The Erdős discrepancy problem over the squarefree and cubefree integers

Abstract

Let g : N → {−1, 1} be a completely multiplicative function and let µ 2 2 (n) be the indicator that n is cubefree. We prove that f = µ 2 2 g has unbounded partial sums. In the squarefree case f = µ 2 g, we provided a necessary condition for bounded partial sums: f pretends to be a real and primitive Dirichlet character χ of condutor q coprime with 6, andThe proofs are built upon Klurman & Mangerel proof of Chudakov conjecture [4], Klurman work on correlations of multiplicative functions [3] and Tao resolution of the Erdős discrepancy problem [6].

Multiplicative functions that are close to their mean

“…In Section 5 we address two more applications of the techniques used in this paper. One of these concerns a problem treated in a recent paper of Aymone [1] on the Erdős discrepancy problem for square-free supported multiplicative sequences. In [1], it is proven among other things, that if for a completely multiplicative function g :…”

“…He conjectured (see Conjecture 1.1 of [1]) that no such functions g exist. The arguments of the present paper allow us to settle this conjecture.…”

“…Then f pretends to be χ , and properties a), b) ,c) are unaffected by replacing χ by χ . This will simplify our work in 1 The same manoeuvre was employed in [11], so there is no loss in consistency at this step with what is done there. what follows.…”

“…Chudakov [4] considered the problem of characterizing completely multiplicative functions with arithmetic constraints whose partial sums behave rigidly, specifically being very close to their mean value. He conjectured in particular that if f : N → C is a completely multiplicative function taking only finitely many values, vanishing at only finitely many primes, and satisfying (1) n≤x…”

“…It is natural to ask if one can relax certain of the hypotheses in Chudakov's problem and still obtain a characterization of those functions that satisfy (1). Ruzsa posed the following question in this direction (see [3,Problem 21]).…”