# Fractional part of rational power arbitrary small

I think that $\{a^n\}$ (where $\{x\}$ is $x \pmod 1$), where $a$ is fixed rational greater than 1 and $n$ is positive integer, is dense in $[0,1]$ is unsolved. However what about $\{a^n\}$ is arbitrary small for some $n$ ($a$ is fixed rational as well).

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I think that the wolfram link cited the first result (Vijayaraghavan 1941) incorrectly where it reports that

$\{ (3/2)^n \}$ has infinitely many accumulation points in both $[0, 1/2)$ and $[1/2, 1]$.

The first result actually is

$\{ (3/2)^n \}$ has infinitely many accumulation points in $[0,1]$.

The second result cited by wolfram is by Flatto, Lagarias, Pollington (1995) which is available here: http://matwbn.icm.edu.pl/ksiazki/aa/aa70/aa7023.pdf. In the introduction of this paper by them, it reports that

Vijayaraghavan later remarked that it was striking that
one could not even decide whether or not $(3/2)^n$ mod $1$ has infinitely
many limit points in $[0, 1/2)$ or in $[1/2, 1)$.

This paper proved that

$$\limsup\{ (3/2)^n \} – \liminf \{(3/2)^n\} \geq \frac13.$$

A few later results are mentioned (Dubickas 2006, 2008) in Yann Bugeaud’s book “Distribution Modulo One and Diophantine Approximation” (for example, p. xi, p. 67 and p. 68).

$\{ (3/2)^n \}$ has a limit point in $[0.238117 . . . , 0.761882 . . .]$ which has length $0.523764 . . .$.

$\{(3/2)^n\}$ has a limit point in $[0, 8/39] \cup [18/39, 21/39] \cup [31/39, 1]$, of total length $19/39$.

I will show that
if $a = 1+\sqrt{2}$
then the limit points of
$\{a^n\}$
are $0$ and $1$.

I know that this doesn’t tell anything
about rational $a$,
but this might be of use.

Note that this method can show this
for $a$ and $b$ roots of
$x^2-2ux-v = 0$
where $u$ and $v$ are
positive integers such that
$v < 2u+1$.
This case is $u=v=1$;
$u=1, v=2$ also works.

If $a = \sqrt{2}+1$
and
$b = 1-\sqrt{2}$
then
$ab = -1$
and
$a+b = 2$.

Therefore $a$ and $b$
are the roots of
$x^2-2x-1 = 0$.

If $u_n = a^n+b^n$,
then
$u_0 = 2, u_1 = 2$.

Since
$a^{n+2} =a^n(a^2) =a^n(2a+1) =2a^{n+1}+a^n$
and similarly for $b$,

$\begin{array}\\ u_{n+2} &=a^{n+2}+b^{n+2}\\ &=2a^{n+1}+a^n+2b^{n+1}+b^n\\ &=2u_{n+1}+u_n\\ \end{array}$

Therefore
$u_n$ is a
positive integer for all $n$.

Since
$|b| < 1$,
$|b^n| \to 0$.

Since
$b < 0$,
$b^{2n} > 0$
and
$b^{2n+1} < 0$.

Therefore,
since
$a^n = u_n – b^n$,
$\{a^{2n}\} =\{u_{2n}-b^{2n}\} =1-b^{2n}$
so
$\{a^{2n}\} \to 1$
and
$\{a^{2n+1}\} =\{u_{2n+1}-b^{2n+1}\} =\{u_{2n+1}+|b^{2n+1}\} =|b^{2n+1}|$
so
$\{a^{2n+1}\} \to 0$.

Therefore
the limits points of
$\{a^n\}$
are $0$ and $1$.