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Let $\varphi(n)$ be the Euler's totient function. It is well know that $\liminf_{n \to \infty} \frac{\varphi(n)}{n / \log \log n} = e^{-\gamma}$, so that for $\varepsilon > 0$ it results $\frac{\varphi(n)}{n} \geq \frac{e^{-\gamma}-\varepsilon}{\log \log n}$ for large $n$. Actually, the "local minima" of $\frac{\varphi(n)}{n}$ are attached for $n = p_1 \cdots p_k$ (the product of the first $k$ primes) and the set of primorial is really sparse. I wonder if it is known a lower bound for $\varphi(n)$ like: "$\varphi(n) / n \geq f(n)$ for all $n$ but a set of null asymptotic density", where $f(n)$ is a function bigger then $\frac{e^{-\gamma}-\varepsilon}{\log \log n}$.

Thank you in advance for your help.

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  • $\begingroup$ For $n/\phi(n)$ "Small values of the Euler function and the Riemann hypothesis Jean-Louis Nicolas" might be related to your question. $\endgroup$
    – joro
    Commented Aug 22, 2013 at 14:21
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    $\begingroup$ 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. $\endgroup$
    – Lucia
    Commented Aug 22, 2013 at 17:08
  • $\begingroup$ @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. $\endgroup$
    – user21706
    Commented Aug 22, 2013 at 19:21
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    $\begingroup$ 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" $\endgroup$
    – user21706
    Commented Aug 22, 2013 at 19:59

2 Answers 2

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Your question has been answered by Lucia already, but you might also be interested in looking up the Erdős-Wintner theorem. A special case (proved already by Schoenberg) is that for each $u \geq 0$, the set of $n$ with $\phi(n)/n \leq u$ has an asymptotic density $D(u)$; moreover, $D(u)$ is continuous and increasing on $[0,1]$.

There are also estimates available for the size of $D(u)$ when $u$ is near zero, and of $1-D(u)$ when $u$ is near $1$. For this, see Erdős's paper "Some remarks about additive and multiplicative functions": http://www.renyi.hu/~p_erdos/1946-11.pdf

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Given $n$, the set of integers $m$ coprime to $n$ has nonzero asymptotic density, thus so does the set $mn$. But then $\phi(n)/n > \phi(mn)/mn$ , so the $\liminf_n\phi(n)/n$ will remain the same off of a set of zero asymptotic density. So there is no such $f(\cdot)$ for asymptotic density. (I assume you mean + epsilon in your formulation.)

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  • $\begingroup$ 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. $\endgroup$
    – user21706
    Commented Aug 22, 2013 at 16:59
  • $\begingroup$ 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. $\endgroup$
    – The Masked Avenger
    Commented Aug 22, 2013 at 18:21

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