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International Journal of Discrete 2017; 2(3): 64-67 http://www.sciencepublishinggroup.com/j/dmath doi: 10.11648/j.dmath.20170203.12

Some Aspects of Certain Form of Near Perfect

Bhabesh Das 1, Helen K. Saikia 2

1Department of Mathematics, B. P. C. College, Nagarbera, Assam, India 2Department of Mathematics, Gauhati University, Guwahati, Assam, India Email address: [email protected] (B. Das), [email protected] (H. K. Saikia) To cite this article: Bhabesh Das, Helen K. Saikia. Some Aspects of Certain Form of Near Perfect Numbers. International Journal of Discrete Mathematics. Vol. 2, No. 3, 2017, pp. 64-67. doi: 10.11648/j.dmath.20170203.12

Received : January 29, 2017; Accepted : March 7, 2017; Published : March 24, 2017

Abstract: It is well known that a positive n is said to be near perfect , if σ(n) = 2n+d where d is a proper divisor of n and function σ(n) is the sum of all positive divisors of n In this paper, we discuss some results concerning with near perfect numbers from known near perfect numbers. Keywords: , , Fermat Prime, , Near Perfect Number

near and deficient hyperperfect numbers [5], near −perfect 1. Introduction numbers. In 2012, P. Pollack and V. Shevelev [10] introduced the notion of near perfect numbers. It is well known that a A positive integer is called perfect number if the sum of positive integer is called near perfect number, if is the all proper divisors of is equal to . Proper divisors of are sum of all of its proper divisors, except for one of them, positive divisors of other than itself. The smallest perfect which is termed as redundant divisor. For example, proper number is since . First four perfect numbers 6 6 = 1 + 2 + 3 divisors of 12 are 1, 2, 3, 4 and 6. We can write are: 6, 28, 496, 8128; which are known from ancient time. 12 = 1 + , which shows that 12 is a near perfect with Well known divisor function is the sum of all positive 2 + 3 + 6 redundant divisor 4. Moreover, using the definition of divisor divisors of . Divisor function is a multiplicative function one can say: a positive integer is near perfect function i.e., satisfies the functional condition number with redundant divisor d if and only if d is a proper = if g.c.d , = 1. For any perfect number , . All known perfect numbers are even. divisor of and = 2 + . Some near perfect numbers = 2 Euler [9] proved that all known perfect numbers are of the are associated with classical perfect numbers. If = 2 M is an even perfect number, then , and P = 2 = 2 form = 2 M P , where both and M P = 2 − 1 are are near perfect numbers with two distinct primes. The primes of the form are called = 2 − 1 = 2 − 1 prime factors. There are infinitely many near perfect numbers Mersenne primes. Up to now (March, 2017), only 49 other than above shapes. In [10], P. Pollack and V. Shevelev Mersenne prime numbers are known which means there are have also defined near perfect number. A positive integer only 49 even perfect numbers that are discovered [14]. Multi- − is called near perfect number [10], if is the sum of all perfect numbers are natural extension of classical perfect of its proper divisors, except numbers of proper divisors. If numbers. A positive integer is called perfect number [3, − is a near perfect number with redundant divisors 4](multi perfect of abundancy or generalized perfect ddd, , ,...., d then we can write number) if , where . All known perfect 1 2 3 k = > 2 − ++ + + + numbers are even. There are only finite number of multi- = 2 ddd1 2 3 ..... d k . perfect numbers are discovered. No single example for an Near perfect numbers are 1 near perfect number. odd perfect number and odd multi perfect number has been From the definition of − perfect number, it is clear that found nor has a proof for their non-existence been there are infinite numbers of − near perfect number. established, although many people have been working on this Another special kind of is Fermat Prime. problem for centuries. m + Odd Prime number of the form Fn = 2 1 is called Fermat In recent time, other possible generalized perfect numbers prime, where must be some power of 2 [2]. There are only are hyper perfect numbers [13], near perfect numbers [10], = = = five known Fermat prime numbers: F0 3 , F1 5 , F2 17 , International Journal of Discrete Mathematics 2017; 2(3): 64-67 65

= F3 = 257 and F4 65537 . an odd prime relatively prime to Fm . If = 2 Fm is a near perfect number with redundant divisor 2, then the prime 15F + 7 2. Main Results must be of the form m . = − Fm 15 We have obtained the following results from known near Proof. Since F is a near perfect number with perfect numbers. = 2 m redundant divisor , so we can write This Proposition 2.1. For any integer ≥ 1, there is no near 2 − 2 = 8. equation implies that (F + 1) F perfect number of the form = 2 . 15 m + 1 − 16 m = 8. Proof. If = 2 is a near perfect number with redundant Solving this equation one can get the form of . divisor 2 , where 0 < < , then from definition of near Example 2.2. n = 17816 = 2. 17.131 is a near perfect perfect numbers = 2 + 2. But the equation number with redundant divisor 2. 2 − 1 = 2 + 2 strictly implies that 2 + 1 = 0, 15F + 7 Here F and therefore m . which is a contradiction. Hence there is no near perfect m = 17 = − = 131 Fm 15 number of the form = 2, where ≥ 1. Proposition 2.8. There is no near perfect number of the Proposition 2.2. For any integer ≥ 1, there is no near form p p , where p and p are distinct odd primes. perfect number of the form = 2 with redundant divisor = 1 2 1 2 where is an odd prime. , Proof. Suppose = p1 p 2 is a near perfect number with Proof. Suppose = 2 is a near perfect number with redundant divisor , then either = p1 or = p2 . If = p1 , redundant divisor , then from definition of near perfect then 2 p p . Therefore (p+ 1)( p + 1) numbers − 2 = . Therefore the equation = 1 2 + 1 2 = , strictly implies that + − 2 − 1 + 1 − 2 = 2 p1 p 2 p 1 , which implies that p2( p 1 1) = 1. This equation 2 − 1 = 2. But the expression 2 − 1 is always odd has solution only for p 2 and p , which contradict for any integer . Therefore there is no near perfect 1 = 2 = 1 ≥ 1 the facts that p and p are odd primes. number of the form = 2 with redundant divisor . 1 2 Proposition 2.9. There is no near perfect number of the Proposition 2.3. If is an odd prime, then any number of form 2 p p , where p and p are distinct odd primes. the form = 2 is a near perfect number if and only if = 1 2 1 2 . = 3 Proof. Suppose = 2 p1 p 2 is a near perfect number with Proposition 2.4. If is an odd prime, then any number of the redundant divisor , then = 4 p1 p 2 +, where possible form = 2 is a near perfect number if and only if = 7. m + values of are = p1 or p2 or 2 p1 or 2 p2 or p1 p 2 . Proposition 2.5. Let Fm = 2 1 is a Fermat prime. If Case I. Suppose p , then 4 p p p , which = 2 Fm is a near perfect number, then = 12 or = 20 . = 1 = 1 2 + 1 gives 3(p + 1) p( p − 2) . From the last equation we Proof. If = 2 Fm is a near perfect number with 2 = 1 2 obtain p p +1 and p − 2 3 . Solving these two redundant divisor , then we have = 2 Fm + and 1 = 2 2 = + − equations, we obtain p 6 and p = 5 , which contradict therefore 7(Fm 1) 8 F m = . This equation strictly implies 1 = 2 − = that p is an odd prime. that 7 Fm R . Since is a positive proper divisor of and 1 Case II. Suppose 2 p , then 4p p+ 2 p , which Fm is a Fermat prime, so the last equation strictly implies = 1 = 1 2 1 gives 3(p + 1) p( p − 1) . From the last equation we get that Fm = 3 and Fm = 5. 2 = 1 2 p p +1and p −1 3 . Solving we obtain p 5 and p Proposition 2.6. Let Fm ≥ 17 is a Fermat prime and is an 1 = 2 2 = 1 = 2 4 , which also contradict that p is an odd prime. odd prime relatively prime to Fm . If = 2 Fm is a near = 2 Case III. Suppose p p , then 4 pp+ pp , perfect number with redundant divisor 2 Fm , then the prime = 1 2 = 12 12 + + 7F + 15 which gives 3(p1 p 2 1) = 2 p1 p 2 . Since the expression must be of the form m . = − p+ p + 1 is always odd, therefore the last equation has no Fm 15 1 2 solution for any odd primes p and p . Proof. If = 2 Fm is a near perfect number with 1 2 Hence the above three cases strictly imply that there is no redundant divisor 2 Fm , then − 2 = 8Fm . This near perfect number of the form 2 p p , where p and equation implies that (F + 1) F F . = 1 2 1 15 m + 1 − 16 m = 8 m p are distinct odd primes. Solving this equation one can obtain the form of 2 . a Example 2.1. Proposition 2.10. If = 2 p1 p 2 , where > 1 and p1 and p2 Suppose = 2. 17. is a near perfect number with are distinct odd primes, then is not a near perfect number redundant divisor 2 . 17 . with redundant divisor p1 p 2 . 7F + 15 m Proof. If = 2 p1 p 2 , then Here Fm then , i.e., = 17, = − = 6 = a+1 Fm 15 σ σ − − 2 = 2 (p1 )( p 2 ) 2 pp 12 + + 2 . 17. 67 is a near perfect number. (2a1− 1)(p + 1)( p +− 1) 2 a 1 pp = 1 2 12 Proposition 2.7. Let Fm ≥ 17 is a Fermat prime and is 66 Bhabesh Das and Helen K. Saikia: Some Aspects of Certain Form of Near Perfect Numbers

+ (2a 1 − 1)(pp + +− 1) pp For , if F is an odd prime, then = 1 2 12 < 2 − m = Suppose is a near perfect number with redundant divisor Fm is a near perfect number with redundant a+1 2 2 − p p , then (2− 1)(p + p + 1) 2 p p m 1 2 1 2 = 1 2 . divisor 2 . a +1 − + + Proof. We have Since the numbers 2 1 and p1 p 2 1 are odd numbers, so L. H. S. of the last equation is an odd number − 2 = 2 − 1 2 − Fm + 1 − 2 2 − Fm a+1 − + + and therefore (2 1)(p1 p 2 1) ≡ 1 (mod 2), but R. H. S. is even and divisible by 2. The of L. H. S. and R. = 2 2 − Fm + 2 − 2 − Fm + 1 − 2 2 − Fm H. S. of the equation do not match, which is a m contradiction. = 2 . Proposition 2.16. If and F 2m + 1 are respectively Proposition 2.11. Let p1 and p2 are distinct odd primes m = 2 perfect number and Fermat prime with g.c.d F , with p1 < p2 . If = 2p1 p 2 is a near perfect number with , m = 1 redundant divisor 2, then . = 650 then = Fm is not a near perfect number. 2 Proof. If = 2 p1 p 2 is a near perfect number with + Proof. We have − 2 = 2 (Fm 1) − 2J Fm = 2J . redundant divisor 2, then But 2 is not a divisor of . 3(p2 + p + 1) (1+p ) − 4 p2 p 2 = − 2 = 1 1 2 1 2 Proposition 2.17. If = 2 − 1 is a Mersenne prime, then =2 +++ −2 + is 2 near perfect number with redundant 3333p1 p 1 p 2 pp 1212 pp 3 = 2 After simple simplification we get divisors and . + ++=2 − Proof. We have (p1 1)(2 pp 12 3 1) pp 12 ( 1) . − 2 = 2 − 1 + + + 1 − 2 We solve this equation in terms of p and p . 1 2 = + . From the last equation we obtain the following three In general we obtain the following proposition possibilities: Proposition 2.18. If = 2 − 1 is a Mersenne prime, then + = − + + = 2 for any , is a near perfect number Case I. If p11 p 2 1 and 2p1 3 p 2 1 p 1 then ≥ 2 = 2 = + − = with redundant divisors , where i=1,2...., k . p2 p 1 2 an d p1( p 1 5) 7 . But the equation − = Proposition 2.19. For each if M 2pi − 1 is a p1( p 1 5) 7 has no solution. = 1, 2; i = − p −1 Mersenne prime, then 2 p1 1 M M is not a near perfect Case II. If p +1 = 2 and 2p+ 3 p + 12 = p 2 then = 1 2 1 2 1 2 1 number. p=2 p + 3 and p2 −4 p − 5 = 0 . Solving these two + 2 2 1 1 1 Proof. We have − 2 = M1 M 1 . equations, we get p = 5 , p =13 and therefore 650 . 2 1 2 = But M1 is not a divisor of . = − Proposition 2.12. If p1 p 2 ... p m , where p1 < p2 < … . < pi 1 Proposition 2.20. If = 2M1 M 2 ... M r , where M i = p are odd primes, then is not a near perfect number with m 2pi − 1 are distinct Mersenne primes, , then is p = 1,2, … . redundant divisor m . not a near perfect number.

Proof. Suppose = p1 p 2 ... p m is a near perfect number with redundant divisor pm , then = 2 + pm and 3. Conclusion + + + therefore pm |, then pm | (1p1 )(1 p 2 )...(1 p m ) . Near perfect numbers are natural extensions of classical p +1 p perfect numbers. In this paper, we discuss near perfect Since p p p , so it is clearly i m for 1 < 2 < … . < m 2 < numbers of certain form. There are very good scopes for studying other form of near perfect numbers. all prime factors of . Therefore + + + pm ∤ (1p1 )(1 p 2 )...(1 p m ) , which is a contradiction. Proposition 2.13. If 2 − 3 is an odd prime, then = References 22 − 3 is a near perfect number with redundant divisors 2. [1] T. M. Apostol, Introduction to Analytic , Proof. We have Springer Verlag, New York, 1976. − 2 = 2 − 12 − 3 + 1 − 2 2 − 3 [2] David M. Burton, Elementary Number Theory, Tata McGraw- Hill, Sixth Edition, 2007. = 22 − 3 + 2 − 2 − 2 − 22 − 3 = 2. [3] R. D. Carmichael, Multiply perfect numbers of three different Proposition 2.14. If 2 − 5 is an odd prime, then = primes, Ann. Math., 8 (1) (1906): 49–56. 22 − 5 is a near perfect number with redundant divisors 4. [4] R. D. Carmichael, Multiply perfect numbers of four different m + primes, Ann. Math., 8 (4) (1907): 149–158. Proposition 2.15. Let Fm = 2 1 is a Fermat prime. International Journal of Discrete Mathematics 2017; 2(3): 64-67 67

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