Some Results on Generalized Multiplicative Perfect Numbers

Home , Superperfect number, Unitary perfect number

Alexandre Laugier Lyc´eeprofessionnel Tristan Corbi`ere 16 rue de Kerveguen - BP 17149 29671 Morlaix cedex France [email protected]

Manjil P. Saikia1 FakultätfürMathematik UniversitätWien Oskar-Morgenstern-Platz 1 1090 Wien Austria [email protected]

Abstract In this article, we deﬁne k-multiplicatively e-perfect numbers and k-multiplicatively e-superperfect numbers and prove some results on them. We also characterize the k- ∗ T0T -perfect numbers deﬁned by Das and Saikia (2013).

1 Introduction

A natural number n is said to be perfect (A000396) if the sum of all aliqout divisors of n is equal to n. Or equivalently, σ(n) = 2n, where σ(k) is the sum of the divisors of k. It is a well known result of Euler-Euclid that the form of even perfect numbers is n = 2kp, where p = 2k+1 − 1 is a Mersenne prime and k ≥ 1. Till date, no odd perfect number is known, and it is believed that none exists. Moreover n is said to be super-perfect if σ(σ(n)) = 2n. It was proved by Suryanarayana-Kanold [3,7] that the general form of such super-perfect numbers are n = 2k, where 2k+1 − 1 is a prime and k ≥ 1. No odd super-perfect numbers are known till date. Unless, otherwise mentioned all n considered in this paper will be a natural number. We also denote by 1, n , the set {1, 2, . . . , n}. J K 1Corresponding author

1 Let T (n) denote the product of all the divisors of n.A multiplicatively perfect (A007422) number is a number n such that T (n) = n2 and n is called a multiplicatively super-perfect if T (T (n)) = n2. S´andor [4] characterized such numbers and also numbers called k-multiplicatively perfect numbers, which are numbers n, such that T (n) = nk for k ≥ 2. In this article we shall give some results on other classes of perfect numbers deﬁned by various authors. For a general introduction to such numbers, we refer the readers to the article [2], which contains various references to the existing literature.

2 k-multiplicatively e-perfect numbers

Sándor[5] studied the multiplicatively e-perfect numbers defined in the previous section. If a1 ar n = p1 ··· pr is the prime factorization of n > 1, a divisor d of n is called an exponential b1 br divisor (or, e-divisor for short) if d = p1 ··· pr with bi | ai for i = 1, . . . , r. This notion is due to Straus and Subbarao [6]. Let σe(n) denote the sum of e-divisors of n, then Straus and Subbarao define n as exponentially perfect (or, e-perfect for short) (A054979) if σe(n) = 2n. They proved that there are no odd e-perfect numbers, and for each r, such numbers with r prime factors are finite. We refer the reader to the article [5] for some historical comments and results related to such e-perfect numbers. Let Te(n) denote the product of all the e-divisors of n. Then n is called multiplicatively 2 2 e-perfect if Te(n) = n and multiplicatively e-superperfect if Te(Te(n)) = n . The main result of Sándor[5] is the following. Theorem 1 (Sándor,[5] Theorem 2.1). n is multiplicatively e-perfect if and only if n = pa, where p is a prime and a is an ordinary perfect number. n is multiplicatively e-superperfect if and only if n = pa, where p is a prime and a is an ordinary superperfect number, that is σ(σ(a)) = 2a. Inspired by the work of others in introducing generalized multiplicative perfect numbers, we now introduce the following two classes of numbers.

k Deﬁnition 2. A natural number n is called k-multiplicatively e-perfect if Te(n) = n , where k ≥ 2. Deﬁnition 3. A natural number n is called a k-multiplicatively e-superperfect number k if Te(Te(n)) = n , where k ≥ 2.

Example 4. There exist 6k-multiplicatively e-perfect numbers with k ∈ N?, which have the α form (p1 · p2 · p3) provided σ(α)d(α)2 = 6kα Indeed, according to Theorem 2.1 of [5], we have

α σ(α)d(α)2 Te((p1 · p2 · p3) ) = (p1 · p2 · p3) Particularly, if d(α) = 6, then αk = 6 · σ(α). Some cases are listed below

2 • α = 2, k = 1 where d(α) = 2 and σ(α) = 3;

• α = 8, k = 5 where d(α) = 4 and σ(α) = 15;

• α = 28, k = 12 where d(α) = 6 and σ(α) = 56;

• α = 18, k = 13 where d(α) = 6 and σ(α) = 39;

• α = 12, k = 14 where d(α) = 6 and σ(α) = 28.

Afterwards, we characterize some of these class of numbers in the following theorems.

Theorem 5. If n = pa, where p is a prime and a is a k-perfect number, then n is k- multiplicatively e-perfect. If n = pa, where p is a prime and a is a k-superperfect number, then n is k-multiplicatively e-superperfect.

Proof. We know from S´andor[5], that if n = pa where p is a prime and a is a non-zero positive integer, then we have σ(a) Te(n) = p . So, if n = pa where p is a prime and a is a k-perfect number, then

σ(a) = ka, and so ka k Te(n) = p = n . Moreover, if n = pa where p is a prime and a is a non-zero positive integer, then we have

σ(a) σ(σ(a)) Te(Te(n)) = Te(p ) = p .

So, if n = pa where p is a prime and a is a k-perfect number, then

σ(σ(a)) = ka, and so ka k Te(Te(n)) = p = n .

In the following, we denote by d(n), the number of positive divisors of n.

α1 αr ∗ Theorem 6. Let p be a prime number and let n = p1 ··· pr , with r ∈ N be the prime ∗ factorisation of an integer n > 1 where pi with i ∈ 1, r are prime numbers and αi ∈ N for all i ∈ 1, r . n is p-multiplicatively e-perfect if andJ onlyK if for each i ∈ 1, r , we have J K J K σ(αi) = gcd(αi, σ(αi))p

3 and Y αi = gcd(αi, σ(αi)) d(αj), j∈ 1,r \{i} J K with r−1 Y 2 ≤ d(αj) < p, j∈ 1,r \{i} J K where we set Y d(αj) = 1. j∈∅

In particular, if r = 1, then α1|σ(α1) and we have

σ(α1) = α1p.

Proof. Before proceeding to the proof, we know from S´andor[5] that

σ(α1)d(α2)···d(αr) σ(αr)d(α1)···d(αr−1) Te(n) = p1 ··· pr . (2.1)

α1 If r = 1, then using Theorem5, n = p1 is p-multiplivatively e-perfect if and only if α1 is a p-perfect number. We now assume that r ≥ 2. We have   σ(α1)d(α2) ··· d(αr) = pα1;  p  Te(n) = n ⇔ ······ (2.2)    σ(αr)d(α1) ··· d(αr−1) = pαr.

Clearly, if for each i ∈ 1, r , we have J K σ(αi) = gcd(αi, σ(αi))p and Y αi = gcd(αi, σ(αi)) d(αj) j∈ 1,r \{i} J K then it can be easily veriﬁed that (2.2) is satisﬁed. Now we notice that if all αi (i ∈ 1, r ) are equal to 1, then (2.2) is consistent only if p = 1. It is impossible since p is a primeJ number.K If at least an αi is equal to 1, say α1 = 1, then from (2.2), we have d(α2) ··· d(αr) = p.

Then one of d(α2), . . . , d(αr) is equal to the prime p and the others are equal to 1. Say

d(α2) = p,

4 and d(α3) = ··· = d(αr) = 1. It implies that α1 = α3 = ··· = αr = 1.

Then the equation σ(α2)d(α1)d(α3) ··· d(αr) = pα2 of (2.2) gives σ(α2) = pα2. But, for all i ∈ 1, r \{1, 2}, the equation σ(αi)d(α1)d(α2) ··· d(αr) = pαi of (2.2) gives σ(αi) = αi whichJ isK not possible since we know that σ(αi) ≥ αi + 1. So, we must have αi ≥ 2 for all i ∈ 1, r . Thus, J K Y r−1 d(αj) ≥ 2 . j∈ 1,r \{i} J K We now prove that if (2.2) is true, then for each i ∈ 1, r , we have J K σ(αi) = gcd(αi, σ(αi))p and Y αi = gcd(αi, σ(αi)) d(αj). j∈ 1,r \{i} J K Let gi = gcd(αi, σ(αi)) for all i ∈ 1, r . So, for each i ∈ 1, r , there exists two non-zero positive integer ai, si such that αi =J giaKi and σ(αi) = gisi withJ K gcd(ai, si) = 1. Notice that si > 1. Otherwise, we would have σ(αi)|αi, a contradiction. For each i ∈ 1, r , the equation J K Y σ(αi) d(αj) = pαi j∈ 1,r \{i} J K of (2.2) now gives Y si d(αj) = pai. j∈ 1,r \{i} J K Since gcd(ai, si) = 1, then from Euclid’s lemma we get Y ai| d(αj), j∈ 1,r \{i} J K and si|p. So, there exists an integer k such that Y d(αj) = kai, j∈ 1,r \{i} J K and p = ksi.

5 Using the fact that p is a prime and si > 1, it results that k = 1 and we get Y d(αj) = ai, j∈ 1,r \{i} J K and p = si.

Therefore σ(αi) = gip and gcd(p, ai) = 1. Moreover, since σ(αi) ≥ αi + 1, we have

gi(p − ai) ≥ 1 implying that p > ai.

α α Example 7. • Let n = p1 · p2 be the prime factorisation of an integer n > 1 where p1 and p2 are prime numbers. Let α = 18. So, we have 18 18 n = p1 · p2 and

d(α) = d(2 · 32) = 6 σ(α) = 1 + 2 + 3 + 6 + 9 + 18 = 39 = 3 · 13 = 3p

where p = 13. Moreover, we have also

gcd(α, σ(α)) = gcd(18, 39) = gcd(3 · 6, 3 · 13) = 3 · gcd(6, 13) = 3 6= α

and gcd(α, σ(α)) · d(α) = 3 · 6 = 18 = α At this stage, notice that Theorem6 can be applied. Let verify that it is well the case. We have

σ(18)·d(18) σ(18)·d(18) 39·6 39·6 3·13·6 3·13·6 18 18 13 p Te(n) = p1 · p2 = p1 · p2 = p1 · p2 = (p1 · p2 ) = n

18 18 So, for all primes p1, p2, integers of the form p1 · p2 are 13-multiplicatively e-perfect numbers.

α α α • Let n = p1 · p2 · p3 be the prime factorisation of an integer n > 1 where p1, p3 and p2 are prime numbers. Let α = 9. So, we have 9 9 9 n = p1 · p2 · p3 and d(α) = d(32) = 3 σ(α) = 1 + 3 + 9 = 13 = 1 · 13 = 1 · p where p = 13. Moreover, we have also

gcd(α, σ(α)) = gcd(9, 13) = 1 6= α

6 and gcd(α, σ(α)) · d(α)2 = 1 · 32 = 9 = α At this stage, notice that Theorem6 can be applied. Let verify that it is well the case. We have

σ(9)·d(9)2 σ(9)·d(9)2 σ(9)·d(9)2 13·9 13·9 13·9 9 9 9 13 p Te(n) = p1 · p2 · p3 = p1 · p2 · p3 = (p1 · p2 · p3) = n

9 9 9 So, for all primes p1, p2 and p3, integers of the form p1 · p2 · p3 are 13-multiplicatively e-perfect numbers. α1 αr Let see how to get the ﬁrst example. According to Theorem6, for r = 2, n = p1 ··· pr is a p-multiplicatively e-perfect number with p a prime number if and only if

α1 = g1d(α2) α2 = g2d(α1)

σ(α1) = g1p σ(α2) = g2p where gi = gcd(αi, σ(αi)) with i = 1, 2. We search a solution such that

a1 a2 α1 = k1 · q1 α2 = k2 · q2 and k1 6 |g1 k2 6 |g2 where ki, qi with i = 1, 2 are prime numbers which are diﬀerent from each other and the ai with i = 1, 2 are integers which are greater than 2. In fact, the assumption ki 6 |gi with i = 1, 2 can be removed as we will see after. But, we will use it in order to derive the result given as an example above. Then we have

0 0 a1 a2 g1 = q1 g2 = q2

0 with ai ∈ 0, ai − 1 (i = 1, 2) and J K d(α1) = 2(a1 + 1) d(α2) = 2(a2 + 1)

a1 a2 σ(α1) = (k1 + 1) · (1 + q1 + ··· + q1 ) = g1p σ(α2) = (k2 + 1) · (1 + q2 + ··· + q2 ) = g2p or equivalently

a1+1 a2+1 q1 − 1 q2 − 1 σ(α1) = (k1 + 1) · = g1p σ(α2) = (k2 + 1) · = g2p q1 − 1 q2 − 1 It comes that

a1 a2 α1 = 2g1(a2 + 1) = k1 · q1 α2 = 2g2(a1 + 1) = k2 · q2

Since we assume that ai ≥ 2, since 2, ki, qi (i = 1, 2) are prime numbers, it results that

k1 = k2 = 2

7 and so a1 a2 g1(a2 + 1) = q1 g2(a1 + 1) = q2 It gives 0 0 a2−a2 a1−a1 a1 = q2 − 1 a2 = q1 − 1 Moreover a1+1 a2+1 q1 − 1 q2 − 1 g1p = 3 · g2p = 3 · q1 − 1 q2 − 1 Since (i = 1, 2) ai qi − 1 6≡ 0 (mod qi) and since (i = 1, 2) 0 ai gi = qi we deduce that 0 0 q1 = q2 = 3 g1 = g2 = 3 a1 = a2 = 1 and 3a1+1 − 1 3a2+1 − 1 p = = 2 2 So a2−1 a1−1 a1 = 3 − 1 a2 = 3 − 1 and a1 = a2 Therefore a1−1 a2−1 a1 = 3 − 1 a2 = 3 − 1

There is only one possibility namely a1 = a2 = 2 and p = 13. It leads to

2 α1 = α2 = 2 · 3 = 18

18 18 meaning that n = p1 · p2 is a 13-multiplicatively e-perfect number. As it was announced before, notice that the case where at least one of gi (i = 1, 2) say g1 has the form (i = 1, 2)

0 ai gi = 2qi is not possible. Indeed, using the same reasoning than above, we will have

a1 α1 = 2g1(a2 + 1) = k1 · q1

0 and so (0 ≤ a1 < a1) 0 2 a1−a1 2 (a2 + 1) = k1 · q1

8 Since k1 is prime, it implies either k1 = 2 or k1 odd prime. If k1 = 2, then the previous 0 equation gives (0 ≤ a1 < a1) 0 a1−a1 2(a2 + 1) = q1 0 Since q1 is prime, if k1 = 2, then it implies that q1 = 2 and so (0 ≤ a1 < a1)

0 a1−a −1 a2 + 1 = 2 1

Since a1+1 q1 − 1 g1p = (k1 + 1) · q1 − 1 if k1 = 2 then we have (q1 = 2)

a0 +1 a +1 2 1 p = 3 · (2 1 − 1)

a0 +1 a +1 It is not possible since 2 1 p is even whereas 3·(2 1 −1) is odd. We reach to a contradiction meaning that k1 is an odd prime. Since

0 2 a1−a1 2 (a2 + 1) = k1 · q1 and since q is prime, if k1 is odd, then we have necessarily q1 = 2 and so

0 a1−a −2 a2 + 1 = k1 · 2 1

0 Notice that it is possible only if a2 is odd and a1 ≥ a1 + 2. Since

a1+1 q1 − 1 g1p = (k1 + 1) · q1 − 1 if k1 is odd then we have (q1 = 2)

0 a +1 a1+1 2 1 p = (k1 + 1) · (2 − 1)

a1+1 Since 2 − 1 is odd, since p is prime and since a1 ≥ 2, it implies that

0 a +1 a1+1 k1 + 1 = 2 1 p = 2 − 1

It follows that 0 a +1 a1+1 k1 = 2 1 − 1 p = 2 − 1

So p is a Mersenne prime and a1 + 1 must be necessarily prime. If

0 a2 g2 = q2 since a2 α2 = 2g2(a1 + 1) = k2 · q2

9 0 then we have (0 ≤ a2 < a2) 0 a2−a2 2(a1 + 1) = k2 · q2

Since a1 + 1 is prime and k2 is prime, either k2 = 2 or k2 = a1 + 1. If k2 = 2, then we have

0 a2−a2 a1 + 1 = q2

0 It is only possible only if a2 − a2 = 1 and q2 = a1 + 1. Since

a2+1 q2 − 1 g2p = (k2 + 1) · q2 − 1 if k2 = 2, then we have a2+1 a1g2p = 3(q2 − 1) 0 a2 a2+1 If g2 = q2 , then since a1 is not prime, since q2 − 1 6≡ 0 (mod q2), it implies that g2 = 0 q2 = 3, a1 = 2, a2 = 1 and a2 = 2. Hence

6p = 3(33 − 1) = 6 · 13

It gives p = 13 which is not a Mersenne prime. Or, if k1 is odd, then p is a Mersenne prime. 0 It results a contradiction. But then k2 = a1 + 1. Or (0 ≤ a2 < a2)

0 a2−a2 2(a1 + 1) = k2 · q2

It gives 0 a2−a2 q2 = 2 0 meaning that q2 = 2 and a2 − a2 = 1. Since

a2+1 q2 − 1 g2p = (k2 + 1) · q2 − 1 we have a2+1 g2p = (a1 + 2)(2 − 1) It implies that 0 a a2+1 a1 + 2 = 2 2 p = 2 − 1 0 a a2+1 a1 + 1 = 2 2 − 1 p = 2 − 1 Or p = 2a1+1 − 1 So a1 = a2

10 0 a2 0 Since a1 + 1 = 2 − 1 and a2 − a2 = 1, we have

a2−1 a2 + 1 = 2 − 1

So, a2 + 1 = a1 + 1 is a Mersenne prime and a2 − 1 must be necessarily prime. Since this equation is not solvable in the set of positive integers, we reach to a contradiction. If

0 a2 g2 = 2q2

Since a2 α2 = 2g2(a1 + 1) = k2 · q2 0 then we have (0 ≤ a2 < a2) 0 2 a2−a2 2 (a1 + 1) = k2 · q2 0 As before, either k2 = 2 or k2 = a1 + 1. If k2 = 2, then we have (0 ≤ a2 < a2)

0 a2−a2−1 a1 + 1 = q2

0 It is only possible if q2 = a1 + 1 and a2 − a2 = 2. Since

a2+1 q2 − 1 g2p = (k2 + 1) · q2 − 1 we have 0 a2 a2+1 2a1q2 p = 3(q2 − 1) 0 It implies that q2 = 3, a2 = 1, a2 = 3, a1 = 2 and so

12p = 3(34 − 1)

p = 20

It is not possible since p is prime. But then k2 = a1 + 1. Since

a2 α2 = 2g2(a1 + 1) = k2 · q2

0 then we have (0 ≤ a2 < a2) 0 2 a2−a2 2 = q2 0 It is only possible if q2 = 2 and a2 − a2 = 2. Since

a2+1 q2 − 1 g2p = (k2 + 1) · q2 − 1 we have 0 a +1 a2+1 2 2 p = (a1 + 2) · (2 − 1)

11 a2+1 Since p is prime, since 2 − 1 6≡ 0 (mod 2) and since a2 ≥ 2, it implies that

0 a +1 a2+1 a1 + 2 = 2 2 p = 2 − 1

Or p = 2a1+1 − 1 So a1 = a2 0 and since a2 − a2 = 2, we have a2−1 a2 + 1 = 2 − 1

So, a2 + 1 = a1 + 1 is a Mersenne prime and a2 − 1 must be necessarily prime. Since this equation is not solvable in the set of positive integers, we reach to a contradiction. It completes the proof that the case where at least one of gi (i = 1, 2) has the form (i = 1, 2)

0 ai gi = 2qi is not possible.

α1 αr ∗ Remark 8. Let n = p1 ··· pr with r ∈ N be the prime factorisation of an integer n > 1 ∗ where pi with i ∈ 1, r are prime numbers and αi ∈ N for all i ∈ 1, r . Let for each J K J K i ∈ 1, r , αi = qi1 qi2 be the prime factorisation of the integer αi > 1 where qi1 , qi2 are distinctJ primeK numbers. We have d(αi) = 4, and

σ(αi) = (1 + qi1 ) · (1 + qi2 ). Let us assume that n is a p-multiplicatively e-perfect number with p a prime number. Ac- cording to Theorem6, we must have

σ(αi) = gip where gi = gcd(αi, σ(αi)), or equivalently

gi = gcd(qi1 qi2 , (1 + qi1 ) · (1 + qi2 )). We also have

(1 + qi1 ) · (1 + qi2 ) = gip.

This implies that p|(1 + qi1 ) or p|(1 + qi2 ). In the following, we will examine the case where one of the prime numbers qi1 , qi2 is equal to 2, say qi1 = 2 and the other, say qi2 , is an odd

12 prime. So, p|3 or p|(1 + qi2 ). If p = 3, then since (1 + qi1 ) · (1 + qi2 ) = 3 · (1 + qi2 ) = gip, we have gi = 1 + qi2 . But then, since gi = gcd(qi1 qi2 , (1 + qi1 ) · (1 + qi2 )), we have

1 + qi2 = gcd(2qi2 , 3 · (1 + qi2 ))

This implies that (1 + qi2 )|2qi2 . From Euclid’s lemma, since gcd(qi2 , 1 + qi2 ) = 1 (qi1 and qi2 + 1 are two consecutive integers) and since 2 is prime, we must have 1 + qi2 = 2 and so qi2 = 1. That is impossible since qi2 is prime. The contradiction stems from the assumption p = 3. So, p 6= 3 and p|(1 + qi2 ). Since 1 + qi2 is even for odd prime qi2 , there exists a non-zero positive integer d such that

1 + qi2 = 2dp. It turns out now, σ(αi) = 3 · 2dp = gip. So

gi = 6d = gcd(2qi2 , 3(1 + qi2 )).

This implies that 6d|2qi2 , and it gives 3d|qi2 and so 3|qi2 . Since qi2 is prime, there is only one possibility, namely qi2 = 3. Then, gi = gcd(6, 12) = 6 and we have αi = 6 and σ(αi) = 6p. Or, σ(αi) = σ(6) = 12 = 2 · 6. At this stage, there is no apparent incompatibility in the reasoning. Nevertheless, if n is a p-multiplicatively e-perfect number, then from Theorem6, Q it implies that j∈ 1,r \{i} d(αj) = 1 since J K Y αi = gi d(αj) j∈ 1,r \{i} J K

Or, d(αi) = 4 for all i ∈ 1, r . So, it leads again to a contradiction (a ﬁrst contradiction was obtained for p = 3) meaningJ K that we cannot have a p-multiplicatively e-perfect number 6 α1 αr n of the form (p1 ··· pr) . There is a generalization of this fact. Indeed, let n = p1 ··· pr ? with r ∈ N be the prime factorisation of an integer n > 1 where pi with i ∈ 1, r are prime ? J K numbers and αi ∈ N for all i ∈ 1, r . If r ≥ 2 and if for all i ∈ 1, r J K J K gcd(αi, σ(αi)) = αi then αi|σ(αi) for all i ∈ 1, r and so there exist r integers qi with i ∈ 1, r such that J K J K σ(αi) = qi · αi Assuming that n is a p-multiplicatively e-perfect number with p a prime number, the set of equation (2.1) becomes   q1 · d(α2) ··· d(αr) = p  ······ (2.3)    qr · d(α1) ··· d(αr−1) = p

13 If at least one of qi (i = 1, . . . , r) is equal to p, say q1, then we have

d(α2) ··· d(αr) = 1 and so α2 = ... = αr = 1 Accordingly σ(α2) = ... = σ(αr) = 1 Since we have σ(αi) = qi · αi it results that q2 = ... = qr = 1

If more than one of qi (i = 1, . . . , r) are equal to p, then there is a contradiction. If only one of qi (i = 1, . . . , r) is equal to p, namely q1, then for each i ∈ 2, r , Equation Q J K qi j∈ 1,r \{i} d(αj) = p of the set of equations (2.3) gives d(α1) = p. If p = 2, then d(α1) = 2. J K It means that α1 must be prime. But then σ(α1) = α1 + 1 which is not divisible by α1. It contradicts the assumtpion done above which implies that σ(α1) = 2α1 (q1 = p = 2). So, p must be necessarily odd. In this case, since d(α1) = p, there exists a prime ρ such that p−1 α1 = ρ . Since σ(α1) = pα1, it would entail that 1 + ρ + ··· + ρp−1 = pρp−1

ρp − 1 = p · (ρ − 1) · ρp−1 (2.4) Notice that ρ = 2 is not possible since 2p −1 is odd whereas p·(ρ−1)·2p−1 is even. Therefore, ρ is an odd prime. From Fermat’s little Theorem, we have

ρp ≡ ρ (mod p) (2.5)

Or p · (ρ − 1) · ρp−1 ≡ p · (ρ − 1) ≡ 0 (mod p) (2.6) Regarding Equations (2.4), (2.5) and (2.6), we must have ρp ≡ 1 (mod p) and so ρ ≡ 1 (mod p). In such a case, there exists a non-zero positive integer k such that ρ = kp + 1. From Equation (2.4), it would imply that

(kp + 1)p − 1 = kp2(kp + 1)p−1 (2.7)

Or, for k ∈ N? and for prime number p, we have kp(p − 1) > 1

Expanding the left side of this inequality and rearranging the diﬀerent terms in this inequality, we have kp2 > kp + 1

14 Multiplying by (kp + 1)p−1 both the sides of the previous inequality, it gives

kp2(kp + 1)p−1 > (kp + 1)p

But, it is not consistent with Equation (2.7). That is a contradiction meaning that all qi (i = 1, . . . , r) are equal to 1. But then σ(αi) = αi for all i ∈ 1, r . It is only possible if all αi (i = 1, . . . , r) are equal to 1. Using the set of equations (2.3J), itK implies that p = 1. That is also impossible since p is prime. Therefore, if an integer n is a p-multiplicatively e-perfect number such that p be a prime number, then the following condition must be fulﬁlled

gcd(αi, σ(αi)) 6= αi

α1 αr ∗ Remark 9. Let n = p1 ··· pr with r ∈ N be the prime factorisation of an integer n > 1 ∗ where pi with i ∈ 1, r are prime numbers and αi ∈ N for all i ∈ 1, r . If all αi are primes, then we have J K J K σ(αi) = αi + 1, and d(αi) = 2.

Notice that in such a case, Theorem6 cannot be applied since σ(αi) is not divisible by αi for all i ∈ 1, r . From (2.1) we have, J K r−1 r−1 (α1+1)·2 (αr+1)·2 Te(n) = p1 ··· pr , and so r Y 2r−1 Te(n) = n pi . i=1

1+α1 If r = 1, then Te(n) = np1 = p . If r ≥ 2 and if αi = 2 for all i ∈ 1, r , then n is a perfect square and we have J K 2 n = (p1 ··· pr) , and 1+2r−2 Te(n) = n . 2 In particular, if r = 2, then Te(n) = n meaning that n is multiplicatively e-perfect number.

∗ 3 k-T0T -perfect numbers A divisor d of n is said to be unitary if gcd(d, n/d) = 1. Let T ∗(n) be the product of unitary divisors of n. Bege [1] has studied the multiplicatively unitary perfect numbers and proved results very similar to S´andor. Das and Saikia [2] introduced the concept of T ∗T -perfect numbers which are numbers n such that T ∗(n)T (n) = n2. They also intorduced k-T ∗T - perfect numbers and characterized both these classes of numbers. They further introduced

15 ∗ ∗ ∗ the concept of T0 T -superperfect and k-T0 T -perfect numbers. A number n is called a T0 T - ∗ 2 ∗ superperfect number if T (T (n)) = n and it is called a k-T0 T -perfect number if T ∗(T (n)) = nk for k ≥ 2. Das and Saikia [2] characterized these classes of numbers. They ∗ ∗ k also introduced the k-T0T -perfect numbers as the numbers n such that T (T (n)) = n ∗ for k ≥ 2. It is our aim in this section to characterize these k-T0T -perfect numbers. a1 ar Let n = p1 ··· pr be the prime factorization of n > 1. Then the number of unitary ∗ r ∗ 2r−1 ∗ divisors of n, τ (n) = 2 and T (n) = n . Das and Saikia [2] mentioned that for k-T0T - perfect number we must have

r r−1 r−1 2 (a12 + 1) ··· (ar2 + 1) = 4k (3.1)

α1 αr for k ≥ 2. In the following results we characterize these class of numbers. Let n = p1 ··· pr ∗ with r ∈ N be the prime factorization of an integer n > 1 where pi with i ∈ 1, r are prime ∗ J K numbers and αi ∈ N for all i ∈ 1, r . J K ∗ 3 Theorem 10. (1) All 2-T0T -perfect numbers have the form n = p1;

∗ 5 (2) All 3-T0T -perfect numbers have the form n = p1;

∗ 7 (3) All 4-T0T -perfect numbers have the form n = p1;

∗ 9 (4) All 5-T0T -perfect numbers have the form n = p1;

∗ 11 (5) All 6-T0T -perfect numbers have the form n = p1 ;

∗ 13 (6) All 7-T0T -perfect numbers have the form n = p1 ;

∗ 15 (7) All 8-T0T -perfect numbers have the form n = p1 ;

∗ 17 (8) All 9-T0T -perfect numbers have the form n = p1 or n = p1p2;

∗ 19 (9) All 10-T0T -perfect numbers have the form n = p1 ; Proof. We will examine in detail the cases where k = 2, 3, 9 leaving to the reader the task to verify the other statements as the proofs are similar. ∗ 3 We ﬁrst prove that all 2-T0T -perfect numbers have the form n = p1. In the following, we will investigate the diﬀerent subcases beginning from r = 1.

• r = 1: (3.1) becomes 2 · (α1 + 1) = 8; it gives α1 = 3.

• r = 2: (3.1) becomes 4·(2α1+1)·(2α2+1) = 8 which is equivalent to (2α1+1)·(2α2+1) = 2; it is not possible since (2α1 + 1) · (2α2 + 1) is odd whereas 2 is even.

• r = 3: (3.1) becomes 8 · (4α1 + 1) · (4α2 + 1) · (4α3 + 1) = 8 which is equivalent to (4α1 + 1) · (4α2 + 1) · (4α3 + 1) = 1; it is not possible since α1, α2, α3 ≥ 1 imply that (4α1 + 1) · (4α2 + 1) · (4α3 + 1) ≥ 125.

16 r−3 r−1 r−1 • r ≥ 4: (3.1) becomes 2 · (α1 · 2 + 1) ··· (αr · 2 + 1) = 1 which is not possible since 2r−3 6 |1 for r ≥ 4.

3 So, only the subcase where r = 1 is valid, which means that if k = 2, then n = p1. ∗ 5 Secondly, we now prove that all 3-T0T -perfect numbers have the form n = p1. In the following, we will investigate the diﬀerent subcases beginning from r = 1.

• r = 1: (3.1) becomes 2 · (α1 + 1) = 12; it gives α1 = 5.

• r = 2: (3.1) becomes 4 · (2α1 + 1) · (2α2 + 1) = 12 which is equivalent to (2α1 + 1) · (2α2 + 1) = 3; it is not possible since for α1, α2 ≥ 1, we have (2α1 + 1) · (2α2 + 1) ≥ 9.

r−2 r−1 r−1 • r ≥ 3: (3.1) becomes 2 · (α1 · 2 + 1) ··· (αr · 2 + 1) = 3 which is not possible since 2r−2 6 |3 for r ≥ 3.

5 So, only the subcase where r = 1 is valid, which means that if k = 3, then n = p1. ∗ 17 Third, we prove that all 9-T0T -perfect numbers have the form n = p1 or n = p1p2. In the following, we will investigate the diﬀerent subcases beginning from r = 1.

• r = 1: (3.1) becomes 2 · (α1 + 1) = 36; it gives α1 = 17.

• r = 2: (3.1) becomes 4 · (2α1 + 1) · (2α2 + 1) = 36 which is equivalent to (2α1 + 1) · (2α2 + 1) = 9; there is a trivial solution which is obtained when α1 = α2 = 1, if at least one of the integers among the integers α1 and α2 is greater than 2, then (2α1 + 1) · (2α2 + 1) > 9 implying that there is no other solution.

r−2 r−1 r−1 • r ≥ 3: (3.1) becomes 2 · (α1 · 2 + 1) ··· (αr · 2 + 1) = 9 which is not possible since 2r−2 6 |9 for r ≥ 3.

So, only the subcases where r = 1 and r = 2 with α1 = α2 = 1 is valid, which means that if 17 k = 9, then either n = p1 or n = p1p2.

∗ 2p−1 Theorem 11. All p-T0T -perfect numbers for a prime p have the form n = p1 . Proof. According to (3.1), we must solve the equation

r r−1 r−1 2 · (α1 · 2 + 1) ··· (αr · 2 + 1) = 4p. (3.2)

In the following, we will investigate the diﬀerent cases beginning from r = 1.

• r = 1: (3.2) becomes 2 · (α1 + 1) = 4p; it gives α1 = 2p − 1.

• r = 2: (3.2) becomes 4·(2α1 +1)·(2α2 +1) = 4p which is equivalent to (2α1 +1)(2α2 + 1) = p; notice that the conditions α1, α2 ≥ 1 imply that (2α1 + 1)(2α2 + 1) ≥ 9 and then (3.2) is not solvable in our setting.

17 r−2 r−1 r−1 • r ≥ 3: (3.2) becomes 2 · (α1 · 2 + 1) ··· (αr · 2 + 1) = p which is not possible r−2 r−2 r−1 r−1 since 2 6 |p for r ≥ 3 for odd prime p and since 2 ·(α1 ·2 +1) ··· (αr ·2 +1) ≥ r−2 r−1 r r2−2 7 2 · (2 + 1) > 2 ≥ 2 = 128 for αi ≥ 1 with i ∈ 1, r and r ≥ 3. 2p−1 J K So, only the case where r = 1 is valid for which n = p1 . 2 ∗ 2p2−1 Theorem 12. All p -T0T -perfect numbers for a prime p have the form n = p1 or (p−1)/2 (p−1)/2 n = p1 p2 . The proof of the above result follows a similar pattern like the proof of Theorem 11. Theorem 13. Let p be a prime, with 2p − 1 being a Mersenne prime. Then 2p−1 · (2p − 1) ∗ is the only even perfect number which is a 3 · (2p − 1)-T0T -perfect number. Proof. According to (3.1), we must solve the equation

r r−1 r−1 2 · (α1 · 2 + 1) ··· (αr · 2 + 1) = 12 · (2p − 1) (3.3) where p is a prime such that 2p − 1 is a Mersenne prime. ∗ First we ﬁnd the possible forms of all 3 · (2p − 1)-T0T -perfect numbers. In the following, we will investigate the diﬀerent cases beginning from r = 1.

• r = 1: (3.3) becomes 2 · (α1 + 1) = 12 · (2p − 1); it gives α1 = 12p − 5.

• r = 2: (3.3) becomes 4 · (2α1 + 1) · (2α2 + 1) = 12 · (2p − 1) which is equivalent to (2α1 + 1) · (2α2 + 1) = 3 · (2p − 1); a trivial solution is given by α1 = 1 and α2 = p − 1 (notice that a particular subcase of this trivial solution is α1 = α2 = 1 which is consistent with (3.3) only if p = 2); more generally, we must have either 2α1 + 1 ≡ 0 (mod 3) or 2α2 +1 ≡ 0 (mod 3); without loss of generality, let us take 2α2 +1 = 3k with k an odd positive integer which divides 2p − 1 so that (3.3) is satisﬁed; then it is not 2p−1 diﬃcult to see that 2α1 + 1 = k and we obtain for α1 and α2, the parametrisation 2p−(k+1) 3k−1 α1(k) = 2k and α2(k) = 2 ; notice that the subcase α1 = p − 1 and α2 = 1 is recovered when k = 1.

r−2 r−1 r−1 • r ≥ 3: (3.3) becomes 2 · (α1 · 2 + 1) ··· (αr · 2 + 1) = 3 · (2p − 1) which is not possible since 2r−2 6 |3 · (2p − 1) for r ≥ 3. 12p−5 So, only the case where r = 1 for which n = p1 and the case where r = 2 for which 2p−(k+1) 3k−1 2k 2 n = p1 · p2 such that k is an odd positive integer which divides 2p − 1 are valid. p−1 In particular, when r = 2, if k = 1, then we get n = p1 · p2. Conversely, when r = 2, if α1(k) = p−1 and α2(k) = 1, then using the parametrisation given above for α1 and α2 when 2p = (k + 1) 3k − 1 r = 2, we have = p − 1 and = 1. This implies that k = 1. Therefore, 2k 2 when r = 2  p−1  α1(k) = p − 1; n = p1 · p2 ⇔ ⇔ k = 1.  α2(k) = 1.

18 ∗ Secondly we see if there exists an even perfect number which is 3 · (2p − 1)-T0T -perfect number. According to the Euclid-Euler theorem, an even perfect number takes the form 2q−1 · (2q − 1), where q is a prime such that 2q − 1 is a Mersenne prime. Accordingly, an even perfect number corresponds to the case where r = 2 which was investigated above. So, α1 α2 n = p1 · p2 is an even perfect number if and only if

α1 α2 q−1 q p1 · p2 = 2 · (2 − 1).

Since the prime factorization of an integer is unique up to the order of the prime factors, q without loss of generality, taking 1 ≤ α2 ≤ α1, since 2 and 2 − 1 are prime, we will have q p1 = 2, p2 = 2 − 1 with α1(k) = q − 1 and α2(k) = 1. Using the parametrization introduced 2p − (k + 1) 3k − 1 above when r = 2, this system is equivalent to = q −1 and = 1. Solving 2k 2 this system for (k, p), it results that k = 1 and p = q. Since when r = 2  p−1  α1(k) = p − 1; n = p1 · p2 ⇔ ⇔ k = 1,  α2(k) = 1.

Let ﬁnish the proof by studying the subcase where both 2α1 + 1 and 2α2 + 1 are divisible by 3. If both 2α1 + 1 and 2α2 + 1 are divisible by 3, then there exist two odd positive integers k1 and k2 such that 2α1 + 1 = 3k1 and 2α2 + 1 = 3k2. It gives 3k − 1 α = 1 1 2 3k − 1 α = 2 2 2 Accordingly, the equation

(2α1 + 1) · (2α2 + 1) = 3 · (2p − 1) becomes 9k1k2 = 3 · (2p − 1) After simpliﬁcation, it gives 3k1k2 = 2p − 1 If n is an even perfect number, then n = 2p−1 ·(2p −1) and without loss of generality, since for α1 α2 p p r = 2, n = p1 · p2 and since 2 − 1 is a Mersenne prime, we can set p1 = 2 and p2 = 2 − 1. Then we have 3k − 1 α = 1 = p − 1 ⇒ 3k = 2p − 1 1 2 1 3k − 1 α = 2 = 1 ⇒ k = 1 2 2 2

19 It is consistent with the equation 3k1k2 = 2p − 1 and the fact that k1, k2 are odd positive p−1 integers. It means that n = p1 p2 as expected. In fact, using the parametrisation 2p − (k + 1) α (k) = 1 2k 3k − 1 α (k) = 2 2 2p−1 and taking by identiﬁcation k = k2 and k1 = 3k , because as well 3k1k2 = 2p − 1 when both 2α1 + 1 and 2α2 + 1 are divisible by 3, this subcase can be recovered. In particular, if k = 1, then it is consistent with the statement of Theorem 13 as it was written in the proof of Theorem 13. Notice that if both 2α1 + 1 and 2α2 + 1 are divisible by 3, then p ≡ 2 (mod 3). Indeed, since k1 is an odd positive integer, there exists an integer m1 such that k1 = 2m1 + 1. Or, 3k1 = 2p − 1. So, 2p = 3k1 + 1 = 6m1 + 4. It gives p = 3m1 + 2. Notice p also that m1 shall be an odd positive integer since p is necessarily prime so that 2 − 1 be a Mersenne prime. We conclude that 2p−1 · (2p − 1) is the only even perfect number which is a 3 · (2p − 1)- ∗ T0T -perfect number.

4 Acknowledgements

The second author is grateful to the excellent facilities provided to him by The Abdus Salam International Centre for Theoretical Physics, Trieste (Italy), where this work was initiated. In particular, he gratefully acknowledges the help recieved from Dr. V. Cappellini in locating some of the references mentioned in this article.

References

[1] A. Bege, On multiplicatively unitary perfect numbers, Seminar on Fixed Point Theory, Cluj-Napoca, 2 (2001), 59–63.

[2] B. Das and H. K. Saikia, On generalized multiplicative perfect numbers, Notes. Numb. Thy. Disc. Math., 19 (2013), 37–42.

[3] H. J. Kanold, Uber¨ super-perfect numbers, Elem. Math., 24 (1969), 61–62.

[4] J. S´andor,On multiplicatively perfect numbers, J. Ineq. Pure Appl. Math., 2 (2001), Art. 3, 6pp.

[5] J. S´andor,On multiplicatively e-perfect numbers, J. Ineq. Pure Appl. Math., 5 (2004), 4pp.

20 [6] E. G. Straus and M. V. Subbarao, On exponential divisors, Duke Math. J., 41 (1974), 465–471.

[7] D. Suryanarayana, Super-perfect numbers, Elem. Math., 24 (1969), 16–17.

2010 Mathematics Subject Classiﬁcation: Primary 11A25; Secondary 11A41. Keywords: perfect numbers, unitary perfect number, multiplicative perfect number, e- perfect number, divisor function.

(Concerned with sequences A000396, A007422, and A054979.)