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Generalized Perfect Numbers COMMENTATIONES MATHEMATICAE Vol. 53, No. 1 (2013), 61-72 D.P. Shukla, Shikha Yadav Generalized Perfect Numbers 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.
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