucian Oct 18 '14 at 10:37