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 𝑔. 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) 𝑔. 푐. 푑. 푛, 휎 푛 = 푛 or (ii) 𝑔. 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 𝑔. 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. But second one is not possible as the LHS is odd and the RHS is even. and (ii) 2푞 = 1 + 푝 + 푝2 + 푝3 and 푝3 = 1 + 푞.Again the second one is not possible as the LHS is odd and the RHS is even .Hence the Result. The above theorem may be generalized as:

Theorem 2.5 There is no perfect number of the form 풑풎풒, where p and q are odd primes and 풎 is an odd numbers. 푚 푖=푚 푖 Proof: Let 푛 = 푝 푞 then 휎 푛 = 푖=0 푝 (1 + 푞). Since we considering the perfect number therefore 푖=푚 푖 푚 휎 푛 = 2푛.That is 푖=0 푝 1 + 푞 = 2 푝 푞 푖=푚 푝푖 1+푞 Or 푖=0 = 2...... (6) 푝푚 푞 푖=푚 푖 푖=푚 푖 푖=푚 푖 Suppose, if possible푝| 푖=0 푝 . Obviously 푝| 푖=1 푝 .Therefore, we have 푝| 푖=0 푝 − 푖=푚 푖 푖=푚 푖 푚 푖=1 푝 . i.e. 푝|1, which is not possible. Hence 푝 ∤. 푖=0 푝 . Therefore the only possibility is 푝 |1 + 푞 and 푖=푚 푖 푚 푖=푚 푖 푞| 푖=0 푝 . 2 being a prime number the possibilities are (i) 2푝 = 1 + 푞 and 푞 = 푖=0 푝 .But second one 푖=푚 푖 푚 is not possible, as the LHS is odd and the RHS is even. (ii) 2푞 = 푖=0 푝 and 푝 = 1 + 푞, again second one is not possible as the LHS is odd and the RHS is even and so the result. Theorem 2.4 is a particular case when m=3. We now state that “There is no harmonic number of the form 풑풎풒, where p and q are odd primes and 풎 is an odd number such that 품. 풄. 풅. 풏, 흈 풏 = 풏. "

III. Some properties of the harmonic numbers having 품. 풄. 풅. 풏, 흈 풏 = 풏 of the form ퟐ풌풑풎풒 In this section we are trying to find the existence of some harmonic numbers of the form 2푘 푝푚 , for small values of m. Theorem 3.1 Let the number be of the form 풏 = ퟐ풌풑 and let 품. 풄. 풅. 풏, 흈 풏 = 풏, then there exists at 흉(풏) least one 풏 such that 푯 풏 = ퟐ 푘 푖=푘 푖 Proof: If 푛 = 2 푝 then 휎 푛 = 푖=0 2 (1 + 푝). Since휎 푛 = 푛푘1, for some number k1 . We have 푖=푘 푖 푛푘1 = 푖=0 2 (1 + 푝) 푘 푖=푘 푖 2 푝푘1 = 푖=0 2 (1 + 푝)

1 + 푝 2푘+1 − 1 푘 = 1 푝 2푘

Let 퐻 푛 be the harmonic mean divisors of n. Then 푘1퐻 푛 = 휏(푛) 1+푝 2푘+1−1 Hence 휏 푛 = 퐻 푛 푝 2푘 1+푝 2푘+1−1 1 + 푘 2 = 퐻 푛 푝 2푘 1+푝 2푘+1−1 1 + 푘 = 퐻 푛 ...... (7) 푝 2푘+1 휏(푛) We observed that if 푝 = 2푘+1 − 1 then 퐻 푛 = . 2 With a small program in MATHEMATICA software it can be easily seen that some of the values of k +1 , for which 푝 = 2푘+1 − 1 is a prime number are {2,3,5,7,13,17,19,31,61,89,107,127,521,607,1279,2203,2281,3217,4253,4423,9689,9941}. However it has been observed that if 푝 = 2푘+1 − 1 is a prime number then 푘 + 1 itself must be a prime number. These

www.iosrjournals.org 41 | Page Some aspects of Harmonic Numbers which divide the sum of its positive divisors numbers are called Mersene primes and the numbers of the form 2푘 (2푘+1 − 1) are called perfect numbers. So far there are 48 Mersene primes . Theorem 3.2 There does not exist any harmonic number of the form 풏 = ퟐ풌풑ퟐ for which 품. 풄. 풅. 풏, 흈 풏 = 풏 ,where 풌 is a positive integer and 풑 is an odd prime. Proof: For 푛 = 2푘 푝2, we have 휎 푛 = 2푘+1 − 1 (1 + 푝 + 푝2). Then 휎(푛) 2푘+1−1 (1+푝+푝2) = 푛 2푘푝2 Now 2푘 ∤ (2푘+1 − 1) and 2푘 ∤ (1 + 푝 + 푝2), as 푝 is an odd prime. Hence 1 + 푝 + 푝2 becomes odd. Hence the theorem. The above theorem is possible as 1 + 푝 + 푝2 becomes odd. So this will happen till 푝푖 is odd for some values of 푖, in fact if 푖 is even. So we can state the following result:

Theorem 3.3 There does not exists any harmonic number of the form ퟐ풌풑ퟐ풙 of which 품. 풄. 풅. 풏, 흈 풏 = 풏 where 풌, 풙 are positive integers and 풑 is an odd prime. Theorem 3.4 There is no harmonic numbers up to ퟏퟎퟕퟔ of the form ퟐ풌풑ퟑ , where 풌 is a positive integer and 풑 is odd prime. Proof: Let 푛 = 2푘 푝3 . Then 휎 푛 = 2푘+1 − 1 1 + 푝 + 푝2 + 푝3 and 휏 푛 = 4(푘 + 1). As

𝑔. 푐. 푑. 푛, 휎 푛 = 푛 we may assume 휎 푛 = 푛푘1 where 푘1 is an integer. 2푘+1−1 (1+푝+푝2+푝3) Then 푘 = is also an integer. 1 2푘 푝3 Now 푝|(푝 + 푝2 + 푝3), if we assume 푝|(1 + 푝 + 푝2 + 푝3) then 푝|(1 + 푝 + 푝2 + 푝3) − (푝 + 푝2 + 푝3) i.e. 푝|1, which is not possible. Hence 푝 ∤ 1 + 푝 + 푝2 + 푝3 i.e. 푝3 ∤ (1 + 푝 + 푝2 + 푝3). Therefore the only possibility is 푝3|(2푘+1 − 1). As . 2푘 ∤ (2푘+1 − 1) so 2푘 |(1 + 푝 + 푝2 + 푝3). Let 푥푝3 = 2푘+1 − 1...... (8) And 1 + 푝 + 푝2 + 푝3 = 푦2푘 ...... (9) for some positive integer 푥, 푦. It is clear that for large value of 2푘+1 − 1 , if 푝 is comparatively very small then the equation (9) is not satisfied for 푦 ≥ 1. We have searched the numbers of the form 2푘+1 − 1 for 1 ≤ 푘 ≤ 251 in MATHEMATICA for which the value of 푝 has been found to be very very small. Hence all these numbers does not satisfy the second equation. Hence there is no number of the form 2푘 푝3 for 푘 ≤ 251.i.e. there is no harmonic numbers up to 1076 of the form 2푘 푝3.

Theorem 3.5 Let the number be of the form 풏 = ퟐ풌풑풒 and let 품. 풄. 풅. 풏, 흈 풏 = 풏 then there exists no 풏 흉(풏) such that 푯 풏 = , 풙 ≥ ퟏ ퟐ풙 푘 푘+1 Proof: For 푛 = 2 푝푞, we have 휎 푛 = 2 − 1 1 + 푝 (1 + 푞) and 휏 푛 = 4(푘 + 1). Let 휎 푛 = 푛푘1 where 푘1a positive integer. Therefore 2푘+1−1 1+푝 (1+푞) 푘 = 1 2푘푝푞 as 푘1퐻 푛 = 휏(푛) 1+푝 1+푞 2푘+1−1 1 + 푘 4 = 퐻 푛 ( )( ) 푝 푞 2푘 1+푝 1+푞 2푘+1−1 Or 1 + 푘 2 = 퐻 푛 ( )( ) 푝 푞 2푘+1 휏(푛) For 퐻 푛 = , 2 1 + 푝 1 + 푞 2푘+1 − 1 = 1 푝 푞 2푘+1 1+푝+푞+푝푞 2푘+1 Or = 푝푞 2푘+1−1 Or ( 2푘+1 − 1) 1 + 푝 + 푞 = 푝푞...... (10)

If 푝 = 푥(2푘+1 − 1) for 푥 > 1 then 푞 is a composite number. Same is true for 푞.Hence the only possibility is 푝 or 푞 must be of the form 2푘+1 − 1. But if 푝 = 2푘+1 − 1 then from the equation (10), we have 휏(푛) 1 + 푝 + 푞 = 푞 푖. 푒. 푝 = −1 is not possible. Same is true for q also. Thus 퐻 푛 = is not possible for all 푝 2 휏(푛) 휎(푛) and 푞 odd primes. Now we can consider the case of 퐻 푛 = , 푥 > 1. Then we have = 1.This implies 2푥 2푥 푛 that

www.iosrjournals.org 42 | Page Some aspects of Harmonic Numbers which divide the sum of its positive divisors

1 + 푝 1 + 푞 2푘+1 − 1 = 1 푝 푞 2푘+푥 1+푝+푞 2푘+1−1 + 1 = 1 푝푞 2푘+푥 1+푝+푞 2푘+1−1 2푘+1−1 = 1 − 푝푞 2푘+푥 2푘+푥 1 + 푝 + 푞 2푘+1 − 1 = 2푘+푥 − 2푘+1 + 1 푝푞 1 + 푝 + 푞 2푘+1 − 1 = 2푘+1 2푥−1 − 1 + 1 푝푞...... (11) For 푥 ≥ 2, 2푘+1 2푥−1 − 1 + 1 > (2푘+1 − 1). Also 푝푞 > 1 + 푝 + 푞 . Hence the RHS of the equation (11) is greater than the LHS , which is not possible. Hence the theorem.

흉(풏) Theorem 3.6 If a number is of the form 풏 = ퟐ풌풑ퟐ풒 with 품. 풄. 풅. 풏, 흈 풏 = 풏 and 푯 풏 = , then 풏 ퟐ does not exist. 푘 2 푘+1 2 Proof: For 푛 = 2 푝 푞 we have 휎 푛 = 2 1 + 푝 + 푝 (1 + 푞) and 휏 푛 = 3(푘 + 1)2. Let 휎 푛 = 푛푘1 where 푘1 is a positive integer. Then we have 푘+1 2 푘 2 2 − 1 1 + 푝 + 푝 1 + 푞 = 2 푝 푞푘1 2푘+1−1 1+푝+푝2 1+푞 Or = 푘 2푘 푝2 푞 1 푛휏(푛) 휏(푛) Again 퐻 푛 = = 휎(푛) 푘1 2푘+1−1 1+푝+푝2 1+푞 휏(푛) Or = 2푘 푝2 푞 퐻(푛) 2푘+1−1 1+푝+푝2 1+푞 Or 퐻(푛) = 3 1 + 푘 2 2푘 푝2 푞 2푘+1−1 1+푝+푝2 1+푞 Or 퐻(푛) = 3 1 + 푘 ...... (12) 2푘+1 푝2 푞 2푘+1−1 1+푝+푝2 1+푞 A necessary condition that 퐻(푛) is an integer. 2푘+1 푝2 푞 Suppose we want to make 퐻 푛 = 3(1 + 푘) 2푘+1−1 1+푝+푝2 1+푞 Then = 1 ...... (13) 2푘+1 푝2 푞 As 2푘+1 − 1, 1 + 푝 + 푝2 are odd numbers so only possibility is 2푘+1|1 + 푞. Let 푞 = 푥2푛 − 1, where 푛 ≥ 푘 + 1 If 푛 > 푘 + 1 then the expression (13) becomes 1+푝+푝2 푥2푦 2푘+1 − 1 = 1 for some 푦 ≥ 0 ...... (14) 푝2 푞 As 푝 and 푞 are odd primes the only possibility is 푦 = 0 i.e. 푛 = 푘 + 1 Hence the expression (14) becomes 1+푝+푝2 푥 2푘+1 − 1 = 1 ...... (15) 푝2 푞

For 푥 ≥ 1, 푞 does not divide 2푘+1 − 1, hence the only possibility is 푞|(1 + 푝 + 푝2) Let 푞푡 = 1 + 푝 + 푝2 for some positive integer 푡, then the expression (15) becomes 푡푥 2푘+1 − 1 = 1 ...... (16) 푝2 As 푥 > 1, and 2푘+1 − 1 > 1, the only possibility is 푡 = 1 and 푥 = 2푘+1 − 1 = 푝 Now 푞 = 1 + 푝 + 푝2 => 푥2푘+1 − 1 = 1 + 푝 + 푝2 => 푝2푘+1 − 푝 = 2 + 푝2 => 푝 2푘+1 − 1 = 2 + 푝2 i.e. 푝2 = 2 + 푝2 , which is not possible. 푡 Now , we consider 푥 = 1. then 푞 = 2푘+1 − 1. Therefore the equation (16) becomes 2푘+1 − 1 = 1 푝2 If 푡 = 1 then 푝2 = 2푘+1 − 1 = 푞.Therefore 푞푡 = 1 + 푝 + 푝2 implies 푞 = 1 + 푝 + 푝2 , which gives 푝 = −1 ,which is not possible. Next we consider that 푡 > 1. Again there are two possibilities, 푡 = 푝 and 푡 = 푝2. If 푡 = 푝2 then 2푘+1 − 1 = 1. This implies that 푘 = 0 which is not acceptable .Lastly we have 푡 = 푝. Then 2푘+1 − 1 = 푝 i.e. 푝 = 푞.So 푞푡 = 1 + 푝 + 푝2 implies 푝2 = 1 + 푝 + 푝2 i.e. 푝 = −1, not possible .Thus there exists no number of 휏(푛) the form 푛 = 2푘 푝2푞 such that H(n) = 3(푘 + 1) = . The above theorem may be generalized on the power of 2 of p and can be written as follows:

www.iosrjournals.org 43 | Page Some aspects of Harmonic Numbers which divide the sum of its positive divisors

Theorem 3.7 Let the number be of the form 풏 = ퟐ풌풑ퟐ풙풒 where 풙 is any positive integer and 풑, 풒 are odd 흉(풏) primes. Let 품. 풄. 풅. 풏, 흈 풏 = 풏 and 푯 풏 = , then 풏 does not exist. ퟐ 휏(푛) Proof: If 퐻 푛 = then we have 2 2푘+1−1 푖=2푥 푝푖 1+푞 푖=0 = 1 ...... (17 2푘+1 푝2푥 푞 푘+1 푖=2푥 푖 푘+1 As 2 − 1 and 푖=0 푝 are odd numbers, so 2 |1 + 푞. 푛 Let 푞 = 푥12 − 1, 푤ℎ푒푟푒 푛 ≥ 푘 + 1 If 푛 > 푘 + 1 then the expression (17) becomes 푖=2푥 푝푖 푥 2푦 2푘+1 − 1 푖=0 = 1 , for some 푦 ≥ 1 ...... (18) 푝2푥 푞 푘+1 푖=2푥 푖 푖=2푥 푖 For 푥 ≥ 1, 푞 does not divide 2 − 1, so the only possibility is 푞| 푖=0 푝 . Let 푞푡 = 푖=0 푝 , then the equation (18) becomes 푡푥 2푘+1 − 1 1 = 1 푝2푥 푘+1 푦1 푦2 푦3 The only possibility is 2 − 1 = 푝 , 푡 = 푝 , 푥1 = 푝 , 푤ℎ푒푟푒 푦1 + 푦2 + 푦3 = 2푥 and 푦푖 ≥ 0, 푖 = 1,2,3 푖=2푥 푖 Now 푞푡 = 푖=0 푝 푘+1 푦2 푖=2푥 푖 => (푥12 − 1)푝 = 푖=0 푝 푦2+푦3 푘+1 푦2+푦3 푦2 푖=2푥 푖 => 푝 2 − 푝 = 1 + 푝 + 푖=1,푖≠푦2+푦3 푝

푘+1 푦2+푦3 푦2 푖=2푥 푖 => 푝 (2 − 1) = 1 + 푝 + 푖=1,푖≠푦2+푦3 푝 푦2 푖=2푥−1 푖 => 0 = 1 + 푝 + 푖=1,푖≠푦2+푦3 푝 ...... (19) This is not possible .Hence the theorem. In theorem 3.6 we have seen that there is no harmonic number of the form 푛 = 2푘 푝2푞 such that 푛 completely divides 휎 푛 .However there are some harmonic numbers of this form such that 푛 does not divide completely 휎 푛 . Let 푞 = 2푘+1 − 1 . Then, we have 퐻 푛 휎(푛) = 휏(푛) 푛 2푘+1−1 1+푝+푝2 1+푞 i.e. 퐻(푛) = 3 1 + 푘 2푘+1 푝2 푞 1+푝+푝2 i.e. 퐻(푛) = 3 1 + 푘 푝2 As 𝑔. 푐. 푑. 푝2, 1 + 푝 + 푝2 = 1, the only possibility is 3 1 + 푘 = 푦(1 + 푝 + 푝2). Taking 푦 = 3 some of the numbers are 21232 213 − 1 , 23052 231 − 1 , taking 푦 = 1 some of the numbers are 2126192 2127 − 1 , 260132 261 − 1 , 21872 219 − 1 .

IV. Some properties of harmonic numbers having 품. 풄. 풅. 풏, 흈 풏 = 풏 of the form 풌 ퟐ 풑ퟏ풑ퟐ … … . 풑풎 풌 풌+ퟏ Theorem 4.1 Let the number be of the form 풏 = ퟐ 풑ퟏ풑ퟐ … … . 풑풎. Let one of them is ퟐ − ퟏ then 흈 풏 ≠ 풌풏, where 풌 = ퟐ풙, 풙 being odd and 풑풊 are odd prime, 풊 = ퟏ, ퟐ, … . . , 풎 푘+1 푚 푘+1 Proof: We have 휎 푛 = 2 − 1 푖=1(1 + 푝)푖 . Let 푝1 = 2 − 1. If possible, let 휎 푛 = 2푥푛 where 푥 is an odd number. We have 푘+1 2 −1 푚 1+푝푖 푘+1 푖=1 = 푥 2 푝푖 푚 1+푝푖 Or 푖=1 = 푥 ...... (20) 푝푖 This is not possible because the numerator part of LHS of (20) is even number and denominator part being odd the LHS becomes even. But the RHS is odd. This theorem can be generalized further to give the following theorem:

풌 Theorem 4.2 Let the number be of the form 풏 = ퟐ 풑ퟏ풑ퟐ … … . 풑풎, where all 풑’s are distinct odd primes, then 흈 풏 ≠ ퟐ풏. Proof: Without loss of generality, we consider that 푝1 < 푝2 … … … … … . 푝푚 . 푘+1 휎(푛) 2 −1 푚 1+푝푖 Now = 푘+1 푖=1 = 푘1(푠푎푦) ...... (21) 푛 2 푝푖 Where 푘is a positive integer. Now 푝푚 ∤ 1 + 푝푖 , ∀ 푖 = 1,2, … . . , 푚 as 푝푚 > 푝푖 , ∀푖 = 1,2,… . . , 푚 − 1. 푘+1 푘+1 Therefore only possibility is 푝푚 |2 − 1. Let 2 − 1 = 푟푚 푝푚, 푟푚 ≥ 1.

www.iosrjournals.org 44 | Page Some aspects of Harmonic Numbers which divide the sum of its positive divisors

푘+1 휎(푛) 2 −1 1+푝푚 푚−1 1+푝푖 So = 푘+1 푖=1 푛 2 푝푚 푝푖 푘+1 푟푚 푝푚 2 −1+푟푚 푚−1 1+푝푖 = 푘 푖=1 2 푟푚 푝푚 푝푖 1 푘+1 1+푝푚 −1 푚−2 1+푝푖 = 푘 (2 − 1 + 푟푚 ) . 푖=1 ...... (22) 2 푝푚 −1 푝푖 Now 푝푚−11 ∤ 1 + 푝푖 , ∀ 푖 = 1,2, … . . , 푚 − 1 as 푝푚−1 > 푝푖 , ∀푖 = 1,2, … . . , 푚 − 2. 푘+1 푘+1 So the possibility is 푝푚−1|2 − 1 + 푟푚 . Let 2 − 1 + 푟푚 = 푟푚−1푝푚−1, .Then the equation (22) can be written as 푘+1 휎(푛) 1 2 −1+푟푚 +푟푚 −1 푚−2 1+푝푖 = 푘 (푟푚−1푝푚−1) 푖=1 푛 2 푟푚 −1푝푚 −1 푝푖 Continuing this process at the 푗th stage , where the remaining primes are 푝1, 푝2 … … … … … . 푝푗 . 2푘+1−1+ 푚 푟 휎(푛) 1 푖=푗 푗 푗 −1 1+푝푖 We have = 푘 (푟푗 푝푗 ) 푖=1 푛 2 푟푗 푝푗 푝푖 푘+1 푚 and 푟푗 −1푝푗 −1 = 2 − 1 + 푖=푗 푟푗 . Therefore at the m-th stage we have 휎(푛) 1 = 2푘+1 − 1 + 푚 푟 ...... (23) 푛 2푘 푖=1 푖 휎(푛) If possible, let = 2, then the equation (23) becomes 2푘+1 − 1 + 푚 푟 =2푘+1 .So 푚 푟 = 1,which is 푛 푖=1 푖 푖=1 푖 not possible .Hence the result.

푘 In fact the above theorem helps us to search numbers of the form 2 푝1푝2 … … . 푝푚 ,where 푝1,푝2, … … . , 푝푚 are distinct odd primes and 푝1<푝2 <, … … . , < 푝푚 (say), which are harmonic of course 𝑔. 푐. 푑. 푛, 휎 푛 = 푛 .Following are the steps: Step 1. We chose the number 2푘+1 − 1 and get its factor.It is desired to have the prime factors whose power is one .However if a prime factor is 2 or ,say 푞푚 , 푚 ≥ 1 occurs which is also a factor of (푘 + 1), then it 푘+1 may be allowed. We take the highest prime say 푝푚 ,as stated in the theorem, let 2 − 1 = 푟푚 푝푚 푘+1 푘+1 Step 2. We calculate the prime factor of 2 − 1 + 푟푚 . In fact the prime factors of 2 − 1 except 푘+1 푝푚 will also be the prime factors of 2 − 1 + 푟푚 .Hence the prime factors are never lost at any stage. Again it is desired to have the prime factors whose power is one. However if a prime factor is 2 or ,say 푞푚 , 푚 ≥ 1 occurs which is also a factor of (푘 + 1),then it may be allowed. Step 3. We repeat the above process and at the j-th stage we consider the prime factors of 2푘+1 − 1 + 푚 푖=푗 푟푖 , again we choose the highest prime provided in the factorization all the prime factor other than prime 푘+1 푚 factors of (k+1) are square free. We end the process when 2 − 1 + 푖=푗 푟푖 has only prime factor 2.Harmonic 푗 mean H(n) of these numbers will be of the form 푞1푞2 … . . 푞푖 2 , for some 푗 where 푞1,푞2, … . . , 푞푖 are some of the prime factors of (k+1). Since the searching process is dependent only on 푘, it takes less time to search the numbers of the said form out of a wide range of numbers. It has been observed that there is no numbers of the said form whose harmonic mean is (푘 + 1)2푥 for some 푥 up to 10100.

References [1] D.Callan, Solution to Problem 6616, Amer.Math.Monthly 99 (1992),783-789,MR1542194 [2] G.L.Cohen, Numbers whose positive divisors have small integral harmonic mean. Mathematics of computation.Vol. 66, No.218,1997, PP 883-891. [3] G.L.Cohen and R.M.Soril. Odd harmonic number exceed 1024, Mathematics of Computation, S 0025- 5718(10)02337-9 [4] T. Goto and S.Shobata . All numbers whose positive divisors have integral harmonic mean up to 300, Mathematics of Computation, Vol. 73, No. 245, PP 473-491 [5] T. Goto and K.Okeya. All Harmonic numbers less than 1014, Japan J. Indust.Appl. Math.24(2007), 275-288 [6] M.Gracia. On numbers with integral harmonic mean, Amer. Math. Monthly, 61(1954), 89-96. MR 15:506d [7] R.K.Guy. Unsolved Problems in Number Theory, 3rd edition. Springer-Verlag, New York, 2004. [8] G.H.Hardy and E.M.Wright. An Introduction to the Theory of Numbers, fourth edition, Oxford (1962). MR0067125(16:673c) [9] O.Ore. On the averages of the divisors of a number, Amer.Math.Monthly, 55 (1948), PP 615-619 [10] C. Pomerance. On a problem of Ore: harmonic numbers, unpublished manuscript (1973) see abstract *709-A5, Notice Amer.Math.soc. 20(1973),a-648 [11] Solution to Problem E3445, Amer. Math.Monthly 99(1992),795 [12] J.Spiess. 1990, Some identities involving harmonic numbers , Math.Comp.55,839-863.