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The Gamma function satisfies:

$$\Gamma(z + 1) = z\Gamma(z)$$ $$\Gamma(z)\Gamma(1 - z) = \frac{\pi}{\sin(\pi z)}$$ $$\prod_{k = 0}^{n-1}\Gamma\left(a+\frac{k}{n}\right) = (2\pi)^{(n-1)/2}n^{-na + 1/2}\Gamma(na)$$

The Rohrlich conjecture states:

Any multiplicative dependence relation of the form $$\pi^{b/2}\prod_{a \in \mathbb{Q}}\Gamma(a)^{m_a} \in \bar{\mathbb{Q}}$$ is a consequence of the relations above.

Do we know special cases where this is valid? For instance, given rationals $a$ and $b$, do we know when $\dfrac{1}{\pi}\dfrac{\Gamma(a)\Gamma(b)}{\Gamma(a + b)}$ is algebraic? And for three rationals?

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    So far as I know, it is unknown whether, say, $\Gamma(1/5)$ is algebraic. – 2017-02-20
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    Wikipedia agrees with me, https://en.wikipedia.org/wiki/Particular_values_of_the_Gamma_function#General_rational_arguments so I suspect very, very little is known about your question. – 2017-02-21
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    Hossein likes this post. Regards! – 2017-03-03
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    Lang strengthened it from monomial to polynomial relations. – 2017-07-24

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At least two of gamma(1/5), gamma(2/5) and pi are algebraically independent. ..

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