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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 function , Euler’s totient function φ and their compositions play a significant role in our results. 𝜎𝜎 Mathematics Subject Classification: 11A25, 11A41, 11Y70.

Keywords: , 2-, super-hyperperfect number, , .

1. INTRODUCTION Let (n) be the sum of positive 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) 𝑑𝑑– 𝑛𝑛n. A n is called a perfect number if T(n) = n or𝜎𝜎 equivalently (n) = 2n. The first few perfect𝜎𝜎 numbers∑ are𝑑𝑑 6, 28, 496, 8128..... 𝜎𝜎 Euclid discovered the𝜎𝜎 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 is a of the form 2 1𝑛𝑛,− where𝑛𝑛 p must also be a prime 1 number. 𝑛𝑛Any even perfect number n is of the form n = 2 (2 1) , where 2𝑝𝑝 1 is a Mersenne− prime. Only 47 Mersenne primes− are known by now (2010[1]), which means𝑝𝑝− that𝑝𝑝 there are 47 known𝑝𝑝 − 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 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. 𝑛𝑛 𝑛𝑛 ∈ For any n 1, Euler’s phi-function and Dedekind’s are given by

≥ 1 1 ɸ(n) = | 1 and ψ(n) = | 1 + respectively,

𝑛𝑛 ∏𝑝𝑝 𝑛𝑛 � − 𝑝𝑝� 𝑛𝑛 ∏𝑝𝑝 𝑛𝑛 � 𝑝𝑝� 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. 𝑘𝑘 𝑑𝑑 𝑛𝑛 An Arithmetic function f is said to be a multiplicative∑ 𝑑𝑑 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. 𝜎𝜎 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, 65536𝜎𝜎 𝜎𝜎, 𝑛𝑛262144.....𝑝𝑝 It− is not known𝑝𝑝 whether there are any odd superperfect numbers. An odd superperfect number n would have to be a − 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 𝑛𝑛 The first few super-hyperperfect numbers are 9, 729, 531441....𝜎𝜎� In𝜎𝜎 [𝑛𝑛1]� the author conjectured that if n = 3 1, where p 3 1 and are primes, then n is a super-hyperperfect number. 𝑝𝑝 − 𝑝𝑝2 − 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 k𝑘𝑘 -hyperperfect𝑘𝑘− 3 1 𝜎𝜎 𝑛𝑛 𝑘𝑘 𝑘𝑘 numbers. For any 2-hyperperfect number n, ( ) = n + . The first few 2-hyperperfect numbers are 21, 2133, 2 2 19521...... 𝜎𝜎 𝑛𝑛 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 quasiperfect𝑘𝑘− 𝑘𝑘 numbers have𝑘𝑘 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 said𝜎𝜎 𝑛𝑛 to be if g.c.d (n, ) = 1. A positive integer n is called if (n) = 2n, where (n) denotes the sum of unitary divisors𝑛𝑛 of n. In [10] the author 𝑑𝑑 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 𝜎𝜎few USP numbers are𝜎𝜎 2, 9, 165, 238,....Thus there are both even and odd USPs [12]. ∗ ∗ σ(n) The function𝜎𝜎 𝜎𝜎 h defined by h(n) = is called index of n. h(n) is a unbounded and . 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 𝑛𝑛 the number n is called deficient number. If h(n) > 2, then the number n is called .

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 (𝑘𝑘2 1) is an even perfect number, then 1 2 2 𝑝𝑝 − ψ(2𝑝𝑝 [2 1]) = 3. 2 − 𝑝𝑝−− 𝑝𝑝 𝑝𝑝− 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 + ) = − 𝑛𝑛 − − 2 2 ≥ 𝑘𝑘 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 𝑘𝑘 Any even perfect− number is of the− form 2 𝑘𝑘1 (2 𝑘𝑘 1), where 2 1 is prime.− So, if𝑘𝑘 k = 2 1 and 2k 1 = 2 1 is a prime, then k(2k 1) is even perfect number𝑝𝑝− and𝑝𝑝 𝑝𝑝 𝑝𝑝− 𝑝𝑝 − − − − − ψ(2 1[2 1]) = 3. 22 2 𝑝𝑝− 𝑝𝑝 𝑝𝑝− 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 𝑝𝑝− 𝑝𝑝 𝑝𝑝 and σ(n) = 2n = 2 (2 1). Since ψ is multiplicative− and (2 , 2 − 1) = 1, 𝑝𝑝 𝑝𝑝 𝑝𝑝 𝑝𝑝 1 so ψ(σ(n)) = ψ(2 [2 −1]) = ψ(2 ) ψ(2 1) = 2 (1+ ) 2 =− 3.22 1 . 2 𝑝𝑝 𝑝𝑝 𝑝𝑝 𝑝𝑝 𝑝𝑝 𝑝𝑝 𝑝𝑝− © 2013, IJMA. All Rights −Reserved − 119 Bhabesh Das*/Some Aspects of Perfect and Generalized Perfect Numbers/ IJMA- 4(6), June-2013.

This implies ψ(σ(n)) = 3.22 1 (i) 𝑝𝑝− From proposition 2.1., ψ(n) = 3. 22 2 . Since (3, 22 2 ) = 1, so 𝑝𝑝− 𝑝𝑝− 1 ψ(ψ(n)) = ψ(3. 22 2) = ψ(3) ψ (22 2) = 4. 22 2 (1+ ) = 3. 22 1. 2 𝑝𝑝− 𝑝𝑝− 𝑝𝑝− 𝑝𝑝− Thus, ψ(ψ(n)) = 3.22 1 (ii) 𝑝𝑝− 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 𝑝𝑝 𝑝𝑝− 𝑝𝑝 4 (n) ( 1) (n) − − then ( ) = and φ( ) = 3( +1) 3 ( ) 𝑚𝑚 ψ 𝑚𝑚 𝑚𝑚− ψ

𝜎𝜎 𝑛𝑛 𝑚𝑚 𝑛𝑛 𝜎𝜎 𝑛𝑛 Proof: Let m = 2 1, then 2 = m + 1 𝑝𝑝 𝑝𝑝 Since n is an even perfect− number, so ( ) = 2n = 2 (2 1) = m (m + 1) 𝑝𝑝 𝑝𝑝 ( ) = m( m + 1) 𝜎𝜎 𝑛𝑛 − (i)

3 3 From the proposition𝜎𝜎 𝑛𝑛 2.1. ψ(n) = 3. 22 2 = 22 = ( + 1)2. This gives 4 4 𝑝𝑝− 𝑝𝑝 3 𝑚𝑚 ψ(n) = ( + 1)2 (ii) 4

𝑚𝑚 ( ) 4 From (i) and (ii), we obtain = . (n) 3( + 1) 𝜎𝜎 𝑛𝑛 𝑚𝑚 ψ 𝑚𝑚 Thus 4 (n) ( ) = (iii) 3( + 1) 𝑚𝑚 ψ

𝜎𝜎 𝑛𝑛 𝑚𝑚 Since φ is a multiplicative function and (2 1, 2 1) = 1, so 𝑝𝑝− 𝑝𝑝 − ( +1) ( 1) 2 1 φ (n) = φ(2 1(2 1)) = φ (2 1) φ (2 1) = 2 1(2 1 1) = . = . 2 2 4 𝑝𝑝− 𝑝𝑝 𝑝𝑝− 𝑝𝑝 𝑝𝑝− 𝑝𝑝− 𝑚𝑚 𝑚𝑚− 𝑚𝑚 − This implies − − − 2 1 φ (n) = (iv) 4 𝑚𝑚 − ( 1) (n) From (iii) and (iv) we obtain φ( ) = 3 ( ) 𝑚𝑚 𝑚𝑚− ψ

𝑛𝑛 𝜎𝜎 𝑛𝑛 Proposition 2.4: Let 3 2 be a prime. If n is a 2-hyperperfect number, then 𝑘𝑘 4 − ψ(n) = ( n + 3 1 ) 3 𝑘𝑘− Proof: From hypothesis, we can take n = 3 1(3 2) . Since ( 3 1, 3k 2) = 1 and ψ is a multiplicative function, 𝑘𝑘− 𝑘𝑘 𝑘𝑘− so ψ(n) = ψ( 3 1(3 2) ) = ψ( 3 1) ψ(3 2) − − 𝑘𝑘− 𝑘𝑘 𝑘𝑘− 𝑘𝑘 1 4 4 = 3 1 (1 + )− (3k 1) = 3 1 . (3 1−) = [ 3 1 (3 2) + 3 1] . 3 3 3 𝑘𝑘− 𝑘𝑘− 𝑘𝑘 𝑘𝑘− 𝑘𝑘 𝑘𝑘− 4 − − − Thus ψ(n) = (n + 3 1) . 3 𝑘𝑘− 8 3 +1 4 Proposition 2.5: If n is a super-hyperperfect number, then ψ(ψ(n)) = , ψ(σ (n)) = and ψ(φ(n)) = . 3 2 3 𝑛𝑛 𝑛𝑛 3 1 𝑛𝑛 Proof: From the definition of super-hyperperfect number, n = 3 1, where p and are primes. 𝑝𝑝2 1 − So, ψ(n) = ψ(3 1) = 3 1(1+ ) = 4. 3 2 . 𝑝𝑝− 3 𝑝𝑝− 𝑝𝑝− 𝑝𝑝− © 2013, IJMA. All Rights Reserved 120 Bhabesh Das*/Some Aspects of Perfect and Generalized Perfect Numbers/ IJMA- 4(6), June-2013.

1 1 Since (3 2, 4) = 1, so ψ(ψ(n)) = ψ(4. 3 2) = ψ(4) ψ(3 2) = 4. 3 2 (1+ )(1+ ) 2 3 𝑝𝑝− 𝑝𝑝− 𝑝𝑝− 𝑝𝑝− 8 = 8. 3 2 = 3 3 1 𝑝𝑝− 𝑛𝑛 Also σ (n) = σ (3 1) = 𝑝𝑝2 𝑝𝑝− − 3 1 3 1 3 +1 3 +1 Thus, ψ(σ (n)) = ψ( ) = + 1 = = 𝑝𝑝2 𝑝𝑝2 𝑝𝑝2 2 − − 𝑛𝑛 1 Since, φ(n) = 3 1(1 ) = 2. 3 2 3 𝑝𝑝− 𝑝𝑝− − 4 So, ψ(φ(n)) = ψ(2. 3 2) = 4. 3 2 = . 3 𝑝𝑝− 𝑝𝑝− 𝑛𝑛 Proposition 2.6: If n is an even superperfect number, then ψ(ψ(n)) = 3n , ψ(σ (n)) = 2n, and φ (σ(σ(n)) = n

Proof: If n is an even superperfect number, then n = 2 1, where 2 1 is Mersenne prime. 𝑝𝑝− 𝑝𝑝 3 Thus ψ(n) = ψ(2 1) = 2 1 =3. 2 2. This implies − 2 𝑝𝑝− 𝑝𝑝− 𝑝𝑝− ψ(ψ(n)) = ψ (3. 2 2)=3. 2 1 = 3n 𝑝𝑝− 𝑝𝑝− Also, σ (n) = σ (2 1) = 2 1 implies ψ(σ (n)) = ψ(2 1 ) = 2 = 2n and 𝑝𝑝− 𝑝𝑝 𝑝𝑝 𝑝𝑝 1 σ(σ(n)) = 2n = 2. 2 1 = 2 . −Hence φ (σ(σ (n)) = φ (2 ) =− 2 (1 ) = 2 1 = n 2 𝑝𝑝− 𝑝𝑝 𝑝𝑝 𝑝𝑝 𝑝𝑝− Proposition 2.7: Let (3 2) be a prime. If n is a 2-hyperperfect number,− then nT(n) is a triangular number. 𝑘𝑘 3 1 Proof: Since n is a 2-hyperperfect− number, so σ(n) = + . This implies 2 2 𝑛𝑛 3 1 1 +1 ( +1) T(n) = σ(n) – n = + – n = + = which gives nT(n) = , a triangular number . 2 2 2 2 2 2 𝑛𝑛 𝑛𝑛 𝑛𝑛 𝑛𝑛 𝑛𝑛 Proposition 2.8: If n is a super-hyperperfect number, then nT(n) is a triangular number.

1 Proof: If n is a super-hyperperfect number, then n= 3 1, where p and (3 1) are primes. 2 𝑝𝑝− 𝑝𝑝 1 1 1 − 1 Then σ(n) = (3 1) = (3n 1). This implies T(n) = σ(n) – n = (3n –1) – n = (n–1). 2 2 2 2 𝑝𝑝 ( 1−) − Thus nT(n) = , a triangular number. 2 𝑛𝑛 𝑛𝑛− Proposition 2.9: If n is an even superperfect number, then T(n) is a triangular number. 2 𝑛𝑛 Proof: If n is an even superperfect number, then n = 2 1 where 2 1 is Mersenne prime. 𝑝𝑝− 𝑝𝑝 So, σ(n) = 2 1 = 2n 1. This gives T(n) = σ(n) – n = 2n 1 n− = n 1 and thus T(n) = (n 1), a triangular 2 2 number. 𝑝𝑝 𝑛𝑛 𝑛𝑛 − − − − − − Proposition 2.10: If n is an even perfect number, then T(n) is a triangular number.

Proof: Since, n is even perfect number, so σ(n) = 2n. Thus T(n) = σ(n) – n = 2n n = n. Every even perfect number is a triangular number, so T(n) = n, is a triangular number. − Proposition 2.11: Let 3 2 be a prime. If n is a 2-hyperperfect number and σ(n) = b, then 2 1 is a prime . 𝑘𝑘 Proof: Since n is a 2-hyperperfect− number, so n = 3 1(3 2); where 3 2 is a prime.√ 𝑏𝑏 − 𝑘𝑘− 𝑘𝑘 𝑘𝑘 Let p = (3 2), then p+1 = (3 1) − − 𝑘𝑘 𝑘𝑘 Thus n = 3 −1(3 2) = 3 1 p −implies © 2013, IJMA.𝑘𝑘− All Rights𝑘𝑘 Reserved𝑘𝑘− 121 − Bhabesh Das*/Some Aspects of Perfect and Generalized Perfect Numbers/ IJMA- 4(6), June-2013.

1 1 1 1 σ(n) = σ(3 1 p) = σ (3 1) σ(p) = (3 1)(p+1) = ( p+1) (p+1) = ( + 1)2. Thus σ (n) = b = ( + 1)2 gives 2 2 2 2 ( + 1)2 =𝑘𝑘 2b.− Therefore,𝑘𝑘− p+1 = ± 2 𝑘𝑘 giving − 𝑝𝑝 𝑝𝑝

p𝑝𝑝 = ± 2 1. Since p = 3 2 is√ a 𝑏𝑏prime, so 2 1 is a prime. 𝑘𝑘 Proposition√ 𝑏𝑏 − 2.12: If n = 2 1−(2 1) is an even√ perfect𝑏𝑏 − number, then 𝑝𝑝− 𝑝𝑝 2. (2 1) + 2 ( 1)+1− 1 2 + 2.(2 1) 2 ( 1) σk(n) = and σ-k (n) = . 𝑘𝑘 𝑝𝑝 2𝑘𝑘 1𝑘𝑘 𝑝𝑝− 𝑝𝑝 −2𝑘𝑘 1−𝑘𝑘 𝑝𝑝− −𝑘𝑘 � 𝑛𝑛 − − − � � − − − 𝑛𝑛 � 𝑘𝑘 𝑘𝑘 1 − − Proof: Since (2 , 2 1) = 1 and σk is a multiplicative function, so 𝑝𝑝− 𝑝𝑝 1 − σk(n) = σk(2 [2 1]) 𝑝𝑝− 𝑝𝑝 1 = σk(2 ) σk(−2 1) 𝑝𝑝− 𝑝𝑝 = (1 + 2 + 22 +−...... + 2 ( 1))(1 + (2 1) ) 𝑘𝑘 𝑘𝑘 𝑘𝑘 𝑝𝑝− 𝑝𝑝 𝑘𝑘 2 ( 1)+1 1 − = . (1 + (2 1) ) 𝑘𝑘 𝑝𝑝2− 1 � − � 𝑝𝑝 𝑘𝑘 𝑘𝑘 − 2 ( 1)+1 – 1 + 2 ( 1)+1 (−2 1) (2 1) = . 𝑘𝑘 𝑝𝑝− 𝑘𝑘 𝑝𝑝2− 1 𝑝𝑝 𝑘𝑘 𝑝𝑝 𝑘𝑘 � − − − � 𝑘𝑘 − 2. (2 1) + 2 ( 1)+1 1 σk(n) = . 𝑘𝑘 𝑝𝑝 2𝑘𝑘 1𝑘𝑘 𝑝𝑝− � 𝑛𝑛 − − − � 𝑘𝑘 − 2. (2 1) + 2 ( 1)+1 1 We have σ-k(n) = σk(n) = 𝑘𝑘 𝑝𝑝 2𝑘𝑘 1𝑘𝑘 𝑝𝑝− −𝑘𝑘 −𝑘𝑘 � 𝑛𝑛 − − − � 𝑘𝑘 − 𝑛𝑛 𝑛𝑛2 + 2.(2 1) 2 ( 1) = 𝑝𝑝 −2𝑘𝑘 1−𝑘𝑘 𝑝𝑝− −𝑘𝑘 � − − − 𝑛𝑛 � 𝑘𝑘 − 3 1 3 Proposition 2.13: If n is a super-hyperperfect number, then σ (n) = , and (n) = , k k 3 𝑘𝑘 1 σ-k 3 −1𝑘𝑘 ℕ 𝑛𝑛 − − 𝑛𝑛 𝑘𝑘 𝑘𝑘 3 1− − ∀ ∈ Proof: n is a super-hyperperfect number, so n=3 1, where p and are primes. 𝑝𝑝2 𝑝𝑝− − 3 ( 1)+1 1 3.3 ( 1) 1 3 1 σ (n) = 1 + 3 + 32 +. . … + 3 ( 1) = = = k 𝑘𝑘 𝑝𝑝3− 1 𝑘𝑘3 𝑝𝑝−1 3 𝑘𝑘 1 𝑘𝑘 𝑘𝑘 𝑘𝑘 𝑝𝑝− − − 𝑛𝑛 − 𝑘𝑘 𝑘𝑘 𝑘𝑘 − − − 3 1 3 We have σ (n) = σk(n) = = -k 3 𝑘𝑘 1 3 −1𝑘𝑘 −𝑘𝑘 −𝑘𝑘 𝑛𝑛 − −𝑛𝑛 𝑘𝑘 𝑘𝑘 − − 𝑛𝑛 𝑛𝑛 2 1 2 Proposition 2.14: If n is an even superperfect number, then σ (n) = , and σ (n) = , k k 2 𝑘𝑘 1 -k 2 −1𝑘𝑘 ℕ 𝑛𝑛 − −𝑛𝑛 𝑘𝑘 𝑘𝑘 − − Proof: Since n is an even superperfect number so n = 2 1, where 2 1 is Mersenne prime. ∀ ∈ 𝑝𝑝− 𝑝𝑝 ( 1)+1 ( 1) 2 ( 1) 2 1 2.2 1 2 −1 σk(n)=1 + 2 + 2 + … . . +2 = = = 𝑘𝑘 𝑝𝑝2− 1 𝑘𝑘2 𝑝𝑝−1 2 𝑘𝑘 1 𝑘𝑘 𝑘𝑘 𝑘𝑘 𝑝𝑝− − − 𝑛𝑛 − 𝑘𝑘 𝑘𝑘 𝑘𝑘 − − − 2 1 2 We have σ-k(n) = σk(n) = = 2 𝑘𝑘 1 2 −1𝑘𝑘 −𝑘𝑘 −𝑘𝑘 𝑛𝑛 − −𝑛𝑛 𝑘𝑘 𝑘𝑘 𝑛𝑛 𝑛𝑛 − − 1 Proposition 2.15: If n is an even superperfect number then index of n is 2 . Moreover n is a .

𝑛𝑛 Proof: If n is an even superperfect number, then n = 2 1, where 2 1 −is Mersenne prime. 𝑝𝑝− 𝑝𝑝 σ(n) = 2 1 = 2n 1 − 𝑝𝑝 σ(n) 2n 1 1 h(n) = = − = 2 − n − 𝑛𝑛 𝑛𝑛 This implies, h(n) < 2, so n is a deficient− number © 2013, IJMA. All Rights Reserved 122 Bhabesh Das*/Some Aspects of Perfect and Generalized Perfect Numbers/ IJMA- 4(6), June-2013.

1 1 Proposition 2.16: If n is a super-hyperperfect number, then index of n is 3 ; moreover n is a deficient number. 2

3 1� − 𝑛𝑛� Proof: If n is a super-hyperperfect number, then n =3 1, where p and are primes. 𝑝𝑝2 𝑝𝑝− − 3 1 3 1 σ(n) = = 𝑝𝑝2 2 − 𝑛𝑛− σ(n) 3 1 1 1 h(n) = = = 3 2 2 𝑛𝑛− 𝑛𝑛 𝑛𝑛 𝑛𝑛 This implies, h(n) < 2 , so n is a deficient number.� − �

3 1 Proposition 2.17: Let 3 2 be a prime. If n is a 2-hyperperfect number, then index of n is + ; Moreover n is a 2 2 deficient number. 𝑘𝑘 𝑛𝑛 − 1 Proof: If n is a 2-hyperperfect number, then n = 3 1(3 2) and σ(n) = (3 + 1) 2 𝑘𝑘− 𝑘𝑘 σ(n) 1 3 1 − 𝑛𝑛 h(n) = = (3 + 1) = + 2 2 2

𝑛𝑛 𝑛𝑛 𝑛𝑛 This implies, h(n) < 2, so n is a deficient𝑛𝑛 number .

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