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?

  • $\begingroup$ I am pretty sure that it is even open whether $\Gamma(1/5)$ is irrational, much less algebraic, but I cannot find a reference. $\endgroup$
    – Boris Bukh
    Dec 3, 2009 at 1:00
  • $\begingroup$ 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.) $\endgroup$ Dec 3, 2009 at 1:10
  • $\begingroup$ -1 until the question is made precise. Is $\Gamma(1/173)\Gamma(4/61)$ morally irrational? Etc. $\endgroup$
    – Boris Bukh
    Dec 3, 2009 at 10:38

6 Answers 6


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

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

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

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

  • 1
    $\begingroup$ Amer. Math. Monthly, November 2010, page 842 $\endgroup$ Aug 20, 2014 at 15:59
  • 3
    $\begingroup$ Problems and Solutions, The American Mathematical Monthly, 117:9 (2010) pp 834-843, doi.org/10.4169/000298910x521724 $\endgroup$
    – David Roberts
    Nov 19, 2021 at 3:08
  • $\begingroup$ $\frac{\Gamma(1/10) \Gamma(9/10)}{\Gamma(3/10) \Gamma(7/10)} = \frac{sin(3\pi / 10)}{sin(\pi / 10)} = \frac{3+\sqrt{5}}{2}$ $\endgroup$
    – jjcale
    Jul 21, 2022 at 17:40

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:

  • Ron Graham, Kevin O'Bryant, A Discrete Fourier Kernel and Fraenkel's Tiling Conjecture, Acta Arith. 118 (2005), no. 3, 283–304, doi:10.4064/aa118-3-4, arXiv:math/0407306,

...but it isn't known if this is all instances of cosecant sums being zero.

  • $\begingroup$ The equality given here is easy to render in terms of geometry. If a regular heptagon ABCDEFG is inscribed in a circle of unit diameter and we apply Ptolemy's Theorem to quadrilateral ABCF within this figure, we find that $\csc(\pi/7)=\csc(2\pi/7)+\csc(3\pi/7)$. This matches the gamma-function relation through the Reflection Formula. $\endgroup$ Jul 22, 2022 at 0:41

All explained here:

  • MR0546622 Pierre Deligne, Valeurs de fonctions $L$ et périodes d'intégrales, 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. link to IAS copy

Rohrlich has conjectured that the multiplicative relations in $\mathbb{C}^\times / \overline{\mathbb{Q}}^\times$ between values of $\Gamma$ at rational numbers are generated by the multiplication formula and the reflection formula. In conceptual terms, Lang says that $\Gamma$ is an odd punctured distribution on $\mathbb{Q}/\mathbb{Z}$, and that conjecturally, it is universal, see Relations de distributions et exemples classiques, Deligne's article mentioned in Felipe Voloch's answer and this MO answer. For more details about distributions, see the book by Kubert and Lang, Modular units (Springer, 1981), Chapter 1.

On the other hand, the first Bernoulli polynomial $B_1(x)=x-\frac12$ also gives rise to a universal distribution. More precisely, the function \begin{equation*} h_1 \colon x \mapsto \begin{cases} B_1(\{x\}) & \textrm{if } x \not\in \mathbb{Z} \\ 0 & \textrm{if } x \in \mathbb{Z}, \end{cases} \end{equation*} where $\{x\}=x-\lfloor x \rfloor$ is the fractional part of $x$, is a distribution on $\mathbb{Q}/\mathbb{Z}$. Following the terminology of Kubert-Lang, the first Bernoulli distribution on $\mathbb{Q}/\mathbb{Z}$ is the Stickelberger distribution associated to $h_1$. It takes values in the direct limit of the group rings $\mathbb{Q}[(\mathbb{Z}/N\mathbb{Z})^\times]$. At each finite level $N$, it is given by \begin{align*} \mathbf{B}_1 \colon \bigl(\frac{1}{N}\mathbb{Z}\bigr)/\mathbb{Z} & \to \mathbb{Q}[(\mathbb{Z}/N\mathbb{Z})^\times] \\ x & \mapsto \sum_{u \in (\mathbb{Z}/N\mathbb{Z})^\times} h_1(ux) [u]. \end{align*} The good thing is that we can show that $\mathbf{B}_1$ is universal (among odd punctured distributions on $\mathbb{Q}/\mathbb{Z}$). This follows from the non-vanishing of the Dirichlet $L$-values $L(\chi,1)$ with $\chi$ odd. Concretely, this means that there are no other linear relations than distribution and parity.

This leads to a conjectural criterion for an arbitrary product of $\Gamma$-values \begin{equation*} \Gamma(x_1)^{n_1} \cdots \Gamma(x_r)^{n_r} \end{equation*} with $x_i \in \mathbb{Q} \backslash \mathbb{Z}$ and $n_i \in \mathbb{Z}$, to be an algebraic number times a power of $\sqrt{\pi}$. Namely, just check whether the divisor $X = \sum_{i=1}^r n_i [x_i]$ on $\mathbb{Q}/\mathbb{Z}$ is in the kernel of the first Bernoulli distribution $\mathbf{B}_1$.

If it is, then using linear algebra, you will be able to write $X$ as a linear combination of the multiplication and reflection relations, because $\mathbf{B}_1$ is universal. Therefore you will be able to compute the $\Gamma$ product as an explicit algebraic number times a power of $\sqrt{\pi}$. This algebraic number will be a cyclotomic number times a product of fractional powers of prime numbers.

If one excepts the possible simplifications of this algebraic number, all this can be made into an algorithm.

In the same article, Lang also asks whether the multiplication and reflection formulas generate the ideal of polynomial relations between $\Gamma$-values over $\overline{\mathbb{Q}}(\sqrt{\pi})$. One could try, similarly as above, to give an explicit conjectural criterion for a given polynomial in $\Gamma$-values to be algebraic.

  • 2
    $\begingroup$ I am not sure that I understand what you mean by being in the kernel of the Bernoulli distribution: for instance the beta function $B(a,b)=\Gamma(a)\Gamma(b)/\Gamma(a+b)$ is certainly not always a power of $\sqrt{\pi}$ times algebraic, but $B_1(a)+B_1(b)-B_1(a+b)$ is always integral (sorry, forgot the $1/2$ but you can square). $\endgroup$ Nov 19, 2021 at 15:30
  • 1
    $\begingroup$ @Henri The terminology "Bernoulli distribution" is ambiguous, indeed it certainly does not suffice to check just one relation. Rather one should consider the Stickelberger distribution associated to $B_1$, which Kubert and Lang also call the Bernoulli distribution. I hope I made it clearer now in the answer. $\endgroup$ Nov 19, 2021 at 17:03

As a consequence of Euler's reflection and Gauss's multiplication formulas, all values $\Gamma(a)$ with $24a\in\mathbb{Z}$ or $60a\in\mathbb{Z}$ can be expressed algebraically in terms of these values: $$ \Gamma\!\left(\frac12\right)=\sqrt{\pi},\quad \Gamma\!\left(\frac13\right), \quad \Gamma\!\left(\frac14\right), \quad \Gamma\!\left(\frac15\right), \quad \Gamma\!\left(\frac25\right), \quad \Gamma\!\left(\frac18\right), $$ $$ \Gamma\!\left(\frac1{15}\right),\quad \Gamma\!\left(\frac1{20}\right), \quad \Gamma\!\left(\frac1{24}\right), \quad \Gamma\!\left(\frac1{60}\right), \quad \Gamma\!\left(\frac7{60}\right). $$

This is worked out in

along with basic expressions for $a\in(0,1)$. For example, $$ \Gamma\!\left(\frac7{10}\right)=\sqrt{\pi}\,2^{3/5}\,\Gamma\left(\frac15\right)^{\!-1}\,\Gamma\!\left(\frac2{5}\right), $$ $$ \Gamma\!\left(\frac1{12}\right)=\frac{3^{3/8}\,\sqrt{\sqrt{3}+1}}{\sqrt{\pi}\,2^{1/4}}\,\Gamma\!\left(\frac13\right)\,\Gamma\!\left(\frac14\right), $$ $$ \Gamma\!\left(\frac{11}{15}\right)=2\pi\cdot 3^{3/10}\,\Gamma\!\left(\frac15\right)\,\Gamma\left(\frac25\right)^{\!-1}\,\Gamma\left(\frac1{15}\right)^{\!-1}, $$ $$ \Gamma\!\left(\frac{11}{20}\right)=2^{1/5}\,\sqrt{5+\sqrt5}\;\Gamma\!\left(\frac15\right)\,\Gamma\!\left(\frac25\right)\,\Gamma\left(\frac1{20}\right)^{\!-1}, $$ $$ \Gamma\!\left(\frac{49}{60}\right)=\frac{\sqrt{\pi}\,\sqrt3\,\sqrt{\sqrt3+1}\,\sqrt{5+\sqrt5}\,\sqrt{\sqrt5+\sqrt3}}{5^{1/24}}\,\Gamma\!\left(\frac13\right)\,\Gamma\!\left(\frac1{60}\right)^{\!-1}. $$ Further algebraic independence of the listed 11 "basic" values (and more general answers) are yet to be settled.


Taking cyclic subgroups seems promising for obtaining identities similar to the ones of Nijenhuis, and so I have looked at products of ${\Gamma\left(\dfrac{a^k }{N}\right)}/{\sqrt{2\pi}}$ with $k$ running over a certain range. The factor $\sqrt{2\pi}$ is motivated by the fact that

- in Nijenhuis' formula as well as in the multiplication formula and even in the reflection formula, a factor $\sqrt{\pi}=\Gamma(\frac12)$ occurs with the same multiplicity as the gamma factors (so my products can be considered in fact as "well-poised" gamma quotients),

  • without including $\sqrt{2}$, those products, whenever they are integers, have a relatively big 2-valuation (see the factor $2^{b(A)}$ in Nijenhuis' formula $$\prod_{x\in A}{\Gamma\left(\frac{x }{2n}\right)}=2^{b(A)} \sqrt{\pi}^{\,\nu(n)},$$ where $A$ is the subgroup of the multiplicative group $\mathbb Z_{2n}^*$ generated by $n + 2$ or any one of its cosets, $\nu(n)$ its order and $b(A)$ the number of elements in $A$ that are larger than $n$).

For brevity, define for coprime integers $a, N$ $$g(a,N):=\prod_{k=1}^{ind_a(N)} \frac{\Gamma\left(\dfrac{a^k \pmod N}{N}\right)}{\sqrt{2\pi}}.$$ Thus the product is taken over the subgroup $\langle a\rangle$ of $\mathbb Z_N^*$ generated by $a$, with all the arguments of the gamma factors between $0$ and $1$. For an integer $b$ coprime to $N$, define similarly the product over the corresponding coset $b\langle a\rangle=\langle a\rangle b$ as $$g_b(a,N):=\prod_{k=1}^{ind_a(N)} \frac{\Gamma\left(\dfrac{a^kb \pmod N}{N}\right)} {\sqrt{2\pi}}.$$

With this notation, the initial identity reads $g(9,14)=\sqrt{2}$, and we further have the somewhat complementary one $g_3(9,14)=1/\sqrt{2}$.

Once you know where to (ask a computer to) search, it is easy to find, in a few minutes, dozens of those products that appear numerically to be algebraic.

I have excluded the self-complementary groups, i.e. the groups for which $\langle a\rangle=\langle N-a\rangle $, because for those, we already know that $g(a,N)$ is algebraic by the reflection formula. Likewise I have excluded subgroups whose members form essentially an arithmetic sequence (more precisely, a set $\{\lambda s+t\}\cap \mathbb Z_N^*$ for given $s,t$), as those can be handled by the multiplication formula and yield again algebraic products. Call the remaining subgroups and the associated gamma products non-trivial.

A systematic search (with reasonably high numerical precision) shows that of the non-trivial algebraic gamma products, most are of the form $q^{u/v}$ with integers $u,v$. Generally, $u/v$ is positive and $q$ a prime dividing $N$. But there are exceptions like $g(24,203)=7^{-1/2}$ or $g(103,420)=(3^3\cdot5^3\cdot7)^{1/6}$.

For $n\le300$, there are $106$ non-trivial algebraic gamma products, $88$ of which can be written in form $q^{u/v}$. (With $v=1$ for 36 of them.) The $18$ remaining ones occur for $N= 60,105,120,140,156,180,220,231,255,285,300$ (note that all these $N$'s have at least three prime divisors) and have minimal polynomials of degrees $2,4,6$ or $8$. The latter holds for all $N\le 1000$.
Moreover so far all of them can be written with radicals, e.g. $g(103,105)=\sqrt[3]{3(1701+166\sqrt{105})}$ or $g(41,156)=2 \sqrt[4]{13}(2\sqrt{3}+\sqrt{13})$ or $g(83,120)=\sqrt{3(\sqrt{3}-1)(\sqrt{30}-5)}$.
A few of these identities, e.g. the one for $g(83,120)$, can be derived by using the standard formulas — see Vidunas's article Expressions for values of the gamma function (thus in the way the OP asks next to the end). But I doubt this is possible where e.g. $\sqrt{13}$ occurs.

You can get more similar identities by taking the cosets of the same groups. I haven't looked systematically at them, but conjecturally, if $g(a,N)$ is algebraic, so is $g_b(a,N)$. Note that generally, $g_b(a,N)/g(a,N)$ does not seem to be algebraic.

The distribution of the non trivial subgroups seems to be at least as irregular as the distribution of the primes and not even very much correlated with the structure of $\mathbb Z_N^*$.


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