Mathworld's discussion of the [Gamma function](http://mathworld.wolfram.com/GammaFunction.html) has the pleasant formula:

$$ \frac{\Gamma(\frac{1}{24})\Gamma(\frac{11}{24})}{\Gamma(\frac{5}{24})\Gamma(\frac{7}{24})} = \sqrt{3}\cdot  \sqrt{2 + \sqrt{3}} $$

This may have been computed algorithmically, according to the page.  So I ask how one might derive this?

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My immediate thought was to look at $(\mathbb{Z}/24\mathbb{Z})^\times
= \big( \{ 1,5,7,11 \big|  13, 17, 19 , 23 \}, \times \big)$ where $1,5,7,11$ are [relatively prime](https://en.wikipedia.org/wiki/Euclidean_algorithm) to 24.  And the other half?  

We could try to use the mirror formula 
$$ \Gamma(z) \Gamma(1-z) = \frac{\pi}{\sin(\pi z)} $$
or the Euler [beta integral](https://en.wikipedia.org/wiki/Beta_function) but nothing has come up yet:
$$ \int_0^1 x^a (1-x)^b \, dx = \frac{\Gamma(a) \Gamma(b)}{\Gamma(a+b)} $$

I am lucky the period integral of some Riemann surface will be the ratio of Gamma functions:
$$ \int_0^1 (x - a)^{1/12}
(x - 0)^{11/12}
 (x - 1)^{-5/12}
(x - d)^{-7/12} \, dx
 $$
these integrals appear in the theory of [hypergeometric function](https://en.wikipedia.org/wiki/Hypergeometric_function)

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In light of comments, I found a paper of [Benedict Gross](https://eudml.org/doc/142544) and the paper of [Selberg and Chowla](https://www.jstor.org/stable/88112?seq=1#page_scan_tab_contents)

$$ F( \tfrac{1}{4},\tfrac{1}{4};1;\tfrac{1}{64}) = \sqrt{\frac{2}{7\pi}} \times \left[\frac{ 
\Gamma(\frac{1}{7})\Gamma(\frac{2}{7})\Gamma(\frac{4}{7})
}{
\Gamma(\frac{3}{7})\Gamma(\frac{5}{7})\Gamma(\frac{6}{7})
}\right]^{1/2} $$

so in our case we are looking at [quadratic residues](https://en.wikipedia.org/wiki/Quadratic_residue) mod 12.  However, however it does not tell us that LHS evaluates to RHS.