Some Results on Generalized Multiplicative Perfect Numbers Alexandre Laugier Lyc´eeprofessionnel Tristan Corbi`ere 16 rue de Kerveguen - BP 17149 29671 Morlaix cedex France [email protected] Manjil P. Saikia1 Fakult¨atf¨urMathematik Universit¨atWien 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 f1; 2; : : : ; ng. 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 S´andor[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 j 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 S´andor[5] is the following. Theorem 1 (S´andor,[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 2 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 2 N be the prime ∗ factorisation of an integer n > 1 where pi with i 2 1; r are prime numbers and αi 2 N for all i 2 1; r . n is p-multiplicatively e-perfect if andJ onlyK if for each i 2 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); j2 1;r nfig J K with r−1 Y 2 ≤ d(αj) < p; j2 1;r nfig J K where we set Y d(αj) = 1: j2; In particular, if r = 1, then α1jσ(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 8 > 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 2 1; r , we have J K σ(αi) = gcd(αi; σ(αi))p and Y αi = gcd(αi; σ(αi)) d(αj) j2 1;r nfig J K then it can be easily verified that (2.2) is satisfied. Now we notice that if all αi (i 2 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 2 1; r n f1; 2g, 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 2 1; r . Thus, J K Y r−1 d(αj) ≥ 2 : j2 1;r nfig J K We now prove that if (2.2) is true, then for each i 2 1; r , we have J K σ(αi) = gcd(αi; σ(αi))p and Y αi = gcd(αi; σ(αi)) d(αj): j2 1;r nfig J K Let gi = gcd(αi; σ(αi)) for all i 2 1; r . So, for each i 2 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)jαi, a contradiction. For each i 2 1; r , the equation J K Y σ(αi) d(αj) = pαi j2 1;r nfig J K of (2.2) now gives Y si d(αj) = pai: j2 1;r nfig J K Since gcd(ai; si) = 1, then from Euclid's lemma we get Y aij d(αj); j2 1;r nfig J K and sijp: So, there exists an integer k such that Y d(αj) = kai; j2 1;r nfig 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; j2 1;r nfig J K and p = si: Therefore σ(αi) = gip and gcd(p; ai) = 1.