Consider $k\geq 2$ biinfinite arithmetic progressions $\mathcal A_i=a_i+b_i\mathbb Z$ (for $i=1,\ldots,k$) in $\mathbb Z$. (We suppose that $a_i$ and $b_i\geq 2$ are strictly positive integers. One can assume $a_i$ and $b_i$ coprime and $a_i$ can be reduced modulo $b_i$ without loss of generality.)

The set $\mathcal P=\prod_{i=1}^k \mathcal A_i$ of all possible products has strictly positive upper and lower densities $\overline{\delta}$ and $\underline{\delta}$ in $\mathbb Z$.

Is it possible that $\overline{\delta}>\underline{\delta}$? Can they be irrational? Are they easy to compute? Fixing $\epsilon>0$ and $k\geq 2$ are there only finitely many such products $\mathcal P$ with upper density at least $\epsilon$?

Examples: $(1+2\mathbb Z)(1+3\mathbb Z)=\mathbb Z\setminus\{0\}$ (and both densities are $1$).

$(1+2\mathbb Z)(1+4\mathbb Z)=1+2\mathbb Z$ (and both densities are equal to $1/2$).

$(1+3\mathbb Z)(1+5\mathbb Z)$ seems to have densities above $.75$.

  • 2
    $\begingroup$ Since the expected number of divisors of $n$ goes to infinity with $n$ (as $\log \log n$) my intuition is that that eventually leaves enough room to group factors together so as to almost always meet whatever decomposition is allowed by considerations modulo $\prod b_i$. In particular the density should be $1$ in your last example. $\endgroup$ Jul 8, 2022 at 11:58
  • $\begingroup$ @YaakovBaruch This seems reasonable. Thanks. $\endgroup$ Jul 8, 2022 at 15:09

1 Answer 1


I think Yaakov Baruch is right in general.

Let $D$ be the l.c.m. of all the $b_i$, and let $S$ be the set of all residue classes in $\mathbb Z/D\mathbb Z$ met by $\mathcal P$. I claim that the density is $|S|/D$.

The upper bound is trivial. In what follows, I assume $(a_i,b_i)=1$, otherwise pass to a (1-dim) sublattice.

Consider any residue class $x+D\mathbb Z\in S$; it is a product of some classes $c_i+D\mathbb Z\subseteq a_i+b_i\mathbb Z$. As far as I understand, most of the integers (in particular, most of those in $x+D\mathbb Z$) have (different) prime divisors $p_i\equiv a_i\pmod {b_i}$. Each such integer $n\in x+D\mathbb Z$ is expanded as $$ n=p_1\dots p_{k-1}\cdot \frac n{p_1\dots p_{k-1}}\in \mathcal P, $$ as the last factor automatically belongs to $c_k+D\mathbb Z$.


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