The Möbius function and the Mertens conjecture

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This blog post was inspired by Holly Krieger's video for Numberphile.

The Möbius function, $\mu(n)$, is related to the factorization of the positive integer $n$ through $$ \mu(n) = \delta_{\Omega(n), \omega(n)} (-1)^{\Omega(n)}, $$ where $\Omega(n)$ is the number of prime factors of $n$, $\omega(n)$ is the number of distinct prime factors of $n$, and $\delta_{i,j}$ is the Kronecker delta function (equal to 1 or 0 according to whether $i=j$ or not).

That is, $\mu(n) = 1$ if $n$ is a square-free and has an even number of factors, $\mu(n) = -1$ if $n$ is a square-free and has an odd number of factors, and $\mu(n) = 0$ if $n$ has a repeated prime factor. For example,

$n$Factorization$\mu(n)$
1$1$
2 = $2$$-1$
3 = $3$$-1$
4 = $2^2$$0$
5 = $5$$-1$
6 = $2\cdot 3$$1$
7 = $7$$-1$
8 = $2^3$$0$
9 = $3^2$$0$
10 = $2\cdot 5$$1$
11 = $11$$-1$
12 = $2^2\cdot 3$$0$

The Mertens function is the cumulative sum of the Möbius function: $$ M(n) = \sum_{k=1}^n \mu(k). $$

This function is the subject of a famous disproven conjecture: that the magnitude of $M(n)$ is always less than the square root of $n$: $$ |M(n)| < \sqrt{n} \quad\quad(not\;true) $$ It was disproved by Andrew Odlyzko and Herman te Riele in 1985; the current upper-bound to a counter-example was shown by Kotnik and Te Riele (2006) to be below $10^{6.91\times 10^{39}}$, but the precise value is not known.

The code below plots the Mertens function and $\pm \sqrt{n}$ for $n \le 10^6$.

Mertens conjecture

import numpy as np
import matplotlib.pyplot as plt

NMAX = 1000000

def get_mu_seq(nmax):
    """Return the values of the Möbius function up to nmax in a list."""

    sqrt_nmax = int(np.sqrt(nmax))
    mu = [1] * (nmax+1)
    # Comopose the numbers from their factors, keeping track of the
    # parity of the number of factors in the sign. If a number is the
    # product of two prime factors set its value of mu to 0.
    for k in range(2, sqrt_nmax+1):
        if mu[k] == 1:
            for j in range(k, nmax+1, k):
                mu[j] *= -k
            for j in range(k*k, nmax+1, k*k):
                mu[j] = 0
    # Set the value of the Möbius function according to the value in mu.
    for k in range(2, nmax+1):
        if mu[k] == k:
            mu[k] = 1
        elif mu[k] == -k:
            mu[k] = -1
        elif mu[k] < 0:
            mu[k] = 1
        elif mu[k] > 0:
            mu[k] = -1
    return mu


mu = get_mu_seq(NMAX)

n = range(1,NMAX+1)
# Plot the Mertens function and its (wrongly) conjectured bounds, ±sqrt(n).
plt.plot(n, np.cumsum(mu[1:]), lw=1)
plt.plot(n, np.sqrt(n), '0.5', lw=1)
plt.plot(n, -np.sqrt(n), '0.5', lw=1)
plt.xlim(0,NMAX+1)
plt.axhline(0, 0, NMAX+1, c='0.5', lw=1)
plt.xlabel('$n$')
plt.ylabel('$M(n)$')
plt.show()
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