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The primes do a nice job of intersecting an arithmetic progression $\{a+dn\}_{n=0}^\infty$ when $a$ and $d$ are coprime (see Dirichlet's theorem).

I would like a set of integers $S$ such that

  • the asymptotic density of $S$ is zero, and
  • for every $a,d>0$, the asymptotic proportion of $S$ in $\{a+dn\}_{n=0}^\infty$ is $1/d$.

Is there a well-known collection of integers which satisfies these properties? Presumably, one may construct such an $S$ with probabilistic techniques, but I am more interested in explicit examples.

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  • $\begingroup$ Sums of two squares almost works. Maybe one can adjoin some numbers to "fill in the holes". $\endgroup$ Commented Nov 15, 2014 at 14:38

1 Answer 1

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Sets of the form $[n^\alpha]$, $\alpha\in(1, \infty)\setminus\mathbb{N}$ are well distributed in arithmetic progressions. More generally we have that for every $\alpha\in(1, \infty)\setminus\mathbb{N}$ and every $\beta\in(0,1)$ the set $\beta n^\alpha\bmod 1$ is equidistributed in $[0,1]$. By Weyl's criterion for equidistribution this is equivalent to the fact that for every $k$ the exponential sum $\sum_{n=1}^N e(k\beta n^\alpha)$ is of magnitude $o(N)$, which follows from the Weyl-van der Corput theory (see "Van Der Corput's Method of Exponential Sums" by Graham and Kolesnik, or "Ten Lectures on the Interface Between Analytic Number Theory and Harmonic Analysis" by Montgomery, or the introduction by Elkies here http://www.math.harvard.edu/~elkies/M259.06/vdc.pdf ).

If you want to have sets of density a lot smaller than $n^\epsilon$, you can take a smooth function $f:\mathbb{R}\rightarrow\mathbb{R}$, which grows faster than any polynomial, but not too fast, and use Vinogradov's estimates in place of Weyl-van der Corput. Filling in the details is probably quite some work, a starting point could be Robert's work on exponent pairs obtained by Vinogradov's method (Robert, Quelques paires d'exposants par la méthode de Vinogradov, Journal de théorie des nombres de Bordeaux 14, 271-285).

A completely different method would be to take a set, which is obviously equidistributed, and stretch. For example the sequence $n!+n$ is equidistributed and extremely thin, but may be considered to be quite artificial.

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