International Journal of Mathematical Archive-4(6), 2013, 118-123 Available online through www.ijma.info ISSN 2229 – 5046 SOME ASPECTS OF PERFECT AND GENERALIZED PERFECT NUMBERS Bhabesh Das* Department of Mathematics, Gauhati University, Guwahati-781014, Assam, India (Received on: 23-04-12; Revised & Accepted on: 20-05-13) ABSTRACT In this paper, some new results for perfect and generalized perfect numbers are derived. Arithmetic functions like the Dedekind’s function ψ, the divisor function , Euler’s totient function φ and their compositions play a significant role in our results. Mathematics Subject Classification: 11A25, 11A41, 11Y70. Keywords: perfect number, 2-hyperperfect number, super-hyperperfect number, superperfect number, triangular number. 1. INTRODUCTION Let (n) be the sum of positive divisors of n, i.e (n) = | and T(n) be the sum of aliquot parts of n , i.e. the positive divisors of n other than n itself, so that T(n) = (n) – n. A natural number n is called a perfect number if T(n) = n or equivalently (n) = 2n. The first few perfect numbers∑ are 6, 28, 496, 8128..... Euclid discovered the first four perfect numbers which are generated by the formula 2 1(2 1) about 300 B.C [8], where (2 1) is a prime. A Mersenne prime is a prime number of the form 2 1, −where p must also be a prime 1 number. Any even perfect number n is of the form n = 2 (2 1) , where 2 1 is a Mersenne− prime. Only 47 Mersenne primes− are known by now (2010[1]), which means− that there are 47 known − even perfect numbers. There is a conjecture that there are infinitely many perfect numbers. The search− for new ones− is to keep on going by search program via the Internet; named GIMPS (Great Internet Mersenne Prime Search). It is not known if any odd perfect number exists, although numbers up to 10300 have been checked without success [2]. There are some results concerning the odd perfect numbers, (i) Any odd perfect number must be of the form 12m + 1 or 36m + 9 (ii) If n is an odd perfect number, it has the form: n = qkp2e... r2f, where q, p,.. ,r are distinct primes and q ≡ k ≡ 1 (mod 4). [5] A positive integer n is said to be a harmonic number if harmonic mean of its divisors is an integer. In 1948 Norwegian mathematician Oystein Ore proved that every perfect number is an Ore harmonic number. ( +1) If positive integer k is a triangular number, then k must be of the form , n . Every even perfect number is a 2 ℕ triangular number. Even perfect numbers end with 6 or 8. ∈ For any n 1, Euler’s phi-function and Dedekind’s Arithmetic function are given by ≥ 1 1 ɸ(n) = | 1 and ψ(n) = | 1 + respectively, ∏ � − � ∏ � � where p runs through the distinct prime divisors of n. Let by convention, φ (1) = 1, ψ(1) = 1. -k Arithmetic function σk(n) is defined as σk(n) = | and σ1(n) = σ (n), and σ-k(n) = n σk(n) for any n ℕ. An Arithmetic function f is said to be a multiplicative∑ function if f(m.n) = f(m)f(n), where g.c.d (m, n) =∈ 1. Functions , φ, σk, and ψ are all multiplicative arithmetic functions. Corresponding author: Bhabesh Das* Department of Mathematics, Gauhati University, Guwahati-781014, Assam, India International Journal of Mathematical Archive- 4(6), June – 2013 118 Bhabesh Das*/Some Aspects of Perfect and Generalized Perfect Numbers/ IJMA- 4(6), June-2013. A positive integer n is called a superperfect number if ( ( ))= 2n. D. Suryanarayana introduced the notation of superperfect number in [11]. Even superperfect numbers are of the form 2 1, where 2 1 is a Mersenne prime. The first few superperfect numbers are 2, 4, 16, 64, 4096, 65536 , 262144..... It− is not known whether there are any odd superperfect numbers. An odd superperfect number n would have to be a square number− such that either n or σ(n) is divisible by at least three distinct primes [7]. 3 1 A positive integer n is called super-hyperperfect number if ( ) = + . 2 2 The first few super-hyperperfect numbers are 9, 729, 531441.... �In [1]� the author conjectured that if n = 3 1, where p 3 1 − and are primes, then n is a super-hyperperfect number. 2 − In [9] Minoli and Bear introduced the concept of k-hyperperfect numbers and they conjectured that there are k- +1 1 hyperperfect numbers for every k. A positive integer n is called k- hyperperfect number if ( ) = n + . A number is a perfect number if and only if it is a 1-hyperperfect number. In [1] A. Beg studied different k -hyperperfect− 3 1 numbers. For any 2-hyperperfect number n, ( ) = n + . The first few 2-hyperperfect numbers are 21, 2133, 2 2 19521....... In [1] the author conjectured that all 2-hyperperfect numbers are of the form n = 3 1(3 2), where 3 2 is prime. A positive integer n is called a quasiperfect number if ( ) = 2n + 1 [6]. No quasiperfect− numbers have been found so far, but if a quasiperfect number exists, it must be an odd square number greater than −1035and have at− least seven distinct prime factors [4]. A divisor d of a number n is said to be unitary divisor if g.c.d (n, ) = 1. A positive integer n is called unitary perfect number if (n) = 2n, where (n) denotes the sum of unitary divisors of n. In [10] the author proved that there are no odd unitary∗ perfect numbers. ∗A positive integer n is said to be a unitary super perfect number (USP) if ( (n)) = 2n. The first few USP numbers are 2, 9, 165, 238,....Thus there are both even and odd USPs [12]. ∗ ∗ σ(n) The function h defined by h(n) = is called index of n. h(n) is a unbounded and multiplicative function. A positive integer n is a perfect number if and only h(n) = 2. The function h takes larger value than 1 for all n > 1. If h(n) < 2, then the number n is called deficient number. If h(n) > 2, then the number n is called abundant number. 2. MAIN RESULTS In this section we prove our main results on perfect and generalized perfect numbers. Proposition 2.1: If k 1, 1 + 2 + 4 + 8 + ... + k = 2k 1, where (2k 1) is a prime and k(2k 1) is a perfect number, then ψ(k[2k 1]) = 3 2. ≥ − − − Moreover, if k(2k 1)− = 2 1 (2 1) is an even perfect number, then 1 2 2 − ψ(2 [2 1]) = 3. 2 − −− − Proof: Let k 1, 1 + 2 + 4 +..... + k =− 2k 1 and (2k 1) is a prime. Since (k, 2k 1) = 1 and Ψ is a multiplicative function , so ψ(k[2k 1]) = ψ(k) ψ(2k 1). From hypothesis k is even and is of the form 2 (n 1). By using the 1 3 definition of Dedekind≥ psi function, we can write− ψ(k) =− k(1 + ) = − − − 2 2 ≥ Since 2k 1 is prime, ψ( 2k 1 ) = 2k. 3 Also, ψ(k[2k− 1]) = ψ(k) ψ(2k− 1) = 2 = 3 2 . This implies ψ(k[2k 1]) = 3 2 2 Any even perfect− number is of the− form 2 1 (2 1), where 2 1 is prime.− So, if k = 2 1 and 2k 1 = 2 1 is a prime, then k(2k 1) is even perfect number− and − − − − − 1 2 2 − ψ(2 [2 1]) = 3. 2 − − Proposition 2.2: For any even perfect number− n, ψ(σ(n)) = ψ(ψ(n)). Proof: If n is a perfect number, then n = 2 1(2 1), where 2 1 is a Mersenne prime − and σ(n) = 2n = 2 (2 1). Since ψ is multiplicative− and (2 , 2 − 1) = 1, 1 so ψ(σ(n)) = ψ(2 [2 −1]) = ψ(2 ) ψ(2 1) = 2 (1+ ) 2 =− 3.22 1 . 2 − © 2013, IJMA. All Rights −Reserved − 119 Bhabesh Das*/Some Aspects of Perfect and Generalized Perfect Numbers/ IJMA- 4(6), June-2013. This implies 2 1 ψ(σ(n)) = 3.2 (i) − From proposition 2.1., ψ(n) = 3. 22 2 . Since (3, 22 2 ) = 1, so − − 2 2 2 2 2 2 1 2 1 ψ(ψ(n)) = ψ(3. 2 ) = ψ(3) ψ (2 ) = 4. 2 (1+ ) = 3. 2 . 2 − − − − Thus, ψ(ψ(n)) = 3.22 1 (ii) − From (i) and (ii), we obtain ψ(σ(n)) = ψ(ψ(n)) . Proposition 2.3: Let m be a Mersenne prime and m = 2 1. If n = 2 1(2 1) is an even perfect number − 4 (n) ( 1) (n) − − then ( ) = and φ( ) = 3( +1) 3 ( ) ψ − ψ Proof: Let m = 2 1, then 2 = m + 1 Since n is an even perfect− number, so ( ) = 2n = 2 (2 1) = m (m + 1) ( ) = m( m + 1) − (i) 3 3 From the proposition 2.1. ψ(n) = 3. 22 2 = 22 = ( + 1)2. This gives 4 4 − 3 2 ψ(n) = ( + 1) (ii) 4 ( ) 4 From (i) and (ii), we obtain = . (n) 3( + 1) ψ Thus 4 (n) ( ) = (iii) 3( + 1) ψ Since φ is a multiplicative function and (2 1, 2 1) = 1, so − 2 1 1 − 1 1 ( +1) ( 1) 1 φ (n) = φ(2 (2 1)) = φ (2 ) φ (2 1) = 2 (2 1) = . = . 2 2 4 − − − − − − This implies − − − 2 1 φ (n) = (iv) 4 − ( 1) (n) From (iii) and (iv) we obtain φ( ) = 3 ( ) − ψ Proposition 2.4: Let 3 2 be a prime.