# A sequence not equidistributed in [0,1]

Let $$a_n=\left(\frac{1+\sqrt{5}}{2}\right)^n.$$ For a real number $r$, denote by $\langle r\rangle$ the fractional part of $r$.

Why is the sequence $$\langle a_n\rangle$$ not equidistributed in $[0,1]$?

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It may be helpful to consider $a_n+1$ vs. $a_n^2$ (: – Nick Alger Dec 11 '10 at 15:40
Have you actually tried computing it? – Qiaochu Yuan Dec 11 '10 at 20:07
I am wondering why $\{ \mbox{frac} (x^n)\}$ is not equidistributed if $\phi^n\to 0\ (\mbox{mod}\ 1)$ holds? – user14538 Nov 15 '12 at 3:59

## 2 Answers

Let $\phi'=(1-\sqrt 5)/2$ denote the Galois conjugate of the golden mean $\phi$. Then $\phi^n+\phi'^{n}$ is an integer for every $n\in\mathbb N$, i.e. $$\phi^n+\phi'^{n}\equiv 0\ (\mbox{mod}\ 1).$$

But $|\phi'|<1$, so $\phi'^{n}\to 0$. This implies that $\phi^n\to 0\ (\mbox{mod}\ 1)$.

The property that the sequence $\{ \mbox{frac} (x^n)\}$ is not equidistributed is shared by other Pisot numbers. There is quite a lot of research publications devoted to them.

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If you just try it, the fractional part becomes very close to 0 or 1 quickly. This is because $\phi ^2=\phi +1$.

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