Journal of Inequalities in Pure and Applied Mathematics

Journal of Inequalities in Pure and Applied Mathematics

http://jipam.vu.edu.au/

Volume 7, Issue 3, Article 99, 2006

A NOTE ON MULTIPLICATIVELYe-PERFECT NUMBERS

LE ANH VINH AND DANG PHUONG DUNG SCHOOL OFMATHEMATICS

UNIVERSITY OFNEWSOUTHWALES

SYDNEY2052 NSW [email protected] SCHOOL OFFINANCE ANDBANKING

UNIVERSITY OFNEWSOUTHWALES

SYDNEY2052 NSW [email protected]

Received 17 October, 2005; accepted 26 October, 2005 Communicated by J. Sándor

ABSTRACT. LetTe(n)denote the product of all exponential divisors ofn. An integernis called multiplicativelye-perfect ifTe(n) =n²and multiplicativelye-superperfect ifTe(Te(n)) =n². In this note, we give an alternative proof for characterization of multiplicativelye-perfect and multiplicativelye-superperfect numbers.

Key words and phrases: Perfect number, Exponential divisor, Multiplicatively perfect, Sum of divisors, Number of divisors.

2000 Mathematics Subject Classification. 11A25, 11A99.

1. INTRODUCTION

Let σ(n)be the sum of all divisors of n. An integern is called perfect if σ(n) = 2n and superperfect ifσ(σ(n)) = 2n. If n =p^α₁¹· · ·p^α_k^k is the prime factorization ofn > 1, a divisor d | n, called an exponential divisor (e-divisor) of n is d = p^β₁¹· · ·p^β_k^k with β_i | α_i for all 1 ≤ i ≤ k. Let T_e(n) denote the product of all exponential divisors of n. The concepts of multiplicativelye-perfect and multiplicativelye-superperfect numbers were first introduced by Sándor in [1].

Definition 1.1. An integern is called multiplicativelye-perfect ifT_e(n) = n² and multiplica- tivelye-superperfect ifT_e(T_e(n)) = n².

In [1], Sándor completely characterizes multiplicatively e-perfect and multiplicatively e- superperfect numbers.

ISSN (electronic): 1443-5756

c 2006 Victoria University. All rights reserved.

313-05

2 LEANHVINH ANDDANGPHUONGDUNG

Theorem 1.1 ([1]).

a) An integernis multiplicativelye-perfect if and only ifn =p^α, wherepis prime andα is a perfect number.

b) An intergernis multiplicativelye-superperfect if and only ifn=p^α, wherepis a prime, andαis a superperfect number.

Sándor’s proof is based on an explicit expression ofT_e(n). In this note, we offer an alternative proof of Theorem 1.1.

2. PROOF OFTHEOREM1.1

a) Suppose thatnis multiplicativelye-perfect; that isT_e(n) =n². Ifnhas more than one prime factor thenn = p^α₁¹· · ·p^α_k^k for some k ≥ 2, α_i ≥ 1andp₁, . . . , p_k are k distinct primes. We have three separate cases.

(1) Suppose thatα₁ = · · · = α_k = 1. Thendis an exponential divisor ofn if and only if d=p₁· · ·p_k=n. HenceT_e(n) =n,which is a contradiction.

(2) Suppose that two ofα1, . . . , αkare greater1. Without loss of generality, we may assume that α1, α2 > 1. Then d1 = p1p^α₂²· · ·p^α_k^k, d2 = p^α₁¹p2p^α₃³· · ·p^α_k^k, d3 = n are three distinct exponential divisors ofn. Hence d1d2d3 | Te(n). However,p^2α₁ ¹⁺¹ | d1d2d3 so T_e(n)6=n²,which is a contradiction.

(3) Suppose that there is exactly one of α₁, . . . , α_k which is greater than 1. Without loss of generality, we may assume that α₁ > 1and α₂ = · · · = α_k = 1. We have that if d is an exponential divisor of n then d = p^β₁¹p2· · ·pk for some β1 | α1. Hence if n has more than two distinct exponential divisors thenp³₂ | T_e(n) =p^2α₁ ¹p²₂· · ·p²_k, which is a contradiction. However, d₁ = p₁p₂· · ·p_k, d₂ = p^α₁¹p₂p₃· · ·p_k are two distinct exponential divisors of n so d₁, d₂ are all exponential divisors of n. Hence T_e(n) = p^α₁¹⁺¹p²₂· · ·p²_k =p^2α₁ ¹p²₂· · ·p²_k. This implies thatα₁ = 1,which is a contradiction.

Thus n has only one prime factor; that is, n = p^α for some prime p. In this case then T_e(n) = p^σ(α). HenceT_e(n) =n² =p^2α if and only ifσ(α) = 2α. This concludes the proof.

b) Suppose thatnis multiplicativelye-superperfect; that isTe(Te(n)) = n². Ifnhas more than one prime factor then n = p^α₁¹· · ·p^α_k^k for some k ≥ 2, αi ≥ 1and p1, . . . , pk are k distinct primes. We have two separate cases.

(1) Suppose that α₁ = · · · = α_k = 1. Thend is an exponential divisor of n if and only if d = p₁· · ·p_k = n. Hence T_e(n) = n and T_e(T_e(n)) = T_e(n) = n which is a contradiction.

(2) Suppose that there is at least one ofα₁, . . . , α_kwhich is greater1. Without loss of gener- ality, we may assume thatα₁ >1. Thend₁ =p₁p^α₂²· · ·p^α_k^k,d₂ =n=p^α₁¹p^α₂²p^α₃³· · ·p^α_k^k, are two distinct exponential divisors of n. Hence d₁d₂ | T_e(n). However, d₁d₂ = p^α₁¹⁺¹p^2α₂ ²· · ·p^2α_k ^k so T_e(n) = p^γ₁¹· · ·p^γ_k^k where γ₁ ≥ α₁ + 1, γ_i ≥ 2α_i ≥ 2 for i = 2, . . . , k. Thus, t₁ = p^γ₁¹p₂p^γ₃³· · ·p^γ_k^k and t₂ = T_e(n) = p^γ₁¹p^γ₂²p^γ₃³· · ·p^γ_k^k are two distinct exponential divisors ofT_e(n). Hencet₁t₂ |T_e(T_e(n)). However,p^2γ₁ ¹ |t₁t₂ andγ₁ > α₁,which is a contradiction.

Thusnhas only one prime factor; that isn =p^αfor some primep. In this case thenT_e(n) = p^σ(α) and Te(Te(n)) = p^σ(σ(n)). HenceTe(Te(n)) = n² = p^2α if and only if σ(σ(α)) = 2α.

This concludes the proof.

Remark 2.1. In an e-mail message, Professor Sándor has provided the authors some more recent references related to the arithmetic functionT_e(n), as well as connected notions on e- perfect numbers and generalizations. These are [2], [3], and [4].

J. Inequal. Pure and Appl. Math., 7(3) Art. 99, 2006 http://jipam.vu.edu.au/

MULTIPLICATIVELY E-PERFECT 3

REFERENCES

[1] J. SÁNDOR, On multiplicativelye-perfect numbers, J. Inequal. Pure Appl. Math., 5(4) (2004), Art.

114. [ONLINE:http://jipam.vu.edu.au/article.php?sid=469]

[2] J. SÁNDOR, A note on exponential divisors and related arithmetic functions, Scientia Magna, 1 (2005), 97–101. [ONLINE http://www.gallup.unm.edu/~smarandache/

ScientiaMagna1.pdf]

[3] J. SÁNDOR, On exponentially harmonic numbers, to appear in Indian J. Math.

[4] J. SÁNDOR AND M. BENCZE, On modified hyperperfect numbers, RGMIA Research Report Collection, 8(2) (2005), Art. 5. [ONLINE: http://eureka.vu.edu.au/~rgmia/v8n2/

mhpn.pdf]

J. Inequal. Pure and Appl. Math., 7(3) Art. 99, 2006 http://jipam.vu.edu.au/