Is it true that
$(\alpha p_k)$ is equidistributed on $[0,1)$ mod 1 (Vinogradov)
$(p_k)$ is equidistributed on $[0,2\pi) $mod $2\pi$ ?
$p_k$ is the kth prime and $\alpha$ is an irrational number.
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I would think the 1st statement implies the second by taking $\alpha=1/2\pi$. The 2nd statement implies the first in the trivial sense that the first is a theorem (I take it that's what you mean when you attribute it to Vinogradov) and everything implies a theorem. But perhaps I misunderstand.