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In this question it is discussed that for a set $$A=\{2^n3^m| n,m=0,1,2,\ldots\}$$ there exists an irrational number $\alpha$ such that the set $\alpha\cdot A$ (being naturally enumerated) is not uniformly distributed modulo 1. For which general conditions on $A$ does such $\alpha$ exist? For example, is it sufficient that $|A\cap [y,y+x]|=O(x^c)$ for all $c>0$ uniformly by $y$ enough? (By Weyl theorem we can not hope for better density condition: the sequence $\alpha\cdot n^d$ is equidistributed modulo 1 for all positive integer $d$ and all irrational $\alpha$. As noted by user42355 in the comment, considering only $y=0$ is not enough, as the set $\sqcup [a_n, a_n+n]$ for rapidly increasing $(a_n)$ shows.)

In this question it is discussed that for a set $$A=\{2^n3^m| n,m=0,1,2,\ldots\}$$ there exists an irrational number $\alpha$ such that the set $\alpha\cdot A$ (being naturally enumerated) is not uniformly distributed modulo 1. For which general conditions on $A$ does such $\alpha$ exist? For example, is it sufficient that $|A\cap [y,y+x]|=O(x^c)$ for every $c>0$ uniformly by $y$? (By Weyl theorem we can not hope for better density condition: the sequence $\alpha\cdot n^d$ is equidistributed modulo 1 for all positive integer $d$ and all irrational $\alpha$. As noted by user42355 in the comment, considering only $y=0$ can not be enough, as the set $\sqcup [a_n, a_n+n]$ for rapidly increasing $(a_n)$ shows.)

In this question it is discussed that for a set $$A=\{2^n3^m| n,m=0,1,2,\ldots\}$$ there exists an irrational number $\alpha$ such that the set $\alpha\cdot A$ (being naturally enumerated) is not uniformly distributed modulo 1. For which general conditions on $A$ does such $\alpha$ exist? For example, is it sufficient that $|A\cap [0,x]|=o(x^c)$ for all $c>0$ enough? (By Weyl theorem we can not hope for more: the sequence $\alpha\cdot n^d$ is equidistributed modulo 1 for all positive integer $d$ and all irrational $\alpha$).

In this question it is discussed that for a set $$A=\{2^n3^m| n,m=0,1,2,\ldots\}$$ there exists an irrational number $\alpha$ such that the set $\alpha\cdot A$ (being naturally enumerated) is not uniformly distributed modulo 1. For which general conditions on $A$ does such $\alpha$ exist? For example, is it sufficient that $|A\cap [y,y+x]|=O(x^c)$ for all $c>0$ uniformly by $y$ enough? (By Weyl theorem we can not hope for better density condition: the sequence $\alpha\cdot n^d$ is equidistributed modulo 1 for all positive integer $d$ and all irrational $\alpha$. As noted by user42355 in the comment, considering only $y=0$ is not enough, as the set $\sqcup [a_n, a_n+n]$ for rapidly increasing $(a_n)$ shows.)

In this question it is discussed that for a set $$A=\{2^n3^m| n,m=0,1,2,\ldots\}$$ there exists an irrational number $\alpha$ such that the set $\alpha\cdot A$ (being naturally enumerated) is not uniformly distributed modulo 1. For which general conditions on $A$ does such $\alpha$ exist? For example, is it sufficient that $A$ has zero upper density, or maybe even that $A$ has upper density less than 1?

In this question it is discussed that for a set $$A=\{2^n3^m| n,m=0,1,2,\ldots\}$$ there exists an irrational number $\alpha$ such that the set $\alpha\cdot A$ (being naturally enumerated) is not uniformly distributed modulo 1. For which general conditions on $A$ does such $\alpha$ exist? For example, is it sufficient that $|A\cap [0,x]|=o(x^c)$ for all $c>0$ enough? (By Weyl theorem we can not hope for more: the sequence $\alpha\cdot n^d$ is equidistributed modulo 1 for all positive integer $d$ and all irrational $\alpha$).