In "M. B. Nathanson – Elementary Methods in Number Theory" is shown (Theorem 7.14) that if $A$ is a set of positive integers such that $\sum_{a \in A} 1 / a$ converges then the set of multiples of $A$ has a natural density (a set of positive integers $S$ has a natural density if exists $\lim_{x \to \infty} |S \cap [1,x]| / x$).
Equivalently, the set of positive integers $n$ such that $a \nmid n$ for all $a \in A$ has a natural density.
I am looking for any reference for the analogous result that if $b$ is some fixed positive integer and $A$ is the same set as above with additional condition that $1\notin A$ then the relative density of primes $p=1+b\cdot m$ where $m$ is not divisible by any $a\in A$ is positive if there exist at least one such prime. I have a proof of this result but I am curious if this result is already in literature.
Thanks in advance for any assistance.
Best Answer
This is too long for a comment and possibly it still answers your question.
Fix a positive integer $b \in \mathbb{N}$ and set $A\subseteq \mathbb{N}$ such that $\sum_{n \in A}1/n<\infty$ and $1\notin A$. Let $\{a_1,a_2,\ldots\}$ be the increasing enumeration of $A$. Let also $\mathbb{P}$ be the set of primes and define $$ X:=\left\{p \in \mathbb{P}: p\equiv 1\bmod{b}\,\, \text{ and }\,\,p\not\equiv 1\bmod{ab} \text{ for all }a\in A\right\}. $$ In other words, $X=\bigcap_{N\ge 1}X_N$, where $$ X_N:=\left\{p \in \mathbb{P}: p\equiv 1\bmod{b}\,\, \text{ and }\,\,p\not\equiv 1\bmod{a_nb} \text{ for all }n=1,\ldots,N\right\} $$ It follows by the main result here (at least for $b=1$) that $X_N$ admits asymptotic density relative to $\mathbb{P}$ and, in addition, such value is at least $$ \frac{1}{\varphi(b)}\prod_{n=1}^N\left(1-\frac{1}{\varphi(a_nb)}\right). $$ With the same proof of Theorem 7.14 given by Nathanson, also the set $X$ admits asymptotic density relative to $\mathbb{P}$. The fact that the latter value is positive should follow by the fact that $$ \lim_{N\to \infty}\lim_{n\to \infty}\frac{|X_N \cap [1,n]|}{|\mathbb{P}\cap [1,n]|} \ge \frac{1}{\varphi(b)}\lim_{N\to \infty}\prod_{n=1}^N\left(1-\frac{1}{\varphi(a_nb)}\right) $$ and since $\varphi(a_nb)\ge \sqrt{a_nb}\to \infty$ as $n\to \infty$, we obtain a lower bound $$ \gg \lim_{N\to \infty}\mathrm{exp}\left(\sum_{n=1}^N \log\left(1-\frac{1}{\varphi(a_nb)}\right)\right) \gg \mathrm{exp}\left(-\lim_{N\to \infty}\sum_{n=1}^N\frac{1}{\varphi(a_nb)} \right); $$ hence to conclude the proof, it is sufficient to show that $\sum_{n=1}^\infty\frac{1}{\varphi(a_nb)}$ is finite. Now $$ \sum_{n=1}^\infty\frac{1}{\varphi(a_nb)}\le \sum_{n=1}^\infty\frac{1}{\varphi(a_n)\varphi(b)}\ll \sum_{n=1}^\infty\frac{1}{\varphi(a_n)}. $$ However, why the latter sum should be finite? I mean, potentially every estimate is optimal if $b=1$ and $A$ has pairwise coprime elements. Still, it is known that there exists infinitely many $n$ such that $$ \varphi(n)<\frac{n}{\log \log n}. $$ Hence, how do you prove that $\sum_n \frac{1}{\varphi(a_n)}<\infty$, assuming that $\sum_n \frac{1}{a_n}<\infty$?