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On Near 3−Perfect Numbers -.:: Natural Sciences Publishing Sohag J. Math. 4, No. 1, 1-5 (2017) 1 Sohag Journal of Mathematics An International Journal http://dx.doi.org/10.18576/sjm/040101 On Near 3−Perfect 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 Jan. 2016, Revised: 18 Sep. 2016, Accepted: 20 Sep. 2016 Published online: 1 Jan. 2017 Abstract: We call a positive integer n be a near 3−perfect number if σ(n) = 3n + d, where σ(n) is the divisor function and d is a α t proper divisor of n. In this paper, we have derived all near 3−perfects of the form 2 p1 p2, where p1 and p2 are distinct odd primes with p1 < p2 and α ≥ 1, 1 ≤ t ≤ 2. There are only ten such numbers. Moreover, we have also obtained some examples of even near 3−perfect numbers with four distinct prime factors. Keywords: Multi perfect number, Hyperperfect number, Near perfect number 1 Introduction The abudancy index I(n) for any positive integers n is associated with the divisor function σ(n) and is defined as σ(n) I(n)= n . Since k−perfect numbers are solutions of the Well known divisor function σ(n) is the sum of all positive equation σ(n)= kn, so I(n)= k. For abundant numbers, divisors of n including 1 and n itself. For any integer k > 1, I(n) > 2 and for deficient numbers, I(n) < 2. a positive integer n is called a k−perfect or multi perfect For any positive integer k > 1, we call a number n be number [3,7], if σ(n)= kn. All known k−perfect numbers a near k− perfect number if σ(n)= kn + d, where d is are even. No odd k−perfect numbers have been found for a redundant divisor of n. Near perfect numbers are near k ≥ 2. In recent years, some properties of odd k−perfect 2−perfect. For any near k−perfect numbers, we have the numbers have been investigated [3],[8]. following result. α α α In particular for k = 2, 2−perfect numbers are Lemma 1.1.If n = p 1 p 2 .....p r is a near k−perfect solutions of the functional equation σ(n)= 2n, which are 1 2 r number with redundant divisor d, where αi are positive also known as perfect numbers. All known perfect integers and pi are distinct primes, then for i = 1,2,....,r, p−1 n n numbers are of the form n = 2 Mp, where both p and σ α σ α p we have pi| ( i ), if pi|d or pi| ( i ) − d, if pi 6 |d. Mp = 2 − 1 are primes. The primes of the form pi pi p Mp = 2 − 1 are called Mersenne primes. Hyperperfect In particular for an even near k−perfect number, using [1], near perfect [6] and near hyperperfect [4] numbers this lemma, one can obtain the following result. are generalization of perfect numbers. Lemma 1.2.Let r, α are positive integers, and let p be i α α i α0 1 2 αr A positive integer n is said to be a near perfect distinct primes. If n = 2 p1 p2 .....pr is an even near number with redundant divisor d if and only if d is a k−perfect number with redundant divisor d, then there proper divisor (divisors excluding 1 and n itself) of n and exists some i with 1 ≤ i ≤ r such that αi is odd if and σ(n)= 2n + d. For example, 12 is a near perfect number only if 2|d. with redundant divisor 4. P. Pollack and V. Shevelev [6] Lemma 1.1 and lemma 1.2 were proved by Y. Li and introduced the concept of near perfect numbers. They Q. Liao [5] in particular for k = 2. derived all even near perfect numbers with exactly two In this paper, we consider near 3− perfect numbers. distinct prime factors and also obtained certain form of For any α ≥ 1, we have derived all the near 3− perfect α α 2 such numbers. Y. Li and Q. Liao [5] derived some numbers of the form 2 p1 p2 and 2 p1 p2, where both p1 examples of even near perfect numbers with three distinct and p2 are odd primes with p1 < p2. We have proved that α t 4 6 7 8 prime factors of the form 2 p1 p2, where p1 < p2 and 2 .3.5, 2 .3.7, 2 .3.7 and 2 .3.7 are the only near α 2 2 α ≥ 1, 1 ≤ t ≤ 2. 3−perfect numbers of the form 2 p1 p2 and 2 .3 .5, ∗ Corresponding author e-mail: [email protected] c 2017 NSP Natural Sciences Publishing Cor. 2 B. Das, H. K. Saikia: On near 3−perfect numbers 23.32.5, 25.32.7, 24.32.13, 25.32.13 and 27.32.23 are the From (1) and(2), it follows that α 2 only near 3− perfect numbers of the form 2 p1 p2. We have also obtained some examples of even near 3−perfect 2k − 3 1 α β numbers with four distinct prime factors. p = 2 + , p = (3.2 −1 p − 2 −1) − 1 (3) 1 2α + 1 2 k 1 2k−3 If p = 3, thenfrom(3), we must have α = 1, which 2 Main Result 1 2 +1 implies that k = 2α−1 + 2 and therefore α α β α β β Near 3−perfect numbers of the form n = 2 p1 p2 8.2 −1−2 −1−2 8.(2 −2+1)−2 −2−9 2 −2+9 p α α 8 α 2 = 2 −1+2 = 2 −2+1 = − 2 −2+1 If n is a near 3−perfect number with redundant (4) σ divisor d, then (n)= 3n + d and therefore abundancy Since p1 < p2 and p1 = 3, therefore (4) also strictly σ(n) d implies that p2 = 5 or p2 = 7. index I(n)= n = 3 + n . Since d is a proper divisor of d α Case I: If p1 = 5, then from (4), we must have n, therefore 1 < < 1. If n = 2 p1 p2, where p1 < p2, is β n 2 −2+9 β −2 α−β α 3 and therefore 2 3.2 1 6. But this a near 3−perfect with redundant divisor d, then 2 −2+1 = ( − )= equation has no solution for non-negative integers α and σ α+1 α (n) = 3n + d =(2 − 1)(p1 + 1)(p2 + 1) = 3.2 p1 p2 + d β. Case II: If p2 = 7, then from (4) we must have d 1 1 1 β Therefore 3 + n = I(n) = (2 − 2α )(1 + p )(1 + p ) < 2 −2+9 β −2 α−β 1 2 α = 1 and therefore 2 (2 − 1) = 8. This 1 1 2 −2+1 2(1 + p )(1 + p ) α β 1 2 α equation has only solution for = 6 and = 5. Thus From the last inequality, we can get if n = 2 p1 p2, n = 26.3.7 is a near 3−perfect number with redundant where p1 < p2, isa near3−perfect number, then the only divisor 25 . α possibilities are p1 = 3 and p2 = 5 or p2 = 7. In fact we (b)If n = 2 p1 p2 is a near 3−perfect number with have the following results. β redundant divisor d = 2 p1, where α ≥ β, then p2 6 |d and Proposition 2.1. Let α be a positive integer. Suppose that proceeding as the proposition 2.1 (a) one can obtain the p1 and p2 are odd primes with p1 < p2, then following two equations, for some positive integer k (a) The only near 3− perfect number of the form n = α β β α+1 2 p1 p2 with redundant divisor 2 , where α ≥ β, is n = 2kp2 + 2 p1 = (2 − 1)(p1 + 1) (5) 26.3.7. (b) The only near 3− perfect numbers of the form n = β α α β 2k(p2 + 1)+ 2 p1 = 3.2 p1 (6) 2 p1 p2 with redundant divisor 2 p1, where α ≥ β, are n = 24.3.5 and n = 27.3.7. From (5) and(6), it follows that (c) The only near 3− perfect number of the form n = α β 2 p1 p2 with redundant divisor 2 p2, where α ≥ β, is n = 28.3.7. 2k − 3 1 α−1 β −1 p1 = 2 + α , p2 = (3.2 p1 − 2 p1 − k) (7) (d) There exists no near 3−perfect number of the form 2 + 1 k α γ α > γ n = 2 p1 p2 with redundant divisor 2 p1 p2,where ≥ α−1 1. If p1 = 3, then from (7), we have k = 2 + 2 and therefore Proof. α If n = 2 p1 p2 is a near 3−perfect number with α β α β β redundant divisor d, then from the lemma 1.2, it follows 8.2 −1−3.2 −1−2 8.(2 −2+1)−3.2 −2−9 3.2 −2+9 p = α = α = 8 − α that all the redundant divisors are even and only possible 2 2 −1+2 2 −2+1 2 −2+1 (8) values of d are 2β , 2β p , 2β p and 2γ p p , where 1 2 1 2 Since p < p and p = 3, therefore (8) also strictly α ≥ β ≥ 1 and α > γ ≥ 1.
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