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# Mertens function

(Redirected from Mertens's function)

The partial sums of the Möbius function give a summatory function, the summatory Möbius function, which is called the Mertens function, named after Franz Mertens. Thus

${\displaystyle M(n):=\sum _{k=1}^{n}\mu (k).\,}$

The following table shows the relationship between the Möbius function and the Mertens function for the integers from 1 to 25.

 ${\displaystyle n}$ 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 ${\displaystyle \mu (n)}$ 1 −1 −1 0 −1 1 −1 0 0 1 −1 0 −1 1 1 0 −1 0 −1 0 1 1 −1 0 0 ${\displaystyle M(n)}$ 1 0 −1 −1 −2 −1 −2 −2 −2 −1 −2 −2 −3 −2 −1 −1 −2 −2 −3 −3 −2 −1 −2 −2 −2

## Asymptotic behavior

### Asymptotic densities

${\displaystyle \lim _{n\to \infty }{\frac {Q(n)}{n}}=\prod _{n=1}^{\infty }{\bigg (}1-{\frac {1}{{p_{n}}^{2}}}{\bigg )}={\frac {1}{\zeta (2)}}={\frac {6}{\pi ^{2}}},\,}$

where ${\displaystyle \scriptstyle Q(n)\,}$ is the summatory quadratfrei function, ${\displaystyle \scriptstyle p_{n}\,}$ is the ${\displaystyle \scriptstyle n\,}$th prime number and ${\displaystyle \scriptstyle \zeta (s)\,}$ is the Riemann zeta function.

Among those, the asymptotic density of numbers with an odd number of prime factors is equal to the asymptotic density of numbers with an even number of prime factors.

Asymptotic densities
${\displaystyle \mu =-1\,}$ ${\displaystyle \mu =0\,}$ ${\displaystyle \mu =+1\,}$
${\displaystyle {\frac {3}{\pi ^{2}}}\,}$ ${\displaystyle 1-{\frac {6}{\pi ^{2}}}\,}$ ${\displaystyle {\frac {3}{\pi ^{2}}}\,}$

### Asymptotic bounds

If we look at the Mertens function ${\displaystyle \scriptstyle M(n)}$ as the result of tossing a fair coin ${\displaystyle \scriptstyle n}$ times and considering only ${\displaystyle \scriptstyle N\,=\,{\frac {1}{\zeta (2)}}\cdot n\,=\,{\frac {6}{\pi ^{2}}}\cdot n}$ independent trials on average (corresponding to the squarefree numbers), we may assume a discrete binomial distribution[1] (asymptotically becoming a normal distribution[2]) with mean 0 and standard deviation of

${\displaystyle \sigma ={\sqrt {Npq}}={\sqrt {{\bigg (}{\frac {6}{\pi ^{2}}}\cdot n{\bigg )}\cdot {\frac {1}{2}}\cdot {\frac {1}{2}}}}={\frac {\sqrt {6n}}{2\pi }}\approx (0.3898484\dots )\cdot {\sqrt {n}},}$

where ${\displaystyle \scriptstyle p\,=\,{\frac {1}{2}}}$ is the probability of getting a head (${\displaystyle \scriptstyle \mu (n)\,=\,+1}$) and ${\displaystyle \scriptstyle q\,=\,{\frac {1}{2}}}$ is the probability of getting a tail (${\displaystyle \scriptstyle \mu (n)\,=\,-1}$).

Now, consider the following table of ranges (in terms of the standard deviation[3]) corresponding to given confidence intervals for the normal distribution.

Normal distribution
Confidence

interval[4]

Range
0.800 ${\displaystyle \scriptstyle \pm 1.28155~\sigma }$
0.900 ${\displaystyle \scriptstyle \pm 1.64485~\sigma }$
0.950 ${\displaystyle \scriptstyle \pm 1.95996~\sigma }$
0.990 ${\displaystyle \scriptstyle \pm 2.57583~\sigma }$
0.995 ${\displaystyle \scriptstyle \pm 2.80703~\sigma }$
0.999 ${\displaystyle \scriptstyle \pm 3.29053~\sigma }$

This does not bode well at all for the von Sterneck conjecture, with range ${\displaystyle \scriptstyle \approx \,{\frac {\pm 0.5}{0.3898484}}\,\approx \,\pm 1.282\,\sigma }$ having a confidence interval of about 80.0%, and not so well either for the Mertens conjecture, with range ${\displaystyle \scriptstyle \approx \,{\frac {\pm 1}{0.3898484}}\,\approx \,\pm 2.565\,\sigma }$ having a confidence interval of about 99.0%. Thus, according to this normal distribution model, the von Sterneck conjecture would asymptotically fail for about 20.0% of the integers, and the Mertens conjecture would asymptotically fail for about 1.0% of the integers. Of course, since the prime factorization of integers is not random (the primes being intricately interdependent,) the Moebius function is not random (the trials are not independent,) so the above confidence intervals may or may not be reflecting the actual behavior of the Mertens function.

#### Riemann hypothesis

The Riemann hypothesis is equivalent to a weaker conjecture than Mertens conjecture on the growth of ${\displaystyle \scriptstyle M(n)}$, namely

${\displaystyle |M(n)|\ll n^{{\frac {1}{2}}+\epsilon }}$

for any ${\displaystyle \scriptstyle \epsilon \,>\,0}$.

More precisely, the Riemann hypothesis is equivalent to

${\displaystyle |M(n)|\ll n^{\frac {1}{2}}~\exp {{\bigg (}c{\frac {\log n}{\log \log n}}{\bigg )}}}$

for some constant ${\displaystyle \scriptstyle c}$.

#### Mertens conjecture

In 1897, Mertens made the following bold conjecture:

Conjecture (Mertens conjecture, 1897). (Franz Mertens)

${\displaystyle \forall n,\,{\Bigg |}{\frac {M(n)}{\sqrt {n}}}{\Bigg |}<1.}$

In 1979, H. Cohen and F. Dress computed the values of ${\displaystyle \scriptstyle M(n)}$ for ${\displaystyle \scriptstyle n\,\leq \,7.8\cdot 10^{9}}$ and found the Mertens conjecture held up to that point.

In 1983 by Hermann te Riele and Andrew Odlyzko disproved the Mertens conjecture. In 1985, Odlyzko found a counterexample near ${\displaystyle \scriptstyle 10^{10^{64.1442}}}$, where

${\displaystyle {\Bigg |}{\frac {M(n)}{\sqrt {n}}}{\Bigg |}>1.06.}$

The smallest ${\displaystyle \scriptstyle n}$ contradicting the Mertens conjecture is estimated to be ${\displaystyle \scriptstyle n\,>\,10^{30}}$. In 1987, J. Pintz showed that a smaller counterexample could be found for ${\displaystyle \scriptstyle n\,<\,10^{65}}$. The smallest ${\displaystyle \scriptstyle n}$ contradicting the Mertens conjecture is still not known. In 1985, Odlyzko and Riele did not expect to find any counterexamples for ${\displaystyle \scriptstyle n\,<\,10^{20}}$.

#### von Sterneck conjecture

In 1897, R. D. von Sterneck made the bolder conjecture

${\displaystyle {\Bigg |}{\frac {M(x)}{\sqrt {x}}}{\Bigg |}<{\frac {1}{2}}.}$

In 1960, Wolgang Jurkat found a counterexample. He discovered that for ${\displaystyle \scriptstyle x\,>\,200}$, the first time the von Sterneck conjecture fails is for

${\displaystyle M(7725030629)=43947>43946.0766992\ldots ={\frac {\sqrt {7725030629}}{2}}.}$

### Asymptotic bias

The first positive value of Mertens function for ${\displaystyle \scriptstyle n\,>\,1\,}$ is for ${\displaystyle \scriptstyle n\,=\,94\,}$. The graph seems to show a negative bias for the Mertens function which is eerily similar to the Chebyshev bias (described in A156749 and A156709). The purported bias seems to be empirically approximated (by looking at the graph).

${\displaystyle -{\frac {1}{\zeta (2)}}\cdot {\frac {\sqrt {n}}{4}}=-{\frac {6}{\pi ^{2}}}\cdot {\frac {\sqrt {n}}{4}}\,}$

where ${\displaystyle \scriptstyle {\frac {1}{\zeta (2)}}\,=\,{\frac {6}{\pi ^{2}}}\,}$ is the asymptotic density of squarefree numbers (the squareful numbers having Möbius mu of 0). This would be a growth pattern akin to the Chebyshev bias.

Here are calculated values of the purported bias with the empirical approximation ${\displaystyle \scriptstyle -{\frac {6}{\pi ^{2}}}\cdot {\frac {\sqrt {n}}{4}}\,}$:

    25: -0.759       50: -1.074    100: -1.519     200: -2.149

1000: -4.806     2000: -6.796   4000: -9.612    8000: -13.593


Could the Mertens conjecture applied to a bias corrected Mertens function (assuming the bias is real, and that we found the actual asymptotic behavior of this purported bias) become true? Obviously, if Mertens conjecture was falsified with too large positive values of the Mertens function, the bias correction would only make things worse, as is the case for the stronger von Sterneck conjecture.

## Sequences

A002321 Mertens's function: ${\displaystyle \scriptstyle \sum _{k=1}^{n}\mu (k),\,}$ where ${\displaystyle \scriptstyle \mu (n)\,}$ is the Moebius function (A008683).

{1, 0, –1, –1, –2, –1, –2, –2, –2, –1, –2, –2, –3, –2, –1, –1, –2, –2, –3, –3, –2, –1, –2, –2, –2, –1, –1, –1, –2, –3, –4, –4, –3, –2, –1, –1, –2, –1, 0, 0, –1, –2, –3, –3, –3, –2, –3, –3, –3, –3, –2, –2, –3, –3, –2, –2, ...}

A084237 ${\displaystyle \scriptstyle a(n)\,=\,M(10^{n})\,}$, where ${\displaystyle \scriptstyle M(n)\,}$ is Mertens's function.

{1, –1, 1, 2, –23, –48, 212, 1037, 1928, –222, –33722, –87856, 62366, 599582, –875575, –3216373, –3195437, ...}

A171096 Numbers ${\displaystyle \scriptstyle n\,}$ for which the Mertens function ${\displaystyle \scriptstyle M(n)\,}$ (A002321) is −1.

{3, 4, 6, 10, 15, 16, 22, 26, 27, 28, 35, 36, 38, 41, 57, 59, 60, 62, 63, 64, 66, 69, 87, 88, 91, 92, 102, 123, 124, 125, 126, 129, 134, 135, 136, 143, 144, 151, 152, 153, 155, 156, 158, 165, 167, 168, 169, 210, ...}

A028442 Numbers ${\displaystyle \scriptstyle n\,}$ such that Mertens's function ${\displaystyle \scriptstyle M(n)\,}$ (A002321) is zero.

{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, ...}

The subset of primes is listed in A100669.

A118684 Numbers ${\displaystyle \scriptstyle n\,}$ for which the Mertens function ${\displaystyle \scriptstyle M(n)\,}$ (A002321) is +1.

{1, 94, 97, 98, 99, 100, 146, 147, 148, 161, 162, 215, 216, 230, 237, 330, 334, 337, 338, 349, 350, 351, 352, 365, 394, 397, 399, 400, 415, 416, 418, 538, 539, 540, 542, 606, 794, 799, 800, 801, 806, 809, ...}

## Notes

1. Weisstein, Eric W., Binomial Distribution, from MathWorld—A Wolfram Web Resource. [http://mathworld.wolfram.com/BinomialDistribution.html]
2. Weisstein, Eric W., Normal Distribution, from MathWorld—A Wolfram Web Resource. [http://mathworld.wolfram.com/NormalDistribution.html]
3. Weisstein, Eric W., Standard Deviation, from MathWorld—A Wolfram Web Resource. [http://mathworld.wolfram.com/StandardDeviation.html]
4. Weisstein, Eric W., Confidence Interval, from MathWorld—A Wolfram Web Resource. [http://mathworld.wolfram.com/ConfidenceInterval.html]