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I want to know when certain expressions of the form

$ {\Gamma(r_1/m) \Gamma(r_2/m) \ldots \Gamma(r_j/m) \over \Gamma(s_1/m) \Gamma(s_2/m) \ldots \Gamma(s_j/m)} $

are algebraic numbers. These ratios of Γ functions occur in the asymptotic enumeration of certain classes of restricted partitions, but I don't think this is relevant. Also, In the partition problems I'm interested in, it's natural to have $r_1 + \ldots + r_j = s_1 + \ldots + s_j$ but this isn't necessary. This seems to happen with some frequency. For example,a note of Albert Nijenhuis (arXiv:0907.1689) shows that $\Gamma(1/14) \Gamma(9/14) \Gamma(11/14) = 4\pi^{3/2}$; the techniques of the same paper show that $\Gamma(3/14) \Gamma(5/14) \Gamma(13/14) = 2\pi^{3/2}$, so the quotient is in fact 2! Similarly, we can get the identity

$ {\Gamma(1/8) \Gamma(5/8) \Gamma(6/8) \over \Gamma(2/8) \Gamma(3/8) \Gamma(7/8)} = \sqrt{2}$

by applying the duplication formula

$ \Gamma(z) \Gamma(z+1/2) = 2^{1-2z} \sqrt{\pi} \Gamma(2z) $

to the first two factors in the numerator and the last two in the denominator. In trying to prove other identities of this type, the duplication formula, its generalization to the "multiplication formula"

$\Gamma(z) \Gamma(z+1/k) \cdots \Gamma(z+(k-1)/k) = (2\pi)^{(k-1)/2} k^{1/2-kz} \Gamma(kz)$

and the reflection formula

$ \Gamma(z) \Gamma(1-z) = \pi \csc(\pi z)$

are the most obvious tools. So this seems to be a problem in combinatorial number theory; given an expression of the form in the first displayed equation, when can we use the multiplication and reflection formulas to reduce it to a rational power of some integer times a product of trig functions of rational multiples of π?

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I am pretty sure that it is even open whether $\Gamma(1/5)$ is irrational, much less algebraic, but I cannot find a reference. – Boris Bukh Dec 3 '09 at 1:00
That's a good point. The question I'm interested in, I suppose, is when we can actually write down a polynomial satisfied by a ratio of gamma functions. (In other words, assume Γ(z) is transcendental for all rational, non-integer z, which is morally true.) – Michael Lugo Dec 3 '09 at 1:10
-1 until the question is made precise. Is $\Gamma(1/173)\Gamma(4/61)$ morally irrational? Etc. – Boris Bukh Dec 3 '09 at 10:38

You should be interested when the solution for Monthly problem 11426 is published. A preview (credit to Albert Stadler):

$$ \frac{\Gamma(1/10)\Gamma(9/10)}{\Gamma(3/10)\Gamma(7/10)} = \frac{3+\sqrt{5}}{2}, $$

$$\frac{\Gamma(1/26)\Gamma(3/26)\Gamma(9/26)\Gamma(17/26) \Gamma(23/26)\Gamma(25/26)}{\Gamma(5/26)\Gamma(7/26)\Gamma(11/26) \Gamma(15/26)\Gamma(19/26)\Gamma(21/26)} = \frac{11+3\sqrt{13}}{2}, $$

$$\frac{\Gamma(1/34)\Gamma(9/34)\Gamma(13/34) \Gamma(15/34)\Gamma(19/34)\Gamma(21/34) \Gamma(25/34)\Gamma(33/34)}{\Gamma(3/34)\Gamma(5/34) \Gamma(7/34)\Gamma(11/34)\Gamma(23/34) \Gamma(27/34)\Gamma(29/34)\Gamma(31/34)} = 1 . $$

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Amer. Math. Monthly, November 2010, page 842 – Gerald Edgar Aug 20 '14 at 15:59

There are other identities that are relevant, but are less systematically understood. For example,

$\Gamma \left(\frac{1}{7}\right) \Gamma \left(\frac{6}{7}\right)=\Gamma \left(\frac{3}{7}\right) \Gamma \left(\frac{4}{7}\right)+\Gamma \left(\frac{2}{7}\right) \Gamma \left(\frac{5}{7}\right).$

There's a known generalization of this with 7 replaced by $2^k-1$ (see my paper with Ron Graham) but it isn't known if this is all instances of cosecant sums being zero.

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All explained here:

Deligne, P. Valeurs de fonctions $L$ et périodes d'intégrales. (French) With an appendix by N. Koblitz and A. Ogus. Proc. Sympos. Pure Math., XXXIII, Automorphic forms, representations and $L$-functions (Proc. Sympos. Pure Math., Oregon State Univ., Corvallis, Ore., 1977), Part 2, pp. 313--346, Amer. Math. Soc., Providence, R.I., 1979. 12A70 (10D15 10D24 10H10)

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Is there a version of that written for mere mortals? – Jacques Carette Jun 14 '10 at 2:28
Maybe the notes of Deligne's lectures at the Arizona Winter School? – Felipe Voloch Jun 14 '10 at 15:31
The original link seems to be dead. Here is an archive copy:… – Vladimir Reshetnikov Sep 2 '14 at 0:05

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