Mertens function

Mertens function to n=10,000
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 μ(k) is the Möbius function. The function is named in honour of Franz Mertens.

Less formally, M(n) is the count of square-free integers up to n that have an even number of prime factors, minus the count of those that have an odd number.

The first 160 M(n) is: (sequence A002321 in OEIS)

 n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 M(n) 1 0 -1 -1 -2 -1 -2 -2 -2 -1 -2 -2 -3 -2 -1 -1 -2 -2 -3 -3 n 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 M(n) -2 -1 -2 -2 -2 -1 -1 -1 -2 -3 -4 -4 -3 -2 -1 -1 -2 -1 0 0 n 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 M(n) -1 -2 -3 -3 -3 -2 -3 -3 -3 -3 -2 -2 -3 -3 -2 -2 -1 0 -1 -1 n 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 M(n) -2 -1 -1 -1 0 -1 -2 -2 -1 -2 -3 -3 -4 -3 -3 -3 -2 -3 -4 -4 n 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 M(n) -4 -3 -4 -4 -3 -2 -1 -1 -2 -2 -1 -1 0 1 2 2 1 1 1 1 n 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 M(n) 0 -1 -2 -2 -3 -2 -3 -3 -4 -5 -4 -4 -5 -6 -5 -5 -5 -4 -3 -3 n 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 M(n) -3 -2 -1 -1 -1 -1 -2 -2 -1 -2 -3 -3 -2 -1 -1 -1 -2 -3 -4 -4 n 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 M(n) -3 -2 -1 -1 0 1 1 1 0 0 -1 -1 -1 -2 -1 -1 -2 -1 0 0

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, ... (sequence A028442 in OEIS).

Because the Möbius function only takes the values −1, 0, and +1, the Mertens function moves slowly and there is no n such that |M(n)| > n. The Mertens conjecture went further, stating that there would be no n where the absolute value of the Mertens function exceeds the square root of n. 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(n), namely M(n) = O(n1/2 + ε). Since high values for M(n) grow at least as fast as the square root of n, this puts a rather tight bound on its rate of growth. Here, O refers to Big O notation.

The above definition can be extended to real numbers as follows:

$M(x) = \sum_{1\le k \le x} \mu(k).$

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

$\psi (x) = M\left( \frac{x}{2} \right) \log(2)+M \left( \frac{x}{3} \right) \log(3) + M \left( \frac{x}{4}\right )\log(4) + \cdots.$

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

$\oint_C F(s)e^{st} \, ds \sim M(e^{t}).$

Assuming that there are not multiple non-trivial roots of $\zeta (\rho)$ we have the "exact formula" by the residue theorem:

$\frac{1}{2 \pi i} \oint_C \frac{x^s}{s \zeta (s)} \, ds = \sum_\rho \frac{x^\rho}{\rho \zeta'(\rho)} - 2+\sum_{n=1}^\infty \frac{ (-1)^{n-1} (2\pi )^{2n}}{(2n)! n \zeta(2n+1)x^{2n}}.$

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.

Titchmarsh(1960), and later J. Garcia provided a Trace formula involving a sum over the Möbius function and zeros of Riemann Zeta in the form

$\sum_{n=1}^{\infty}\frac{\mu(n)}{\sqrt{n}} g \log n = \sum_t \frac{h(t)}{\zeta'(1/2+it)}+2\sum_{n=1}^\infty \frac{ (-1)^{n} (2\pi )^{2n}}{(2n)! \zeta(2n+1)}\int_{-\infty}^{\infty}g(x) e^{-x(2n+1/2)} \, dx,$

where 't' sums over the imaginary parts of nontrivial zeros, and (g, h) are related by a 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

$M(n)= \sum_{a\in \mathcal{F}_n} e^{2\pi i a}$   where   $\mathcal{F}_n$   is the Farey sequence of order n.

This formula is used in the proof of the Franel–Landau theorem.[1]

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.

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 an increasing range of n.

 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

The Mertens function for all integer values up to N may be computed in O(N2/3+ε) time, while better methods are known. Elementary algorithms exist to compute isolated values of M(N) in O(N2/3*(ln ln(N))1/3) time.

See for values of M(N) at powers of 10.

Notes

1. ^ Edwards, Ch. 12.2