14

We know that Ramanujan's constant $e^{\pi \sqrt{163}}$ 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.

138 Aspen
  • 1
  • 13
    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. – Theo Bendit Dec 17 '22 at 01:12
  • 2
    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 $163$. – anon Dec 17 '22 at 02:43
  • Some interesting comments can be found in A069014. – Gary Dec 27 '22 at 07:13
  • 1
    It is certainly true, but how to prove it? – Ataulfo Dec 27 '22 at 10:35
  • For all $a\in[0,1]$ one has $\pi\sqrt x=a$ with $x=\dfrac{a^2}{\pi^2}$, so we can deduce that the fractional part $\pi\sqrt x-\lfloor{\pi\sqrt x}\rfloor$ is dense in $[0,1]$. Thus the set of fractional parts of the exponents of $e$ is dense in $[0,1]$.From this, I think someone can deduce the asked result. – Ataulfo Dec 27 '22 at 11:33
  • 1
    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... – NeitherNor Dec 31 '22 at 19:45
  • Related post on MathOverflow: Why are powers of $\exp(\pi\sqrt{163})$ almost integers? – 138 Aspen Mar 04 '24 at 01:17

0 Answers0