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Some Aspects of Harmonic Numbers Which Divide the Sum of Its Positive Divisors

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Some Aspects of Harmonic Numbers Which Divide the Sum of Its Positive Divisors IOSR Journal of Mathematics (IOSR-JM) e-ISSN: 2278-5728,p-ISSN: 2319-765X, Volume 7, Issue 3 (Jul. - Aug. 2013), PP 39-45 www.iosrjournals.org Some aspects of Harmonic Numbers which divide the sum of its Positive divisors. 1 2 Pranjal Rajkhowa , Hemen Bharali 1(Department of Mathematics),Gauhati University. India 2(Department of Basic Sciences),Don Bosco College of Engineering and Technology ,Guwahati ,India Abstract: In this paper some particular harmonic numbers n, where n divides σ(n) completely have been 푚 푘 푚 푘 taken. Three types of numbers prime factorised as say, 푝 푞, 2 푝 푞, 2 푝1푝2 … … 푝푚 have been discussed and some propositions have been developed to understand the properties of these type of numbers. It has been observed that harmonic numbers n having . 푐. 푑. 푛, 휎 푛 = 푛 of the form 푝푞, 푝2푞, 푝푚 푞 does not exist if p,q both of them are odd primes. Further some useful properties of the harmonic numbers of the form 2푘 푝푚 푞, for some values of m have been discussed with the help of some propositions. After that an algorithm has been 푘 proposed to get the numbers of the form 2 푝1푝2 … … 푝푚 , where n divides σ(n) completely. Keywords: Harmonic number, Mersene prime, Ore’s conjecture, Perfect number I. Introduction In 1948 Ore [9] introduced the concept of Harmonic numbers, and these numbers were named as Ore’s harmonic numbers(some 15 years later) by Pomerance[10].In general, the harmonic mean of positive numbers 푎1, 푎2, … … . , 푎푘 is defined by −1 푘 1 푖=1 푎푖 푛휏(푛) A positive integer n is said to be harmonic if the harmonic mean of its positive divisors 퐻 푛 = is 휎(푛) an integer, where 휏(푛) denotes the number of positive divisors and σ(n) denotes the sum of the positive divisors of n. We call 1 the trivial harmonic number. A list of harmonic numbers less than 2.109 have been given by Cohen [2]. This have been extended by Goto and Okeya [5] which is up to 1014. If n is a harmonic divisor number then there are two possibilities (i) . 푐. 푑. 푛, 휎 푛 = 푛 or (ii) . 푐. 푑. 푛, 휎 푛 ≠ 푛.(i) implies that σ(n)=kn, where k is a positive integer. In fact if σ(n)=2n, then the numbers are called perfect numbers. Ore [9] proved that every perfect number is harmonic. But the converse is not true. For example, 140 is not perfect, but H(140)=5. Ore [4] conjectured that all harmonic numbers other than 1 must be even. In the present paper , we focus on those harmonic numbers which satisfies the property (i). Some examples of this kind of numbers are 6,28,496,672,8128,30240, … ….In section 2, some properties of the numbers of the form 푝푚 푞 where p,q are any prime numbers have been discussed. In section 3, the type of numbers have been extended to the form 2푘 푝푚 푞. In section 4, the numbers have been further extended to the 푘 form 2 푝1푝2 … … 푝푚 . In fact in this section a search technique have been obtained to get the numbers of the 푘 form . 푐. 푑. 푛, 휎 푛 = 푛, where 푛 = 2 푝1푝2 … … 푝푚 . II. Some properties of harmonic numbers having 품. 풄. 풅. 풏, 흈 풏 = 풏 of the form 풑풎풒 Theorem 2.1 The only harmonic divisor number of the form 풑풒, where 풑 and 풒 are prime numbers and 품. 풄. 풅. 풏, 흈 풏 = 풏, is ퟔ. Proof: Let 푛 = 푝푞, where 푝, 푞 are two prime numbers. Now 휎 푛 = 1 + 푝 + 푞 + 푝푞. If n divides σ(n) completely then 1 + 푝 + 푞 + 푝푞 = 푘푝푞, for some number k. Since n is a harmonic divisor number so 푘|휏 푛 = 4. So the possible values of k are 1,2,4 . Since 1 + 푝 + 푞 ≠ 0 therefore 푘 ≠ 1. The other possible 1+푞 1+푞 equations are 1 + 푝 + 푞 = 푝푞 or 1 + 푝 + 푞 = 3푝푞. The possible values of 푝 are , . Now 1 + 푞 > 푞−1 3푞−1 3푞 − 1 => 2푞 < 2, which is not possible. Hence the only possible solutions are 2,3. Therefore 6 is the only harmonic divisor number of the form 푝푞. Theorem 2.2 The only harmonic divisor number of the form 풑ퟐ풒, where 풑 풂풏풅 풒 are prime numbers such that 품. 풄. 풅. 풏, 흈 풏 = 풏 is ퟐퟖ. www.iosrjournals.org 39 | Page Some aspects of Harmonic Numbers which divide the sum of its positive divisors Proof: Here 휎 푛 = 1 + 푝 + 푝2 + 푞 + 푝푞 + 푝2푞 and 푛 = 6 . To satisfy the condition 푛|휎(푛) we have 1+푝+푝2 푞 = , where휎(푛) = 푛푘. The possible values of k are 1,2,3,6. Clearly k cannot be 1.As 휎(푛) = 푛푘 −1−푝−푝2+푘푝2 implies 휎 푛 = 푛 푖. 푒. 1 + 푝 + 푝2 + 푞 + 푝푞 + 푝2푞 = 푝2푞, 푖. 푒. 1 + 푝 + 푝2 + 푞 + 푝푞 = 0, which is not possible as the LHS is greater than 0. Let 푘푝2 − 1 + 푝 + 푝2 = 푡 … … … … … … 1 and 1 + 푝 + 푝2 = 푞푡 … … … … … … … … … . 2 Where t is some positive integer. Adding (1) and (2), 푘푝2 = 푞 + 1 푡. Firstly, we consider that both 푝 푎푛푑 푞 are odd primes. From (2) we have 푞 푎푛푑 푡 are odd numbers as 푞푡 must be odd, which is because, 1 + 푝 + 푝2 is odd. Now from (1) 푘푝2 must be even as the RHS is odd. As 푝 is an odd prime, 푘 must be even. Hence the possible values of 푘 are 2,6 Case (i). 푘 = 2 We have 푘푝2 = 푞 + 1 푡 ⇒ 2푝2 = 푞 + 1 푡 ⇒ 2. 푝. 푝 = 푞 + 1 푡 Then the possible equations are 푞 + 1 = 2푝2, 푡 = 1 or 푞 + 1 = 2푝, 푡 = 푝 If 푞 + 1 = 2푝2, 푡 = 1 , then equation (2) implies 1 + 푝 + 푝2 = 2푝2 − 1 . 1 ⇒ 푝2 − 푝 − 2 = 0 ,which has no prime solutions. .Similarly, if 1 + 푝 + 푝2 = (2푝 − 1)푝 then 푝2 − 2푝 − 1 = 0 , which does not give any prime values of p. Therefore 푘 cannot be 2. Case (ii). 푘 = 6 We have 푘푝2 = 푞 + 1 푡 ⇒ 6푝2 = 푞 + 1 푡 ⇒ 2.3. 푝. 푝 = 푞 + 1 푡 The possible values of 푞, 푡 are 2,3푝2 , 6, 푝2 , 2푝, 3푝 , 2푝2, 3 , 6푝, 푝 , 6푝2, 1 . Clearly any of the above pair does not satisfy the equation (2) for odd prime 푝. Next we assume that 푝 is an even prime. From equation (2), we have 1 + 푝 + 푝2 = 푞푡 ⇒ 푞푡 = 7. Since 푞 is prime the only possibility is 푞 = 7. Hence the number becomes 푛 = 22. 7.Also we have 휏 푛 = 6 and 휎 푛 = 56.Therefore 휎 푛 = 2푛 and 퐻 푛 = 3. Similarly if we consider 푞 is even then equation (2) is not satisfied as the LHS is odd and the RHS is even. Hence the theorem. In fact the above theorem may be generalised to some extent and may be stated as follows: Theorem 2.3 If 풏 is a harmonic number of the form 풑풎풒, where 풑, 풒 are prime numbers and 풎 is a 흉(풏) positive integer such that 품. 풄. 풅. 풏, 흈 풏 = 풏 then 푯 풏 = ퟐ Proof: Since 푛|휎(푛) and 푛 = 푝푚 푞 , therefore 푞 can be expressed as 푖=푚 푖 푖=0 푝 푞 = 푚 푖=푚 푖 푘푝 − 푖=0 푝 푚 푖=푚 푖 Let 푘푝 − 푖=0 푝 = 푡 … … … … … … … … … (3) 푖=푚 푖 And 푖=0 푝 = 푞푡 … … … … … … … … … … … . (4) Where t is a particular positive integer .Adding (3) and (4), we have 푘푝푚 = 푞 + 1 푡 Again 푖=푚 푖 푖=0 푝 푞 = 푚 푖=푚 푖 > 1 푘푝 − 푖=0 푝 푖=푚 푖 푚 ⇒ 2 푖=0 푝 > 푘푝 푝푚 +1−1 ⇒ 2 > 푘푝푚 푝−1 2(푝푚 +1−1) ⇒ 푘 < 푝푚 (푝−1) 1 2(푝− 푚 ) ⇒ 푘 < 푝 (푝−1) 푝 1 ⇒ 푘 < 2 − 푝−1 푝푚 (푝−1) ⇒ 푘 ≤ 2 So 푘 is either 1 or 2. If 푘 = 1, then from equation (3),we have 푚 푖=푚 푖 푝 − 푖=0 푝 = 푡 푚 푖=푚−1 푖 푚 푝 − 푖=0 푝 − 푝 = 푡 푖=푚−1 푖 − 푖=0 푝 = 푡 휏(푛) Which is not possible as the LHS is negative and RHS is positive. Therefore 푘 must be 2. i.e.퐻 푛 = . 2 Hence the theorem is proved. www.iosrjournals.org 40 | Page Some aspects of Harmonic Numbers which divide the sum of its positive divisors In fact, it has been observed by Pomerance [10] and Callan [7] that the only harmonic numbers of the form 푝푎 푞푏 are perfect numbers. Theorem 2.4 There is no perfect number of the form 풑ퟑ풒, where 풑, 풒 are odd primes. Proof: Let 푛 = 푝3푞 then 휎 푛 = 1 + 푝 + 푝2 + 푝3 1 + 푞 . Since we are considering the perfect number ,therefore 휎 푛 = 2푛.That is 1 + 푝 + 푝2 + 푝3 1 + 푞 = 2푝3푞. 1+푝+푝2+푝3 1+푞 Or = 2 … … … … … … … … … . (5) 푝3 푞 Suppose if possible 푝| 1 + 푝 + 푝2 + 푝3 . Obviously 푝| 푝 + 푝2 + 푝3 . Therefore, we have 푝|( 1 + 푝+푝2+푝3−푝+푝2+푝3). i.e. 푝|1, which is not possible. Hence 푝∤(1+푝+푝2+푝3). Therefore the only possibility is 푝3| 1 + 푞 and 푞| 1 + 푝 + 푝2 + 푝3 . 2 being a prime number the only possibilities are (i) 2푝3 = 1 + 푞 and 푞 = 1 + 푝 + 푝2 + 푝3.
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