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Let $X \subset \mathbb{N}$ and say that $X$ is super-equidistributed if for all $\alpha \in \mathbb{R} \setminus \mathbb{Z}$ there exists $C(\alpha) > 0$ such that for all $N$ $$ \left| \sum_{x \in X, x \leq N} \exp(2\pi i \alpha x) \right| < C(\alpha). $$

My conjecture is the following: If $X$ is super-equidistributed, then either $X$ is finite or $\mathbb{N} \setminus X$ is finite.

Why should this be true? Well there isn't too much evidence but one can show that if $X$ is super-equidistributed, then for all integers $a,q$ we have $$ \#\{x \in X: x \leq N, x \equiv a \pmod{q}\} = \frac{|X \cap [1,N]|}{q} + O_q(1). $$

This is a strong condition on $X$, but it is not enough. For example a friend of mine suggested the set $S = \bigcup_{n \geq 1} [n!, 2 \cdot n!) \cap \mathbb{Z}$. This satisfies this property mod $q$ for all $q$ but is neither finite nor is $\mathbb{N} \setminus S$. However if we pick $\alpha = e$, one can show that $\sum_{x \in S, x \leq N} \exp(2 \pi i \alpha x)$ is unbounded. To see this one essentially uses that the fractional part of $e \cdot k!$ is very well behaved for any $k$.

Also note that any set $X$ such that $X$ or $\mathbb{N} \setminus X$ is finite is super-equidistributed since $\sum_{x \geq 0} \exp(2 \pi i \alpha x) = (1 - \exp(2 \pi i \alpha))^{-1}$.

I would appreciate any thoughts on this problem. Possible counter-examples, solutions or consequences are all welcome.

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    $\begingroup$ Of possible interest is W Narkiewicz, Uniform distribution of sequences of integers in residue classes, in the Springer Lecture Notes series. $\endgroup$ – Gerry Myerson Jan 10 at 0:36
  • $\begingroup$ On a first glance it seems there are results about when a multiplicative sequence is uniformly distributed. Maybe one can try to establish the result for X multiplicative and somehow reduce to this case? $\endgroup$ – 1213 Jan 10 at 0:51
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    $\begingroup$ One could consider also a version where we replace $X$ with a $\pm 1$-valued sequence $x_n$ and demand $\sum_{n=1}^N x_n \exp(2 \pi i \alpha n)$ bounded for all $\alpha$, even zero. $\endgroup$ – Will Sawin Jan 10 at 1:15
  • $\begingroup$ Does there exist such a $\pm 1$ sequence? The erdos discrepancy problem tells us that there doesn't exists such a sequence for which $\sum_{n=1}^N x_n \exp(2\pi i \alpha n)$ is bounded independently of $\alpha$. $\endgroup$ – 1213 Jan 10 at 1:26
  • $\begingroup$ Does the Thue-Morse sequence (or rather, $X$ being the set of indices of $1$s in the Thue-Morse sequence) provide a counterexample? It's similar to the $mod 2^n$ set, but seems like it should work for dyadic roots of unity. $\endgroup$ – user44191 Jan 10 at 1:48
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Not quite an answer, but maybe it points to one:

Ivan Niven, Uniform distribution of sequences of integers, Compositio Mathematica 16 (1964) 158-160, defines a sequence $A=(a_1,a_2,\dots)$ of integers to be uniformly distributed if $$A(n,j,m)={n\over m}+{\rm o}(n)$$ for every integer $m\ge2$ and every $j$, $1\le j\le m$, where $A(n,j,m)$ is the number of terms among $a_1,a_2,\dots,a_n$ satisfying $a_i\equiv j\bmod m$. He gives as an example the sequence $[\theta],[2\theta],\dots$ of integer parts of the multiples of $\theta$, where $\theta$ is any real irrational. He cites a result of Uchiyama to the effect that $A$ is uniformly distributed if and only if $$\sum_{k=1}^Ne^{2\pi iha_k/m}={\rm o}(N)$$ for all positive integers $m$ and $h$, $1\le h\le m-1$. He gives some applications.

The Uchiyama reference is On the uniform distribution of sequences of integers, Proc Jap Acad 37 (1961) 605-609. There is also an earlier paper of Niven on the topic, Uniform distribution of sequences of integers, Trans Amer Math Soc 98 (1961) 52-61.

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