Let $S$ be the set of integers with largest prime factor bounded by a given positive integer $k$. Is there a formula for the asymptotic density of such a set $S$?
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1$\begingroup$ Please use a high-level tag like "nt.number-theory". I added this tag now. $\endgroup$– GH from MOCommented Aug 9, 2023 at 13:42
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4$\begingroup$ These are called "smooth" or "friable" numbers, and there is a vast literature about them. $\endgroup$– Joshua StuckyCommented Aug 9, 2023 at 22:58
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$\begingroup$ @JoshuaStucky : I would think you're talking about sets of numbers being "smooth" or "friable" or whatever the adjective is, rather than about the numbers themselves being "smooth", etc., since every number belongs to such a set. $\endgroup$– Michael HardyCommented Aug 11, 2023 at 20:37
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$\begingroup$ @MichaelHardy No, I am talking about numbers themselves being smooth, e.g. en.wikipedia.org/wiki/Smooth_number $\endgroup$– Joshua StuckyCommented Aug 13, 2023 at 0:48
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2 Answers
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If $k$ is fixed, then the simple bound $|S\cap[1,x]|\leq(\log_2 x)^{\pi(k)}$ shows that the aymptotic density of $S$ is zero. For stronger bounds, see Chapter III.5 in Tenenbaum: Introduction to analytic and probabilistic number theory.
answered Aug 9, 2023 at 13:41
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You're saying no prime factors bigger than the $n$th prime number appear. So we're looking at the set $$ \{ p_1^{e_1} \cdots p_n^{e_n} : (e_1,\ldots,e_n) \in \{0,1,2,3,\ldots\}^n \}. $$ There is a one-to-one correspondence between such numbers and the $n$-tuples $(e_1,\ldots,e_n),$ by uniqueness of prime factorizations.
If the sum of the reciprocals of these numbers is finite, then the asymptotic density must be zero. The sum of the reciprocals is $$ \left( 1 + \frac1{p_1} + \frac1{p_1^2} + \cdots \right) \cdots \left( 1 + \frac1{p_n} + \frac1{p_n^2} + \vphantom{ \frac1{p_1^2} + {} } \cdots \right), $$ and that is finite.
