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Let the Mertens function $$M(x) = \sum_{n \le x} \mu(n)$$ I assume (perhaps foolishly) that it is known that $M(x)$ changes sign infinitely often. If that's true, the question is a quantitative version :

How many sign changes of $M(x)$ are there between $1$ and $y$ (asymptotically) ?

**ADDITION* GH from MO cites a result which gives a logarithmic number of changes. This, while better than nothing, is not (empirically the truth): for $N=1000000,$ you get around $5500$ sign changes, for $N=10000000,$ around $12000,$ and here is the graph of the total number of sign changes.This looks square-rootish. Now, what is even more interesting is that for a symmetric random walk, the number of returns to the origin is asymptotic to $\frac{2}{\pi} \sqrt{N},$ which is much smaller than this. (see https://math.stackexchange.com/questions/1338097/expected-number-of-times-random-walk-crosses-0-line for deviation). The difference is even more striking, if you remember that a lot of numbers are non-square-free.

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    $\begingroup$ It does change of sign infinitely often, otherwise the RH would be trivial since $\frac{1}{\zeta(s)}$ would have a singularity at its abscissa of convergence. What you want is a smooth real function $f(x)$ such that $\int_1^\infty M(x) \sin( f(x)) x^{-s-1}dx$ has a singularity at its abscissa of convergence. Look at $x-\psi(x)= \sum_\rho \frac{x^\rho}{\rho}+\mathcal{O}(1)$ then $M(x)=\sum_\rho \frac{x^\rho}{\rho\zeta'(\rho)} + \mathcal{O}(1)$ $\endgroup$
    – reuns
    Commented Jul 6, 2017 at 17:14
  • $\begingroup$ Intuitively, the more sign changes, the less is $ M(x) $ cause the elementary increasing is at most 1 ( $ \vert M(n+1)-M(n)\vert\leq 1 $ ). As RH is equivalent to $M(x)\ll_{\varepsilon}x^{1/2+\varepsilon} $ and $ M(x)<x^{1/2} $ is known to be false, one can expect the number of sign changes below $ x $ to be at most $ x^{1/2+\varepsilon} $ too. $\endgroup$
    – Sylvain JULIEN
    Commented Jul 7, 2017 at 9:57
  • $\begingroup$ Let $ S(x) $ be the number of sign changes of the Mertens function between 1 and $ x $, $ I(x) $ the average length of the intervals below x on which the sign of the Mertens function is constant and $m(x) : =\sup_{n\leq x}\vert M(n)\vert $. Maybe one can try to prove that 1) $ S(x)m(x)\ll x $ and 2) $ I(x)m(x)\asymp x^{1+o(1)} $ . $\endgroup$
    – Sylvain JULIEN
    Commented Jul 7, 2017 at 14:53

2 Answers 2

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Let $\gamma_1=14.1347251\dots$ be the imaginary part of the first $\zeta$-zero. It was proved by Kaczorowski and Pintz (Acta Math. Hungar. 48 (1986), 173-185, doi: 10.1007/BF01949062) that $M(x)$ has at least $(\gamma_1/\pi-o(1))\log y$ sign changes in $[1,y]$. See Corollary 4 in their paper (for $a=0$). The paper contains several other interesting and relevant results, such as an effective version of the quoted bound (see Corollary 5), or information on sign changes and oscillation in "shortish" intervals (see (2.2) and (2.3)).

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  • $\begingroup$ See the addition to the post - comments welcome. $\endgroup$
    – Igor Rivin
    Commented Jul 7, 2017 at 1:16
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You may also find quantitative results concerning error term from prime number theorem in the article S. B. Stechkin, A. Yu. Popov, “The asymptotic distribution of prime numbers on the average”, Uspekhi Mat. Nauk, 51:6(312) (1996), 21–88, english translation is published in Russ. Math. Surv. 51 (1996), pp. 1025-1092.

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