In number theory, the Mertens function is defined for all positive integers n as
M(n)=
n | |
\sum | |
k=1 |
\mu(k),
where
\mu(k)
M(x)=M(\lfloorx\rfloor).
Less formally,
M(x)
The first 143 M(n) values are
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
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, 428, ... .
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,
x | |
\sum | |
n=1 |
\mu(n)\exp(i2\pin\theta)=O\left(
x | |
loghx |
\right)
uniformly in
\theta
\theta=0
M(x)=O\left(
x | |
loghx |
\right) .
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
\sqrt{x}
The true rate of growth of M(x) is not known. An unpublished conjecture of Steve Gonek states that
0<\limsupx
|M(x)| | |
\sqrt{x |
(logloglogx)5/4
Probabilistic evidence towards this conjecture is given by Nathan Ng.[2] In particular, Ng gives a conditional proof that the function
e-y/2M(ey)
\nu
R
f
\limY\toinfty
1 | |
Y |
Y | |
\int | |
0 |
f(e-y/2M(ey))dy=
infty | |
\int | |
-infty |
f(x)d\nu(x),
if one assumes various conjectures about the Riemann zeta function.
Using the Euler product, one finds that
1 | |
\zeta(s) |
=\prodp(1-p-s)=
infty | |
\sum | |
n=1 |
\mu(n) | |
ns |
,
where
\zeta(s)
1 | |
2\pii |
c+iinfty | |
\int | |
c-iinfty |
xs | |
s\zeta(s) |
ds=M(x),
where c > 1.
Conversely, one has the Mellin transform
1 | |
\zeta(s) |
=s
infty | |
\int | |
1 |
M(x) | |
xs+1 |
dx,
which holds for
\operatorname{Re}(s)>1
A curious relation given by Mertens himself involving the second Chebyshev function is
\psi(x)=M\left(
x | |
2 |
\right)log2+M\left(
x | |
3 |
\right)log3+M\left(
x | |
4 |
\right)log4+ … .
Assuming that the Riemann zeta function has no multiple non-trivial zeros, one has the "exact formula" by the residue theorem:
M(x)=\sum\rho
x\rho | |
\rho\zeta'(\rho) |
-2+
infty | |
\sum | |
n=1 |
(-1)n-1(2\pi)2n | |
(2n)!n\zeta(2n+1)x2n |
.
Weyl conjectured that the Mertens function satisfied the approximate functional-differential equation
y(x) | |
2 |
-
N | |
\sum | |
r=1 |
B2r | |
(2r)! |
2r-1 | |
D | |
t |
y\left(
x | |
t+1 |
\right)+x
x | |
\int | |
0 |
y(u) | |
u2 |
du=x-1H(logx),
where H(x) is the Heaviside step function, B are Bernoulli numbers, and all derivatives with respect to t are evaluated at t = 0.
There is also a trace formula involving a sum over the Möbius function and zeros of the Riemann zeta function in the form
infty | |
\sum | |
n=1 |
\mu(n) | |
\sqrt{n |
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
2\pig(x)=
infty | |
\int | |
-infty |
h(u)eiuxdu.
Another formula for the Mertens function is
M(n)=-1+\suma\inl{Fn}e2\pi,
where
l{F}n
This formula is used in the proof of the Franel–Landau theorem.[3]
M(n) is the determinant of the n × n Redheffer matrix, a (0, 1) matrix in whichaij is 1 if either j is 1 or i divides j.
M(x)=1-\sum21+\underset{ab\leqx}{\suma\sumb
This formulation expanding the Mertens function suggests asymptotic bounds obtained by considering the Piltz divisor problem, which generalizes the Dirichlet divisor problem of computing asymptotic estimates for the summatory function of the divisor function.
From [4] we have
n | |
\sum | |
d=1 |
M(\lfloorn/d\rfloor)=1 .
Furthermore, from [5]
n | |
\sum | |
d=1 |
M(\lfloorn/d\rfloor)d=\Phi(n) ,
\Phi(n)
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 | |
von Sterneck | 1901 | 5 | |
von Sterneck | 1912 | 5 | |
Neubauer | 1963 | 108 | |
Cohen and Dress | 1979 | 7.8 | |
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 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 time; a further improvement by Harald Helfgott and Lola Thompson in 2021 improves this to,[12] and an algorithm by Lagarias and Odlyzko based on integrals of the Riemann zeta function achieves a running time of .[13]
See for values of M(x) at powers of 10.
Ng notes that the Riemann hypothesis (RH) is equivalent to
M(x)=O\left(\sqrt{x}\exp\left(
C ⋅ logx | |
loglogx |
\right)\right),
for some positive constant
C>0
\begin{align}|M(x)|&\ll\sqrt{x}\exp\left(C2 ⋅ (log
| ||||
x) |
\right)\ |M(x)|&\ll\sqrt{x}\exp\left(\sqrt{logx}(loglogx)14\right).\end{align}
Known explicit upper bounds without assuming the RH are given by:[14]
\begin{align} |M(x)|&<
12590292 ⋅ x | |
log236/75(x) |
, forx>\exp(12282.3)\ |M(x)|&<
0.6437752 ⋅ x | |
logx |
, forx>1.\end{align}
It is possible to simplify the above expression into a less restrictive but illustrative form as:
\begin{align} M(x)=O\left(
x | |
log\pi(x) |
\right).\end{align}
\pi(x)
\mu(n)
\pi(x)