0
$\begingroup$

Is it true that

$(\alpha p_k)$ is equidistributed on $[0,1)$ mod 1 (Vinogradov)

$\Leftrightarrow$

$(p_k)$ is equidistributed on $[0,2\pi) $mod $2\pi$ ?

$p_k$ is the kth prime and $\alpha$ is an irrational number.

$\endgroup$
1
  • $\begingroup$ sorry. p_k is the kth prime and alpha is an irrational. $\endgroup$ – daniel Nov 27 '11 at 22:53
2
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Oof. Thanks, yes that's what I meant. "Everything implies a theorem" will do in this case. $\endgroup$ – daniel Nov 27 '11 at 23:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.