for which $A\subset \mathbb{N}$ does there exist irrational $\alpha$ for which $\alpha\cdot A \pmod 1$ is not uniformly distributed?

Question

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.)

Answer 1

Consider a random set $A$ where $n\in A$ with probability $f(n)/n$ for $f$ a function going to $\infty$ arbitrarily slowly. Then I think $A$ has $\alpha \cdot A$ uniformly distributed for all nonzero $\alpha$ with probability $1$. Equivalently $$\sum_{ n \in A, n \leq N} e( n \alpha ) = o ( f(N) \log N) $$ for all $\alpha \neq 0$.

The idea is that $$\sum_{ n \in A, n \leq N} e( n \alpha )$$ has mean $0$ and variance $f(N) \log N$ so it can probably be well-approximated by a Gaussian with mean $0$ and variance $f(N) \log N$, in which case the probability that it is $> \delta f(N) \log N$ is $$\leq e^{ - \frac{ \delta (f(N) \log N )^2}{ 2f(N) \log N } } =e^{ - \frac{ \delta f(N)}{2} \log N} = N^{- \frac{\delta f(N)}{2}}$$ which choosing $\delta$ a bit smaller than $2/f(N)$ is $N^{-1-\epsilon}$.

Now break $[0,1]$ into $N^{1+\epsilon}$ intervals of length $N^{-1 -\epsilon}$. Changing $\alpha$ from one point in the interval to the other changes the value of the sum $\sum_{ n \in A, n \leq N} e( n \alpha )$ by at most $$N^{-\epsilon} f(N) \log N = o(f (N)\log N)$$ so it suffices to check that $$\sum_{ n \in A, n \leq N} e( n \alpha ) = o ( f(N) \log N)$$ for $\alpha$ the midpoint of each interval. There are $N^{1+\epsilon}$ such $\alpha$ and each one has probability $N^{-1-\epsilon}$ so we win by the union bound.

If this exists then even very sparse sets with no hidden structure may not have an irrational $\alpha$ with this property.

Answer 2

Perhaps the most elementary example of an $(\alpha,A)$ pair like this is $A=\{ F_0,F_1,\ldots \}$, the Fibonacci numbers, $\alpha=\frac{1+\sqrt{5}}{2}$, the golden ratio. Then $\alpha\ast A \pmod{1}$ has a single limit point (or two, if you think 0 and 1 are different limits modulo 1).

In Representing Ordinal Numbers with Arithmetically Interesting Sets of Real Numbers, Joel Hamkins, Dakota Blair, and I considered this problem under the additional constraint that we considered only the order type of $\alpha\ast A$. In a nutshell, we found that for any countable order types $t_1,\dots,t_k$ and any irrationals $x_1,\dots,x_k$ there is a single set $A$ of positive integers with $x_i\ast A$ having order type $t_i$. There are some other relevant results in that paper. For example, if $A$ has positive density and $\alpha$ is irrational, then $\alpha\ast A$ has an interval of limit points.