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:

$$\#\{n\leq x : a_{n}>1\ \mathrm{and} \ (a_n, \mathcal{P})=1\},$$ 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

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

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?

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?