Bounds on the size of rough numbers

A positive integer $$n$$ can be described as $$B$$-rough if all of the prime factors of $$n$$ strictly exceed $$B$$.

The first five 2-rough numbers are 1, 3, 5, 7, 9. We always include 1 by convention.

It appears to be true that the $$k$$th $$B$$-rough number will never exceed $$Bk$$.

Answers to this question, for example, show why it is true for finite ranges of $$k$$ and $$B$$. Is it possible to show that it is true in general without a major breakthrough in number theory?

• For k larger than something like $2^{\pi(B)}\log B$ this is clear from counting numbers coprime to the primes at most $B$. For better estimates one can use bounds on Jacobsthal's function to show it holds for k greater than g(P), where P is the product of the $m$ many primes at most P and g(P) is bounded above by $m^{3+3.81\log\log m}$ (and likely also by $m^2$). Gerhard "Don't Know About Small K" Paseman, 2019.02.11. – Gerhard Paseman Feb 11 at 17:09
• One can also use existing estimates on $\pi(2N)-\pi(N)$ for $N=B,2B, 4B$, etc. to bump up $k$. However I do not know how large a value of $k$ you can achieve with this. Gerhard "Less Sure About Middling K" Paseman, 2019.02.11. – Gerhard Paseman Feb 11 at 17:36
• @Gerhard Paseman: The two ranges actually overlap. To get explicit results I only use Chebychev type inequalities, not the prime number theorem. Looking only at primes we find that the claim is true if $\pi(Bk)>k+B$, which is true for $B>2\log k$. On the other hand the first approach works for $k>3^{B/\log B}\log B$. Thus the claim is true for all $B<30$ and all $k<5\cdot 10^8$. Counting the number of 30-rough integers below $5\cdot 10^8$ greatly reduces these bounds. Then check the small cases individually. – Jan-Christoph Schlage-Puchta yesterday