Is the set $A:= \{ \sin(p): \text{$p$ is prime} \}$ a dense subset of $[-1,1]$? Is it known whether or not $-1$, $0$ or $1$ are limit points of $A$?

I would imagine so, otherwise this means the primes are related to $\pi$ in some special way, which I have not heard of. I'm not sure how to prove the affirmative though: I think the unpredictable nature of the primes makes this a challenge. All I know is that $\{ \sin(n):n\in\mathbb{N}\}$ is a dense subset of $[-1,1]$. For example, see here. There are many known inequalities and bounds for the prime counting function, for example, see here. However, I'm not sure these are useful for answering the above question.

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    $\begingroup$ There's probably some result about the uniform distribution of the fractional part of $p\alpha$ where $\alpha$ is a fixed irrational and $p$ runs through the primes, probably somewhere in the Kuipers and Niederreiter book. math.stackexchange.com/questions/2272248/… also looks relevant. $\endgroup$ Feb 11, 2023 at 23:09
  • $\begingroup$ Found it. Page 22 of K & N, Uniform Distribution of Sequences, it says, "... it is known that $(p_n\theta)$, $n=1,2,\dots$, is u.d. mod $1$, where $p_1=2$, $p_2=3,\dots,p_n,\dots$ is the sequence of primes arranged in ascending order." Here $\theta$ is an arbitrary (real) irrational, and $(x)$ is the fractional part of $x$. References are given to work by Vinogradov and by Hua, e.g., Chapter 11 of Vinogradov's book, The Method of Trigonometric Sums in the Theory of Numbers. $\endgroup$ Feb 12, 2023 at 23:10

1 Answer 1


Apparently I. M. Vinogradov proved it's dense in $1935.$ Though not equally distributed.

This is a theorem in Algebraic number theory.

That $\{-1,0,1\}$ are limit points is a trivial consequence.

  • $\begingroup$ To be clear: Vinogradov proved only that the set is dense, not that it is uniformly distributed. Nevertheless it almost certainly is uniformly distributed as well. Correct? $\endgroup$
    – mjqxxxx
    Dec 30, 2023 at 19:02
  • $\begingroup$ My guess is that he proved that it is not uniformly distributed. $\endgroup$ Dec 30, 2023 at 19:20
  • $\begingroup$ @mjqxxxx Intuitively, rather than a uniform distribution, it's more likely we would have density near $x$ proportional to $1/\sqrt{1-x^2}$, since that's what we would get if $p$ were uniformly distributed in $[0,2\pi]$. $\endgroup$
    – aschepler
    Dec 30, 2023 at 19:44
  • $\begingroup$ Yes, good point. So it’s most likely that the distribution is the same as if $p$ were uniformly random. $\endgroup$
    – mjqxxxx
    Dec 31, 2023 at 3:41

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