# Is the decimal part of $\exp (\pi \sqrt{n})$ dense on the interval $(0,1)$?

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.

• 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. Dec 17, 2022 at 1:12
• 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, 2022 at 2:43
• Some interesting comments can be found in A069014.
– Gary
Dec 27, 2022 at 7:13
• It is certainly true, but how to prove it? Dec 27, 2022 at 10:35
• 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... Dec 31, 2022 at 19:45