6
$\begingroup$

Let $p_n$ be the $n$th prime and $\alpha$ an irrational number. Vinogradov proved that the sequence $\{p_n \alpha\}$ is equidistributed. Is it known whether the sequence $\{p_n^2 \alpha \}$ is equidistributed?

$\endgroup$
12
$\begingroup$

Yes - this follows from a general theorem of Bergelson, Kolesnik, Madritsch, Son, and Tichy (Theorem 2.1 in https://people.math.osu.edu/bergelson.1/BKMS_PrimePowers.pdf):

Let $\xi(x)=\sum_{j=1}^m\alpha_j x^{\theta_j}$ be a polynomial with real coefficients $\alpha_i\in\mathbb{R}$ such that $0<\theta_1<\cdots <\theta_m$ and either

a) at least one of the $\theta_j$ is not at integer, or

b) at least one of the $\alpha_j$ is irrational.

If $\xi$ satisfies at least one of these conditions then $(\xi(p))$ is uniformly distributed.

Your special case $\xi(x)=\alpha x^2$ was probably known earlier, but I can't find a reference for that at the moment.

EDIT: This theorem in the case $\theta_j$ all integers was proved by Rhin in 1973 using Vinogradov's method, see https://mathscinet.ams.org/mathscinet-getitem?mr=323731. According to that Mathscinet review, this result was also implicit in Vinogradov's book.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ No, then - if all the powers are integers then the additional constraint that at least one $\alpha_j$ is irrational must be imposed. If at least one is not an integer, then it doesn't matter what the coefficients are, and the result holds even if they're all rational. $\endgroup$ – Thomas Bloom Dec 23 '18 at 13:41
  • 1
    $\begingroup$ I have rephrased the statement to hopefully make the condition a little clearer. $\endgroup$ – Thomas Bloom Dec 23 '18 at 14:03

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.