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I have a sequence of positive integers $a_n$ which assumes infinitely many times the value $1$. I want to estimate the cardinality of the following set:

$$\#\{n\leq x : a_n>1 \text{ and } (a_n, \mathcal{P})=1\},$$ where $P=\prod_{p\leq y}p$ is the product of all the primes less than $y$. Here we can assume that $y<x$ are sufficiently large.

I was wondering if there exists a method to approach this problem. Obviously, if $a_{n}$ took only finitely many times the value $1$, we could apply Sieve Theory results.

A possibility might be to separate according to the least prime diving $a_n$ and obtain that

$$\#\{n\leq x : a_n>1 \text{ and } (a_n, \mathcal{P})=1\}\leq \sum_{p>y}\#\{n\leq x : p\mid a_ n \text{ and } (a_n/p, \mathcal{P})=1\}$$

and then to apply Sieve Theory to the sequence $b_n=a_n/p$. The problem is that $b_n$ is not anymore an integer sequence.

Can anyone help me, please?

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  • $\begingroup$ I don't see how the condition (or the lack of it) that infinitely many $a_n$'s are equal to $1$ changes anything. The idea of sieve theory is to upper bound and lower bound the divisor sum $\sum_{d\mid m}\mu(d)$ by replacing $\mu(d)$ in it by other functions. Starting from such bounds, you can estimate your quantity, which is $\sum_{\substack{n\leq x\\a_n>1}}\sum_{d\mid(a_n,\mathcal{P})}\mu(d)$. $\endgroup$
    – GH from MO
    Commented Apr 30, 2018 at 16:08
  • $\begingroup$ Yes, you are right. The problem is that I don't have any information about the number of integers $n\leq x$ such that $a_n>1$. Therefore, I don't see how can I find a bound using your formula. $\endgroup$
    – The Number Theorist
    Commented Apr 30, 2018 at 17:33
  • $\begingroup$ If you really have no information, it could be that $a_n$ is always $1$, and then you can't do better than the trivial bound. $\endgroup$
    – Denis Chaperon de Lauzières
    Commented Apr 30, 2018 at 19:18
  • $\begingroup$ Actually, what I expect (and what I would like to prove) is that the set $\{n : a_n=1\}$ has positive natural density. But in order to reach this, I need to estimate the cardinality stated above. $\endgroup$
    – The Number Theorist
    Commented Apr 30, 2018 at 19:30
  • $\begingroup$ Notice the use of \text{} and \mid in my edits to the question. Both affect spacing. $\endgroup$
    – Michael Hardy
    Commented May 1, 2018 at 17:19

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