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After playing around with transforms of a certain parametric integral, I am inclined to think that the linear combination $$f(n):=\dfrac1{n-2}\left({\,}_2F_1(\dfrac{n-2}{4n},\dfrac12;\dfrac{5n-2}{4n};-1)\right)+\dfrac1{n+2}\left({\,}_2F_1(\dfrac{n+2}{4n},\dfrac12;\dfrac{5n+2}{4n};-1)\right)$$ has a closed form for integer $n$. I know for example that $f(3)=\dfrac{1}{12^{3/4}}\dfrac{\Gamma(\frac14)^2}{\sqrt{\pi}}$. Any ideas?

*Edit*: putting $a:=\frac14-\frac1{2n}$, we can define $g(a):= \frac{8}{1-4a}f(\frac{8}{1-4a})$ to get arguments closer to the “standard” notation used in formula collections. The question then becomes:

Which rational values of $a$, other than $a= \frac1{12}$, allow a closed form for$$g(a)=\dfrac1{a}\left({\,}_2F_1(a,\frac12;a+1;-1)\right)+\dfrac1{\frac{1}{2}-a}\left({\,}_2F_1(\dfrac{1}{2}-a,\frac12;\frac32-a;-1)\right)?$$

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I have found two instances of $_2F_1(a,b;c;z), z=-1$ in

A. Erdelyi, *Higher Transcendental Functions*, Vol. 1 (and particularly, Sec. 2.8), Krieger Publishing.

that may be of help to you.

$$

_2F_1(a,b;1+a-b;-1)=2^{-a}\frac{\Gamma(1+a-b)\Gamma(1/2)}{\Gamma(1-b+a/2)\Gamma(1/2+a/2)}, \quad 1+a-b\ne0,-1,-2,…

$$

$$

(a+1)_2F_1(-a,1;b+2;-1)+(b+1)_2F_1(-b,1;a+2;-1)=2^{a+b+1}\frac{\Gamma(a+2)\Gamma(b+2)}{\Gamma(a+b+2)}, \quad a,b\ne-2,-3,-4,…

$$

And this one from the *NIST Handbook of Mathematical Functions*, which is a variation of the first one above

$$

_2F_1(a,b;1+a-b;-1)=\frac{\Gamma(1+a-b)\Gamma(a/2+1)}{\Gamma(1-b+a/2)\Gamma(1+a)}

$$

There was one additional relation for $_2F_1(a,b;c;z), z=-1$, but it was in terms of the digamma function.

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