thumb|right|Mertens function to n=10,000 thumb|right|Mertens function to n=10,000,000 In number theory, the Mertens function is defined for all  positive integers n as
M(n) = \sum_{k=1}^n \mu(k)
where \mu(k) is the Möbius function.
The function is named in honour of Franz Mertens.
This definition can be extended to positive real numbers as follows:
M(x) = M(\lfloor x \rfloor).
Less formally, M(x) 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) is:
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.
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}, 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An unpublished conjecture of Steve Gonek states that
0 < \limsup_{x \to \infty} \frac{|M(x)|}{\sqrt{x} (\log \log \log x)^{5/4}} < \infty.
Probabilistic evidence towards this conjecture is given by Nathan Ng.Ng In particular, Ng gives a conditional proof that the function e^{-y/2} M(e^y) has a limiting distribution \nu on \mathbb{R}.
That is, for all bounded Lipschitz continuous functions f on the reals we have that
\lim_{Y\rightarrow\infty} \frac{1}{Y} \int_0^{Y} f\left(e^{-y/2} M(e^y)\right) dy = \int_{-\infty}^{\infty} f(x) d\nu(x).
Representations
As an integral
Using the Euler product one finds that
\frac{1}{\zeta(s) }= \prod_{p} (1-p^{-s})= \sum_{n=1}^{\infty} \frac{\mu(n)}{n^s}
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{1}{2\pi i}\int_{c-i\infty}^{c+i\infty} \frac{x^{s}}{s\zeta(s)} \, ds = M(x)
where c > 1.
Conversely, one has the Mellin transform
\frac{1}{\zeta(s)} = s\int_1^\infty \frac{M(x)}{x^{s+1}}\,dx
which holds for  \mathrm{Re}(s)>1.
A curious relation given by Mertens himself involving the second Chebyshev function is
\psi (x) = M\left( \frac{x}{2} \right) \log(2)+M \left( \frac{x}{3} \right) \log(3) + M \left( \frac{x}{4}\right )\log(4) + \cdots.
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 \frac{x^\rho}{\rho \zeta'(\rho)} - 2+\sum_{n=1}^\infty \frac{ (-1)^{n-1} (2\pi )^{2n}}{(2n)! n \zeta(2n+1)x^{2n}}.
Weyl conjectured that the Mertens function satisfied the approximate functional-differential equation
\frac{y(x)}{2}-\sum_{r=1}^N \frac{B_{2r}}{(2r)!}
D_t^{2r-1} y \left(\frac{x}{t+1}\right) + x\int_0^x \frac{y(u)}{u^{2}} \, du = x^{-1}H(\log x)
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
\sum_{n=1}^{\infty}\frac{\mu(n)}{\sqrt{n}} g(\log n) = \sum_{\gamma} \frac{h(\gamma)}{\zeta'(1/2+i\gamma)}+2\sum_{n=1}^\infty \frac{ (-1)^{n} (2\pi )^{2n}}{(2n)! \zeta(2n+1)}\int_{-\infty}^{\infty}g(x) e^{-x(2n+1/2)} \, dx,
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 \pi g(x)= \int_{-\infty}^{\infty}h(u)e^{iux} \, du.
As a sum over Farey sequences
Another formula for the Mertens function is
M(n)= -1+ \sum_{a\in \mathcal{F}_n} e^{2\pi i a}   where    \mathcal{F}_n   is the Farey sequence of order n.
This formula is used in the proof of the Franel–Landau theorem.Edwards, Ch.
12.2 As a determinant
M(n) is the determinant of the n × n Redheffer matrix, a (0,1) matrix in which aij is 1 if either j is 1 or i divides j.
As a sum of the number of points under n-dimensional hyperboloids{{citation needed|date=July 2018}}
M(x) = 1 - \sum_{2 \leq a \leq x} 1 + \underset{ab \leq x}{\sum_{a \geq 2} \sum_{b \geq 2}} 1 - \underset{abc \leq x}{\sum_{a \geq 2} \sum_{b \geq 2} \sum_{c \geq 2}} 1 + \underset{abcd \leq x}{\sum_{a \geq 2} \sum_{b \geq 2} \sum_{c \geq 2} \sum_{d \geq 2}} 1 - \cdots
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.
Calculation
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.
The Mertens function for all integer values up to x may be computed in O(x log log x) time.
Combinatorial based algorithms can compute isolated values of M(x) in O(x2/3(log log x)1/3) time, and faster non-combinatorial methods are also known.
See  for values of M(x) at powers of 10.
Known upper bounds
Ng notes that the Riemann hypothesis (RH) is equivalent to
M(x) = O\left(\sqrt{x} \exp\left(\frac{C \cdot \log x}{\log\log x}\right)\right),
for some positive constant C>0.
Other upper bounds have been obtained by Maier, Montgomery, and Soundarajan assuming the RH including
\begin{align}  |M(x)| & \ll \sqrt{x} \exp\left(C_2 \cdot (\log x)^{\frac{39}{61}}\right) \\  |M(x)| & \ll \sqrt{x} \exp\left(\sqrt{\log x} (\log\log x)^{14}\right).
\end{align}
Other explicit upper bounds are given by Kotnik as
\begin{align} |M(x)| & < \frac{x}{4345},\ \text{ for } x>2160535 \\  |M(x)| & < \frac{0.58782 \cdot x}{\log^{11/9}(x)},\ \text{ for } x>685.
\end{align}
See also
Perron's formula
Liouville's function
Notes
References
Deléglise, M. and Rivat, J. "Computing the Summation of the Möbius Function."
Experiment.
Math.
5, 291-295, 1996.
https://projecteuclid.org/euclid.em/1047565447
Nathan Ng, "The distribution of the summatory function of the Möbius function", Proc.
London Math.
Soc. (3)
89 (2004) 361-389.
http://www.cs.uleth.ca/~nathanng/RESEARCH/mobius2b.pdf
