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1 Introduction Received: 09.11.2016 Year: 2017, Pages: 13-26 Published: 30.11.2017 Original Article** The Abundancy Index of Divisors of Spoof Odd Perfect Numbers Jose Arnaldo1;* <[email protected]> Bebita Dris2 <[email protected]> Department of Mathematics and Physics, Far Eastern University, Manila, Philippines Abstaract − We call n a spoof odd perfect number if n is odd and n = km for two integers k; m > 1 such that σ(k)(m + 1) = 2n, where σ is the sum-of-divisors function. In this paper, we show how results analogous to those of odd perfect numbers could be established for spoof odd perfect numbers (otherwise known in the literature as Descartes numbers). In particular, we predict that an analogue of the Descartes-Frenicle-Sorli conjecture for odd perfect numbers holds for spoof odd perfect numbers. Furthermore, we conjecture that the quasi-Euler prime of a spoof odd perfect number is also its largest quasi-prime factor. Keywords − Abundancy index, spoof odd perfect number, Descartes number. 1 Introduction If N is a positive integer, then we write σ(N) for the sum of the divisors of N. A number N is perfect if σ(N) = 2N. A number M is almost perfect if σ(M) = 2M − 1. It is currently unknown whether there are innitely many even perfect numbers, or whether any odd perfect numbers (OPNs) exist. On the other hand, it is known that powers of two (including 1) are almost perfect, although it is still an open problem to rule out other possible forms of even almost perfect numbers, and also, if 1 will be the only odd almost perfect number to be discovered". Ochem and Rao recently proved [1] that, if N is an odd perfect number, then N > 101500 and that the largest component (i.e., divisor pa with p prime) of N is bigger than 1062. This improves on previous results by Brent, Cohen and te Riele [2] in 1991 (N > 10300) and Cohen [3] in 1987 (largest component pa > 1020). An odd perfect number N = qrt2 is said to be given in Eulerian form if q is prime with q ≡ r ≡ 1 (mod 4) and gcd(q; t) = 1. (The number q is called the Euler prime, while the component qr is referred to as the Euler factor. Note that, since q is prime and q ≡ 1 (mod 4), then q ≥ 5.) We denote the abundancy index I of the positive integer x as σ(x) I(x) = : x If w = z is the only solution to the equation I(w) = I(z), then z is said to be solitary. Greening showed that the condition gcd (z; σ(z)) = 1 is sucient (but not necessary) to show that z is solitary. In this paper, we will refer to this result as Greening's Theorem. If the equation I(x) = y **Edited by A. D. Godase (Editor-in-Chief). *Corresponding Author. Indian Journal in Number Theory, 2017, 13-26 14 has no integral solution x for a given rational number y, then the rational y is called an abundancy outlaw [4]. In his Ph. D. thesis, Sorli [5] conjectured that r = 1. In the M. Sc. thesis [6], it was conjectured that the divisors qr and t are related by the inequality qr < t. This conjecture was made on the basis of the result I(qr) < I(t). In a recent preprint [7], Brown shows that the inequality q < t is true, and that qr < t holds in many cases". In the paper [8], it is shown that t < q is sucient for the Descartes-Frenicle-Sorli conjecture that r = 1 [9]. If proved, the inequality t < q would also imply that q is the largest prime factor of the odd perfect number N = qt2. In this paper, we will focus on spoof odd perfect numbers (otherwise known as Descartes numbers in the literature). We will discuss the analogue of the Descartes-Frenicle-Sorli conjecture for odd perfect numbers (i.e., r = 1), for the case of spoof odd perfect numbers. We will also consider the possibility that the quasi-Euler prime of a spoof odd perfect number is also its largest quasi-prime factor. For recent papers on Descartes/spoof odd perfect numbers, we refer the interested reader to [9] and [10]. 2 Preliminaries A spoof odd perfect number (hereinafter abbreviated as spoof") is an odd integer n = km > 1 such that σ(k)(m + 1) = 2n. Notice that the following statements follow from this denition. Lemma 2.1. Let n = km > 1 be a spoof. • k must be an odd square. • m satises the congruence m ≡ 1 (mod 4). • m j σ(k). • (m + 1) j 2k. σ(k) (This is from [9].) • m = 2k − σ(k) = D(k) Remark 2.2. We will call m the quasi-Euler prime of the spoof n. In this paper, we will restrict m to be composite, since otherwise if m is prime, then the equation σ(k)(m + 1) = 2n = 2k · m1 will imply that n = km is in fact an odd perfect number, whose Euler prime m has exponent 1. Thus, since m ≡ 1 (mod 4) by Lemma 2.1, a lower bound for m is given by m ≥ 9. The following result follows directly from Lemma 2.1. Lemma 2.3. If n = km > 1 is a spoof, then gcd (k; σ(k)) = 2k − σ(k): Proof. Let n = km > 1 be a spoof. By the denition of spoofs, we have σ(k)(m + 1) = 2n = 2km σ(k) = m · (2k − σ(k)) : Note that we can also rewrite this equation as m + 1 2k = · σ(k) = (m + 1) · (2k − σ(k)) ; m where we have used the equation σ(k)=m = 2k − σ(k) from Lemma 2.1. Indian Journal in Number Theory, 2017, 13-26 15 Consequently, we obtain m + 1 gcd (k; σ(k)) = gcd · (2k − σ(k)) ; m · (2k − σ(k)) : 2 But we know that gcd ((m + 1)=2; m) = gcd(m + 1; m) = 1 since m and m + 1 are consecutive (positive) integers. Thus, we get gcd (k; σ(k)) = (2k − σ(k)) · gcd ((m + 1)=2; m) = (2k − σ(k)) · 1 = 2k − σ(k); and we are done. Remark 2.4. Since the equation gcd (k; σ(k)) = 2k − σ(k) holds if n = km > 1 is a spoof (by Lemma 2.3), and since σ(k) = 2k − σ(k) m by Lemma 2.1, then we have σ(k) gcd (k; σ(k)) = = 2k − σ(k): m In particular, if we can show that σ(k) = m, then this would imply that σ(k) gcd (k; σ(k)) = = 2k − σ(k) = 1; m which would further imply that the odd integer k > 1 is an almost perfect number and also a solitary number (by Greening's Theorem [11]). Note that σ(k) = m implies that, since k > 1, we obtain σ(k) m 1 < = ; k k from which it would follow that k < m. We pattern our approach to a study of spoofs via an analogous method for odd perfect numbers described in the M. Sc. thesis [6], the main results of which were published in [8]. Remark 2.5. To begin with, note that we do not have the divisibility constraint gcd(m; k) = 1 for spoofs (as compared to odd perfects). (Wikipedia's denition for Descartes numbers in http://en.wikipedia.org/wiki/Descartes_number does specify gcd(m; k) = 1. However, this assumption was not used in the papers [9] and [10]. A reference request has been posted online via http://math.stackexchange.com/q/1178386 to aid in clarifying this. Lastly, in http://oeis. org/wiki/Descartes_number, Descartes numbers are dened as odd numbers n = km with k > 1, m > 1 and m - k such that σ(k) · (m + 1) = 2n = 2km.) Consequently, we will be needing the following lemma, whose proof may be found in any standard textbook on (elementary) number theory, and which we will omit in this paper. Lemma 2.6. Let a and b be any two positive integers. Then the following chain of inequalities hold: aσ(b) ≤ σ(ab) ≤ σ(a)σ(b): We now give our rst formal result in this section. Theorem 2.7. Let n = km > 1 be a spoof. The following chain of inequalities hold: 2k m+1 10 9 σ(k) . • 1 < σ(k) = m ≤ 9 < 5 ≤ k < 2 p σ( k) • p3 < p < 2. 5 k Indian Journal in Number Theory, 2017, 13-26 16 Proof. The rst inequality m + 1 1 < m is trivial. The inequality m + 1 1 10 = 1 + ≤ m m 9 as m ≥ 9 follows from Remark 2.2. Similarly, the inequality 9 σ(k) ≤ 5 k follows from σ(k) 2m m 1 1 9 = = 2 = 2 1 − ≥ 2 1 − = : k m + 1 m + 1 m + 1 10 5 The inequality σ(k) < 2 k follows from σ(k) m = < 1: 2k m + 1 (In other words, k is decient.) The inequality p 3 σ( k) p < p 5 k follows from Lemma 2.6, since it implies that r p σ(k) σ( k) < p : k k (Note that, by Lemma 2.1, k is an odd square.) Lastly, the inequality p σ( k) σ(k) p < k k p p follows from Lemma 2.1 (since k > 1 is an odd square), and because then k j k and k < k. Remark 2.8. Note the rational approximations 10 ≈ 1:11111 9 and 3 p ≈ 1:34164; 5 so that we obtain p m + 1 10 3 σ( k) ≤ < p < p : m 9 5 k This inequality is an analogue of a similar result for odd perfect numbers, as detailed in [6].
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