Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have a set S of prime numbers and I would like to find the size (in some sense, ideally some nice asymptotic expression) of the set of positive integers which are the product of with all prime divisors in S. (That is, for each prime $p$ dividing such a number, $p\in S$.)

What are good methods for going about this? The trivial cases are when $S$ is finite or cofinite (in the set $\mathcal{P}$ of primes):

  • If $S$ is finite a Mertens-like product would give a relative density.
  • If $\mathcal{P}\setminus S$ is finite

Some particular cases of interest:

  • $S$ is the set of primes in a finite collection of arithmetic progressions: $S=\mathcal{P}\cap\left(\bigcup_{i=1}^k(a_i+b_in)_{n\in\mathbb{N}}\right)$.
  • $S$ is not known, but $s_n$, the n-th term of $S,$ obeys $f(n)\le s_n\le g(n)$ for sufficiently nice function $f,g.$ Example: $n^3\le s_n\le n^4$ for $n>100.$

These include very nice problems like "how common are numbers which are the sum of two squares" and "how dense are abundant numbers".

An asymptotic would be ideal, but at this point I'll take what I can get.

share|cite|improve this question
What do you mean by " I would like to find the size of the set of positive integers with all prime divisors in S." It will always be infinite, and its density will always be $0$ when $S$ does not have density $1$ in the primes. I am not sure I understand your question, are you looking for something like the summatory function? You could possibly give bounds for the elements in $S$, or the summatory function of the set $S$. – Eric Naslund Mar 21 '12 at 19:15
@EricNaslund: I left it open intentionally, in case there is some method that works better than others. Ideal would be an asymptotic "a(n) ~ k n^a log^b n ..." where a(n) is the n-th S-smooth number. A big-Theta would be almost as good. But if less is possible/convenient I would accept whatever mathematical technology permits. – Charles Mar 21 '12 at 20:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.