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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.

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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ @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
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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.

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  • $\begingroup$ 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

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