How many numbers below $N$ are sieved by a certain fraction of primes?

Question

Suppose we have an infinite sequence of primes which have an asymptotic density $1/a$ among all the primes. For convenience let $P=\{p_a,p_{2a},...,p_{na},...\}$ be the primes we are interested in.
Can we approximate the number of natural numbers below $N$ which are divisible by at least one of the primes $\in P$ ?
It would be safe to conjecture that $N/a$ numbers are "sieved" by those primes but I am not sure if this is correct and I have not any idea how to calculate something more precise.
Any ideas?

Answer 1

The fraction of numbers divisible by at least one of your primes tends toward 100%. For any finite set of primes $P = \{p_a, p_{2a},\ldots,p_{ka}\}$ the fraction of numbers which are divisible by none of them is $$ c_P = \prod_{p\in P}1 - \frac1p = \exp\sum_{p\in P}\log(1-\frac1p) \approx \exp-\sum_{p\in P}\frac1p $$ and in particular there are $(1 - c_P)x + O(1)$ such numbers up to $x$.

But your primes are $1/a$ of primes, so the sum over all the primes in $P$ up to $x$ is asymptotically $$ \frac{\log\log x}{e^\gamma a}(1+o(1)) $$ and since this tends to $\infty$ the constant $c_P$ is less than any positive number and hence equal to 0.