Inequalities for the Arithmetical Functions of Euler and Dedekind
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Czechoslovak Mathematical Journal
11 pp
Online first
INEQUALITIES FOR THE ARITHMETICAL FUNCTIONS OF EULER AND DEDEKIND Horst Alzer, Waldbröl, Man Kam Kwong, Hong Kong Received December 2, 2018. Published online January 27, 2020.
Abstract. For positive integers n, Euler’s phi function and Dedekind’s psi function are given by Y Y 1 1 ϕ(n) = n 1− and ψ(n) = n 1+ , p p p|n p prime
p|n p prime
respectively. We prove that for all n > 2 we have and
1−
1 n−1 1 n+1 ϕ(n) ϕ(n) ψ(n) ψ(n) 1+ 6 n n n n
ϕ(n) ψ(n) ψ(n) ϕ(n) n
n
1 n−1 1 n+1 1+ 6 1− . n n
The sign of equality holds if and only if n is a prime. The first inequality refines results due to Atanassov (2011) and Kannan & Srikanth (2013). Keywords: Euler’s phi function; Dedekind’s psi function; inequalities MSC 2010 : 11A25
1. Introduction In this paper, we are concerned with two classical arithmetical functions: Euler’s phi (or totient) function ϕ and Dedekind’s psi function ψ. If n ∈ N, then ϕ(n) is the number of positive integers up to n which are relatively prime to n. The product formula Y 1 ϕ(n) = n 1− p p|n p prime
DOI: 10.21136/CMJ.2020.0530-18
1
is an important tool for calculating ϕ(n). Dedekind’s psi function given by ψ(n) = n
Closely related to the ϕ-function is
Y
p|n p prime
1+
1 . p
Both functions can be expressed in terms of the Möbius function µ which is defined for n ∈ N by ( (−1)k if n is square-free and k is the number of prime factors of n, µ(n) = 0 if n is not square-free. We have the representations Xn ϕ(n) = µ(d) d
and ψ(n) =
d|n
Xn d|n
d
|µ(d)|.
These functions have remarkable applications in various mathematical and physical problems. Recently, Solé and Planat in [8] proved that a certain inequality involving the Dedekind psi function is equivalent to the famous Riemann hypothesis. For more information on this subject we refer the reader to Apostol, see [1], Mitrinovi´c et al., see [4], Sándor, see [5], Sándor and Crstici, see [7] and the references cited therein. Our work is motivated by two interesting papers published by Atanassov, see [2] and Kannan and Srikanth, see [3]. Atanassov proved that for n > 2, n2n < ϕ(n)ϕ(n) ψ(n)ψ(n) .
(1.1)
Kannan and Srikanth claimed that if Θ(n) =
ϕ(n) + ψ(n) , 2n
then for n > 2, n2nΘ(n) 6 ϕ(n)ϕ(n) ψ(n)ψ(n) .
(1.2)
Since Θ(n) > 1 (n > 2), it follows that (1.2) refines (1.1). Unfortunately, the proof given in [3] is incorrect. For instance, the inequality in the last line of page 20 is false. A correct proof of (1.2) was given by Sándor, see [6], page 51, who actually offered an improvement of (1.2): (1.3) 2
n2nΘ(n) 6
ϕ(n) + ψ(n) 2nΘ(n) 2
< ϕ(n)ϕ(n) ψ(n)ψ(n)
(n > 2). Online first
The referee pointed out that a slight modification of the proof of (1.3) leads to the following chain of inequalities: ϕ(n)ψ(n) ψ(n)ϕ(n)
2).
From (1.3) and (1.4) we conclude that the elegant inequalities (1.5)
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