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S Jul 23, 2021 at 12:24 history suggested vitamin d CC BY-SA 4.0
Optimzing latex expressions.
Jul 23, 2021 at 11:45 review Suggested edits
S Jul 23, 2021 at 12:24
Aug 22, 2013 at 18:21 comment added The Masked Avenger With the loglog n factor, I'm not sure. I do know for every n and every z.d. E, there is m coprime to n such that mn is not in E. Then phi(n)/n > phi(nm)/nm and so lim inf phi(n)/n is the same with or without E. I was addressing your question about f and phi(n)/n.
Aug 22, 2013 at 16:59 comment added user21706 Sorry, but I do know understand you answer. How do you prove that if $E$ is a set of null asymptotic density then $\liminf_{E \not\ni n \to \infty} \varphi(n) / (n / \log\log n) = e^-\gamma$ ? Thanks.
Aug 22, 2013 at 16:28 history answered The Masked Avenger CC BY-SA 3.0