Is $n=2$ the only root of $M(n!)$...?

Question

Wolfram can help till $n=9$, but are there other value larger than $2$ for which $$ M(n!)=0, $$ where $M(n)$ is Merten's function.

Answer 1

If $M(n!)=0$ then $n=2$ or $n\ge18$, see A087989. Probably a few more terms could be computed using the Deléglise-Rivat algorithm, but the chance of finding other terms is remote.

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Let's treat $\mu(n)$ as a random function which is 0 if $n$ is divisible by a square and $\pm1$ with probability 1/2 each otherwise. Then $M(x)$ is essentially $\sim y=:\frac{6}{\pi^2}x$ Bernoulli trials and so the 'probability' that $M(n)=0$ is $2^{-y}{y\choose y/2}\sim2^{-y}\cdot\frac{2^y}{\sqrt{\pi y/2}}=1/\sqrt{\pi y/2}=1/\sqrt{3x/\pi}$. So the 'chance' that $M(n!)=0$ is sharply decreasing with $n$, and in fact $$ \int_{17.5}^\infty\frac{dx}{\sqrt{3\Gamma(x+1)/\pi}}<2\cdot10^{-8} $$ so the chance that another even exists is small. (Note the 17.5 comes from continuity correction; actually this choice is optimistic, with a more realistic value leading to a yet smaller 'probability'.)