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My friend and I were talking earlier today and he posed the following problem, that he does not know the answer to: Take a real number, $a$, and look at consecutive powers of $a$: $a,a^{2},a^{3},...$ and look at the fractional part of the powers, i.e. $a^k - \lfloor a^k \rfloor$. What values can the fractional part of the powers converge to, if any? Obviously $0$ is one fractional part, but he believes that all rationals can be. He then extended his answers to all numbers that are not transcendental. Just thought it was a nice question and I'm curious as to what the answer is myself.

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I guess you are assuming that $a$ is not in $(-1,1)$, as that's the boring case... – J. M. Nov 18 '11 at 6:09
We don't have to assume that, it just doesn't help answer the question; if $a$ is in (-1,1), the fractional part will be 0, which we already know is possible. I'm curious as to what other possible values we can get. – Rebecca Saramosa Nov 18 '11 at 6:18
Some years ago I considered a similar question in the context of the collatz-problem. For some rational a=b/c I expressed the powers as irregular fractions and plotted the fractional part against the denominator and got some nice pictures. For the collatz-problem it was relevant, whether the dots for 1.5^N are all below the main diagonal. But I found some nice structures for some common irrational numbers like golden ratio and pi so perhaps you'll find it interesting, too. Here it is:… – Gottfried Helms Nov 18 '11 at 6:52

In case you had not observed it, I should mention that you can get $1$ as the limit. For example, let $a=3+2\sqrt{2}$. It is not hard to verify that $$(3+2\sqrt{2})^n +(3-2\sqrt{2})^n$$ is an integer for every non-negative integer $n$. One way is to expand each power using the Binomial Theorem, and observe that the terms that involve odd powers of $\sqrt{2}$ cancel.

For large $n$, $(3-2\sqrt{2})^n$ is positive and close to $0$. It follows that $(3+2\sqrt{2})^n$ is slightly below an integer, and therefore $$(3+2\sqrt{2})^n -\lfloor(3+2\sqrt{2})^n\rfloor$$ is nearly equal to $1$.

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Or note that $a_n=(3+2\sqrt{2})^n +(3-2\sqrt{2})^n$ satisfies the recursion relation $a_{n+1}=6a_n-a_{n-1}$, with initial conditions $a_0=2, a_1=6$. – J. M. Nov 18 '11 at 13:23

It's pretty well-studied. See this classic by Hardy and Littlewood, and this series of papers by Vijayaraghavan. In particular, it is known that the sequence $\{x^k\},\quad x > 1$ is equidistributed for most numbers, with a few exceptions like $x=1+\sqrt 2$ and $x=\phi$. See the MathWorld link for the interesting behavior when $x=\frac32$.

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