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 $.75$.



Density for products of arithmetic progressions
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 $c\_k+D\mathbb Z$.


