MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
For $n/\phi(n)$ "Small values of the Euler function and the Riemann hypothesis Jean-Louis Nicolas" might be related to your question. – joro Aug 22 '13 at 14:21
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. – Lucia Aug 22 '13 at 17:08
@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. – user21706 Aug 22 '13 at 19:21
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" – user21706 Aug 22 '13 at 19:59

Your question has been answered by Lucia already, but you might also be interested in looking up the Erd\H{o}s--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 Erd\H{o}s's paper "Some remarks about additive and multiplicative functions":

share|cite|improve this answer

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 lim inf of phi(n)/n will remain the same off of a set of zero asymptotic density. So there is no such f() for asymptotic density. (I assume you mean + epsilon in your formulation.)

share|cite|improve this answer
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. – user21706 Aug 22 '13 at 16:59
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. – The Masked Avenger Aug 22 '13 at 18:21

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.