Let $(u_n)_{n\in\mathbb N}$ be the sequence of $3$-smooth numbers (that is whole number that can be written as $2^a3^b$ with $a,b\in\mathbb N$) sorted by increasing order. I am looking for the principal term of the asymptotic of $\log(u_n)$ as well as the error term, for $n\to+\infty$. Any idea to achieve that?

Thanks in advance.

  • 2
    $\begingroup$ Closely related: the number of smooth numbers up to a given bound: math.stackexchange.com/questions/1182775/… $\endgroup$ – Wojowu Jun 25 '18 at 11:44
  • $\begingroup$ Related, yes, but does it give an answer to my problem? $\endgroup$ – joaopa Jun 25 '18 at 11:50
  • 2
    $\begingroup$ You can deduce from there $\log u_n\sim\sqrt{2\log 2\log 3}\sqrt{n}$. If you figure out an error term for the number of smooth numbers, you can translate it to an error term to your problem. $\endgroup$ – Wojowu Jun 25 '18 at 12:22
  • $\begingroup$ How do you obtain this equivalent? I can not s the link between the number of $u_k$ lesser than $n$ and $u_n$. $\endgroup$ – joaopa Jun 25 '18 at 12:27
  • 1
    $\begingroup$ @joaopa: Let $N(x)$ denote the number of $u_n$'s up to $x$. Then, according to the link above, $N(x)\sim(\log x)^2/(2\log 2\log 3)$. Hence $n=N(u_n)\sim(\log u_n)^2/(2\log 2\log 3)$, which gives $\log u_n\sim\sqrt{(2\log 2\log 3)n}$. $\endgroup$ – GH from MO Jun 25 '18 at 18:05

3-smooth numbers are tabulated at http://oeis.org/A003586 and there you will also find the asymptotic, $${1\over\sqrt6}e^{\sqrt{2(\log2\log3)n}}$$ (attributed to Benoit Cloitre) as well as several links to the literature.

  • $\begingroup$ How do you obtain such an equivalent? $\endgroup$ – joaopa Jun 25 '18 at 14:07
  • $\begingroup$ Have you gone to that webpage, joaopa, and followed the links given there? $\endgroup$ – Gerry Myerson Jun 25 '18 at 22:51
  • $\begingroup$ I did not find a proof in the links on the oeis page. $\endgroup$ – joaopa Jun 30 '18 at 2:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.