Generalized Perfect Numbers

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) σ ^p (n) = qn,

where

σ ^p (n) = σ(σ(...(σ

| {z } (n)...))

p −times

).

In this paper we will discuss one more type of modified perfect numbers called (p, q)+perfect numbers.

A natural number n is called (p, q)+perfect numbers if

(1.2) σ ₊ ^p (n) = qn,

where

σ ^p ₊ (n) = σ + (σ + (...(σ +

| {z } (n)...))

p−times

).

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 = 2 ^k .N, k ­ 1 with N odd, then σ + (n) = 2(2 ^k − 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

σ ^p ₊ (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 σ ₊ ^p q × n

1 6 -

2 8 2×4

3 14 -

4 16 4×4

5 30 -

6 48 12×4

7 120 30×4

8 336 84×4

9 960 240×4

10 3024 756×4

11 9600 2400×4

12 31496 7874×4

13 57344 14336×4

14 131056 32764×4 15 245760 61440×4 16 786384 196596×4 17 2130180 532545×4 18 5507712 1376928×4 19 18068544 4517136×4 20 58705920 14676480×4 21 110046720 27511680×4

p σ ^p ₊ q × n

1 8 -

2 14 -

3 16 -

4 30 5×6

5 48 8×6

6 120 20×6

7 336 56×6

8 960 160×6

9 3024 504×6

10 9600 1600×6

11 31496 -

12 57344 -

13 131056 -

14 245760 40960×6

15 786384 131064×6

16 2130180 355030×6

17 5507712 917952×6

18 18068544 3011424×6

19 58705920 9784320×6

20 110046720 18341120×6

21 423843840 70640640×6

Let n = 8 Let n = 10

Table -1.3 Table-1.4

p σ ₊ ^p q × n

1 14 -

2 16 2×8

3 30 -

4 48 6×8

5 120 15×8

6 336 42×8

7 960 120×8

8 3024 378×8

9 9600 1200×8

10 31496 3937×8

11 57344 7168×8

12 131056 16382×8 13 245760 30720×8 14 786384 98298×8 15 2130180 2662725×8 16 5507712 688464×8 17 18068544 2258568×8 18 58705920 7338240×8 19 110046720 13755840×8 20 423843840 52980480×8

p σ ^p ₊ q × n

1 12 -

2 24 -

3 56 -

4 112 -

5 240 24×10

6 720 72×10

7 2340 234×10

8 6552 -

9 20384 −

10 49476 -

11 122880 12288×10

12 393168 -

13 983040 98304×10 14 3145680 314568×10 15 10866960 1086696×10 16 40360320 4036032×10 17 189719712 -

18 711065600 71106560×10 19 2080358912 -

20 4286578688 -

Let n=12 Let n=14

Table=1.5 Table = 1..6

p σ ₊ ^p q × n

1 24 2×12

2 56 -

3 112 -

4 240 20×12

5 720 60×12

6 2340 195×12

7 6552 546×12

8 20384 -

9 49476 4123 × 12

10 122880 10240×12 11 393168 32764×12 12 983040 81920×12 13 3145680 262140×12 14 10866960 905580×12 15 40360320 3363360×12 16 189719712 15809976×12 17 711065600 -

18 2080358912 - 19 4286578688 -

p σ ^p ₊ q × n

1 16 -

2 30 -

3 48 -

4 120 -

5 336 24×14

6 960 -

7 3024 216×14

8 9600 -

9 31496 −

10 57344 4096×14

11 131056 -

12 245760 -

13 786384 -

14 2130180 -

15 5507712 393408×14 16 18068544 -

17 58705920 4193280×14

18 110046720 7860480×14

19 423843840 30274560×14

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 =2 ^k. m, with m is odd.

Now

σ + (σ + (2 ^k .m)) = σ + (2(2 ^k − 1)σ(m)) ­ σ(m)σ ⁺ (2(2 ^k − 1)) (using 1.5).

So, using (1.3) and (1.5), we have

σ(m)σ + (2(2 ^k − 1) = 2σ(m)σ(2 ^k − 1) ­ 2mσ(2 ^k − 1) ­ 2m2 ^k = 2n.

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) σ ₊ ⁴ (n) ­ 4n.

Similarly, using inequation (1.9) and theorem 1.1

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

So,

(1.10) σ ₊ ⁶ (n) > 6n.

Therefore by induction we can prove that

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

Theorem 1.3. n is (2, 2)+ perfect number if and only if it has the form n = 2 ^k , where 2 ^k − 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 = 2 ^k , where 2 ^k − 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) = X

d/n

d,

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

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

σ + (n) = 2(2 ^k − 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,

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 = 2 ^k N 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 = 2 ^k N with N odd & N ­ 1, where 2 ^k − 1 is prime, is a (p, q)+ super perfect number then q ­ (p + 1). Further, n = 2 ^k is the solution for (1,2)+super perfect numbers.

Proof. Let n = 2 ^k N be an even positive integer, if n is a (p, q)+super perfect number then using (2.5), we obtain

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

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

σ + (p2(2 ^k − 1)σ(N)) ­ pσ ⁺ (2(2 ^k − 1)σ(N)) ­ p.σ(N)σ ⁺ (2(2 ^k − 1)).

Further, using (1.4) and Lemma (2.1), we obtain

p.σ(N )σ + (2.(2 ^k − 1)) ­ p.Nσ ⁺ (2(2 ^k − 1)) = p.N.2σ(2 ^k − 1) = 2p.N.2 ^k , where 2 ^k − 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

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 = 2 ^k , where 2 ^k − 1 is a prime.

n = 2 ^k .

Hence n = 2 ^k 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 σ ^p,0 (n) denotes the sum of all divisors of n which are ≡ 0( mod p). For all prime p, σ ^p,0 (n) is defined as

(3.1)

σ p,0 (n) = X

1¬d¬k

Q/t

p ^d Q for n = p ^k t

= 0 for n = 1

 



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

(3.2) σ p,0 (n) = 2n.

For p = 2, σ 2,0 (n) is defined as the sum of all even divisors of n. So one can write σ 2,0 (n) = σ + (n). Hence +perfect numbers are also called (2, 0)-perfect numbers.

For p = 3, σ 3,0 (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

Lemma 3.1. σ p,0 (n) = p. ^(p _(p−1)

⁻¹⁾ σ(t) for n = p ^k t and p - t.

Proof. Using (3.1), we get

σ p,0 (n) = X

1¬d¬k

Q/t

p ^d Q = (p + p ² + ...p ^k )σ(t)

= p. (p ^k − 1)

(p − 1) σ(t).

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,

(2) n = p ^k 

p

−1 p −1

 s for k odd where p - q and

σ

 p ^k − 1 p − 1 s



= 2p ^k−1 s,

(3) n = p ^k 

p

−1 2(p−1)

 s

for k even where p - q

and.

σ

 p ^k − 1 2(p − 1) s ⁰



= p ^k−1 s ⁰ .

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

(3.3) (p ^k − 1)σ(t) = 2(p − 1)p ^k−1 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 = (p ^k − 1)q and it is clear that (p − 1)||(p ^k − 1). So there exist positive integers q and s such that q = 2s and t = 

p

−1 p−1

 s. Hence using (3.3), we have

σ

 p ^k − 1 p − 1 s



= 2p ^{k −1} s and

n = p ^k

 p ^k − 1 p − 1

 s.

Case III. If k is even then 2(p − 1)/(p ^k − 1), so

t = p ^k − 1 2(p − 1) s ⁰ . Hence

n = p ^k

 p ^k − 1 2(p − 1)

 s ⁰ and

σ

 p ^k − 1 2(p − 1) s ⁰



= p ^{k −1} s ⁰ .

For n = p ^k t, 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 σ p,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 σ p,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 σ p,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 σ p,0 (n)

6 66 132

28 308 616

496 5456 10912

8128 89408 178816

33550336 369053696 738107392

Let n = p ^k t, 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(4s ⁰ ) = 7s ⁰

and

σ(4s ⁰ ) ­ s ⁰ σ(4) = 7s ⁰ ,

with equality holds for only s

= 1. so t = 4 and n = 7 ² .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.

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.

Proof. Let n = p 1 .p 2 ....p r , where p 1 , p 2 , ..., p r are of the form 2 ^k − 1 is a Mersenne prime.

Now

σ(σ(n)) = σ(σ(p ₁ p 2 ...p r )) = σ(σ(p ₁ )σ(p ₂ )...σ(p r ))

= σ(2 ^k

.2 ^k

....2 ^k

) = 2 ^k

^+k

^+...+k

⁺¹ − 1

= 2.(2 ^k

^+k

^+...+k

) − 1 = 2σ(p ¹ p 2 ...p r ) − 1

= 2σ(n) − 1.

Similarly, if n = p, where p = 2 ^k − 1 is a Mersenne prime, then σ(σ(n)) = σ(σ(p)) = σ(2 ^k ) = 2 ^k+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 contem-

porary Mathematics, Vol. 16 (2008), No.1, 34-45.

[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)