On Unitary Analogue of f g −perfect numbers and Ψ s −perfect numbers

D.P.Shukla, Shikha Pandey

On Unitary Analogue of f g −perfect numbers and Ψ s −perfect numbers

Abstract. In this paper unitary analogue of f

−Perfect numbers and some proper- ties of Dedekind’s function and all the Ψ

−perfect numbers have been discussed.

2000 Mathematics Subject Classification: 11A25.

Key words and phrases: f

−Perfect Numbers , unitary divisors, arithmetic functions, Dedekind’s function, g-perfect numbers, Ψ

−perfect numbers.

1. Introduction. We shall discuss some arithmetic functions and notations.

• N is the set of all positive integers. If n > 1 and n ∈ N, then n has the form

n = Y k i=1

p ^α _i

that is called a connonical factorization of n,

where k, α 1 , α 2 ...α k ∈ N, p ¹ , p 2 ....p k are different primes, p 1 < p 2 .... < p k .

• w(n) is the number of all different divisors of n ∈ N, n > 1 and w(1) = 0

• d(n) is the number of all different divisors of n ∈ N and

(1) d(n) =

Y k i=1

(α _i + 1) for n > 1

and

d(1) = 1.

Also

d(n) = 2 ^w(n) ∀ n = Y k i=1

p i .

• A divisor d of n is called unitary if d/n and d, ⁿ d

= 1 and it is denoted by d||n.

• Let E ^k (n) = n ^k and E(n) = n, e(n) = 1 ∀ n ­ 1, k ­ 1, a positive integer.

• µ ^∗ (n) is the unitary analogue of M ^.. obius function µ(n) and we have µ ^∗ (n) = (−1) ^w(n) ∀ n ­ 1, a positive integer.

µ ^∗ (n) is a multiplicative function and we get X

d ||n

µ ^∗ (d) =

 1 , n = 1 0 , n > 1

• d ^∗ (n) is the number of all unitary divisors of n and (2) d ^∗ (n) = 2 ^k = 2 ^w(n) if n = p ^α ₁

p ^α ₂

...p ^α _k

it is a multiplicative function.

• Let φ ^∗ (n) denotes the unitary analogue of the Euler totient function, that is φ ^∗ (n) represents the number of positive integers m ¬ n with (m, n) ∗ = 1.

Then it is easy to see that

(3) φ ^∗ (n) = X

d ||n

dµ ^∗  n d



so φ ^∗ (n) is multiplicative. Further,

(4) φ ^∗ (n) =

 



  Y k i=1

(p ^α

− 1) , n = Y k i=1

p ^α _i

1 , n = 1

φ ^∗ (n) is called unitary totient function.

• Let σ ^∗ (n) denotes the unitary analogue of the divisor function, that is σ ^∗ (n) is the sum of all unitary divisors of n. Then it is easy to see that

(5) σ ^∗ (n) = X

d||n

d = X

d||n

E(d)e  n d



so σ ^∗ (n) is multiplicative function. Further,

(6) σ ^∗ (n) =

 



  Y k i=1

(p ^α

+ 1) , n = Y k i=1

p ^α _i

1 , n = 1

σ ^∗ (n) is called unitary divisor function.

• Ψ is dedekind’s function which is multiplicative and for n > 1 and n ∈ N.

(7) Ψ(n) = n

Y k i=1

 1 + 1

p i



and

Ψ(1) = 1

• Function Ψ ^s is the generalization of function Ψ,where s is a real number. It is a multiplicative function. So many properties of Ψ s functions are discussed in [4]·

for n > 1 and n ∈ N,

Ψ s (n) = n ^{s k} π

i=1

 1 + 1

p ^s _i



and

(8) Ψ s (1) = 1

Function Ψ ₁ coincides with Ψ.

while

(9) Ψ 0 (n) = 2 ^ω(n) for n ∈ N

For n = p ^m p being a prime number.

(10) Ψ s (p ^m ) = p ^(m−1)s (p ^s + 1) for all real s

Let f and g are two arithmetic functions then the unitary convolution ””is defined by

(11) (f  g)(n) = X

d ||n

f (d)g  n d



where d||n denotes that d/n and d, ⁿ d

= 1

Let F is the set of arithmetic functions then (F, +, ) is a commutative ring with unity elements.

On the other hand, one has for the unitary totient, unitary number of divisors and unitary sum of divisors,

(12) σ ^∗ = E  e, φ ^∗ = µ ^∗  E, d ^∗ = e  e

In M.V. Vassilev-Missana and K.T. Atanassov [5] f g −perfect numbers has been

defined as-

A number n ∈ N is said to be f ^g −perfect number iff (g ∗ f)(n) = 2f(n) holds,

where ”∗”denotes the Dirichlet product of two arithmetical functions f and g given as

(g ∗ f)(n) = X

d |n

g  n d

 f (d)

Let g be a fixed arithmetic function. A natural number n is said to be g-additive perfect number if and only if

(13) X

d/n

g(d) = 2g(n)

In this paper will study unitary analogue of f g −perfect numbers called as unitary f g −perfect numbers which is defined as given below in section 2.

Recently Mladen V. Vassilev- Missana and Krassimir T. Atanassov [5] have proved that there is no Ψ 0 − perfect numbers.

In view of this we will discuss some Ψ s −perfect numbers for a real numbers s 6= 0 in section 3.

2. Unitary f g −Perfect Number.

Definition 2.1 A number n ∈ N is said to be unitary f ^g −perfect number iff (g  f)(n) = 2f(n), holds.

where ””dentoes the unitary convolution of two arithmetical function f and g, given by (11)

so,

(14) X

d ||n

g  n d

 f (d) = 2f (n)

Let us consider unitary linear convolution operators D ^∗ and L ^∗ that is given by D ^∗ g = µ ^∗  g

L ^∗ g = E o  g

These operators act on the set of all arithmetic function g. It is easy to see that D ^∗ (L ^∗ g) = g

and

L ^∗ (D ^∗ g) = g

Since unitary convolution is commuative and assosiative. Therefore

D ^∗ (g  f) = (D ^∗ g)  f = g  (D ^∗ f )

and

L ^∗ (g  f) = (L ^∗ g)  f = g  (L  f )

with the help of the above relations we will prove the following lemma.

Lemma 2.2 (φ ^∗  σ ^∗ )(n) = (σ ^∗  φ ^∗ )(n) = (E 1 d ^∗ )(n) = n.d ^∗ (n) ∀ n ∈ N.

Proof Using (3) and (5) we can write

φ ^∗ = D ^∗ E and σ ^∗ = L ^∗ E now

φ ^∗  σ ^∗ = (D ^∗ E)  (L ^∗ E)

= D ^∗ (E  L ^∗ E)

= D ^∗ L ^∗ (E  E)

= E  E so

(φ ^∗  σ ^∗ )(n) = (E  E)(n)

= X

d ||n

E  n d

 E(d)

= X

d ||n

 n d

 d

= n. X

d||n

d

= n.d ^∗ (n) Hence,

(φ ^∗  σ ^∗ )(n) = n.d ^∗ (n) Similary we can also show that

(σ ^∗  φ ^∗ )(n) = ((L ^∗ E)  (D ^∗ E))(n)

= (L ^∗ D ^∗ (E  E))(n)

= (E  E)(n) = n.d ^∗ (n) _

Proposition 2.3 Unitary φ ^∗ _σ∗ −perfect numbers do not exist.

Proof Let n ∈ N is an unitary φ ^∗ σ∗ −perfect numbers, so using definition 1, we have

(σ ^∗  φ ^∗ )(n) = 2φ ^∗ (n)

Using lemma 1 we get

n.d ^∗ (n) = 2φ ^∗ (n).

Since

φ ^∗ (n) < n, therefore

n.d ^∗ (n) < 2n,

this implies that d ^∗ (n) < 2. But d ^∗ (n) ­ 2, so it arises a contradiation. Hence n is

not an unitary φ ^∗ _{σ ∗} −perfect number. 

Theorem 2.4 The only unitary σ _φ ^∗ _∗ −perfect number is n=6

Proof Let n ∈ N is an unitary σ ^∗ φ ∗ −perfect number, then using defintion 1. We have

(φ ^∗  σ ^∗ )(n) = 2σ ^∗ (n)

using lemma 1. n ∈ N is an unitary σ φ ^∗ ∗ −perfect number iff it holds that

(15) σ ^∗ (n)

d ^∗ (n) = n 2

In order to prove (15) we will consider following cases- (i) w(n) = 1

(ii) w(n) = 2 (iii) w(n) ­ 3

Case 1. If w(n) = 1,then n = p ^α , p being a prime and α ­ 1, α ∈ N.Let n ∈ N is a unitary σ _φ ^∗ _∗ −perfect number then (15) holds true.

Now using (2) and (6) in (15), we have σ ^∗ (n)

d ^∗ (n) = p ^α + 1 2 = p ^α

2

this imples that p ^α +1 = p ^α , which is impossible. So it arises a contradiction. Hence n is not an unitary σ _φ ^∗ _∗ −perfect number.

Case 2. If w(n) = 2 then n = p ^α q ^β where p and q are different primes such that 2 ¬ p < q and α, β ∈ N, for n is an unitary σ ^∗ φ∗ −perfect number (15) holds true.

So,

σ ^∗ (n) d ^∗ (n) = n

2 using (2) and (6) we have

(p ^α + 1)(q ^β + 1)

4 = p ^α q ^β

2 ,

this implies that  1 + 1

p ^α

  1 + 1

q ^β



= 2 Let 2 ¬ p < q with min (α, β) ­ 2 then

 1 + 1

p ^α

  1 + 1

q ^β



¬

 1 + 1

2 ^α

  1 + 1

3 ^β



¬ 25 18 . it gives 2 ¬ ²⁵ ₁₈ , which is impossible.

Further, Let min(α, β) = 1, then two conditions are possible (a) min(α, β) = α

(b) min(α, β) = β

Let (a) holds ture then α = 1 and β ­ α. So

 1 + 1

p ^α

  1 + 1

q ^β



¬

 1 + 1

2

  1 + 1

3 ^β



¬ 2

with equality for α = β, q = 3 and p = 2.

Let (b) holds true then β = 1 and α ­ β.So

 1 + 1

p ^α

  1 + 1

q ^β



¬

 1 + 1

2 ^α

  1 + 1

3



¬ 2

with equality for α = β, q = 3 and p = 2

So for both the cases we can see that (15) holds true for p = 2, q = 3 and α = β = 1. Hence n = 6 is the solution of unitary σ _φ ^∗ _∗ −perfect number.

Case 3. Take w(n) ­ 3, then n = Y k i=1

p ^α _i

where i = 1, 2, 3, ...k and α i ∈ N with p i ­ 2 and p ¹ < p 2 < p 3 .... < p

Let min(α 1 , α 2 , ...α k ) ­ 2, then n is a squarefull number. Now

σ ^∗ (n)

n =

Y k i=1

 1 + 1

p ^α _i



¬ Y k i=1

 1 + 1

p ² _i



< π  1 − _p ¹

i



π  1 − _p ¹

i



= ζ(2) ζ(4) < 2 Since

ζ(2)

ζ(4) = 1.52135989...

This imples that ^σ

_n ⁽ⁿ⁾ < 2, but if n is an unitary σ _φ ^∗ _∗ −perfect number then according to (15) we have

σ ^∗ (n)

n = d ^∗ (n)

2 = 2 ^{w (n)−1} ,

where w(n) ­ 3 it gives, 2 ^{w (n)−1} < 2, which is impossible for w(n) ­ 3.So it arises a contradiction.Thus our assumption is wrong.

Hence n is not an unitary σ _φ ^∗ _∗ −perfect number.

Let min(α ₁ , α 2 , ...α k ) = 1 with α ₁ = α ₂ ... = α k = 1,then n = p ₁ , p 2 ...p k . Since σ ^∗ (n) = Ψ(n) for n =

Y k i=1

p i , therefore

σ ^∗ (n)

n = Ψ(n) n . In J. Sandor and E. Egri [4] it has been proved that

2 ^w(n) ¬ Ψ(n)d(n)

σ(n) ¬ 2 ^Ω(n) ,

where Ω(n) = α 1 + α 2 + ...α k ∀ n = Y k i=1

p ^α _i

But for n = Y k i=1

p i , we have w(n) = Ω(n). So using (15) we have

2 ^{w (n)−1} = σ ^∗ (n)

n = Ψ(n)

n = 2 ^w(n) σ(n) n.d(n) , using (1), this implies that

σ(n) d(n) = n

2

But according to the inequality which has been proved in Mitrinovic, D. M. popadic [1]

σ(n) d(n) ¬

 3 4

 w(n)

.n <

 3 4

 3

.n < n 2 for w(n) ­ 3, it gives

σ(n) d(n) < n

2

So it arises a contradiction. Thus our assumption is wrong. Hence n is not unitary σ _φ ^∗ _∗ −perfect number.

Let min(α 1 , α 2 , ...α k ) = 1 where all α ⁰ _i s are not equal to 1, then n = M N = Y

p

|M

p i

Y

p

|N

p ^α _j

where p ⁰ _i s and p ⁰ _j s are different prime α j ­ 2 and M = Y

p

|M

p i , N = Y

p

|N

p ^α _j

with

1 ¬ w(M) < w(n) and 1 ¬ w(N) < w(n), also w(M) + w(N) = w(n).

Now,

σ ^∗ (n) d ^∗ (n) =

Y

p

|M

(p i + 1) Y

p

|N

(p ^α _j

+ 1)

2 ^w(n) < 2 ^w(M)+1 πp i πp ^α _j

2 ^w(n) since, p + 1 < 2p and π(p ^α _j

+ 1) < 2πp ^α _j

∀ α ^j ­ 2

therefore,

σ ^∗ (n)

d ^∗ (n) < M N

2 ^{w (n)−w(M)−1} < n 2 ; if

w(n) − w(M) > 1 Now, if w(n) − w(M) = 1, then

σ ^∗ (n)

n = σ ^∗ (MN)

n = σ ^∗ (M)σ ^∗ (N)

M N < 2σ ^∗ (M)

M = 2σ(M)

M Since

σ ^∗ (N) < 2N ∀ N = Y

p

|M

p ^α _j

, α j ­ 2, σ ^∗ (M) = σ(M) and

d(M ) = d ^∗ (M) ∀ M = Y

p

|M

p i

using inequality which has been proved in [5]

2 ^{w (n)−1} < 2σ(M) M < 2.

 3 4

 w(M)

d(M ) for w(M ) ­ 3 it gives

2 ^{w (n)−1} < 2.

 3 4

 w(M)

.2 ^w(M) = 2 ^w(n)

 3 4

 w (n)−1

, this implies that

1 2 <

 3 4

 w (n)−1

, which is impossible for w(n) ­ 4.

Thus it arises a contradiction. So our assumption is wrong for w(M) ­ 3.

Let w(M) = 2 then w(n) = 3 and w(N) = 1 Now, M = p 1 .p 2 and N = p ^α , α ­ 2

σ ^∗ (n)

n =

 1 + 1

p 1

  1 + 1

p 2

  1 + 1

p ^α



< 2

 1 + 1

p 1

  1 + 1

p



so,

2 ^{w (n)−1} < 2

 1 + 1

p 1

  1 + 1

p 2



,

this implies that 2 < 

1 + _p ¹

 

1 + _p ¹

 , but

 1 + 1

p 1

  1 + 1

p 2



< 2 ∀ 2 ¬ p ¹ < p 2

Thus it arises a contradiction. So our assumption is wrong. Hence n is not an

unitary σ ^∗ _{φ ∗} −perfect number. 

3. Solutions of Ψ s −Perfect number. Since Ψ ^s (n) = 1 for all real numbers therefore n=1 is not a Ψ s − Perfect number.

Now we will discuss about some solutions of Ψ s −Perfect numbers for all s 6= 0, a real number.

Theorem 3.1 If n = p ^m , where p is prime number and m ­ 1,a positive integer, then there do not exist any solution of Ψ s −Perfect numbers for all s < 0 and s > 1.

If 0 < s < 1, then for s = ¹ ₂ , n = 4 is the only solution for Ψ s −Perfect number.

Proof Let n = p ^m where p is a prime number and m ­ 1,a positive integer is a Ψ s − perfect number for a real number s 6= 0.

Case 1. Take s < 0 using, (10) we get

(16) Ψ s (p ^m ) = p ^(m−1)s (p ^s + 1)

X

d/p

Ψ s (d) = Ψ s (1) + Ψ s (p) + Ψ s (p ² ) + Ψ s (p ³ ) + ... + Ψ s (p ^m )

= 1 + (p ^s + 1) + p ^s (p ^s + 1) + p ^2s (p ^s + 1) + ... + p ^(m−1)s (p ^s + 1)

= 1 + (p ^s + 1)(1 + p ^s + p ^2s + ... + p ^(m−1)s ) (17)

Let s = −c, where c > 0, a real number.

using equation (17), we get X

d/p

Ψ s (d) = 1 +

 1 p ^c + 1

  1 + 1

p ^c + 1

p ^2c + ... + 1 p ^(m−1)c



= 1 + (p ^c + 1) p ^c

1 − p

¹

1 − p ¹

!

so

(18) X

d/p

Ψ s (d) = 1 + (p ^c + 1)(p ^mc − 1)

p ^mc (p ^c − 1)

and

(19) Ψ s (p ^m ) = (p ^c + 1)

p ^mc

Since we have supposed that n is a Ψ− perfect number therefore using, (18) and (19) we get,

P

d/p

Ψ _s (d) = 2Ψ _s (p ^m ), where s = −c, c > 0 a real numer.

Thus

(20) 1 + (p ^c + 1)(p ^mc − 1)

p ^mc (p ^c − 1) = 2(p ^c + 1) p ^mc By solving equation (20), we get

(21) 1 = (p ^c + 1)(2p ^c − p ^mc − 1) p ^mc (p ^c − 1)

Further, by solving equation (21) we get an equation of degree m + 1 given as

(22) 2p ^(m+1)c − 2p ^2c − p ^c + 1 = 0

Let x = p ^c , then, (22) becomes,

(23) 2x ^m+1 − 2x ² − x + 1 = 0

In general equation, (23) can be written as,

(24) (x − 1)(2x ^m + 2x ^{m −1} + 2x ^{m −2} + ... + 2x ² − 1) = 0 Let m = 1, then from equation (24) we get,

x − 1 = 0, which implies, p ^c = 1. But it is impossible that p ^c = 1, for c > 0, a real number and a prime p.

Let m = 2, then equation (24) becomes

(x − 1)(2x ² − 1), which implies either p ^c = 1 or p ^c = ± ^{√ 1} ₂ .

Now, it is clear that p ^c = 1 is not possible for any c > 0, a real number and a prime p. Now for p ^c = √ ¹

2 , c = − ¹ 2 and p = 2, but c > 0. So p ^c = √ ¹

2 is not possible.

Since p ^c can not be negative for all c > 0 and a prime p, therefore p ^c = − ^{√ 1} ₂ is also not possible. Further, for m > 2 equation (24) has only one solution, x = 1.

This implies p ^c = 1 which is not possible for c > 0, a real number and a prime p.

Thus we can not get any value of p and c for which equation (22) is solvable for

all m ­ 1, a positive integer. Hence, there do not exist any solution of Ψ ^s − perfect

number of the form n = p ^m , where m ­ 1, a positive integer and p is a prime for

s < 0.

Case 2. Take s > 0 using, (10) we get X

d/p

Ψ s (d) = Ψ s (1) + Ψ s (p) + Ψ s (p ² ) + Ψ s (p ³ ) + ... + Ψ s (p ^m )

= 1 + (p ^s + 1) + p ^s (p ^s + 1) + p ^2s (p ^s + 1) + ... + p ^(m−1)s (p ^s + 1)

= 1 + (p ^s + 1)(1 + p ^s + p ^2s + ... + p ^(m−1)s )

= 1 + (p ^s + 1) (p ^ms − 1) (p ^s − 1) (25)

and

(26) Ψ s (p ^m ) = p ^(m−1)s (p ^s + 1) Since n = p ^m , is a Ψ s − perfect number therefore,

X

d/p

Ψ s (d) = 2Ψ s (p ^m )

using equation (25) and (26), we get

(27) 1 + (p ^s + 1)(p ^ms − 1)

(p ^s − 1) = 2p ^(m−1)s (p ^s + 1) By solving equation (27), we get

(28) 1 = (p ^s + 1)(p ^ms − 2p ^(m−1)s + 1) (p ^s − 1)

Further by solving equation (28) we get an equation of degree m + 1, given as (29) p ^(m+1)s − p ^ms − 2p ^(m−1)s + 2 = 0

Let x = p ^s then equation (29) becomes

(30) x ^m+1 − x ^m − 2x ^m−1 + 2 = 0

In general equation (30) can be written as

(31) (x − 1)(x ^m − 2x ^{m −2} − 2x ^{m −3} .... − 2x − 2) = 0

Let m = 1, then from equation (31), we get x − 1 = 0, which implies p ^s = 1 is not possible for s > 0, a real number and a prime p.

Letm = 2 then from equation (31),

(x − 1)(x ² − 2) = 0, which implies either p ^s = 1 or p ^s = ± √ 2. Now it is clear that p ^s = 1 is not possible for any s > 0, a real number and a prime p. Now for p ^s = √ 2, s = ¹ ₂ and p = 2 which is possible. Since p ^s can not be negative for any s > 0, a real number and a prime p, therefore p ^s = − √ 2 is not possible.

Hence for m = 2, p ^s = √ 2 is the only of equation (29).

Since for p ^s = √

2, s = ¹ ₂ and p = 2 therefore n = 2 ² = 4. So n = 4 is a Ψ s − perfect number for s = ¹ ₂ .

Further for m > 2 equation (31) has only solution x = 1. This implies p ^s = 1, which is not possible for s > 0, a real number a prime p.

Thus we cannot get any value of p and s for which equation (29) is solvable for m = 1 and m > 2. For m = 2, p ^s = √ 2 is the only solution of equation (29).

Hence from case 1 and case 2 n = p ^m is not the solution of Ψ s − perfect number for s < 0 and s > 1. If 0 < s < 1 then n = 4 is the only solution of Ψ s − perfect

number for s = ¹ ₂ . _

4. Conclusion. The problem formulated in Mladen V. Vassilev Missana and Krassimir T. Atanassov [5] viz.

”For a given real number s 6= 0 discribe all Ψ ^s − perfect numbers”is still open for the numbers other than discussed in Theorem 3.1.

References

[1] D. Mitrinovic and M. Popadic, inequalites in Number theory, Univ. of Nis., Nis., 1978.

[2] J. Pe, On a generalization of perfect numbers. J. Rec. Math. Vol.31, No.3, (2002-2003), 168- 172.

[3] J. Sandor, On Dedekind’s arithmetical function, Seminarul de t. Structurilor No. 51, 1988, 1-15, Univ. Timisoara, Romania.

[4] J. Sandor and E. Egri, Arithmetic functions in algebra, gemoetry and analysis, Advnaced studies in Centemporary Mathematics, Vol.14, No.2, (2007), 163-213.

[5] M.V. Vassilev-Missana and K.T. Atanassov, A new point of view on perfect and other similar numbers, Advanced studies in contemporary Mathematics, Vol.5, No.2 (2007), 153-169.

[6] M.V. Vassilev-Missana and K.T. Atanassov, Modification of the concept of the perfect num- bers. Proceedigns of thirty first spring conference of the union of Bulgarian Mathematicians, Borovets, 3-6 April, (2002), 221-224.

D.P.Shukla

Department of Mathematics and Astronomy, Lucknow University Lucknow 226007, India

E-mail: dpshukla3@gmail.com Shikha Pandey

Department of Mathematics and Astronomy, Lucknow University Lucknow 226007, India

(Received: 6.05.2011)