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PROBLEM. Show that the sequence $\,a_n=\lfloor \mathrm{e}^n\rfloor$ contains infinitely many odd and infinitely many even terms.

It suffices to show that the terms of the sequence $$\,b_n=\mathrm{e}^n\,\mathrm{mod}\, 2,\,\,\,n\in\mathbb N,$$ are dense in $[0,2]$.

Unfortunately, Weyl's Theorem does not look helpful in this case.

EDIT. As Chris Culter said, the claim that the terms of the sequence $$\,b_n=\mathrm{e}^n\,\mathrm{mod}\, 2,\,\,\,n\in\mathbb N,$$ are dense in $[0,2]$ is (or might be) an open problem. Nevertheless, this does not imply that the claim that the sequence $\,a_n=\lfloor \mathrm{e}^n\rfloor$ contains infinitely many odd and infinitely many even terms is necessarily an open problem as well. It is also noteworthy that it is relatively easy to construct an irrational $\alpha$ with the property that the sequence $\,\alpha^n\,\mathrm{mod}\, 2,\,\,n\in\mathbb N,$ is NOT dense in $[0,2]$.

Yiorgos S. Smyrlis
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    If $e^n$ is dense modulo $2$, then it is also dense modulo $1$, yet the latter is an open problem, at least circa 2003 per [this paper](https://www.researchgate.net/publication/283609413_Salem_numbers_and_uniform_distribution_modulo_1). – Chris Culter Jan 26 '18 at 07:59
  • @MathematicsStudent1122 Does "look" helpful, but not really helpful. – Sungjin Kim Jan 26 '18 at 22:27
  • What's the source of this question, please? – Gerry Myerson Nov 11 '19 at 08:34
  • I had found it in a Greek site with Mathematical problems for university students. – Yiorgos S. Smyrlis Nov 11 '19 at 08:38
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    Exercises for university students? Seems hard. The sequence is tabulated at http://oeis.org/A000149 but I don't think anything at that page helps to resolve the question. – Gerry Myerson Nov 11 '19 at 08:45
  • For any $a\in\mathbb N$, $\lfloor a^n\rfloor$ will have the same parity as $a$. For $0 < a < 1$ we always have $\lfloor a^n\rfloor = 0$. But for any other values of $a$ I don't see obvious way to prove or disprove the property from the problem. Maybe it has nothing to do specifically with $e$ and a proof would cover a class of reals (potentially all non-integers $>1$ ?). – Cristian Gratie Apr 06 '22 at 17:08

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