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Aug 22, 2013 at 22:10 answer added so-called friend Don timeline score: 3
Aug 22, 2013 at 19:59 comment added user21706 I got. The average value of $n / \varphi(n)$ is $315\zeta(3)/(2\pi^4)$. "R. Sitaramachandrarao. On an error term of Landau II, Rocky Mountain J. Math. 15 (1985), 579-588"
Aug 22, 2013 at 19:21 comment added user21706 @Lucia Thank you for your answer! However I can't find a reference for the average value of $n / \varphi(n)$, I know that average value of $\varphi(n) / n$ is $6 / \pi^2$, but $n / \varphi(n)$ I don't know.
Aug 22, 2013 at 17:08 comment added Lucia Since the average value of $n/\phi(n)$ is bounded, it follows that for any function $f(n)$ tending to zero as $n$ tends to infinity one has $\phi(n)/n \ge f(n)$ except on a set of zero density.
Aug 22, 2013 at 16:28 answer added The Masked Avenger timeline score: 0
Aug 22, 2013 at 14:21 comment added joro For $n/\phi(n)$ "Small values of the Euler function and the Riemann hypothesis Jean-Louis Nicolas" might be related to your question.
Aug 22, 2013 at 12:43 history asked user21706 CC BY-SA 3.0