On Abundancy Index of Some Special Class of Numbers

Sohag J. Math. 4, No. 2, 37-40 (2017) 37 Sohag Journal of Mathematics An International Journal http://dx.doi.org/10.18576/sjm/040202 On Abundancy Index of Some Special Class of Numbers Bhabesh Das1,∗ and Helen K. Saikia2 1 Department of Mathematics, B.P.C.College, Assam, 781127, India 2 Department of Mathematics, Gauhati University, Assam, 781014, India Received: 10 Dec. 2015, Revised: 12 Sep. 2016, Accepted: 23 Sep. 2016 Published online: 1 May 2017 σ(n) σ Abstract: For any positive integer n, abundancy index I(n) is defined as I(n) = n , where (n) is the sum of all positive divisors of n. In this paper, we have discussed non trivial lower and upper bounds of I(n) for some special class of numbers like Quasi perfect numbers, Super hyperperfect numbers, Near perfect numbers and Hyperperfect numbers. Keywords: Quasi perfect number, Super hyperperfect number, Near perfect number, Hyperperfect number p−1 1 Introduction of n. If m = 2 Mp is an even perfect number, where p p both p and Mp = 2 − 1 are primes, then n = 2m, n = 2 m We known divisor function σ(n) is the sum of all positive and n = (2p − 1)m are near perfect numbers. There are divisors of n , including 1 and n itself. The abundancy finitely many near perfect numbers other than these three p index I(n) for any positive integer n is associated with the shapes. The primes of the form Mp = 2 − 1 are called divisor function σ(n) and is defined as Mersenne primes. Definition 1.3. If a positive integer n is a k− hyperperfect σ(n) I(n)= number, then n n = 1 + k[σ(n) − n − 1] Since perfect numbers are solutions of the equation σ(n) = 2n, therefore I(n) = 2. For abundant numbers, for some positive integer k. This equation can also be I(n) > 2 and for deficient numbers, I(n) < 2. written as k + 1 k − 1 We first recall some well known definitions. σ(n)= n + k k Definition 1.1. A positive integer n, which is a solution of D. Minoli and R. Bear [6] introduced the notion of the functional equation σ(n) = 2n + 1, is called quasi k−hyperperfect numbers. They conjectured that there are perfect number. No single example for quasi perfect k−hyperperfect numbers for every natural number k. number has been found, nor has a proof for their non Perfect numbers are 1−hyperperfect numbers. In 2000, J. existence been established. It is still an open question to S. McCranie [5] computed all hyperperfect numbers less determine existence (or otherwise) for a quasi perfect than 1110. If n isa2−hyperperfect number, then number. Interest to this problem has produced many analogous notions. If a quasi perfect number exists, then 3 1 σ(n)= n + it must be an odd square number and have at least seven 2 2 10 distinct prime factors and also greater than 35 [4]. For A. Bege and K. Fogarasi [2] have given list of k− more details on quasi perfect numbers see [1,3]. hyperperfect numbers for some positive integer k and also Definition 1.2. P. Pollack and V. Shevelev [7] introduced they conjectured that all 2−hyperperfect numbers are of the concept of near perfect numbers and classified all the the form n = 3k−1(3k − 2), where 3k − 2 are primes. near perfect numbers with two distinct prime factors. Definition 1.4. A positive integer n, which satisfies the Near perfect numbers are solutions of the equation equation σ(n) = 2n + d , where d is a proper divisor (divisors 3 1 σ(σ(n)) = n + excluding 1 and n itself) of n, known as redundant divisor 2 2 ∗ Corresponding author e-mail: [email protected] c 2017 NSP Natural Sciences Publishing Cor. 38 B. Das, H. K. Saikia: On abundancy index of some special... is called super hyperperfect number. A. Bege and K. Proposition 2.3. A positive integer n is quasi perfect Fogarasi [2] conjectured that solutions of this equation number if and only if the following inequality holds: p are of the form 3p−1, where both p and 3 −1 are primes. 2 2n + 3 2n + 4 In this paper, we have determined non trivial lower < I(n) < and upper bounds of I(n) for quasi perfect, near perfect, n + 1 n + 1 super hyperperfect and k−hyperperfect numbers. Moreover, quasi perfect, near perfect and k−hyperperfect Proof. First we assume that n is a quasi perfect number, then from the proposition 2.2, we can write I n < 2n+4 . numbers can also be defined in terms of abundancy index ( ) n+1 2n+3 I(n). Next we prove n+1 < I(n). On the contrary suppose that 2n+3 ≥ 2n2+3n ≥ σ n+1 I(n), then n+1 (n). But 2 2n +3n = 2n(n+1)+n = 2n + n = 2n + 1 − 1 . Therefore 2 Main Result n+1 n+1 n+1 n+1 the last inequality becomes − 1 ≥ σ 2n + 1 n+1 (n) = 2n + 1, which is clearly a Proposition 2.1. If n is a quasi perfect number, then contradiction. This contradiction strictly implies that 2n+3 2n + 4 n+1 < I(n). Hence if n is a quasi perfect number, then 2 < I(n) < 2n+3 2n+4 n + 1 n+1 < I(n) < n+1 holds. Conversely, suppose that n is a positive integer which Proof. If n is a quasi perfect number, then σ(n)= 2n + 1. satisfies the following inequality σ Therefore abundancy index I(n)= (n) = 2 + 1 > 2. Next n n 2n + 3 2n + 4 we prove I(n) < 2n+4 . On the contrary suppose that < I(n) < n+1 n + 1 n + 1 ≥ 2n+4 I(n) n+1 , then from definition of I(n), one obtains 2 Then from definition of I(n), we can write σ(n) ≥ 2n +4n . But n+1 2n2+3n 2n2+4n 2 < σ(n) < . This inequality implies that 2n +4n = 2n(n+1)+2n = 2n + 2n = 2n + 2 − 2 . n+1 n+1 n+1 n+1 n+1 n+1 2n 1 − 1 < σ n < 2n 2 − 2 . Since σ n is Therefore our assuming inequality becomes + n+1 ( ) + n+1 ( ) σ(n) ≥ 2n + 2 − 2 . Since n is a quasi perfect number, always a positive integer, therefore the last inequality n+1 implies that σ(n) = 2n + 1. Hence n is a quasi perfect σ ≥ − 2 therefore 2n + 1 = (n) 2n + 2 n+1 , this is clearly a number. contradiction for any positive integer n > 1. This obstacle 2n+4 Proposition 2.4. If n is a super hyperperfect number, then implies that I(n) < n+1 . Remark 2.1. Converse of the proposition 2.1 is also true, 3n + 2 1 < I(n) < i.e., the inequality obtained in proposition 2.1 for n, is also 2(n + 1) necessary for n to be a quasi perfect number. Proposition 2.2. If n is a positive integer which satisfies Proof. If n is a super hyperperfect number, the n is a σ σ 3 1 p−1 the inequality solution of the equation ( (n)) = 2 n + 2 . If 3 are 2n + 4 the only solutions [2] of this equation, where both p and 2 < I(n) < 3p−1 n + 1 2 are primes, then for super hyperperfect numbers, , then n is a quasi perfect number. σ(n) are always primes and therefore σ σ σ Proof. Let n be a positive integer, and suppose that n ( (n)) = (n)+ 1. From the last equation, we obtain σ 3n − 1 satisfies the inequality 2 < I(n) < 2n+4 , then from (n) = 2 2 . Therefore abundancy index, n+1 3 1 2 − σ 2n +4n I(n) = 2 2n . Clearly I(n) > 1. Moreover definition of I(n), one obtains 2n < (n) < n+1 . But 3n+2 3 1 3 1 n 2 = − = − (1 − ) > I(n). 2n +4n = 2n + 2 − 2 , therefore the inequality becomes 2(n+1) 2 2(n+1) 2 2 n+1 n+1 n+1 For super hyperperfect number, the lower of I(n) can σ − 2 σ 2n < (n) < 2n + 2 n+1 . Since divisor function (n) is also be improved. always a positive integer for any n > 1, therefore the last Proposition 2.5. Let n be a positive integer. Then n is a inequality strictly implies that σ(n)= 2n + 1. Hence n is super hyperperfect number if a quasi perfect number. Remark 2.2. From proposition 2.1 and proposition 2.2, 3n + 1 3n + 2 < I(n) < we can define quasi perfect numbers in terms of 2(n + 1) 2(n + 1) abundancy index I(n). We can say that positive integer n is a quasi perfect number if and only if the inequality Remark 2.3. The converse of proposition 2.4 and 2n+4 2 < I(n) < n+1 holds. The number 2 is the lower bound proposition 2.5 are not true in general if n is a super of I(n), when n is a quasi perfect number. Moreover, this hyperperfect number. Any numbers of the form 3k, where ≥ σ 3n − 1 lower bound of I(n) can be expressed in terms of rational k 1, satisfy the equation (n)= 2 2 , but all such function of n. Following proposition gives the more type of numbers do not satisfy the equation σ σ 3 1 improve lower bound for I(n). ( (n)) = 2 n + 2 .

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