The **Rohrlich-Lang Conjecture for polynomial relations in Gamma values** predicts that all polynomial relations between Gamma values over $\mathbb Q$ come from the functional equations satisﬁed by the Gamma function. An other statement, somewhat narrower: if a quotient of products of gamma values over $\mathbb Q$ is an algebraic number, then it can be evaluated by using only the well-known reflection and multiplication formulae.


*Lang, S. – Relations de distributions et exemples classiques. Séminaire
Delange-Pisot-Poitou, 19e année: 1977/78, Théorie des nombres,
Fasc. 2, Exp. N° 40, 6 p. Collected Papers, vol. III, Springer
(2000), 59–65.*


I am wondering what that means for the following. Define 
$$A:=\prod\_{k=0}^5{\Gamma(\frac{49^k \pmod {78}}{78})}= {\Gamma(\frac{1}{78})\Gamma(\frac{25}{78})\Gamma(\frac{43}{78})\Gamma(\frac{49}{78})\Gamma(\frac{55}{78})\Gamma(\frac{61}{78})} $$ and


$$B:=\prod\_{k=0}^5 {\Gamma(\frac{7\cdot49^k\pmod {78}}{78})}= \Gamma(\frac{7}{78})\Gamma(\frac{19}{78})\Gamma(\frac{31}{78})\Gamma(\frac{37}{78})\Gamma(\frac{67}{78})\Gamma(\frac{73}{78}).$$
Then numerically, it appears that $$\frac AB=4.$$ Equivalently, as $AB=(2\pi)^6\;\sqrt[3\ \ \ ]{13}$ by the multiplication formula, we'd have to show $$A= 16\pi^3\;\sqrt[6\ \ \ ]{13}\ \ \text{ and / or }\ \ B=4\pi^3\;\sqrt[6\ \ \ ]{13}.$$ So the product of the $12$ Gamma terms $\Gamma(\frac{6j+1}{78})$ with $j=0,1,3,...,12$ (excluding $j=2$ where $\frac{13}{78}=\frac16$) splits into two halves, one $4$ times as big as the other.


How to "get rid of" the numerators, which are all $\equiv1\pmod6$? Applying the multiplication formula with step $\frac16$ to each term as suggested in [this comment](https://math.stackexchange.com/questions/3041736/how-to-derive-the-closed-form-of-this-gamma-quotient/3041759?noredirect=1#comment6272475_3041759) has the same effect (in terms of producing algebraic factors that can then be "ignored") as applying the reflection formula to each term: applying it for instance to $A$ will leave us with $\Gamma(\frac{77}{78})\Gamma(\frac{53}{78})\Gamma(\frac{35}{78})\Gamma(\frac{29}{78})\Gamma(\frac{23}{78})\Gamma(\frac{17}{78})$ in the denominator, no gain. 


I am aware that [there can be surprises like the fact that $\frac{ \Gamma\left(\frac{11}{42}\right)\Gamma\left(\frac{12}{42}\right)}{\Gamma\left(\frac{21}{42}\right)\Gamma\left(\frac2{42}\right)} $ and $\frac{ \Gamma\left(\frac{18}{42}\right)\Gamma\left(\frac{4}{42}\right)}{\Gamma\left(\frac{21}{42}\right)\Gamma\left(\frac1{42}\right)} $ are algebraic](https://math.stackexchange.com/questions/2804457/simplify-gamma-left-frac27-right-gamma-left-frac1142-right-gamma-l) and I would not dare to claim that the identity $\frac AB=4$ provides a counterexample to the Rohrlich-Lang conjecture. It is just that the simplicity of the result, being an integer, is particularly intriguing. 


Is there any other way to evaluate this, using only the functional equations of the Gamma function? Has anybody ever written a program for that (which is rather a combinatorial issue that should not be awfully hard)?



A tantalizing Gamma quotient to challenge the Rohrlich-Lang Conjecture
It turns out that triplication is not needed here: 
the recursion $\Gamma(z+1) = z \Gamma(z)$, the reflection formula
$$
\Gamma(z) \Gamma(1-z) = \frac\pi{\sin \pi z},
$$
and the duplication formula
$$
\Gamma(z) \Gamma\bigl(z+\frac12\bigr) = 2^{1-2z} \sqrt\pi \Gamma(2z)
$$
suffice to reduce the gamma product to a trigonometric identity.


Apply duplication to $z = \{2^n/39\}$ with $0 \leq n < 12$
(where $\{\cdot\}$ is the fractional part) and multiply the resulting
$12$ identities, using the recursion when $z>1/2$ 
to write $\Gamma(z+\frac12)$ and $\Gamma(2z)$ as 
rational multiples of $\Gamma(z-\frac12)$ and $\Gamma(2z-1)$.
Then the factors $\Gamma(\{2^n/39\})$ cancel out. 
For the remaining factors $\Gamma(\{ \frac{2^n}{39} + \frac12 \})$,
apply reflection when $n$ is odd, we get a formula for $A/B$.


