Generalized Perfect Numbers

COMMENTATIONES MATHEMATICAE Vol. 53, No. 1 (2013), 61-72

D.P. Shukla, Shikha Yadav

Abstract. In this paper a modified form of perfect numbers called (p, q)+ perfect numbers and their properties with examples have been discussed. Further properties of σ+ arithmetical function have been discussed and on its basis a modified form of perfect number called (p, q)+ super perfect numbers have been discussed. A modified form of perfect number called (p, 0)-perfect and their characterization has been stu- died. In the end of this paper almost super perfect numbers have been introduced. 2000 Mathematics Subject Classification: 11A25. Key words and phrases: Prime numbers, perfect numbers, super perfect numbers, Arithmetical Functions .

1. (p, q)+perfect numbers. In J. Sandor and K. Atanassov [4] (p, q) perfect numbers with their properties and examples have been discussed. G.L. Cohen H.J.J. te Ride [1] called the natural number n , (p, q) perfect number if

(1.1) σp(n) = qn, where σp(n) = σ(σ(...(σ(n)...))).

p times − In this paper we will discuss one more| {z type} of modiﬁed perfect numbers called (p, q)+perfect numbers. A natural number n is called (p, q)+perfect numbers if

p (1.2) σ+(n) = qn, where p σ+(n) = σ+(σ+(...(σ+(n)...))). p times − | {z } 62 Generalized Perfect Numbers

Here, σ+ denotes the sum of all even divisors of n. Since, σ+(n) denotes the sum of all even divisors of n therefore, σ+(n) = 0 if n is odd. Hence (p, q)+ perfect numbers are deﬁned for only even numbers. Further, if n is even, n = 2k.N, k 1 with N odd, then σ (n) = 2(2k 1)σ(N). + −

Let n = 2, as σ+(2) = 2 so, σ+(σ+(2)) = 2 . Now take p = 3 σ+(σ+(σ+(2))) = σ+(σ+(2)) = 2,

for p = 4, σ+(σ+(σ+(σ+(2)))) = 2. By induction, we can see that

σp (2) = 2 = 1 2, for all p 1 + × Hence n = 2 is the solution for (p, 1)+ perfect numbers Some examples of (p, q)+ perfect numbers for p 21 have been given in the tables as follows ¬

Let n = 4 Let n = 6 Table -1.1 Table -1.2 p p p σ+ q n p σ+ q n 1 6 - × 1 8 - × 2 8 2 4 2 14 - 3 14 - × 3 16 - 4 16 4 4 4 30 5 6 5 30 - × 5 48 8×6 6 48 12 4 6 120 20× 6 7 120 30×4 7 336 56×6 8 336 84×4 8 960 160× 6 9 960 240× 4 9 3024 504×6 10 3024 756×4 10 9600 1600× 6 11 9600 2400× 4 11 31496 - × 12 31496 7874×4 12 57344 - 13 57344 14336× 4 13 131056 - 14 131056 32764×4 14 245760 40960 6 15 245760 61440×4 15 786384 131064× 6 16 786384 196596× 4 16 2130180 355030×6 17 2130180 532545×4 17 5507712 917952×6 18 5507712 1376928× 4 18 18068544 3011424× 6 19 18068544 4517136×4 19 58705920 9784320×6 20 58705920 14676480× 4 20 110046720 18341120× 6 21 110046720 27511680×4 21 423843840 70640640×6 × × D.P. Shukla, S. Yadav 63

Let n = 8 Let n = 10 Table -1.3 Table-1.4 p p p σ+ q n p σ+ q n 1 14 - × 1 12 - × 2 16 2 8 2 24 - 3 30 - × 3 56 - 4 48 6 8 4 112 - 5 120 15× 8 5 240 24 10 6 336 42×8 6 720 72×10 7 960 120× 8 7 2340 234× 10 8 3024 378×8 8 6552 - × 9 9600 1200× 8 9 20384 10 31496 3937×8 10 49476 -− 11 57344 7168×8 11 122880 12288 10 12 131056 16382× 8 12 393168 - × 13 245760 30720×8 13 983040 98304 10 14 786384 98298×8 14 3145680 314568× 10 15 2130180 2662725× 8 15 10866960 1086696× 10 16 5507712 688464 ×8 16 40360320 4036032×10 17 18068544 2258568× 8 17 189719712 - × 18 58705920 7338240×8 18 711065600 71106560 10 19 110046720 13755840× 8 19 2080358912 - × 20 423843840 52980480×8 20 4286578688 - × Let n=12 Let n=14 Table=1.5 Table = 1..6

p p p σ+ q n p σ+ q n 1 24 2 ×12 1 16 - × 2 56 - × 2 30 - 3 112 - 3 48 - 4 240 20 12 4 120 - 5 720 60×12 5 336 24 14 6 2340 195× 12 6 960 - × 7 6552 546×12 7 3024 216 14 8 20384 - × 8 9600 - × 9 49476 4123 12 9 31496 10 122880 10240× 12 10 57344 4096− 14 11 393168 32764×12 11 131056 - × 12 983040 81920×12 12 245760 - 13 3145680 262140× 12 13 786384 - 14 10866960 905580×12 14 2130180 - 15 40360320 3363360× 12 15 5507712 393408 14 16 189719712 15809976× 12 16 18068544 - × 17 711065600 - × 17 58705920 4193280 14 18 2080358912 - 18 110046720 7860480×14 19 4286578688 - 19 423843840 30274560× 14 × 64 Generalized Perfect Numbers

From the tables given above it is clear that all (2,2)+ perfect numbers coincide with + super perfect numbers and (1,2)+ perfect numbers coincide with + perfect numbers. Before discussing the properties of (p, q)+ perfect numbers let us discuss some properties of σ and σ+ given as follows For all positive integers n 2 one has (1.3) σ(n) n + 1, with equality only for n =prime. Similarly

(1.4) σ(n) n, with equality only for n = 1. For all positive integers m, n

(1.5) σ (mn) mσ (n), + + with equality only for m = 1, m and n both odd.

(1.6) σ (mn) σ (m)σ (n), + + + with equality only for m and n both odd. For an even positive integer n

(1.7) σ (n) n, + with equality only for n = 2, if n > 2 then,

(1.8) σ+(n) > n + 1.

Theorem 1.1. For every n 4 σ (σ (n)) 2n + +

Proof. Let n = 2, then σ (σ (2)) = 2 < 2 2 = 4. + + × Take n 4. Since, σ+(n) = 0 for n is odd therefore, n is always even.

If n is even then n =2k.m, with m is odd. Now k k k σ+(σ+(2 .m)) = σ+(2(2 1)σ(m)) σ(m)σ+(2(2 1)) (using 1.5). So, using (1.3) and (1.5),− we have −

σ(m)σ (2(2k 1) = 2σ(m)σ(2k 1) 2mσ(2k 1) 2m2k = 2n. + − − − D.P. Shukla, S. Yadav 65

Hence σ (σ (n)) 2n, for every n 4. + + Now using theorem 1.1, we have

σ (σ (σ (σ (n)))) 2σ (σ (n)) 2.2n = 4n. + + + + + + So,

(1.9) σ4 (n) 4n. + Similarly, using inequation (1.9) and theorem 1.1

σ6 (n) = σ (σ (σ (σ (σ (σ (n)))))) 4(σ (σ (n)) 4.2n = 8n > 6n. + + + + + + + + + So,

6 (1.10) σ+(n) > 6n. Therefore by induction we can prove that

Theorem 1.2. For every even positive integer p 2 and n 4 σp (n) pn +

Theorem 1.3. n is (2, 2)+ perfect number if and only if it has the form n = 2k, where 2k 1 is a prime. Proof. It− is clear that (2, 2)+ perfect numbers coincide with the + super perfect numbers. In J. Sandor and E. Egri [2] it has been proved that n is + super perfect if and only if it has the form n = 2k, where 2k 1 is a prime. Hence theorem is proved. −

2. (p, q)+ super perfect numbers. J. Sandor and K. Atanssov [3] have deﬁned (p, q) super perfect numbers as follows- Given two positive integers p and q, a number n N is said to be (p, q) super perfect if it satisﬁes ∈

(2.1) σ(pσ(n)) = qn,

where σ(n) is a divisor function which is the sum of diﬀerent divisors of n. n N, n > 1. ∈ σ(1) = 1 and

(2.2) σ(n) = d, Xd/n 66 Generalized Perfect Numbers for all n 1, σ(n) n and σ(n) n + 1 for n 2,with equality holds only for n is prime. Let σ+(n) denotes the sum of all even divisors of n. We say that a positive integer n is called + perfect number if

(2.3) σ+(n) = 2n.

J. Sandor and E. Egri [2] have proved a result deﬁned as follows

k Lemma 2.1: If n is odd, then σ+(n) = 0. If n is even, n = 2 N, k 1 with N odd then

σ (n) = 2(2k 1)σ(N). + − A positive integer n is called + super perfect if

(2.4) σ+(σ+(n)) = 2n, for two positive integers p and q we have deﬁned (p, q)+super perfect numbers as follows Given two positive integers p and q a number n N is said to be (p, q)+ super perfect number if it satisﬁes. ∈

(2.5) σ+(pσ+(n)) = qn.

Theorem 2.2 The single (1, 1)+ super perfect number is n = 2 Proof.Let n is (1, 1)+super perfect number then by equation, (2.5)

σ+(σ+(n)) = n, using (1.7)

σ (σ (n)) σ (n). + + + So,

(2.6) n σ (n). + Further, using (1.7) we obtain

(2.7) σ (n) n. + From (2.6) and (2.7) we get, D.P. Shukla, S. Yadav 67

σ+(n) = n. Since equality in (1.7), holds only for n = 2. Hence n = 2 is single (1,1)+super perfect number.

Theorem 2.3. If n is (p, q)+ super perfect number then n = 2kN be an even positive integer where N is odd positive integer and k 1 Proof. Let n is an odd (p, q)+ super perfect number. Then By (2.5), we have

σ+(pσ+(n)) = qn.

By Lemma (2.1), σ+(n) = 0. So, pσ+(n) = 0, which contradicts the fact that σ+is an arithmetical function. Therefore, our assumption is wrong. Hence n is an even (p, q)+ super perfect number.

Theorem 2.4. Let n = 2kN with N odd & N 1, where 2k 1 is prime, is a (p, q)+ super perfect number then q (p + 1). Further, n = 2k −is the solution for (1,2)+super perfect numbers. Proof. Let n = 2kN be an even positive integer, if n is a (p, q)+super perfect number then using (2.5), we obtain

k k σ+(pσ+(2 N)) = q.2 .N, using Lemma (2.1), we have

σ (pσ (2kN)) = σ (p2(2k 1)σ(N)), + + + − so, using (1.5), we obtain

σ (p2(2k 1)σ(N)) pσ (2(2k 1)σ(N)) p.σ(N)σ (2(2k 1)). + − + − + − Further, using (1.4) and Lemma (2.1), we obtain

p.σ(N)σ (2.(2k 1)) p.Nσ (2(2k 1)) = p.N.2σ(2k 1) = 2p.N.2k, + − + − − where 2k 1 is odd. So, − q 2p > p + 1 Hence q p + 1 Further, if p = 1 and q = 2 then using (2.5), we get

σ+(σ+(n)) = 2n, using (2.4), we have 68 Generalized Perfect Numbers

n is a +super perfect number. Therefore, by following theorem, we get

Theorem 2.5. n is + super perfect if and only if it has the form n = 2k, where 2k 1 is a prime. − n = 2k. Hence n = 2k is the solution for (1,2) +super perfect numbers.

Remark 2.6 All + super perfect numbers are (1,2) + super perfect numbers. For example n = 4 and n = 8...... are (1,2) + super perfect numbers. as

σ+(σ+(4)) = σ+(6) = 8 = 2.4 and

σ+(σ+(8)) = σ+(14) = 16 = 2.8. Similarly, solution for (2,3); (3,4); (5,6)...... + super perfect numbers is n = 2

3. (p, 0) perfect number. If p is a prime number then σp,0(n) denotes the sum of all divisors− of n which are 0( mod p). For all prime p, σ (n) is deﬁned as ≡ p,0 d k σp,0(n) = p Q for n = p t (3.1) 1 d k . ¬XQ/t¬  = 0 for n = 1 

A positive integer n is called (p, 0)-perfect number if 

(3.2) σp,0(n) = 2n.

For p = 2, σ2,0(n) is deﬁned as the sum of all even divisors of n. So one can write σ2,0(n) = σ+(n). Hence +perfect numbers are also called (2, 0)-perfect numbers. For p = 3, σ3,0(n) is deﬁned as the sum of all divisors of n which are 0( mod 3).In J. Sandor and E.Egri [2] properties of (2,0)-perfect numbers and (3,0)-≡ perfect numbers have been given. In this section we will discuss some properties of (p, 0)-perfect numbers for a prime p given as following

k (p 1) k Lemma 3.1. σp,0(n) = p. (p −1) σ(t) for n = p t and p - t. Proof. Using (3.1), we get−

d 2 k σp,0(n) = p Q = (p + p + .....p )σ(t) 1 d k ¬XQ/t¬ (pk 1) = p. − σ(t). (p 1) − D.P. Shukla, S. Yadav 69

Theorem 3.2. A positive integer n is a (p, 0)-perfect number if (1) n = pt, where t is a perfect number and p - t, k pk 1 (2) n = p p −1 s for k odd where p - q and − k p 1 k 1 σ − s = 2p − s, p 1 − k k p 1 0 0 (3) n = p 2(p−1) s for k even where p - q and. − k p 1 k 1 σ − s0 = p − s0. 2(p 1) −

Proof. Let n = pkt, where p - t is a (p, 0)-perfect number. Using Lemma 3.1, we have

k k 1 (3.3) (p 1)σ(t) = 2(p 1)p − t. − − Case I. Let k = 1, then using (3.3)

σ(t) = 2t.

Hence t is a perfect number not divisible by p. Case II. Let k 2. If k is odd, using (3.3), we must have 2(p 1)t = (pk 1)q and it is clear that (p 1) (pk 1). So there exist positive integers− q and s−such p−k 1 || − that q = 2s and t = p −1 s. Hence using (3.3), we have − k p 1 k 1 σ − s = 2p − s p 1 − and pk 1 n = pk − s. p 1 − Case III. If k is even then 2(p 1)/(pk 1), so − − pk 1 t = − s0. 2(p 1) − Hence k k p 1 n = p − s0 2(p 1) − and k p 1 k 1 σ − s0 = p − s0. 2(p 1) − 70 Generalized Perfect Numbers

For n = pkt, where t 3355036 a perfect number, k 1, a positive integer, 3 p 11, a prime number,¬ some (p, 0) perfect numbers have been described in the¬ following¬ tables- Table 3.1. Let p = 3 t n = pt σp,0(n) 6 18 36 28 84 168 496 1488 2976 8128 24384 48768 33550336 100651008 201302016 Table 3.2. Let p = 5 t n = pt σp,0(n) 6 30 60 28 140 280 496 2480 4960 8128 40640 81280 33550336 167751680 335503360 Table 3.3. Let p = 7 t n = pt σp,0(n) 6 42 84 28 196 392 496 3472 6944 8128 56896 113792 33550336 234852352 469704704 Table 3.4. Letp = 11 t n = pt σp,0(n) 6 66 132 28 308 616 496 5456 10912 8128 89408 178816 33550336 369053696 738107392 Let n = pkt, where p - t. Let p = 3, then for k = 2 and t = 2, n = 18. From table 3.1 it can be seen that n = 18 is the only solution of (3,0)-perfect number. Let p = 7 and k = 2, using (3), of Theorem 3.2, we get

σ(4s0) = 7s0

and

σ(4s0) s0σ(4) = 7s0, with equality holds for only s0 = 1. so t = 4 and n = 72.4 = 196. Hence for k = 2, n = 196 is the only solution of (7,0)-perfect number. It has been deﬁned in J. Sandor and E. Egri [2] that for k > 2 even or odd however, the study of (3,0)-perfect numbers remains open. The problem of solution of (7,0)-perfect numbers is still open for k > 2. Further this problem is also open for all other primes and k 2. D.P. Shukla, S. Yadav 71

4. Almost Super Perfect numbers . Mladen V. Vassilev -Missana and Krassimir T. Atanassov [5] introduced almost perfect numbers defined as follows A number n N is called almost perfect number if ∈ σ(n) = 2n 1. − It is also known as a łeast deficientór Ślightly defective numbers”. The only known almost perfect numbers are the power of 2 i.e. 1,2,4...... Now we have defined almost super perfect numbers as follows- A number n N is called almost super perfect number”if ∈ , σ(σ(n)) = 2σ(n) 1. −

Theorem 4.1: If n is a Mersenne prime or product of distinct Mersenne primes then n is almost super perfect number. k Proof. Let n = p1.p2....pr, where p1, p2, ..., pr are of the form 2 1 is a Mersenne prime. − Now

σ(σ(n)) = σ(σ(p1p2...pr)) = σ(σ(p1)σ(p2)...σ(pr)) = σ(2k1 .2k2 ....2kr ) = 2k1+k2+.....+kr +1 1 − = 2.(2k1+k2+.....+kr ) 1 = 2σ(p p ...p ) 1 − 1 2 r − = 2σ(n) 1. − Similarly, if n = p, where p = 2k 1 is a Mersenne prime, then − σ(σ(n)) = σ(σ(p)) = σ(2k) = 2k+1 1 = 2σ(n) 1. − − Hence n is an almost super perfect number.

Remark 4.2 If n N is product of distinct Mersenne primes then it is almost super perfect, M perfect,∈ modiﬁed e-perfect and e-harmonic of both types. Further, n = 8 is a nobly deﬁcient number, almost perfect number and multiplicatively perfect number.

References

[1] G. H. Hardy and E. M. Wright, An introduction to the theory of numbers, Oxford at the Clarendon Press, 1979.

[2] J. Sandor and E. Egri, Arithmetical functions in algebra, geometry and analysis, Advanced studies in contemporary Mathematics, Vol.14 (2007), No.2, 163-213.

[3] J. Sandor and K. Atanassov, On (m, n)-super perfect numbers, Advanced studies in contemporary Mathematics, Vol. 16 (2008), No.1, 34-45. 72 Generalized Perfect Numbers

[4] J. Sandor and K. Atanassov, On a modiﬁcation of perfect numbers, Advanced studies in Contemporary mathematics 17(2008), No.2, pp. 249-255.

[5] Mladen V. Vassilev- Missana and Krassimir T. Atanassov, A new point of view on perfect and other similar numbers, Advanced studies in contemporary mathematics 15 (2007), No.2, pp.153-169.

D.P. Shukla Department of Mathematics & Astronomy, Lucknow University Lucknow 226007 Shikha Yadav Department of Mathematics & Astronomy, Lucknow University Lucknow 226007

(Received: 28.03.2013)