Mertens function

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In number theory, the Mertens function is

[M(n) = \sum_ \mu(k)]

where μ(k) is the Möbius function. The function is named in honour of Franz Mertens.

Because the Möbius function has only the return values -1, 0 and +1, it's obvious that the Mertens function moves slowly and that there is no x such that M(x) > x. The Mertens conjecture goes even further, stating that there is no x where the absolute value of the Mertens function exceeds the square root of x. The Mertens conjecture was disproven in 1985. However, the Riemann hypothesis is equivalent to a weaker conjecture on the growth of M(x), namely [M(x) = o(x^)]. Since high values for M grow at least as fast as the square root of x, this puts a rather tight bound on its rate of growth. Here, [o] refers to little-o notation.

Integral representations

Using the Euler product one finds that

[ \frac= \prod_ (1-p^)= \sum_^\mu (n)n^ ]

where [\zeta(s)] is the Riemann zeta function and the product is taken over primes. Then, using this Dirichlet series with Perron's formula, one obtains:

[ \frac\oint_ds \frac}=M(x) ]

where "C" is a closed curve encircling all of the roots of [\zeta(s).]

Conversely, one has the Mellin transform

[\frac = s\int_1^\infty \frac}\,dx]

which holds for [Re(s)>1].

A good evaluation, at least asymptotically, would be to obtain, by steepest descent method, an inequality:

[ \oint_dsF(s)e^ \sim M(e^) ]

References

F. Mertens, "Über eine zahlentheoretische Funktion", Akademie Wissenschaftlicher Wien Mathematik-Naturlich Kleine Sitzungsber, IIa 106, (1897) 761-830.
A. M. Odlyzko and H.J.J. te Riele, "[Disproof of the Mertens Conjecture]", Journal für die reine und angewandte Mathematik 357, (1985) pp. 138-160.
, [Mertens function] at MathWorld.
- Values of the Mertens function for the first 50 n are given by [SIDN A002321]
- Values of the Mertens function for the first 2500 n are given by [PrimeFan's Mertens Values Page]
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