# The sequence $\,a_n=\lfloor \mathrm{e}^n\rfloor$ contains infinitely many odd and infinitely many even terms

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 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 $[0,2]$.

• 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. – Chris Culter Jan 26 '18 at 7:59
• When you say "Weyl's theorem does look helpful", you write that it is "[unfortunate]". Why would that be unfortunate?.... – MathematicsStudent1122 Jan 26 '18 at 9:01
• @MathematicsStudent1122 Does "look" helpful, but not really helpful. – Sungjin Kim Jan 26 '18 at 22:27