Suppose that the sequence $(a_{j})_{j \in \mathbb{N}}$ is an increasing sequence of positive integers that satisfies $$(1)\text{ }\text{ }\text{ }\text{ }\text{ }\text{ } d | a_{d}$$ and $$ (2)\text{ }\text{ }\text{ }\text{ } |\{v| v\in (a_{j})_{j \in \mathbb{N}}, v \leq N \}| \geq C\log(N).$$ for a given constant $C > 0$ and every $N \in \mathbb{N}$. For a given $x \in \mathbb{R}$, do we have

$$(3)\text{ }\text{ }\text{ }\text{ }\text{ } \lim_{N \rightarrow \infty}\min_{1 \leq j \leq N}\|a_{j}x\| = 0$$

(Here $\|y\|$ refers to the distance from $y$ to the nearest integer).

Note that if the sequence $(a_{j})_{j \in \mathbb{N}}$ satisfies the property $(3)$ then it is a Heilbronn sequence. https://en.wikipedia.org/wiki/Heilbronn_set

A Heilbronn sequence must contain multiples of every number, furthermore, this is not a sufficient condition. If the underlying sequence is sparse like $(j!)_{j \in \mathbb{N}}$ then it is not a Heilbronn sequence.

This conjecture is related to Dirichlet's approximation theorem, the paper "Small values of $n^2 \alpha$ $\mod 1$" and "Analytischer Beweis des Satzes uber die Verteilung der Bruchteile eines ganzen Polynoms" where they consider the case where $a_{j}$ is some fixed power of $j$.

  • 1
    $\begingroup$ I don't think so. You can fix an irrational $\theta$ and define the sequence greedily subject to $||a_d\theta|| > 1/3$, i.e., let $a_d$ be the smallest positive integer greater than $a_{d-1}$ that satisfies $d \mid a_d$ and $||a_d\theta|| > 1/3$. I think the resulting sequence will indeed have $\ge c\log N$ elements in $\{1,\dots,N\}$ for all (large) $N$. $\endgroup$ May 13, 2022 at 12:47
  • $\begingroup$ Maybe a nice $\theta$ to consider would be an algebraic number like $\frac{1+\sqrt{5}}{2}$ and consider your approach. $\endgroup$ May 13, 2022 at 12:57
  • $\begingroup$ Go ahead. Easy to code, for example. $\endgroup$ May 13, 2022 at 14:44

1 Answer 1


I will expand upon @mathworker21's comment.

Let $\theta = \frac{1 + \sqrt{5}}{2}$, and define the sequence $a_d$ greedily to be the smallest positive integer greater than $a_{d - 1}$ such that $d | a_d$ and $\|a_d \theta\| > \frac{1}{3}$.

Let $m$ be the smallest multiple of $d$ which is greater than $a_{d - 1}$. Clearly, $m \leq a_{d - 1} + d$. To find $a_d$, we take the number $m \theta$, and repeatedly add $d \theta$ until we get a number whose fractional part is between $\frac{1}{3}$ and $\frac{2}{3}$.

By Dirichlet's approximation theorem, for some $1 \leq k \leq 3$ we have $$\| kd \theta \| \leq \frac{1}{3}$$ and it is well known (for this specific value of $\theta$) that we have $$\| kd \theta \| \geq \frac{1}{3 k^2 d^2} \geq \frac{1}{27 d^2}.$$ Therefore, if we take $m \theta$ and add $kd \theta$ repeatedly, after at most $27 d^2$ steps we will get a number with fractional part between $\frac{1}{3}$ and $\frac{2}{3}$, which shows that $$a_d \leq a_{d - 1} + 30 d^3.$$ By induction, we have $$a_d \leq 30 d^4$$ which is sufficient for condition (2).


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