COMMENTATIONES MATHEMATICAE Vol. 51, No. 2 (2011), 141-153
D.P.Shukla, Shikha Pandey
On Unitary Analogue of fg −perfect numbers and
Ψs −perfect numbers
Abstract. In this paper unitary analogue of fg −Perfect numbers and some properties of Dedekind’s function and all the Ψs −perfect numbers have been discussed.
2000 Mathematics Subject Classification: 11A25.
Key words and phrases: fg −Perfect Numbers , unitary divisors, arithmetic functions,
Dedekind’s function, g-perfect numbers, Ψs −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=
k
Y
1
pα
i
i=1
that is called a connonical factorization of n,
where k, α1 , α2 .....αk ∈ N, p1 , p2 ....pk are different primes, p1 < p2 .... < pk .
• 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) =
k
Y
(αi + 1)
for n > 1
i=1
and
d(1) = 1.
Also
w(n)
d(n) = 2
∀
n=
k
Y
i=1
pi .
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On Unitary Analogue of fg −perfect numbers and Ψs −perfect numbers
• A divisor d of n is called unitary if d/n and d, nd = 1 and it is denoted by
d||n.
• Let Ek (n) = nk and E(n) = n, e(n) = 1 ∀ n ­ 1, k ­ 1, a positive integer.
..
• µ∗ (n) is the unitary analogue of Mobius 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
1 , n=1
∗
µ (d) =
0 , n>1
d||n
• d∗ (n) is the number of all unitary divisors of n and
(2)
αk
1 α2
if n = pα
1 p2 ...pk
d∗ (n) = 2k = 2w(n)
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
n
X
(3)
φ∗ (n) =
dµ∗
d
d||n
so φ∗ (n) is multiplicative. Further,

k

 Y αi
(pi − 1) ,
(4)
φ∗ (n) =
i=1


1
,
n=
k
Y
i
pα
i
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
n
X
X
(5)
σ ∗ (n) =
d=
E(d)e
d
d||n
d||n
so σ ∗ (n) is multiplicative function. Further,

k

 Y αi
(pi + 1) ,
∗
(6)
σ (n) =
i=1


1
,
σ ∗ (n) is called unitary divisor function.
n=
k
Y
i=1
i
pα
i
n=1
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D.P.Shukla, S. Pandey
• Ψ is dedekind’s function which is multiplicative and for n > 1 and n ∈ N.
k Y
1
Ψ(n) = n
1+
pi
i=1
(7)
and
Ψ(1) = 1
• Function Ψs is the generalization of function Ψ,where s is a real number. It
is a multiplicative function. So many properties of Ψs functions are discussed
in [4]·
for n > 1 and n ∈ N,
1
1+ s
i=1
pi
k
Ψs (n) = ns π
and
(8)
Ψs (1) = 1
Function Ψ1 coincides with Ψ.
while
Ψ0 (n) = 2ω(n)
(9)
for n ∈ N
For n = pm p being a prime number.
(10)
Ψs (pm ) = p(m−1)s (ps + 1)
for all real s
Let f and g are two arithmetic functions then the unitary convolution ””is
defined by
n
X
(11)
(f g)(n) =
f (d)g
d
d||n
where d||n denotes that d/n and d, nd = 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] fg −perfect numbers has been
defined as-
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On Unitary Analogue of fg −perfect numbers and Ψs −perfect numbers
A number n ∈ N is said to be fg −perfect number iff
(g ∗ f )(n) = 2f (n) holds,
where ” ∗ ”denotes the Dirichlet product of two arithmetical functions f and g given
as
X n
f (d)
(g ∗ f )(n) =
g
d
d|n
Let g be a fixed arithmetic function. A natural number n is said to be g-additive
perfect number if and only if
X
(13)
g(d) = 2g(n)
d/n
In this paper will study unitary analogue of fg −perfect numbers called as unitary
fg −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 Ψ0 − perfect numbers.
In view of this we will discuss some Ψs −perfect numbers for a real numbers
s 6= 0 in section 3.
2. Unitary fg −Perfect Number.
Definition 2.1 A number n ∈ N is said to be unitary fg −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,
X n
(14)
g
f (d) = 2f (n)
d
d||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 )
145
D.P.Shukla, S. Pandey
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) = (E1 d∗ )(n) = n.d∗ (n)
∀ n ∈ N.
Proof Using (3) and (5) we can write
φ∗ = D ∗ E
σ ∗ = L∗ E
and
now
φ∗ σ ∗
=
=
=
=
(D∗ E) (L∗ E)
D∗ (E L∗ E)
D∗ L∗ (E E)
EE
so
(φ∗ σ ∗ )(n)
=
=
=
=
(E E)(n)
X n
E(d)
E
d
d||n
X n
d
d
d||n
X
n.
d
d||n
=
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)
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On Unitary Analogue of fg −perfect numbers and Ψs −perfect numbers
Using lemma 1 we get
Since
therefore
n.d∗ (n) = 2φ∗ (n).
φ∗ (n) < n,
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 σφ∗
−perfect number is n=6
∗
Proof Let n ∈ N is an unitary σφ∗
−perfect number, then using defintion 1. We
have
(φ∗ σ ∗ )(n) = 2σ ∗ (n)
∗
using lemma 1. n ∈ N is an unitary σφ∗
−perfect number iff it holds that
(15)
n
σ ∗ (n)
=
∗
d (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 = pα , p being a prime and α ­ 1, α ∈ N.Let n ∈ N
∗
is a unitary σφ∗
−perfect number then (15) holds true.
Now using (2) and (6) in (15), we have
σ ∗ (n)
pα + 1
pα
=
=
d∗ (n)
2
2
this imples that pα + 1 = pα , which is impossible. So it arises a contradiction. Hence
∗
n is not an unitary σφ∗
−perfect number.
Case 2. If w(n) = 2 then n = pα q β where p and q are different primes such
∗
that 2 ¬ p < q and α, β ∈ N, for n is an unitary σφ∗
−perfect number (15) holds
true.
So,
σ ∗ (n)
n
=
∗
d (n)
2
using (2) and (6) we have
(pα + 1)(q β + 1)
pα q β
=
,
4
2
147
D.P.Shukla, S. Pandey
this implies that
1
1+ α
p
1
1+ β =2
q
Let 2 ¬ p < q with min (α, β) ­ 2 then
1
1
1
1
25
1+ α
1+ β ¬ 1+ α
1+ β ¬
.
p
q
2
3
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
1
1
1+ α
1+ β ¬ 1+
1+ β ¬2
p
q
2
3
with equality for α = β, q = 3 and p = 2.
Let (b) holds true then β = 1 and α ­ β.So
1
1
1
1
1+ α
1+ β ¬ 1+ α
1+
¬2
p
q
2
3
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 σφ∗
−perfect number.
k
Y
i
Case 3. Take w(n) ­ 3, then n =
pα
i where i = 1, 2, 3, ...k and αi ∈ N with
pi ­ 2 and p1 < p2 < p3 .... < pk
i=1
Let min(α1 , α2 , ...αk ) ­ 2, then n is a squarefull number. Now
Y
π 1−
k k Y
1
1
σ ∗ (n)
1+ 2 < 1 + αi ¬
=
n
pi
pi
π 1−
i=1
i=1
=
Since
This imples that
to (15) we have
ζ(2)
<2
ζ(4)
1
p4i
1
p2i
ζ(2)
= 1.52135989.....
ζ(4)
σ ∗ (n)
n
∗
< 2, but if n is an unitary σφ∗
−perfect number then according
d∗ (n)
σ ∗ (n)
=
= 2w(n)−1 ,
n
2
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On Unitary Analogue of fg −perfect numbers and Ψs −perfect numbers
where w(n) ­ 3 it gives, 2w(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 σφ∗
−perfect number.
Let min(α1 , α2 , ...αk ) = 1 with α1 = α2 ... = αk = 1,then n = p1 , p2 ...pk . Since
k
Y
∗
σ (n) = Ψ(n) for n =
pi , therefore
i=1
Ψ(n)
σ ∗ (n)
=
.
n
n
In J. Sandor and E. Egri [4] it has been proved that
2w(n) ¬
where Ω(n) = α1 + α2 + ...αk
But for n =
k
Y
Ψ(n)d(n)
¬ 2Ω(n) ,
σ(n)
∀ n=
k
Y
i
pα
i
i=1
pi , we have w(n) = Ω(n). So using (15) we have
i=1
2w(n)−1 =
Ψ(n)
2w(n) σ(n)
σ ∗ (n)
=
=
,
n
n
n.d(n)
using (1), this implies that
n
σ(n)
=
d(n)
2
But according to the inequality which has been proved in Mitrinovic, D. M. popadic
[1]
w(n)
3
σ(n)
3
3
n
¬
.n <
.n <
d(n)
4
4
2
for w(n) ­ 3, it gives
n
σ(n)
<
d(n)
2
So it arises a contradiction. Thus our assumption is wrong. Hence n is not unitary
∗
σφ∗
−perfect number.
Let min(α1 , α2 , ...αk ) = 1 where all αi0 s are not equal to 1, then
n = MN =
Y
pi |M
pi
Y
α
pj j
pj |N
where p0i s and p0j s are different prime αj ­ 2 and M =
Y
pi |M
pi , N =
Y
pj |N
1 ¬ w(M ) < w(n) and 1 ¬ w(N ) < w(n), also w(M ) + w(N ) = w(n).
α
pj j with
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D.P.Shukla, S. Pandey
Now,
σ ∗ (n)
=
d∗ (n)
Y
(pi + 1)
pi |M
Y
α
(pj j + 1)
pj |N
2w(n)
α
α
<
2w(M )+1 πpi πpj j
2w(n)
α
since, p + 1 < 2p and π(pj j + 1) < 2πpj j ∀ αj ­ 2
therefore,
σ ∗ (n)
MN
n
< w(n)−w(M )−1 < ;
d∗ (n)
2
2
if
Now, if w(n) − w(M ) = 1, then
w(n) − w(M ) > 1
σ ∗ (n)
σ ∗ (M N )
σ ∗ (M )σ ∗ (N )
2σ ∗ (M )
2σ(M )
=
=
<
=
n
n
MN
M
M
Since
σ ∗ (N ) < 2N
∀ N=
and
Y
pj |M
d(M ) = d∗ (M )
α
pj j , αj ­ 2, σ ∗ (M ) = σ(M )
∀ M=
using inequality which has been proved in [5]
2w(n)−1 <
it gives
w(n)−1
2
this implies that
2σ(M )
< 2.
M
pi
pi |M
w(M )
3
d(M ) for w(M ) ­ 3
4
w(M )
w(n)−1
3
3
w(M )
w(n)
< 2.
.2
=2
,
4
4
1
<
2
so,
Y
w(n)−1
3
,
4
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 = p1 .p2 and N = pα , α ­ 2
σ ∗ (n)
1
1
1
=
1+
1+
1+ α
n
p1
p2
p
1
1
< 2 1+
1+
p1
p2
w(n)−1
2
1
1
<2 1+
1+
,
p1
p2
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On Unitary Analogue of fg −perfect numbers and Ψs −perfect numbers
this implies that 2 < 1 +
1
p1
1+
1
p2
, but
1
1
1+
1+
<2
p1
p2
∀ 2 ¬ p1 < p2
Thus it arises a contradiction. So our assumption is wrong. Hence n is not an
∗
unitary σφ∗
−perfect number.
3. Solutions of Ψs −Perfect number. Since Ψs (n) = 1 for all real numbers
therefore n=1 is not a Ψs − Perfect number.
Now we will discuss about some solutions of Ψs −Perfect numbers for all s 6= 0,
a real number.
Theorem 3.1 If n = pm , where p is prime number and m ­ 1,a positive integer,
then there do not exist any solution of Ψs −Perfect numbers for all s < 0 and s > 1.
If 0 < s < 1, then for s = 21 , n = 4 is the only solution for Ψs −Perfect number.
Proof Let n = pm where p is a prime number and m ­ 1,a positive integer is a
Ψs − perfect number for a real number s 6= 0.
Case 1. Take s < 0
using, (10) we get
Ψs (pm ) = p(m−1)s (ps + 1)
(16)
X
Ψs (d)
=
Ψs (1) + Ψs (p) + Ψs (p2 ) + Ψs (p3 ) + ... + Ψs (pm )
=
1 + (ps + 1) + ps (ps + 1) + p2s (ps + 1) + ... + p(m−1)s (ps + 1)
=
1 + (ps + 1)(1 + ps + p2s + ........ + p(m−1)s )
d/pm
(17)
Let s = −c, where c > 0, a real number.
using equation (17), we get
X
1
1
1
1
+1
1 + c + 2c + ..... + (m−1)c
Ψs (d) = 1 +
pc
p
p
p
d/pm
!
1
(pc + 1) 1 − pmc
= 1+
pc
1 − p1c
so
(18)
X
d/pm
Ψs (d) = 1 +
(pc + 1)(pmc − 1)
pmc (pc − 1)
D.P.Shukla, S. Pandey
151
and
Ψs (pm ) =
(19)
(pc + 1)
pmc
Since we have supposed that n is a Ψ− perfect number therefore using, (18) and
(19) we get,
P
Ψs (d) = 2Ψs (pm ), where s = −c, c > 0 a real numer.
d/pm
Thus
(20)
1+
(pc + 1)(pmc − 1)
2(pc + 1)
=
mc
c
p (p − 1)
pmc
By solving equation (20), we get
(21)
1=
(pc + 1)(2pc − pmc − 1)
pmc (pc − 1)
Further, by solving equation (21) we get an equation of degree m + 1 given as
(22)
2p(m+1)c − 2p2c − pc + 1 = 0
Let x = pc , then, (22) becomes,
(23)
2xm+1 − 2x2 − x + 1 = 0
In general equation, (23) can be written as,
(24)
(x − 1)(2xm + 2xm−1 + 2xm−2 + ..... + 2x2 − 1) = 0
Let m = 1, then from equation (24) we get,
x − 1 = 0, which implies, pc = 1. But it is impossible that pc = 1, for c > 0, a
real number and a prime p.
Let m = 2, then equation (24) becomes
(x − 1)(2x2 − 1), which implies either pc = 1 or pc = ± √12 .
Now, it is clear that pc = 1 is not possible for any c > 0, a real number and a
prime p. Now for pc = √12 , c = − 21 and p = 2, but c > 0. So pc = √12 is not possible.
Since pc can not be negative for all c > 0 and a prime p, therefore pc = − √12 is
also not possible. Further, for m > 2 equation (24) has only one solution, x = 1.
This implies pc = 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 Ψs − perfect
number of the form n = pm , where m ­ 1, a positive integer and p is a prime for
s < 0.
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On Unitary Analogue of fg −perfect numbers and Ψs −perfect numbers
Case 2. Take s > 0 using, (10) we get
X
Ψs (d) = Ψs (1) + Ψs (p) + Ψs (p2 ) + Ψs (p3 ) + ... + Ψs (pm )
d/pm
(25)
=
1 + (ps + 1) + ps (ps + 1) + p2s (ps + 1) + ... + p(m−1)s (ps + 1)
=
1 + (ps + 1)(1 + ps + p2s + ........ + p(m−1)s )
(pms − 1)
1 + (ps + 1) s
(p − 1)
=
and
Ψs (pm ) = p(m−1)s (ps + 1)
(26)
Since n = pm , is a Ψs − perfect number therefore,
X
Ψs (d) = 2Ψs (pm )
d/pm
using equation (25) and (26), we get
(27)
1+
(ps + 1)(pms − 1)
= 2p(m−1)s (ps + 1)
(ps − 1)
By solving equation (27), we get
(28)
1=
(ps + 1)(pms − 2p(m−1)s + 1)
(ps − 1)
Further by solving equation (28) we get an equation of degree m + 1, given as
(29)
p(m+1)s − pms − 2p(m−1)s + 2 = 0
Let x = ps then equation (29) becomes
(30)
xm+1 − xm − 2xm−1 + 2 = 0
In general equation (30) can be written as
(31)
(x − 1)(xm − 2xm−2 − 2xm−3 .... − 2x − 2) = 0
Let m = 1, then from equation (31), we get x − 1 = 0, which implies ps = 1 is
not possible for s > 0, a real number and a prime p.
Letm = 2 then from equation (31),
√
(x − 1)(x2 − 2) = 0, which implies either ps = 1 or ps = ± 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
ps = 2, s = 21 and p = 2 which is possible. Since ps can not be negative for any
√
s > 0, a real number and a√prime p, therefore ps = − 2 is not possible.
Hence for m = 2, ps = 2 is the only of equation (29).
D.P.Shukla, S. Pandey
153
√
Since for ps = 2, s = 12 and p = 2 therefore n = 22 = 4. So n = 4 is a Ψs −
perfect number for s = 21 .
Further for m > 2 equation (31) has only solution x = 1. This implies ps = 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, ps = 2 is the only solution of equation (29).
Hence from case 1 and case 2 n = pm is not the solution of Ψs − perfect number
for s < 0 and s > 1. If 0 < s < 1 then n = 4 is the only solution of Ψs − perfect
number for s = 21 .
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 Ψs − 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), 168172.
[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 numbers. 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: [email protected]
Shikha Pandey
Department of Mathematics and Astronomy, Lucknow University
Lucknow 226007, India
(Received: 6.05.2011)