Less formally, is the count of square-free integers up to x that have an even number of prime factors, minus the count of those that have an odd number.
The first 143 M(n) values are (sequence A002321 in the OEIS)
M(n)
+0
+1
+2
+3
+4
+5
+6
+7
+8
+9
+10
+11
0+
1
0
−1
−1
−2
−1
−2
−2
−2
−1
−2
12+
−2
−3
−2
−1
−1
−2
−2
−3
−3
−2
−1
−2
24+
−2
−2
−1
−1
−1
−2
−3
−4
−4
−3
−2
−1
36+
−1
−2
−1
0
0
−1
−2
−3
−3
−3
−2
−3
48+
−3
−3
−3
−2
−2
−3
−3
−2
−2
−1
0
−1
60+
−1
−2
−1
−1
−1
0
−1
−2
−2
−1
−2
−3
72+
−3
−4
−3
−3
−3
−2
−3
−4
−4
−4
−3
−4
84+
−4
−3
−2
−1
−1
−2
−2
−1
−1
0
1
2
96+
2
1
1
1
1
0
−1
−2
−2
−3
−2
−3
108+
−3
−4
−5
−4
−4
−5
−6
−5
−5
−5
−4
−3
120+
−3
−3
−2
−1
−1
−1
−1
−2
−2
−1
−2
−3
132+
−3
−2
−1
−1
−1
−2
−3
−4
−4
−3
−2
−1
The Mertens function slowly grows in positive and negative directions both on average and in peak value, oscillating in an apparently chaotic manner passing through zero when n has the values
Because the Möbius function only takes the values −1, 0, and +1, the Mertens function moves slowly, and there is no x such that |M(x)| > x.
H. Davenport[1] demonstrated that, for any fixed h,
uniformly in . This implies, for that
The Mertens conjecture went further, stating that there would be no x where the absolute value of the Mertens function exceeds the square root of x. The Mertens conjecture was proven false in 1985 by Andrew Odlyzko and Herman te Riele. However, the Riemann hypothesis is equivalent to a weaker conjecture on the growth of M(x), namely M(x) = O(x1/2 + ε). Since high values for M(x) grow at least as fast as , this puts a rather tight bound on its rate of growth. Here, O refers to big O notation.
The true rate of growth of M(x) is not known. An unpublished conjecture of Steve Gonek states that
Probabilistic evidence towards this conjecture is given by Nathan Ng.[2] In particular, Ng gives a conditional proof that the function has a limiting distribution on . That is, for all bounded Lipschitz continuous functions on the reals we have that
There is also a trace formula involving a sum over the Möbius function and zeros of the Riemann zeta function in the form
where the first sum on the right-hand side is taken over the non-trivial zeros of the Riemann zeta function, and (g, h) are related by the Fourier transform, such that
Neither of the methods mentioned previously leads to practical algorithms to calculate the Mertens function.
Using sieve methods similar to those used in prime counting, the Mertens function has been computed for all integers up to an increasing range of x.[6][7]
Person
Year
Limit
Mertens
1897
104
von Sterneck
1897
1.5×105
von Sterneck
1901
5×105
von Sterneck
1912
5×106
Neubauer
1963
108
Cohen and Dress
1979
7.8×109
Dress
1993
1012
Lioen and van de Lune
1994
1013
Kotnik and van de Lune
2003
1014
Hurst
2016
1016
The Mertens function for all integer values up to x may be computed in O(x log log x) time. A combinatorial algorithm has been developed incrementally starting in 1870 by Ernst Meissel,[8]Lehmer,[9]Lagarias-Miller-Odlyzko,[10] and Deléglise-Rivat[11] that computes isolated values of M(x) in O(x2/3(log log x)1/3) time; a further improvement by Harald Helfgott and Lola Thompson in 2021 improves this to O(x3/5(log x)3/5+ε),[12] and an algorithm by Lagarias and Odlyzko based on integrals of the Riemann zeta function achieves a running time of O(x1/2+ε).[13]
See OEIS: A084237 for values of M(x) at powers of 10.
^Davenport, H. (November 1937). "On Some Infinite Series Involving Arithmetical Functions (Ii)". The Quarterly Journal of Mathematics. Original Series. 8 (1): 313–320. doi:10.1093/qmath/os-8.1.313.
^Nathan Ng (October 25, 2018). "The distribution of the summatory function of the Mobius function". arXiv:math/0310381.
Mertens, F. (1897). ""Über eine zahlentheoretische Funktion", Akademie Wissenschaftlicher Wien Mathematik-Naturlich". Kleine Sitzungsber, IIA. 106: 761–830.