Approximate how the Numbers $n$ such that Mertens' function is zero grow. Is it possible to approximate how the "Numbers $n$ such that Mertens' function is zero" grow?

 A: This paper looked at zeros $n$ small enough to (at present day) compute $M(n)$ at. All zeros less than $10^{16}$ were computed.
In this range, the number of zeros less than $x$ seems to grow as $\Theta(x^{1/2+\varepsilon})$.
Here's the relevant excerpt from section 6 in the paper:

All zeros of $M(n)$ for $n \leq 10^{16}$ were recorded.
     A natural question to ask is for any $x$, how many zeros are less than $x$? Defining $V(x)$ to be the number of zeros less than $x$, a theorem of Landau states $V(x) = \Omega(\log x)$. This however is expected to be a weak lower bound.
Treating $M(n)$ as a random walk with probability of staying stationary $1-6/\pi^2$ and with both probabilities of moving up and down $3/\pi^2$, it would follow that $V(x) = \sqrt{\pi x/3} + o(\sqrt{x}\,)$.
      In practice however $M$ cannot be modeled as a random walk because there is regularity, e.g. $M(4n+3) = M(4n+4)$, etc.
      Nonetheless, the data suggest $V(x) = \Theta(x^{1/2 + \varepsilon})$.
      In fact $3.5 \sqrt{x}$ or even $\sqrt{x} \log \log x$ seem like good approximations.


