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The Non-Euler Part of a Spoof Odd Perfect Number Is Not Almost Perfect

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The Non-Euler Part of a Spoof Odd Perfect Number Is Not Almost Perfect Received: 09.11.2017 Year: 2018, Pages: 06-12 Published: 22.06.2018 Original Article** THE NON-EULER PART OF A SPOOF ODD PERFECT NUMBER IS NOT ALMOST PERFECT Jose Arnaldo B. Dris1;* <[email protected]> Doli-Jane T. Lugatiman2 <[email protected]> 1PhD Student, University of the Philippines - Diliman, Quezon City, Philippines 2Mindanao State University, Fatima, General Santos City, 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 prove that k is not almost perfect. Keywords − Abundancy index, spoof odd perfect number, Descartes number 1 Introduction This article is an elucidation of some of the recent discoveries/advances in the paper [1], as applied to the case of spoof odd perfect numbers, with details in the older version of this paper titled The Abundancy Index of Divisors of Spoof Odd Perfect Numbers [2]. Recall that 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 = 2km. The number m is called the quasi-Euler prime of the spoof n, while k is referred to as the non-Euler part of n. For some other recent papers on spoofs (otherwise known as Descartes numbers), we refer the interested reader to [3] and [4]. 2 Preliminaries We begin with the following very useful lemmas. (In what follows, we take σ(x) to be the sum of the divisors of x, and denote the abundancy index of x 2 N as I(x) = σ(x)=x, where N is the set of natural numbers or positive integers.) Lemma 2.1. Let n = km be a spoof odd perfect number. Then p p σ(m) σ( k) m k p + is bounded () p + is bounded : k m k m Remark 2.2. Note that, in Lemma 2.1, we dene σ(m) := m + 1: **Edited by A. D. Godase (Editor-in-Chief). *Corresponding Author. Indian Journal in Number Theory, 2018, 06-12 7 Proof. The claimed result follows from the inequalities m < σ(m) = m + 1 < 2m p p p k < σ( k) < 2 k p (since k is a proper divisor of the decient number k). Consequently, m σ(m) m p < p < 2 · p k k k p p p k σ( k) k < < 2 · m m m from which it follows that p p p m k σ(m) σ( k) m k p + < p + < 2 · p + : k m k m k m We therefore conclude that p p σ(m) σ( k) m k p + is bounded () p + is bounded ; k m k m as desired. Remark 2.3. In general, the function 1 f(z) := z + z is not bounded from above. (It suces to consider the cases z ! 0+ and z ! +1.) This means that we do not expect the sum p σ(m) σ( k) p + k m to be bounded from above. Corollary 2.4. Let n = km be a spoof odd perfect number. Then the following conditions hold: p 1. σ( k) 6= m + 1 = σ(m) p 2. σ( k) 6= m Proof. (a) Suppose that n = km is a spoof satisfying the condition p σ( k) = σ(m): It follows that p σ( k) σ(m) p = p k k and p σ( k) σ(m) = m m from which we obtain p p σ(m) σ( k) σ( k) σ(m) p + = p + : k m k m But we also have p σ( k) σ(m) σ(k) σ(m) m + 1 10 28 p + < + = I(k) + < 2 + = : k m k m m 9 9 Finally, we get p σ(m) σ( k) 28 p + < ; k m 9 Indian Journal in Number Theory, 2018, 06-12 8 which contradicts Lemma 2.1 and Remark 2.3. We conclude that p σ( k) 6= m + 1 = σ(m): (b) Suppose that n = km is a spoof satisfying the condition p σ( k) = m: It follows that p σ( k) = 1: m But p p σ( k) σ(m) σ( k) σ(m) 1 < p · = · p < 2: k m m k This implies that σ(m) 1 < p < 2: k In particular, p σ( k) σ(m) σ(m) + p = 1 + p < 1 + 2 = 3: m k k This contradicts Lemma 2.1 and Remark 2.3. We conclude that p σ( k) 6= m: Lemma 2.5. Let a; b 2 N. 1. If I(a) + I(b) < σ(a)=b + σ(b)=a, then a < b () σ(a) < σ(b) holds. 2. If σ(a)=b + σ(b)=a < I(a) + I(b), then a < b () σ(b) < σ(a) holds. 3. If I(a) + I(b) = σ(a)=b + σ(b)=a holds, then either a = b or σ(a) = σ(b) is true. Proof. We refer the interested reader to a proof of part (2) in page 7 of this paper [1]. The proofs for parts (1) and (3) are very similar. p Remark 2.6. Note that if we let and in Lemma 2.5, and if we make the additional a = m b = k p assumption that , then case (3) is immediately ruled out, as gcd(pa; b) = gcd(m; k) = 1 p gcd(m; k) = 1 implies that m 6= k. Additionally, note that σ(m) 6= σ( k) per Corollary 2.4 (a). Likewise, note that Lemma 2.1 and Remark 2.3 rules out case (2), as it implies that p σ(m) σ( k) p m + 1 10 28 p + < I(m) + I( k) < + I(k) < + 2 = ; k m m 9 9 a contradiction. Hence, we are left with the scenario under case (1): p p p m + 1 σ( k) m + 1 σ( k) I(m) + I( k) = + p < p + ; m k k m which per Lemma 2.5 implies that p p m < k () m + 1 < σ( k): The considerations in Remark 2.6 prove the following proposition. Theorem 2.7. Let n = km be a spoof odd perfect number. Then the series of biconditionals p p p m + 1 σ( k) m < k () m + 1 < σ( k) () p < k m hold. Indian Journal in Number Theory, 2018, 06-12 9 Proof. The proof is trivial. Remark 2.8. Note that p m + 1 σ( k) p 6= k m p is in general true under the assumption gcd(m; k) = 1, since it follows from the fact that p p m + 1 σ( k) m + 1 σ( k) 1 < p · = · p < 2: k m m k p p p Also, note that since m 6= k (because gcd(m; k) = 1), and m + 1 = σ(m) 6= σ( k) (by Corollary 2.4, (a)), then equivalently, we have the series of biconditionals p p p σ( k) m + 1 k < m () σ( k) < m + 1 () < p : m k 3 Main Results Remark 3.1. Let us double-check the ndings of Theorem 2.7 (and Remark 2.8) using the only spoof that we know of, as a test case. In the Descartes spoof, m = 22021 = 192 · 61 and p k = 3 · 7 · 11 · 13 = 3003 so that we obtain m + 1 = 22022 = 2 · 11011 and p σ( k) = (3 + 1) · (7 + 1) · (11 + 1) · (13 + 1) = 5376 = 28 · 3 · 7: Notice that we then have p k < m; p σ( k) < m + 1; and p σ( k) 5376 22022 m + 1 = < 1 < = p ; m 22021 3003 k in perfect agreement with the results in Theorem 3. Remark 3.2. Using Theorem 3, we list down all allowable permutations of the set n p p o m; m + 1; k; σ( k) (following the usual ordering on N) subject to the biconditional p p p m + 1 σ( k) m < k () m + 1 < σ( k) () p < k m and the constraints p p m < m + 1; k < σ( k): We have the following cases to consider: p p Case A: m < m + 1 :=pσ(m) < k < σ( k) Under this case, m < k < k which is true if and only if k is not an odd almost perfect number (see page 6 of this preprint [2]). That is, p σ(k) σ( k) D(k) := 2k − σ(k) = > > 1; m m whence there is no contradiction. Indian Journal in Number Theory, 2018, 06-12 10 p p Case B: k < σ( k) < m < m + 1 Noting that, for the Descartes spoof, we have D(k) = D(30032) = 2(3003)2 − σ(30032) = 819 > 1; so that k is not almost perfect, it would seem prudent to try to establish the inequality m < k for Case B. To do so, one would need to repeat the methodology starting with Lemma 2.5, but this time with a = m and b = k. Again, we are left with the scenario under case (1): m + 1 σ(k) m + 1 σ(k) I(m) + I(k) = + < + ; m k k m which per Lemma 2.5 implies that m < k () m + 1 < σ(k): Thus we see that we have the following corollary to Theorem 2.7. Corollary 3.3. Let n = km be a spoof odd perfect number. Then the series of biconditionals m + 1 σ(k) m < k () m + 1 < σ(k) () < k m hold.
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