# Density for products of arithmetic progressions

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 $$.75$$.

• 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. Jul 8, 2022 at 11:58
• @YaakovBaruch This seems reasonable. Thanks. Jul 8, 2022 at 15:09

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 $$c_k+D\mathbb Z$$.