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The title says the problem. What are the values of $\gamma$ that satisfy equidistribution? As far as I know, $\forall x\in (\mathbb{Q}\cup {ne|n\in \mathbb{Z}})$, the sequence is not equidistributed.

Ginger88895
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  • Do you have any reason to think there's any hope of answering this question? I doubt one can prove anything for $n!\sqrt2$, or $n!\pi$. – Gerry Myerson Sep 27 '16 at 12:34
  • I believe the same argument as for multiples of $e$ shows that the sequence is also not equidistributed for integer multiples of $e^{-1}$. – Dejan Govc Sep 27 '16 at 12:43
  • @DejanGovc You are right :) By the way I have such thought: I think the fractional part of $\gamma$ can be expanded like $\sum\limits_{n=2}^\infty \frac{a_n}{n!}$, where $a_n\in N\land a_n<n$(just like binary, hexadecimal numbers...) I hope I can get more people to answer this question by posting this thought :D – Ginger88895 Sep 27 '16 at 12:57

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Korobov, Concerning some questions of uniform distribution, Izvestiya Akad. Nauk SSSR. Ser. Mat. 14 (1950) 215–238, MR0037876 (12, 321a), according to the review by Lowell Schoenfeld, proves that $\alpha n!$ is uniformly distributed if $\alpha=\sum_1^{\infty}[k^{1+\lambda}]/k!$ and $0<\lambda<1$. Schoenfeld goes on to write,

...other specializations of these results are also given. While the uniform distribution of these functions for almost all values of $\alpha$ (in the sense of measure 0) has been known since Weyl's work [Math. Ann. 77, 313–352 (1916)], not a single value of $\alpha$ has been known for which these functions actually are uniformly distributed.

Korobov's paper was in Russian. I don't know whether it has been translated into other languages. Several later papers cite this paper. The review by Olaf Stackelberg of Salat, Zu einigen Fragen der Gleichverteilung (mod 1), Czechoslovak Math. J. 18 (93) 1968 476–488, MR0229586 (37 #5160), gives more details on Korobov's result.

Gerry Myerson
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  • Thanks for providing the information about this article. Though I also have trouble reading Russian, from the formulas listed within I think the proof is reasonable and such construct satisfies equidistribution. – Ginger88895 Sep 28 '16 at 07:18