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Let $\alpha$ be any positive irrational and $\beta$ be any positive real. We have the following results.

H. Weyl (1909): The fractional part of the sequence $\alpha n$ is equidistributed modulo 1.

I. Vinogradov (1935): The fractional part of the sequence $\alpha p_n$ is equidistributed modulo 1 where $p_n$ is the $n$-th prime.

E. Hlawka (1975): The fractional part of the sequence $\beta \gamma_n$ is equidistributed modulo 1 where $\gamma_n$ is the imaginary part of the $n$-th zero the Riemann zeta function.

The common thing in each of the above three celebrated results is that the sequences are of the form $as_n$ where $a$ is a positive real and $s_n$ has the property that the sequence

$$ \frac{s_1}{s_n}, \frac{s_2}{s_n}, \ldots , \frac{s_{n-1}}{s_n} $$

approaches equidistribution modulo 1 as $n \to \infty$.

Question: I would like a nontrivial counterexample of a positive real $a$ and a sequence $s_n$ such that the fractional part of the sequence $as_n$ is equidistributed modulo 1 but the sequence of the ratios $s_i/s_n$ do not approach equidistribution modulo 1 as $n \to \infty$.

Edit: I am explaining what I mean by nontrivial because the example given by Noam indicates that it is necessary to explain it explicitly. The examples of Weyl, Vinogradov and Hlawka are nontrivial because there is no assumption on the normality constant. If we take the constant to be normal, we indirectly already assume what we want to prove and so we can construct many artificial examples.

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I still don't think you've nailed down exactly what you mean by "nontrivial". For example, in Weyl's and Vinogradov's results, you have an assumption on $\alpha$, namely that is irrational. Why is "irrational" okay but "normal in base 2" not okay? –  Greg Martin May 19 '13 at 0:50
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6 Answers 6

Let $s_n = 2^n$ and choose for $a$ any real number that's normal in base $2$.

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@Noam: While your example is correct, it is highly trivial because if we take $a$ to be normal in base $b$, then by the very definition of normality, we can construct similar example in that base $b$. Please note that in my examples of Weyl, Vonogradov had Hlawka, there is no such assumption on the normality constant which makes them non trivial. My question is can we have such a non trivial example such that $s_i/s_n$ does approach equidistribution modulo 1. –  Nilotpal Sinha May 18 '13 at 8:31
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Take $s_n = n^2$ and $\alpha$ irrational.

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But this is nothing but the extension of the Equidistributuion Theorem (first result) to polynomials, which was done my Weyl himself. So not just $n^2$, but any polynomial $s_n=f(n)$ is trivial. –  Nilotpal Sinha May 18 '13 at 11:17
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I agree. But the quotients $s_1/s_n,\dots, s_{n-1}/s_n$ do not tend to equidistribution as $n\to\infty$. –  Sean Eberhard May 18 '13 at 11:25
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Take $ s_n=\sqrt n $ and $a \ne 0.$

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J. F. Koksma proved in 1935 that for Lebesgue-almost-every $\beta>1$ the sequence $\beta^n$ is equidistributed modulo 1. I am not sure whether or not you would consider this to be nontrivial. Explicit examples of $\beta$ such that this property holds are not known, but it is widely thought that $\beta:=\frac{3}{2}$ has this property, as has been occasionally alluded to in answers to some other questions.

Koksma's article also contains a range of other results of this type: for example, Satz 5 states that if $a \colon \mathbb{N} \to \mathbb{Z}$ is an arbitrary injection then $\theta \cdot a(n)$ is equidistributed modulo 1 for Lebesgue-almost-every $\theta \in (0,1)$. It is easy to then construct an example of a repetition-free sequence of integers $s_n$ such that the sequence of measures $\frac{1}{n}\sum_{i=1}^n \delta_{s_is_n^{-1} \mod 1}$ does not converge to Lebesgue measure on the interval and deduce that an appropriate constant $a>0$ exists for this sequence.

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Let $S$ be the semi-group generated by $2$ and $3$. The interest in this semi-group is that it is non-lacunary, meaning $s_{n+1}/s_{n} \rightarrow 1$.

A famous theorem due to Furstenberg will tell you that that for every irrational $a$ the sequence $S\cdot a$ is dense modulo $1$.

As Furstenberg himself observed, this sequence not necessarily equidistribute, because of some Liouville-type numbers in base $6$.

Ergodicity will tell you that Lebesgue almost-every number equidistribute.

But as long as one does say impose some diophantine conditions on $a$, one can get effective equidistribution, this is called the effective Roduloph-Johnson theorem proved by Bourgain-Lindenstrauss-Michel-Venkatesh.

One can certainly take non-normal numbers here (say, as long as they generate a positive-dimensional Cantor set).

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Consider all possible pairs $(a,s_n)$ of a positive irrational $a$ and an integer sequence $s_n$ with exponential growth. It seems likely (maybe even easy to show) that it is almost always true that the sequence of fractional parts $[r_n]=\{s_na\}|_{n \in \mathbf{N}}$ is equally distributed in $[0,1).$ For example for any fixed sequence $s_n$ (or even all the sequences $b^n$) the set of appropriate $a$ will have density $1$.

You've already rejected this, but take $s_n=10^{n^{2}}.$ then given a desired sequence $r_n^'$ in $[0,1)$ we can easily describe an explicit $a$ such that $|\{s_na\}-r_n^'| \lt 10^{1-2n}$ (use the first $2j-1$ digits of $r_j'$ followed by the first $2j+1$ of $r^'_{j+1}$ etc. for $j \ge 1$.)

You will probably be as displeased with this, but I'll mention it anyway: Pick any positive irrational $a$ you wish and any desired sequence $r_n^'$ in $[0,1).$ Then we can also pick an integer sequence $s_n$ one term at a time so that for each $n$, $|\{s_na\}-r_n^'| \lt 10^{-n}$ and $\frac{s_1}{s_n} \lt \frac{s_2}{s_n} \lt \ldots \lt \frac{s_{n-1}}{s_n} \lt10^{-n}$: Suppose $s_{k-1}$ has been determined and consider in order $s_k=10^ks_{k-1},1+10^ks_{k-1},2+10^ks_{k-1},\cdots .$ From the theorem of Weyl, you are sure to eventually arrive at a choice with $\{s_ka\}$ differing from $r^'_k$ by no more than $10^{-k}.$

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