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I did some numerical experiments about $\{ \frac{a}{b}: a,b \in A, \;a < \, b\}$ for various integer sets $A$. Does anyone recognize these densities?

Here is $A = \{ p: \text{prime}\}$

enter image description here

Here is $A = \mathbb{Z}$

enter image description here

Here is if $\{ (a,b): a < 2 \, b, \; \gcd(a,b) = 1 \}$. Does this uniform distribution surprise everybody?

enter image description here

Literature on each of these could be interesting. I know there is discussion about the primes case being dense already:

By default I can look for the Weyl Equidistribution Theorem. We are told to evaluate sums:

$$ c_N(m) = \sum_{0 < a < b < N} e^{2\pi i m \frac{a}{b}} $$

These are the sums of Ramanujan, $c_q(n)$. This paper of Huxley states in the affirmative, and - in usual style - the result is so obvious he does not prove it.

In another place Huxley works it out. He shows (unconditionally) just reasoning about fractions:

$$ \sum_{n=1}^N e^{ 2\pi i \, m \frac{a}{b}} \ll d(m) \sqrt{N} $$ and this is Weyl's criterion. The other lemma I like is: Let $0 < x_n < 1$ be the Farey fractions between $0$ and $1$:

$$ \sum_{n=1}^N e(m x_n) = \sum_{d|m} d \left[ M( \tfrac{Q}{d}) \right] $$

where $M(x) = \sum_{m \leq x} \mu(m)$. No effort was taken to optimize exponent. This is already equivalent to $M(x) = o(x)$.

This is taken from Chapter 1 Huxley's book Area, Lattice Points, and Exponential Sums

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  • $\begingroup$ I think it's related to Farey sequences, from which one can get a criterion for the truth of RH. $\endgroup$
    – Sylvain JULIEN
    Commented Aug 14, 2016 at 15:49
  • $\begingroup$ Quotients of Primes David Hobby and D. M. Silberger $\endgroup$
    – john mangual
    Commented Aug 14, 2016 at 15:49
  • $\begingroup$ I have no idea what the graphs are meant to be. I don't even know what the axes are. But I know there's much discussion of the equidistribution of Farey sequences in the Kuipers & Niederreiter book. $\endgroup$
    – Gerry Myerson
    Commented Aug 14, 2016 at 23:25
  • $\begingroup$ There is some work by Boca, Cobeli, and Zaharescu that I think is quite closely related to your question. (see also the Hall distribution). $\endgroup$
    – Anthony Quas
    Commented Aug 15, 2016 at 7:54
  • $\begingroup$ @AnthonyQuas I googled Hall distribution and found this paper by Jayadev Athreya on the Gaps of Farey Fractions: arXiv:1210.0816 Gap distributions and homogeneous dynamics $\endgroup$
    – john mangual
    Commented Aug 16, 2016 at 23:03

1 Answer 1

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The equidistribution in the last graph is equivalent to the prime number theorem, the optimal speed of equidistribution is equivalent to the Riemann hypothesis. The way you get there is by taking Fourier transform of the empirical measure, and when the smoke clears, the decay of coefficients is equivalent to the fact that $\sum_{n<x} \mu(n) = o(x).$

EDIT A nice exposition can be found in this University of Edinburgh senior project.

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  • $\begingroup$ Are you sure? It looks to me like the simpler fact that $\sum_{n<x} \mu(n)=o(x^2)$. $\endgroup$
    – Will Sawin
    Commented Aug 14, 2016 at 18:34
  • $\begingroup$ @WillSawin Did you note that $Q \asymp \sqrt{N}$ ? I have that $\sum_{n<x} \mu(n)=O(x)$ which is somewhat less than the PNT. $\endgroup$
    – john mangual
    Commented Aug 16, 2016 at 16:46
  • $\begingroup$ @johnmangual The equidistribution follows from $\sum_{d|m} d M(Q/d) = o(N)$, which follows from $M(x) = o(x^2)$, because then $M(Q/d)=o(Q^2/d^2)=o(Q^2)=o(N)$. We can assume $m=O(1)$ so $d=O(1)$ and the sum is finite. $\endgroup$
    – Will Sawin
    Commented Aug 16, 2016 at 17:34
  • $\begingroup$ @WillSawin The book uses $\sum_{n < x}\mu(n) \leq x$ which even I can verify. Unfortunately, this is definitely not the PNT. Even if I use $\zeta(1+it) \neq 0$ I still need the Wiener-Ikehara (or something) to prove $\sum_{n < x} \mu(n) = o(x)$. At least, if I find some way to get better than $N^{1/2}$ on these Weyl averages, it's equivalent to PNT. $\endgroup$
    – john mangual
    Commented Aug 16, 2016 at 18:40
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    $\begingroup$ @IgorRivin I am not being sarcastic, I'm just trying to make a very slight correction but not communicating very clearly. I think your first statement is wrong because the Prime Number Theorem does not follow quickly from the equidistribution statement. The bound that the equidistribution statement is equivalent to is $\sum_{n < x} \mu(n)= o(x^2)$, which is trivial because, as you note, $|\mu(n)| \leq 1$. I agree that optimal equidistribution is equivalent to the Riemann hypothesis. $\endgroup$
    – Will Sawin
    Commented Aug 17, 2016 at 0:02

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