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S. Kim, P. Q. Nguyen Res. Number Theory (2025) 11:3
https://doi.org/10.1007/s40993-024-00603-9
RESEARCH
On counterexamples to the Mertens
conjecture
Seungki Kim1* and Phong Q. Nguyen2
*Correspondence:
seungki.math@gmail.com
1Department of Mathematical
Sciences, University of Cincinnati,
4199 French Hall West, 2815
Commons Way, Cincinnati, OH
45221-0025, USA 2Department
of Computer Science, Ecole
Normale Supérieure, 45 rue
d’Ulm, 75005 Paris, France
Full list of author information is
available at the end of the article
Abstract
We use state-of-art lattice algorithms to improve the upper bound on the lowest
counterexample to the Mertens conjecture to ≈exp(1.96 ×1019), which is significantly
below the conjectured value of ≈exp(5.15 ×1023) by Kotnik and van de Lune (Exp
Math 13:473–481, 2004).
Keywords: Mertens conjecture, Mertens function, Lattice-point enumeration
1 Introduction
The Mertens conjecture [17], dating back to 1897, is a statement about the growth rate of
the Mertens function
M(x):=
1≤n≤x
μ(n),
where μ(n) is the Möbius function
μ(n)=⎧
⎨
⎩
(−1)kif nis squarefree, and has kdistinct prime factors,
0ifnis not squarefree.
The size of M(x) is of interest in number theory, since it is closely related to the size of the
real parts of the zeroes of the Riemann zeta function ζ(s). For example, a short argument
(see e.g. [19, Sect. 2]) shows that if M(x)=O(xθ), then the Riemann zeta function ζ(s)has
no zeroes on the half-plane Re s>θ.Forθ=1/2+εfor arbitrarily small ε>0, the latter
statement is the famous Riemann hypothesis. The Mertens conjecture is a much bolder
claim that
|M(x)|<x1/2for all x>1.(1.1)
It took nearly a century for this conjecture to be disproved, by Odlyzko and te Riele [19].
Their argument consisted of certain insights from classical analytic number theory, and,
perhaps surprisingly, the use of a lattice reduction algorithm. No alternative (dis)proof
that does not rely on lattice reduction is known to this day.
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