Advances in Theoretical and Applied Mathematics ISSN 0973-4554 Volume 12, Number 1 (2017), pp. 39-50 © Research India Publications http://www.ripublication.com

Perfect if and only if Triangular

Tilahun Muche, Mulatu Lemma, George Tessema and Agegnehu Atena

Department of Mathematics, Savannah State University, Savannah, GA, 31404, U.S.A.

Abstract

A number 푛 is perfect when 휎(푛) = ∑0≤푑≤푛 푑 = 2푛. It was Euclid who 푑|푛 proved that if (2푘 − 1) is a prime number, Mersenne prime, then 푁 = 2푘−1(2푘 − 1) is an even perfect number. Moreover, if 푁 is an even perfect

number then 푁 = 푇푚 for some 푚 ∈ ℕ and 푚 ≥ 3 is a triangular number 푚 where 푇푚 = ∑푖=1 푖 . In this paper we proved the necessary and sufficient condition for an even 푘−1 푘 triangular number 푇푚 to be a perfect number N= 2 (2 − 1)besides 푇푚 ≢ 4 푚표푑 10 and 푇푚 ≢ 2 푚표푑 10 .

Keywords: Perfect Numbers, Triangular Numbers and Mresenne Primes.

Mathematical subject Classification: 11B72, MSC 2010

INTRODUCTION A Perfect Number is a positive integer with the property that it coincides with the sum of all its positive divisors other than the number itself [1]. Thus, an integer 푛 ≥ 1 is a perfet number if

∑ 푑 = 푛 0≤푑<푛 푑|푛

40 Tilahun Muche, Mulatu Lemma, George Tessema and Agegnehu Atena

The nth triangular number is the number of dots composing a triangle with 푛 dots on a side, and is equal to the sum of the n natural numbers from 1 to n. [2]

푛 푛(푛 + 1) ∑ 푖 = 푖=1 2

Example 1: Triangular Numbers: 1, 3, 6,10,15,21,28,36,45. . . Perfect Numbers: 6, 28,496, 8126, 33550336, 8589869056,137438691328, 2305843008139952128, 2658455991569831744654692615953842176,... The number 6 is unique in that 6 = 1 + 2+ 3 where 1, 2 and 3 are all of the proper divisors of 6. The number 28 also shares this property, for 28 = 1+ 2 +4 +7 + 14. These perfect numbers have been a great deal of mathematical study- indeed, many of the basic theorems of numbers theory stem from the investigation of the Greeks into the problem of perfect and Pythagorean numbers. The Pythagoreans introduced the name perfect and there are speculations that there could be religious or astrological origins because the earth was created in 6 and the moon needs 28 days to circle the earth, mystical associations are natural. The early Hebrews also studied perfect numbers [3].

Definition 1: The sum of divisors is the function 휎(푛) = ∑푑|푛 푑,where 푑 runs over the positive divisors of 푛 including 1 and 푛 itself.

Definition 2: The number n is called perfect if 휎(푛) = 2푛, when 휎(푛) < 2푛 we say 푛 is deficient, 휎(푛) > 2푛 we say 푛 is abundant.

Example 2: 6 and 28 are perfect as 휎(6) = 1 + 2 + 3 + 6 = 12 = 2(6) and 휎(28) = 1 + 2 + 4 + 7 + 14 + 28 = 56 = 2(28).

Perfect if and only if Triangular 41

Euclid was the first mathematician who categorized even perfect numbers. He noticed that 6= 21. 31 = 21(22 − 1) 28 = 22. 7 = 22( 23 − 1) 496 = 16 . 31 = 24(25 − 1) 8126 = 64 . 127 = 26(27 − 1)

Theorem 3 (Euclid)[4,9]: If (2푛 − 1) is prime then 푁 = 2푛−1(2푛 − 1) is perfect. Proof: The only prime divisors of N are (2푛 − 1) and 2. Since (2푛 − 1) occurs as a single prime, we have that 휎(2푛 − 1)= (1 + (2푛 − 1))= 2푛, and thus 2푛−1 휎(푁) = 휎 (2푛−1)휎 (2푛 − 1) = ( ) 2푛 = 2푛(2푛 − 1) = 2. 2푛−1(2푛 − 1) = 2푁 2−1 So N is perfect.

Mersenne primes: Monk Martin Mersenne, a colleague of Descartes, Fermat and Pascal created with investigating these unique primes as early as 1644. He knew (2푛 − 1) is prime for n = 2, 3, 5, 7, 11, 13,17 and 19. [5, 6]

푃푛 Definition 4: A Mersenne prime is a prime number of the form 푀푛 = 2 − 1 where 푃푛 is a prime number.

Proposition 5:[5] (Cateldi – Fermat) If (2푛 − 1) is prime, then 푛 itself is prime. Proof: 푥푛 − 1= (푥 − 1) (푥푛−1+ . . . +푥 + 1). Suppose we can write 푛 = 푟푠 where 푟, 푠 > 1. Then 2푛 − 1 = (2푟)푠 − 1 = (2푟 − 1)((2푟)푠−1+ . . . + 2푟 + 1) so that (2푟 − 1)|(2푛 − 1) which is prime, a contradiction. ∎

Theorem 6: If 푁 is an even perfect number, then 푁 = 2푛−1(2푛 − 1) where(2푛 − 1) is prime. Proof: Since 2푛푚 = (2푛 − 1)휎(푚) , every prime divisor of (2푛 − 1) must also divide 푚, for it is odd and cannot divide 2푛 . So, suppose 푝훼 divides (2푛 − 1) with 푝 prime. 42 Tilahun Muche, Mulatu Lemma, George Tessema and Agegnehu Atena

휎(푎) 휎(푏) From the fact that if 푎|푏 , then ≤ where equality holds only if 푎 = 푏 we 푎 푏 have, 휎(푚) 휎(푝훼) 1+푝+...+ 푝∝ 푝∝−1+ 푝∝ 1+푝 ≤ = ≥ = . Hence 푚 푝훼 푝∝ 푝∝ 푝 휎(푁) 휎(2푛−1)휎(푚) (2푛−1)(1+푝) (2푛−1)−푝 1 = = ≥ = 1 + . 2푁 2푛푚 2푛푝 2푛푝 This is only satisfied when the fraction on the right is zero, so that 푝 = (2푛 − 1), ∝ = 1 and 푚 = 푝. Hence 푁 = 2푛−1(2푛 − 1). ∎

Proposition 7: Even perfect number ends in either 6 or 8. Proof: Every prime number 푝 ≥ 2 is of the form 푝 = 4푚 + 3 or 푝 = 4푚 + 1 . In the former case, 푁 = 2푛−1(2푛 − 1) = 24푚(24푚+1 − 1) = (16)푚(2. (16)푚 − 1) ≡ 6푚(2(6)푚 − 1) ≡ 6(푚표푑10) Since by induction one can show that 6푚 ≡ 6(푚표푑10) for all 푚 . Similarly in the latter case, 푁 = 2푛−1(2푛 − 1) = 24푚+2(24푚+3 − 1) = 4(16)푚(8. (16)푚 − 1) ≡ (4)(6)(8(6) − 1) ≡ 4(8 − 1) ≡ 8(푚표푑10). Finally, if 푛 = 2, 푁 = 22−1(22 − 1) = 6 and so we have the result that even perfect number ends in either 6 or 8. ∎

MAIN RESULTS 푚 푖 푛 푖 Let 퐴(푥) = ∑푖=0 푎푖 푥 and 퐵(푥) = ∑푖=0 푏푖 푥 . Then 퐴(푥)퐵(푥) = 퐶(푥) = 푛+푚 푘 ∑푘=0 푐푘 푥 where 푘 퐶푘 = ∑푖=0 푎푖 푏푘−푖 for 0 ≤ 푘 ≤ 푚 + 푛 .

The decimal expansion of a positive integer 푁, 0 ≤ 푎푘 < 10 where 푎0 is the unit digit of N is given by 푚 푚−1 1 0 푁 = 퐴(10) = 푎푚10 + 푎푚−110 + . . . + 푎110 + 푎010

= (푎푚푎푚−1푎푚−2 . . . 푎1푎0)10 푚 푖 = ∑푖=0 푎푖10

Perfect if and only if Triangular 43

푚 Theorem 8 [7]: A triangular number 푇푚 = ∑푖=1 푖 is even if and only if 푚 = (4푘 − 1) or 푚 = 4푘 for some 휖 ℤ+ .

Theorem 9: Even triangular numbers 푇푚 end not with 2 or 4. That is neither 푇푚 ≡

4 푚표푑 10 nor 푇푚 ≡ 2 푚표푑 10 .

Proof: Suppose 푇푚 is an even triangular number. Then either 푚 = (4푘 − 1)or = 4푘 . a) Suppose 푚 = 4푘 − 1. 4푘−1 This implies 푇푚 = 푇4푘−1 = ∑푖=1 푖 (4푘−1)( 4푘) = = 2푘( 4푘 − 1). 2 푚 푖 Let퐴 (10) = 2푘 = ∑푖=0 푎푖10 = (푎푚푎푚−1푎푚−2 . . . 푎1푎0)10 be decimal expansion of the factor(2푘) of an even triangular number 푇4푘−1 where the unit digit 푎0 ∈ { 0 , 2 , 4, 6, 8}.

Let 푏0 be the unit digit of the factor 퐵(10) = (4푘 − 1) of 푇푚 where the decimal expansion is 푛 푖 퐵(10) = 4푘 − 1 = ∑푖=0 푏푖10 = (푏푛푏푛−1 . . . 푏1푏0)10 .

Consider 푇푚 = 푇4푘−1 = (2푘)(4푘 − 1) = (2푘)( 2(2푘) − 1) = 퐴(10) 퐵(10) 푚 푖 푛 푖 = (∑푖=0 푎푖10 ) (∑푖=0 푏푖10 ) = 퐶(10) 푛+푚 푖 = ∑푖=0 푐푖 10 = (푐푚+푛푐푚+푛−1 … 푐1푐0)10

The constant term 푐0 of 퐶(10) = 푇푚 = 푇4푘−1 is 푐0 = 푎0푏0 .

We consider each unit digit 푎0 ∈ { 0 , 2 , 4, 6, 8} of 2푘 = 퐴(10) to determine unit digits 푏0 of

퐵(10) = (4푘 − 1) and 푐0 of 퐶(10).

1) 푎0= 0 ⇒ 푐0 = 0

2) 푎0 = 2, 푏0 = 2*2 – 1 = 3 ⇒ 푐0 = 6 1 3) 푎0 = 4, 푏0 = 2*4 – 1= 7 ⇒ 푎0푏0 = 푐0 = 8 (Because (4)*(7) = 28 = 2 ∗ 10 + 8 ∗ 100 ) 1 0 4) 푎0= 6, 푏0 = 1, because 2*6 – 1 = 11 = 1 ∗ 10 + 1 ∗ 10 ⇒ 푐0 = 6 1 0 5) 푎0= 8, the unit digit 푏0 = 5 because 2*8 – 1 = 1 5 = 1 ∗ 10 + 5 ∗ 10 ⇒ 푐0 = 0.

44 Tilahun Muche, Mulatu Lemma, George Tessema and Agegnehu Atena

b) For푚 = 4푘, in similar approach one can show that an even triangular number 푇푚 has the following sequence of unit digits.

Hence if a triangular number is even, then its unit digit is either 0, 6 or 8 but not 2 and 4. This implies

푇푚 ≢ 4 푚표푑 10 and 푇푚 ≢ 2 푚표푑 10 . ∎

Proposition 10 [8]: Every even perfect number ends in either 6 or 8.

Theorem 11: An even triangular number 푇4푘 for each 푘 ≥ 1cannot be written in the form of 2푛−1(2푛 − 1) for any 푛 ≥ 2 .

Perfect if and only if Triangular 45

4푘( 4푘+1) Proof: Suppose 푇 = = 2푘( 4푘 + 1) = 2푛−1(2푛 − 1) for some 푛 ≥ 4푘 2 2 and 푘 ≥ 1 . Then, 2푘( 4푘 + 1) = 2푛−1(2푛 − 1) iff 4푘2 + 푘 = 2푛−2(2푛 − 1) ⟺ 4푘2 + 푘 - 2푛−2(2푛 − 1) = 0 ⟺ (4푘 − ( 2푛 − 1))(푘 + 2푛−2)= 0 ⟺ 4푘 = 2푛 − 1 or 푘 + 2푛−2 = 0 2푛−1 ⟺ 푘 = (not an integer) or 푘 = − 2푛−2(not a positive integer) 22 ⇒ 푘 휖 ∅ → ← 푛−1 푛 Hence 푇4푘 ≠ 2 (2 − 1) for any 푘 ≥ 1 and 푛 ≥ 2 . ∎

Corollary 12: If an even triangular number 푇푚 is perfect, then 푚 = (4푘 − 1) for some 푘 ≥ 1 .

6 10 28 36 66 78 120 136 190 210 276 300 378 406

2*3 2*5 4*7 4*9 6*11 6*13 8*15 8*17 10*19 10*21 12*23 12*25 14*27 14*29

푡3 푡4 푡7 푡8 푡11 푡12 푡15 푡16 푡19 푡20 푡23 푡24 푡27 푡28 24−1 22−1 23−1 ∗ (24 ∗ (22 ∗ (23 − 1) − 1) − 1)

496 528 630 666 780 820 946 990 1128 1176 1326 1378 1540 1540

16*31 16*33 18*35 18*37 20*39 20*41 22*43 22*45 24*47 24*49 26*51 26*53 28*55 28*57

푡31 푡32 푡33 푡36 푡35 푡40 푡37 푡44 푡39 푡48 푡41 푡52 푡43 푡56 25−1 *(25 − 1)

Table III: Even Triangular Numbers with some in ퟐ풏−ퟏ(ퟐ풏 − ퟏ) form. 46 Tilahun Muche, Mulatu Lemma, George Tessema and Agegnehu Atena

푡 Theorem 13: An even triangular number 푇푚 is perfect if and only if 푚 = (2 − 1) for some prime number 푡 .

Proof: A triangular number 푇푚 is even if and only if either 푚 = (4푘 − 1) or 푚 = 4푘 for some 푘 ≥ 1 and an every prime number 푝 > 2 is of the form 푝 = (4푙 + 3) or 푝 = 4푙 + 1. [7]

Suppose 푇푚 is perfect. 푛−1 푛 푛 푛 This implies 푇푚 = 2 (2 − 1) and (2 − 1) is prime. But (2 − 1) prime only if 푛 is prime number. By (Theorem 11) and the later remark above the only choice for 푚 is 푚 = (4푘 − 1) = 4(푘 − 1) + 3 = 4푙 + 3 but not 푚 = 4푘. 푚 4푘−1 푛−1 푛 Hence, 푇푚 = ∑푖=1 푖 = ∑푖=1 푖 = (2푘)(4푘 − 1) = 2 (2 − 1) and 푛−1 푛 푛−1 푛 푇푚 = 2 (2 − 1) ⟺ ( 4푘 − 1)(2푘) = 2 (2 − 1) (1) It is easy to show that gcd(2푘 , 4푘 − 1) = gcd (2푛−1,, 2푛 − 1) = 1 = gcd(2푘 ,2푛 − 1) = gcd(2푛−1 , 4푘 − 1). (2) We use (2) to show that (4푘 − 1 ) = (2푛 − 1). Because (2푛 − 1)is prime, either gcd (4푘 − 1, 2푛 − 1) = 1 or (2푛 − 1)| (4푘 − 1 ). If the former is true, incorporating with what we have in (2), gcd(2푘 , 4푘 − 1) = gcd (2푛−1,, 2푛 − 1) = 1 = gcd(2푘 ,2푛 − 1) = 1 = gcd(2푛−1 , 4푘 − 1) = gcd (4푘 − 1, 2푛 − 1) = 1 and is clear to see that (4푘 − 1) = (2푛 − 1) and then 2푘 = 2푛−1. If the later holds, then there exists 푑 ∈ ℤ+ such that (4푘 − 1) = 푑(2푛 − 1). (3) As (4푘 − 1) and (2푛 − 1) are both odd, this implies 푑 is an odd integer too. Substituting (3) into (1) we have, ( 4푘 − 1)(2푘) = (2k) (2푛 − 1) 푑=2푛−1(2푛 − 1). This implies (2k) 푑 =2푛−1 and either 2푛−1 2푛−1 2푛−1 푑 = ∈ ℚ ⋀ 푑 = ∉ ℤ+or 푑 = ∈ ℤ+ and is an even integer for the only 2푘 2푘 2푘 factors of 2푛−1 are multiples of 2, which is a contradiction to 푑 is an odd integer.

Perfect if and only if Triangular 47

Hence(2푛 − 1)| (4푘 − 1 ) only when 푑 = 1 and hence (4푘 − 1) = (2푛 − 1) and 2푘 = 2푛−1.

Consequently, (2푘)( 4푘 − 1) = 2푛−1(2푛 − 1) ⟺ ( 4푘 − 1) = (2푛 − 1)and 2푘 = 2푛−1 ⟺ 4푘 = 2푛 and 푘 = 2푛−2 ⟺ 푘 = 2푛−2 for some prime 푛 . 2 푛−2 푛 Thus 푚 = 4푘 − 1 = 2 2 − 1 = (2 − 1), and if an even Triangular number 푇푚 is perfect, 푚 = (2푡 − 1) for some prime number 푡. ∎ Alternative proof: (2푘)(4푘 − 1) = 2푛−1(2푛 − 1) ⟺ 8푘2 − 2푘 = 2푛−1(2푛 − 1) ⟺ 4푘2 − 푘 = 2푛−2(2푛 − 1) ⟺ 4푘2 − 푘 - 2푛−2(2푛 − 1) = 0⟺ 4푘2 − 푘 - 2푛−2(2푛 − 1) = 0 ⟺ 4푘2 − 2푛푘 + (2푛 − 1)푘 - 2푛−2(2푛 − 1) = 0 ⟺ 4푘(푘 − 2푛−2) + (2푛 − 1)(푘 - 2푛−2) = 0 ⟺ (푘 − 2푛−2)(4푘 + 2푛 − 1) = 0 ⟺ 푘 = 2푛−2 or 4푘 = 1 − 2푛 ⟺ 푘 = 2푛−2 or 푘 = 22 − 2푛−2 = 2−2(1 − 2푛) 1−2푛 ⟺ 푘 = 2푛−2 or 푘 휖 ∅ (because (1 − 2푛) < 0, ∀ 푛 ≥ 3 and 푘 = ∉ ℤ+). 22 ⟹ 푘 = 2푛−2 . Consequently, 푚 = 4푘 − 1 = 4(2푛−2) − 1= (2푛 − 1), where 푛 is prime. ∎ 푡 Conversely, suppose 푇푚 is an even triangular number where 푚 = ( 2 − 1) for some prime number 푡 . 2푡−1 Then 푇푚 = 푇2푡−1 = ∑푖=1 푖 (2푡−1 )(2푡−1+1) = = (2푡−1)(2푡 − 1 ) = N and which is perfect. ∎ 2

Corollary 14: An even triangular 푇푀 is perfect if and only if 푀 is a Mersenne prime.

Theorem 15: An even triangular number 푇4푘−1 is not perfect if 푘 ≡ −1 푚표푑 5 or 푘 ≡ 0 푚표푑 5.

Proof: Consider an even triangular number 푇4푘−1. Suppose 푘 ≡ −1 푚표푑 5 or

48 Tilahun Muche, Mulatu Lemma, George Tessema and Agegnehu Atena

푘 ≡ 0 푚표푑 5 . Then 푘 ≡ −1 푚표푑 5 if and only if 푘 ≡ 4 푚표푑 5 if and only if 5 | (푘 − 4) if and only if 푘 = 4 + 5푡 for some 푡 ∈ ℤ+. ( 15+20푡)( 20푡+16) Hence 푇 = 푇 = 푇 = 4푘−1 4( 4+5푡)−1 15+20 푡 2 20(3+4푡)(4+5푡) = = 10( 3 + 4푡)( 4 + 5푡) and 10 | 푇 . 2 4푘−1

Consequently 푇4푘−1 is not a perfect number. (Proposition 10).

Similarly if 푘 ≡ 0 푚표푑 5 one can show that 10 | 푇4푘−1 which implies it is not perfect. ∎

Theorem 16 : If a triangular number 푇푚 is perfect , then 푛−1 (2 2 ) 푛 2 3 2 3 푇(2 −1) = 푇 푛+1 - 2 푇 푛−1 = ∑푘=1 (2푘 − 1) . (2 2 ) (2 2 )

푛 Proof: If a triangular number Tm is perfect then 푚 = (2 − 1) where 푛 is a prime number (Theorem 13).

푛(푛+1) 2 푇2 =∑푛 푖3 = ( ) . 푛 푖=1 2

푛+1 푛+1 2 푛+1 푛+1 2 (푛+1) ( ) (2 2 ) ( ) ( ) 2 (2 2 +1) 2 3 2 2 (2 2 +1) ⇒ 푇 푛+1 = ∑푖=1 푖 = ( ) = = (2 2 ) 2 4

푛+1 2 2푛−1 (2 2 + 1) and (4)

푛−1 푛−1 2 푛−1 푛−1 2 (푛−1) ( ) (2 2 ) ( ) ( ) 2 (2 2 +1) 2 3 2 2 (2 2 +1) 푇 푛−1 = ∑푖=1 푖 = ( ) = = (2 2 ) 2 4

푛−1 2 ( ) ( ) 2 푛−3 (2 2 + 1)

푛−1 2 푛−1 2 3 2 3 (푛−3) ( ) 푛 ( ) ⇒ 2 푇 푛−1 = 2 . 2 (2 2 + 1) = 2 (2 2 + 1) (5) (2 2 )

Perfect if and only if Triangular 49

Combining (4) and (5) we have,

푛+1 2 푛−1 2 2 2 (푛−1) ( ) 푛 ( ) 푇 푛+1 - 푇 푛−1 = 2 (2 2 + 1) - 2 (2 2 + 1) (2 2 ) (2 2 )

푛+1 2 푛−1 2 ( ) ( ) = (2푛−1) ((2 2 + 1) − 2 (2 2 + 1) )

푛+1 푛−1 ( ) ( ) = (2푛−1) ((2푛+1 + 2. 2 2 + 1) − 21( 2푛−1 + 2. 2 2 + 1))

푛+3 푛+1 ( ) ( ) = (2푛−1) ((2푛+1+. 2 2 + 1 ) - 21( 2푛−1 + 2 2 + 1)) 푛+3 푛+3 ( ) ( ) = (2푛−1) (2푛+1 + 2 2 + 1 − 2푛 − 2 2 − 2) = (2푛−1)( 2푛+1 − 2푛 − 1) = (2푛−1)( 2.2푛 − 2푛 − 1) 푛 푛 푛−1 푛 2 (2 −1) = (2 )( 2 − 1) = = 푇 푛 ∎ (6) 2 (2 −1)

푛−1 (2 2 ) 푛 3 Next we show that, 푇(2 −1) = ∑푖=1 (2푖 − 1) . 푛+1 푛−1 (2 2 ) (2 2 ) 2 2 푛 3 3 3 3 푇(2 −1) = 푇 푛+1 - 2 푇 푛−1 = ∑푖=1 (푖) - 2 ∑푖=1 (푖) (2 2 ) (2 2 )

푛+1 3 푛−1 3 ( ) ( ) = (12 + 23 + 33 + … + (2 2 ) ) - 23 (12 + 23 + 33 + … + (2 2 ) )

푛+1 3 푛+1 3 ( ) ( ) = (12 + 23 + 33 + … + (2 2 ) ) - (23 + 43 + 63 + … + (2 2 ) )

푛+1 3 푛+1 3 푛+1 3 ( ) = (12 + 23 + 33 + 43 + 53 + 63 + 73 + 83 + ⋯ + (2 2 − 2) + (2 2 − 1) + (2 2 ) )

푛+1 3 푛+1 3 - (23 + 43 + 63 + 83 … + (2 2 − 2) + (2 2 ) )

푛+1 3 푛−1 3 ( ) ( ) = (12 + 33 + 53 + 73 + ⋯ + (2 2 − 1) ) = (12 + 33 + 53 + 73 + ⋯ + (2. 2 2 − 1) )

푛−1 ( ) = (12 + 33 + 53 + 73 + ⋯ + (2푘 − 1)3) where 푘 = 2 2 푛−1 (2 2 ) 3 = ∑푘=1 (2푘 − 1) 푛−1 ( ) 2 2 3 This implies 푇2푛−1 = ∑푘=1 (2푘 − 1) . ∎ (7) 50 Tilahun Muche, Mulatu Lemma, George Tessema and Agegnehu Atena

From (6) and (7) it follows, 푛−1 (2 2 ) 푛 2 3 2 3 푇(2 −1) = 푇 푛+1 - 2 푇 푛−1 = ∑푘=1 (2푘 − 1) . ∎ (2 2 ) (2 2 )

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