3
$\begingroup$

Let $S$ be the set of integers with at least one prime factor in the arithmetic progression $km+d$, $(m, d)=1$. I am looking for results on the density of $S$. I found this post which talked about the density of the set of integers whose prime factors are all in the arithmetic progression, though I cannot find the citation from Landau.

If anyone can point me to results of this type that would be great. Thanks!

Improve this question
$\endgroup$
1
  • 3
    $\begingroup$ I suspect the density is $1$. Out near infinity, numbers have lots of prime factors, it's hard for all of them to avoid one given congruence class. $\endgroup$
    – Gerry Myerson
    Commented Jul 28, 2022 at 5:58

1 Answer 1

Reset to default
10
$\begingroup$

Let $P$ be any non-empty finite set of primes. By the Chinese Remainder Theorem, the density of integers not having a prime factor in $P$ is $\prod_{p\in P}(1-1/p)$, which is smaller than $\exp(-\sum_{p\in P}1/p)$. It is known that the sum of reciprocals of primes in the OP's arithmetic progression $A$ diverges (see Serre: A course in arithmetic), hence for any $\epsilon>0$ one can choose $P\subset A$ such that the above density is less than $\epsilon$. It follows that the density of $S$ exceeds $1-\epsilon$ for any $\epsilon>0$, hence it equals one.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Does this imply that for any prime $p$, there are infinitely many $n$ such that $p|r_{0}(n)$ where $r_{0}(n):=\inf\{r>0,(n-r,n+r)\in\mathbb{P}^{2}\}$ assuming $p$ belongs to some given such arithmetic progression? $\endgroup$
    – Sylvain JULIEN
    Commented Jul 29, 2022 at 9:49

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .