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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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sorry. p_k is the kth prime and alpha is an irrational. – daniel Nov 27 '11 at 22:53
up vote 2 down vote accepted

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.

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Oof. Thanks, yes that's what I meant. "Everything implies a theorem" will do in this case. – daniel Nov 27 '11 at 23:35

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