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Jun 15, 2020 at 7:27 history edited CommunityBot
Commonmark migration
Aug 14, 2015 at 19:04 comment added Eric Naslund @DavidZhang: It means that it is both $\ll$ and $\gg$. (The symbol $\sim$ means that the quotient tends to $1$)
Aug 14, 2015 at 18:17 comment added David Zhang Excuse me, but what does $\asymp$ denote here? That the quotient of the two quantities tends to $1$ as $N \to \infty$?
Aug 14, 2015 at 17:35 history edited GH from MO CC BY-SA 3.0
improved spelling
Aug 14, 2015 at 13:10 comment added Eric Naslund @FedorPetrov: Thank you for the correction - it is now fixed.
Aug 14, 2015 at 13:10 history edited Eric Naslund CC BY-SA 3.0
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S Jun 27, 2013 at 20:18 history suggested Vít Tuček CC BY-SA 3.0
fixed math formatting: \\{ ---> \lbrace
Jun 27, 2013 at 20:14 review Suggested edits
S Jun 27, 2013 at 20:18
Jun 25, 2013 at 8:03 comment added Anthony Quas @jernej: There's a very nice small book by Mark Kac about the number of prime factors of integers. Otherwise, try Hardy and Wright.
Oct 7, 2012 at 8:27 vote accept Kamalakshya
Oct 6, 2012 at 8:08 comment added Jernej @Kevin P. Costello Do you happen to know a book or survey paper covering these types of results as the one of Hardy and Ramanujan about almost all integers and their prime factors?
Oct 6, 2012 at 4:48 history edited Eric Naslund CC BY-SA 3.0
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Oct 6, 2012 at 4:44 comment added Kevin P. Costello If all you're looking to show is $o(n^2)$, then I believe Erdos' original argument can get you there somewhat faster. As a very rough sketch, the idea is: 1. By a result of Hardy and Ramnujan, almost all integers between $1$ and $n$ have roughly $\ln \ln (n^2)=\ln \ln n+O(1)$ prime factors. 2. Almost all pairs $(x,y)$ have approximately $2\ln \ln n$ prime factors dividing $xy$ (since $x$ and $y$ usually won't share many factors). 3. Since most products lie in only a small subset of $\{1,…,n^2\}$ (the numbers having an unusually large number of factors), most of the rest must remain uncovered.
Oct 6, 2012 at 1:09 history edited Eric Naslund CC BY-SA 3.0
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Oct 5, 2012 at 17:40 comment added Felipe Voloch Presumably $n=N$.
Oct 5, 2012 at 17:29 history answered Eric Naslund CC BY-SA 3.0