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On F− Amicable Pairs OCTOGON MATHEMATICAL MAGAZINE Vol. 17, No.2, October 2009, pp 627-635 627 ISSN 1222-5657, ISBN 978-973-88255-5-0, www.hetfalu.ro/octogon On f− amicable pairs J´ozsefS´andorand Mih´alyBencze10 ABSTRACT.P For an arithmetical function f : N → R define F : N → R by F (n) = f (d) , n ∈ N. We say that the numbers a and b are f− amicable, if d|n F (a) = F (b) = f(a) + f(b). We will study this notion for various particular arithmetical functions f. 1. INTRODUCTION It is well-known that two positive integers a and b are called amicable, if one has: σ (a) = σ (b) = a + b, (1) P where σ (n) = d denotes the sum of distinct positive divisors of n. For d|n a = b we reobtain from (1) the perfect numbers, as σ (a) = 2a (2) The least pair of amicable numbers a 6= b are a = 220 and b = 284, known also from the Bible. For the history and survey of old or recent results on amicable numbers, we quote the monograph [10]. The aim of this paper is to introduce a generalization of the notion of amicable numbers; and to study some particular cases. As we will see, in certain cases we will be able to settle completely the problem; in other cases 10Received: 21.09.2009 2000 Mathematics Subject Classification. 11A25. Key words and phrases. Arithmetic functions; amicable numbers; inequalities. 628 J´ozsefS´andorand Mih´alyBencze only particular solutions will be obtained, along with many open problems-which is similar in certain sense with the classical case. Let f : N → R be an arithmetical function, and define its summatory function by F : N → R, X F (n) = f (d) , n ∈ N (3) d|n where the sum runs through all distinct positive divisors d of n. In what follows, we shall say that the pair of numbers a and b are f− amicable, if F (a) = F (b) = f (a) + f (b) (4) When a = b, then we reobtain a notion of f - perfect number a, given by F (a) = 2f (a) (5) studied in certain papers in the References (along with similar or variations of notions) We note then we could define, in place of (4) a and b to be almost-amicable, if F (a) = F (b) = f (a) + f (b) − 1 (6) or quasi-amicable, if F (a) = F (b) = f (a) + f (b) + 1 (7) but these will be considered in another places. Clearly, when f (n) = E (n) = n, (n ∈ N) , from (3) we get F (n) = σ (n) , so (4) contains the classical case of amicable numbers, in the sense of (1). Similarly, from (6) and (7) we get the classical cases of almost-amicable and quasi-amicable numbers (see [10]). On f− amicable pairs 629 2. MAIN RESULTS In this section we will consider separately some particular cases of (4). 1. I-amicable pairs. Let I (n) = 1, ∀n ∈ N. The following result is true. Theorem 1. The only I- amicable numbers a, b are given by a = p and b = q; where p, q are arbitrary prime numbers. P Proof. By (3) we get F (n) = 1 = d (n) = number of divisors of n. Then d|n equation (4) becomes d (a) = d (b) = 2 It is well-known that d (n) = 2 if and only if n = prime, so the result follows. 2. ϕ-amicable pairs. Let ϕ (n) be Euler‘s totient. We have: Theorem 2. All ϕ- amicable numbers a and b are given by a = b = 2k (k ≥ 1, integer) P Proof. As F (n) = ϕ (d) = n (by Gauss‘ identity), we get from (4) that d|n a = b = ϕ (a) + ϕ (b) , so 2ϕ (a) = a; i.e. a ϕ (a) = (8) 2 It is well-known (see e.g. [11]) all solutions of (8) are a = 2k, k ≥ 1. This finishes the proof of Theorem 2. 3. 1/E - AMICABLE PAIRS 1 Let E : N → N, E (n) = n. Then f (n) = n , and P 1 1 P n 1 P σ(n) F (n) = d = n d = n d = n . d|n d|n d|n Therefore (4) gives σ (a) σ (b) 1 1 = = + (9) a b a b 630 J´ozsefS´andorand Mih´alyBencze a As from (9), at one part σ (a) = 1 + b , so b divides a; and from another part, b σ(a) 2 σ (b) = 1 + a , so a divides b, we get b = a. This implies by (9) that a = a , so σ (a) = 2 (10) As for a ≥ 2 one has σ (a) ≥ a + 1 ≥ 3; and σ (1) = 1, equation (10) has no solutions. We have obtained: 1 Theorem 3. There are no E − amicable pairs. 4. k - AMICABLE PAIRS Let in what follows introduce the arithmetical function ½ n, if n = prime k (n) = (11) 1, if n = composite or n = 1 Then F (1) = 1 and for n > 1 one has0 1 P P P P P F (n) = d+ 1 = p+@ 1 − 1A = B (n)+d (n)−ω (n) , d|n d|n p|n d|n d|n d prime d composite d prime where ω (n) denotes the number of distinct prime divisors of n, while B is a much studied arithmetical function (see [10]) X B (n) = p, (12) p|n where p runs through the prime divisors of n. Let a, b > 1. Then (4) leads to B (a) + d (a) − ω (a) = B (b) + d (b) − ω (b) = k (a) + k (b) As k (a) + k (b) = a + b if a and b are prime; a + 1 if a = prime, b = composite; 2 if a, b are composite; and since B (a) = a if a = prime; > a if a = composite; clearly only the case a = prime; b = composite can be allowed. As ω (a) = 1, d (a) = 2, a + 1 = a + 1 but B (b) + d (b) − ω (b) = a + 1 only if B (b) + d (b) − ω (b) − 1 = a is a prime number (13) On f− amicable pairs 631 Theorem 4. All k- amicable numbers a > 1 and b > 1 are given by the equation (13), where B is given by (12). Remark. There are infinitely many solutions to (13). Put e.g. b = pk, where p is a prime. Then (13) gives p + k − 1 = prime. Thus is true for any k = q − p + 1, where q > p are primes. For another example, let b = p · p0, where p · p0 are primes. Then we have that p + p0 + 1 = q is a prime. This is possible, e.g. for p = 3, p0 = 7, where q = 11; or p = 5, p0 = 11, when q = 17; etc. 5. δ+ - AMICABLE PAIRS Let d+ (n) denote the nuber of even divisors of n. (This function has been studied e.g. in [6], [8]). k Clearly, d+ (n) = 0 if n is odd; while for neven, written in the for n = 2 N, one has (with N ≥ 1, odd) d+ (n) = kd (N) (14) ½ 1, n even (see e.g. [6]). let δ (n) = . + 0, n odd Theorem 5. If a and b are δ+− amicable numbers, then a = 4 or a = 2p and b = 4 or b = 2q; whereP p and q Pare prime numbers. Proof. Since F (n) = δ+ (d) = 1 = d+ (n) , we get d|n d|n, even F (n) = d+ (n) (15) Now, equation (4) gives ½ 2, if a, b are even d (a) = d (b) = δ (a) + δ (b) = + + + + 1, if a and b are of distinct parity As d+ (a) = 2 by (14) only if kd (N) = 2, and d (N) = 1 ⇔ N = 1 we get m 2 that d+ (a) = 2 only if a = 2 A (m ≥ 1,A odd) has the form a = 2 · 1 or a = 21 · p. Similarly for b, so Theorem 5 follows. 632 J´ozsefS´andorand Mih´alyBencze 6. ON + AMICABLE PAIRS Let ½ n, if n = even f (n) = 0, if n = odd P Then F (n) = d = σ+ (n) = sum of even divisors of n. Therefore, d|n d even equation (4) leads in this case to: σ+ (a) = σ+ (b) = a + b; a, b even (16) We will call the numbers a and b as + amicable numbers. Theorem 6. All + amicable numbers a = 2kN and b = 2sM (N, M odd) must satisfies the equations ³ ´ 2k − 1 σ (N) = (2s − 1) σ (M) = 2k−1N + 2s−1M (17) Proof. As it is shown in [6], one has ³ ´ ³ ´ k k σ+ (a) = σ+ 2 N = 2 2 − 1 σ (N) (18) By using (18), and defintion (4), relation (17) follows Remarks. 1). When a = b = 2kN, then (17) leads to the + perfect - numbers, which all were determined in [6]. For + superperfect or related numbers, see [8]. 2). The determination of all dolutions to (17) is however, an open problem. 7. ϕ − I− AMICABLE PAIRS Let I (n)P = 1, so f (n) = Pϕ (n) − 1. PThen F (n) = (ϕ (d) − 1) = ϕ (d) − 1 = n − d (n) ; so by (4) we get the d|n d|n d|n equations: a − d (a) = b − d (b) = ϕ (a) + ϕ (b) − 2 (19) Theorem 7. All ϕ − I− amicable numbers are given by equation (19). If a is prime, then a = 2 and b ∈ {1, 2} . There are no ϕ − I− amicable numbers a, b such that a, b ≥ 3 and a − d (a) is odd. On f− amicable pairs 633 Proof. If a is prime, then d (a) = 2 and ϕ (a) = a − 1, so (19) implies ϕ (b) = 1; giving b ∈ {1, 2} .
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