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The sequence $\{h_n \alpha\}$ is equidistributed mod $1$ if $\alpha$ is irrational and $h_n = n$. Is there a generalization of the equidistribution theorem with some sequence $h_n$ (other than the type $h_n = \lfloor \beta + \gamma n\rfloor$) that would still work? I am looking for a sequence $h_n$ such that $h_n/n \rightarrow \infty$ if at all possible. I need something like this in a specific context, see section 3.1. in this article.

I thought I read that a sequence satisfying $\sqrt{h_{n+1}}-\sqrt{h_n} \rightarrow 0$ would do (one such sequence is $h_n$ being the $n$-th prime number), but I asked the question on MSE (here) and the answer is negative in general.

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Weyl proved in 1916 that a polynomial sequence with at least one irrational coefficient (besides the constant coefficient) is equidistributed modulo $1$. In particular, $\alpha n^2$ works, so $h_n=n^2$.

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  • $\begingroup$ This is VERY useful (in connection with my problem.) Thank you. $\endgroup$ Aug 30, 2019 at 4:59

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