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# Talk:Totient summatory function

## A. Walfisz (reference)

Thanks for correcting the German text of the reference on both the wiki and the main OEIS, as I had copied it verbatim from the main OEIS, assuming the correctness of it. I will adopt the practice of doing at least a quick Google search on content I want to copy from the main OEIS to the wiki. Is there an english translation of (I did a Google search without success)

A. Walfisz, Weylsche Exponentialsummen in der neueren Zahlentheorie, VEB Deutscher Verlag der Wissenschaften, Berlin 1963.

Daniel Forgues 23:05, 31 December 2011 (UTC)

## Asymptotic behavior of totient summatory function

$\Phi (n)={\frac {3n^{2}}{\pi ^{2}}}+O(n\log n)={\frac {n^{2}}{2\zeta (2)}}+O(n\log n).\,$ Is there a relation with the asymptotic behavior of the summatory quadratfrei function or is it a coincidence? — Daniel Forgues 23:43, 31 December 2011 (UTC) — Daniel Forgues 03:52, 4 May 2013 (UTC)

$\lim _{n\to \infty }{\frac {\Phi (n)}{n^{2}}}={\frac {3}{\pi ^{2}}}={\frac {1}{2}}\lim _{n\to \infty }{\frac {Q(n)}{n}}=\lim _{n\to \infty }{\frac {\sum _{i=1}^{n}[\mu (i)=-1]}{n}}=\lim _{n\to \infty }{\frac {\sum _{i=1}^{n}[\mu (i)=+1]}{n}}.\,$ The summatory quadratfrei function is defined as

$Q(n):=\sum _{i=1}^{n}q(i)=\sum _{i=1}^{n}|\mu (i)|,\,$ where $q(n)\,:=\,|\mu (n)|\,$ is the quadratfrei function (characteristic function of squarefree numbers) and $\mu (n)\,$ is the Moebius function.

The asymptotic density of squarefree numbers corresponds to the probability that two randomly chosen integers are coprime

$\lim _{n\to \infty }{\frac {Q(n)}{n}}=\prod _{n=1}^{\infty }{\bigg (}1-{\frac {1}{{p_{n}}^{2}}}{\bigg )}={\frac {1}{\zeta (2)}}={\frac {6}{\pi ^{2}}},\,$ where $p_{n}\,$ is the $n\,$ th prime number, and $\zeta (s)\,$ is the Riemann zeta function.

The asymptotic density of squarefree numbers with an odd number of prime factors is equal to the the asymptotic density of squarefree numbers with an even number of prime factors, i.e.

$\lim _{n\to \infty }{\frac {\sum _{i=1}^{n}[\mu (i)=-1]}{n}}=\lim _{n\to \infty }{\frac {\sum _{i=1}^{n}[\mu (i)=+1]}{n}}={\frac {1}{2\zeta (2)}}={\frac {3}{\pi ^{2}}},\,$ where $[\cdot ]\,$ is the Iverson bracket.