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Under what conditions on $a,b,c$ a hypergeometric function $ \text{ }_2F_1\left(a,b ;c;z\right)$ be a rational function?

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closed as off-topic by Cameron Williams, Sharkos, José Carlos Santos, Namaste, Shailesh Jan 29 '18 at 1:29

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  • $\begingroup$ It happens only occasionally (for instance $\phantom{}_2 F_1(3,1,1,z)=\frac{1}{(1-z)^3}$), since the radius of convergence of a $\phantom{}_2 F_1$ is $1$ while the radius of convergence of $\frac{p(x)}{q(x)}$ depends on the zero of $q(x)$ which is closest to the origin. $\endgroup$ – Jack D'Aurizio Jan 28 '18 at 21:21
  • $\begingroup$ So, there are not any general theorems about the rationality of hypergeometric functions? $\endgroup$ – Leox Jan 28 '18 at 21:40
  • $\begingroup$ what about if we change the last parameter $z$ to some function of $z?$ $\endgroup$ – Leox Jan 28 '18 at 21:43
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    $\begingroup$ Well, a function is a $\phantom{}_2 F_1$ iff its Taylor coefficients fulfill a certain recurrence relation, and if $q(x)f(x)$ is a polynomial then the Taylor coefficients of $f(x)$ have to fulfill an additional recurrence relation. It might be useful to know which kind of hypergeometric functions you are dealing with, in order to prove/disprove they are rational functions or elementary functions in some other sense. $\endgroup$ – Jack D'Aurizio Jan 28 '18 at 21:57
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    $\begingroup$ There is an expository paper by Beukers called "Hypergeometric Functions, How Special Are They?" which addresses exactly this question. $\endgroup$ – Nemo Jan 29 '18 at 6:23