I am trying to approximate the number of integers less than $x$ that have no prime factors less than $B$. Is such an approximation known? If so, how good is it?

Note that this should not be confused with approximating $x - \Psi(x, B)$, since that would approximate the numbers less than $x$ that have at least one prime factor greater than or equal to $B$.

  • $\begingroup$ Perhaps you are thinking about smooth numbers, in which case, see: en.wikipedia.org/wiki/Smooth_number $\endgroup$ – BlackAdder Feb 17 '14 at 1:54
  • $\begingroup$ See my note: it is not so simple as subtracting from x the number of B-smooth numbers less than x. $\endgroup$ – Gnosis_X Feb 18 '14 at 4:48

Hint: How many primes are there in between B and X ? How many of their products are less then X ? For the first question, see prime-counting function $\pi(n)\simeq\dfrac n{\ln n}$ . For the second, $p_k^2\in[B^2,x]$ would be a necessary, though insufficient, first condition.

  • $\begingroup$ Did you mean to put $=$ there? $\endgroup$ – Matthew Conroy Feb 17 '14 at 2:06
  • $\begingroup$ No. I meant that, apart from the actual primes in between B and X, the smallest composites are of the form $p_jp_k$, with both $p_j$ and $p_k$ in between B and X. But that first prime $\ge B$ also has to be $\le\sqrt x$ in order for such composites to even exist... $\endgroup$ – Lucian Feb 17 '14 at 2:28
  • $\begingroup$ I meant the $\pi(n)=n/\ln n$ statement. $\endgroup$ – Matthew Conroy Feb 17 '14 at 2:37
  • $\begingroup$ Oh! Sorry. :-) I've fixed it now. $\endgroup$ – Lucian Feb 17 '14 at 2:42
  • $\begingroup$ I am not sure how to fully and correctly express this in mathematical notation, but is the solution: $(\pi(x)-\pi(B)) + (\pi(x^{\frac{1}{2}})-\pi(B))^2 + (\pi(x^{\frac{1}{3}})-\pi(B))^3 + ...$ until the first term in which $\pi(x^{\frac{1}{i}})-\pi(B)$ is negative? $\endgroup$ – Gnosis_X Feb 18 '14 at 4:42

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