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Define harmonic numbers for a complex argument $z$ as $H_z=\frac{\Gamma'(z+1)}{\Gamma(z+1)}-\Gamma'(1)$.

For $n\in\mathbb{N}$, $H_n$ are usual harmonic numbers $\sum^n_{k=1} k^{-1}$ . They are obviously rational and are known (Taeisinger 1915) to be non-integers for $n>1$.

Question: Is there a non-integer rational $q$ such that $H_q\in\mathbb{Q}$?

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  • $\begingroup$ Historical details: (1) first published proof of this is by Theisinger (Monatshefte für Mathematik und Physik 26, 1915, S. 132–134), not 'Taeisinger' (2) this is misspelling is very widespread (e.g., as of 2020-05-16 it is still misspelled by the English Wikipedia in en.wikipedia.org/wiki/Harmonic_number#cite_note-5, while the German version has it right; (3) I'm of course not the first to note (2); (4) Theisinger used Bertrand's postulate, not the more elementary 2-adic reasoning. $\endgroup$ – Peter Heinig May 16 '20 at 20:03
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The answer is "no". Your function $H_z$ which is the same as $\psi(z+1)+\gamma$, where $\psi$ is the digamma function, and $\gamma$ is the Euler-Mascheroni constant, takes transcendental values at non-integer rationals. This is a theorem of M. Ram Murty and N. Saradha, "Transcendental values of the digamma function".

Notice that at rational values the digamma function has an explicit evaluation given by Gauss's formula.

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