# Numbers $n$ such that Mertens' function is zero.

OEIS (A028442) lists the Numbers n such that Mertens' function
$$M(n)=\sum_{k=1}^n\mu(k)$$ is zero:

2, 39, 40, 58, 65, 93, 101, 145, 149, 150, 159, 160, 163, 164, 166, 214, 231, 232, 235, 236, 238, 254, 329, 331, 332, 333, 353, 355, 356, 358, 362, 363, 364, 366, 393, 401, 403, 404, 405, 407, 408, 413, 414, 419, 420, 422, 423, 424, 425, 427,...

Do these numbers have a deeper significance other than: The set of numbers below $n$ is split into $2$ equally large sets with $\mu(m\le n)=\pm 1$ (with asymptotic density each $\frac{3}{\pi^2}$) and the set $\mu(m\le n)=0$ (with asymptotic density $1-\frac{6}{\pi^2}$)?

I mean, does the fact that $\lim_{n\to\infty}M(n)=0$ play a role (whatever that is in the infinte case) in a finite case as well? I'm especially interested in the case where $n$ is even.

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–  joriki Oct 13 '12 at 20:25
It's not a fact that $M(n) \to 0$, obviously (since $\mu(n) = 1$ infinitely often). –  Erick Wong Jan 8 '13 at 8:36
@ErickWong so it's not sufficient that the asymptotic density equal each other? And why should there be more numbers $n$ with an even number of factors $(\mu(n_e)=1)$ compared to an odd number of factors$(\mu(n_o)=-1)$? –  draks ... Jan 8 '13 at 11:49
@draks... You're confusing $M(n) \to 0$ with $M(n) = o(n)$. The latter is true (and somewhat deep, being equivalent to PNT), the former is certainly false (by the most trivial test of divergence). –  Erick Wong Jan 8 '13 at 17:37
@ErickWong I found this one: *... while the last one is obvious if we allow generalized summation $\sum_{n=1}^{\infty}\mu(n) = \frac{1}{\zeta(0)} = -2$ * (see here). What do you think? –  draks ... Jan 10 '13 at 21:36