14
$\begingroup$

We know that $e^{\pi \sqrt{163}}$ is very close to an integer(see wikipedia), and there is no closer to an integer than $163$ when $n$ is less than $1$ million.

Can we prove such a strengthened proposition:

the decimal part of $l_n = e^{\pi \sqrt{n}}, n \in \mathbb{Z}$ is dense on $(0, 1)$.

In this way, $l_n$ can approach $\alpha\in (0, 1)$, so we can know that $l_{163}$ is not the one closest to integers.

$\endgroup$
6
  • 11
    $\begingroup$ I suspect this is a hard problem. The question of the density of the decimal part of $x^n$ (where $x > 1$ is not an integer) is already tricky, even in the case $x = 3/2$. It's possible that the $x^{\sqrt{n}}$ variant is easier somehow, but I'd be surprised. $\endgroup$
    – Theo Bendit
    Dec 17, 2022 at 1:12
  • 2
    $\begingroup$ Even if $e^{\pi\sqrt{163}}$ is not the closest to an integer, if the fractional parts are (e.g.) "uniformly distributed" in $(0,1)$ then $l_{163}$ can be much closer than we would have expected from a sample size of $163$. $\endgroup$
    – anon
    Dec 17, 2022 at 2:43
  • $\begingroup$ Some interesting comments can be found in A069014. $\endgroup$
    – Gary
    Dec 27, 2022 at 7:13
  • 1
    $\begingroup$ It is certainly true, but how to prove it? $\endgroup$
    – Piquito
    Dec 27, 2022 at 10:35
  • 1
    $\begingroup$ I wonder if it would help to concentrate on subsequences with $n=m^2$, $m\in\mathbb{Z}$. Then we look on $e^{m\pi}\ mod\ 1$, which seems at least a bit simpler... $\endgroup$
    – NeitherNor
    Dec 31, 2022 at 19:45

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .