k-generalized FIBONACCI NUMBERS CLOSE TO THE FORM 2 a + 3 b + 5 c 1. Introduction

Transcription

1 Acta Math. Uiv. Comeiaae Vol. LXXXVI, 2 (2017), pp k-generalized FIBONACCI NUMBERS CLOSE TO THE FORM 2 a + 3 b + 5 c N. IRMAK ad M. ALP Abstract. The k-geeralized Fiboacci sequece { F (k) } is defied as the sum 0 of the k proceedig terms ad iitial coditios are 0,..., 0, 1 (k terms). I this paper, we solve the diophatie equatio F (k) = 2 a + 3 b + 5 c + δ, where a, b, c ad δ are oegative itegers with max{a, b} c ad 0 δ 5. This work geeralizes a recet Marques [9] ad the first author, Szalay [6] results. 1. Itroductio Let k 2 be a iteger. The k-geeralized Fiboacci sequece { F (k) } by the followig recurrece relatio = k, (k 2) 0 is defied with the iitial coditios (k 2) = (k 3) = = 0 = 0 ad k 1 = 1. Naturally, the case k = 2 turs to well-kow Fiboacci sequece {F }. If k = 3, the it gives the Triboacci sequece {T }. The problem of fidig differet type of the umbers amog the terms of liear recurrece has a log history. Oe of the results by Bugeaud, Migotte ad Siksek [1] is that oly 0, 1, 8, 144 i Fiboacci umbers ad 1, 4 i Lucas umbers ca be writte i the form y t, where t > 1. Szalay ad Luca showed that there are oly fiitely quadruples (, a, b, p) such that F = p a ± p b + 1, where p is a prime umber i [7]. Recetly, Bravo ad Luca [2] solved the diophatie equatio F (k) = 2 m for positive itegers, k, m with k 2. The paper of Marques ad Togbe [8] determies the Fiboacci umbers ad the Lucas umbers of the form 2 a + 3 b + 5 c where Lucas sequece is defied by relatio L = L 1 + L 2 for 2, together with L 0 = 2 ad L 1 = 1. Recetly, Marques [9] solved the diophatie equatio = 2 a + 3 b + 5 c with max{a, b} c. The first author ad Szalay showed that there are 22 solutios to the diophatie equatio 0 T 2 a 3 b 5 c 10, where T is the Triboacci sequece. Received May 6, 2016; revised March 22, Mathematics Subject Classificatio. Primary 11B39, 11J86. Key words ad phrases. k-geeralized Fiboacci sequece; liear forms i logarithms; reductio method. This project is supported by Niğde Uiversity s Project No: FEB 2013/30 BAGEP.

2 280 N. IRMAK ad M. ALP (1) I this paper, we solve the equatio = 2 a + 3 b + 5 c + δ, where a, b, c ad δ are oegative itegers with max{a, b} c ad 0 δ 5. There is a differece from the paper [9]. We add a diameter to the equatio. If we take k = 3 ad 0 δ 5 i the equatio (1), the we obtai the result of the paper [6]. Takig δ = 0 i the equatio (1) yields the paper [9]. Our result is the followig theorem. Theorem 1.1. For k + 2, the solutios of the equatio are give i the followig tables k a b c δ k a b c δ = 2 a + 3 b + 5 c + δ k a b c δ k a b c δ k a b c δ where a, b ad c are positive itegers with max{a, b} c ad the sequece { F (k) } is the k-geeralized Fiboacci sequece. Whe we take δ = 0, it coicides with the Marques [9] results. Moreover, we ote that the solutios (, k, a, b, c, δ) = (7, 4, 0, 1, 2, 0) ad (9, 7, 0, 0, 3, 0) are ot observed i [9]. 2. Auxiliary Results Before proceedig further, we recall some facts ad tools which will be used ext. Dresde ([4, Theorem 1]) gave the Biet-type formula of the terms of the sequece { (k)} F as follows (2) = k i=1 α i (k + 1) (α i 2) α 1 i,

3 k-generalized FIBONACCI NUMBERS 281 for α 1,..., α k, the roots x k x k 1 1 = 0. Also, it was prove i the same paper that F (k) g (α, k) α 1 1 (3) < 2 where α is the domiat root of the characteristic equatio x k x k 1 1 = 0 ad the otatio g (α, k) := (α 1)/(2 + (k + 1) (α 2)). Also, Bravo ad Luca [2] proved that α 2 α 1 for all 1. Aother tool to prove our theorem is a lower boud for liear forms i logarithms of algebraic umbers, give by Matveev [10]. The first oe is the followig lemma. Lemma 2.1. Let K be a umber field of degree D over Q, γ 1, γ 2,... γ t be positive real umbers of K, ad b 1, b 2,... b t be ratioal itegers. Put ad B max { b 1, b 2,..., b t } Λ := γ b1 1 γbt t 1. Let A 1,..., A t be real umbers such that A i max {Dh (γ i ), log γ i, 0.16}, i = 1,..., t. The, assumig that Λ 0, we have Λ > exp ( t+3 t 4.5 D 2 (1 + log D) (1 + log B) A 1... A t ). As usual, i the above lemma, the logarithmic height of algebraic umber η is defied as h (η) = 1 ( d ( log a 0 + max { η (i), 1 })) d i=1 where d is the degree of η over Q, ( η (i)) are the cojugates of η over Q, 1 i d ad a 0 is the positive leadig coefficiet of the miimal polyomial of η over the itegers. The applicatio of Matveev theorem gives a large upper boud. I order to reduce this boud, we use the followig lemma ([5, Lemma 2.2]). Lemma 2.2. Suppose that M is a positive iteger. Let p/q be a coverget of the cotiued fractio expasio of the irratioal umber γ such that q > 6M ad ε = µq M γq, where µ is a real umber ad deotes the distace from the earest iteger. If ε > 0, the there is o solutio to the iequality i positive itegers m ad with 0 < mγ + µ < AB m log(aq/ε) log B m < M. We use also the followig lemma from the paper [2].

4 282 N. IRMAK ad M. ALP Lemma 2.3. For every positive iteger 2, we have Proof of Theorem 1.1 The formula (2) together with the diophatie equatio yields that The we have k i=1 α i (k + 2) (α i 2) α 1 i = 2 a + 3 b + 5 c + δ. g (α, k) α 1 5 c 1 = 2a + 3 b ξ + δ 5 c, where 1 < α = α 1 R is the domiat root of the characteristic equatio x k x k 1 1 = 0 ad ξ = k i=2 g (α i, k) α 1 i is the real umber whose absolute value is less tha 1. Therefore, g(α,k)α > 0. Cosequetly, we get c (4) g (α, k) α 1 5 c 1 < 2a + 3 b + δ 5 c < c sice 2 < 5, 3 < ad c max{a, b}. I order to apply the Lemma 2.1, we take t := 3 ad γ 1 := g (α, k), γ 2 := α, γ 3 := 5, together with the expoets b 1 := 1, b 2 := 1 ad b 3 := c. For this choise, we have D = k, A 2 = k log 5, A 3 = 0.7, ad by the paper [2, page 73], A 1 = 4k log k. By the iequality (3), 5 c < F (k) < α 1 gives that 1 c > log 5 log α > 1. Hece, B = max {1, 1, c} = 1. Whe we compare the upper ad lower bouds for Λ, we get e T (1+log k)(1+log( 1))(4k log k)(k log 5)0.7 < c, where T = k 2. After takig logarithm of both sides ad some simplificatios together with 1 + log ( 1) < 2 log ( 1) ad 1 + log k < 2 log k, we obtai Sice α 2 < yields that c log 1.6 log 2 < k 4 (log k) 2 log ( 1). < 3 5 c < 5 c+1, the ( 2) log α log 5 1 < c. So, the iequality ( ( 2) log α log 5 1 ) log 1.6 < k 4 (log k) 2 log ( 1) 1 log ( 1) < k 4 (log k) 2.

5 k-generalized FIBONACCI NUMBERS 283 Sice the fuctio x x/ log x is icreasig for all x > e, it is easy check that the iequality x < A yields x < 2A log A. log x Thus, takig A := k 4 (log k) 2, we have 1 < 2 ( k 4 (log k) 2) log ( k 4 (log k) 2) < k 4 (log k) 2 (log log log k + 2 log (log k)) < k 4 (log k) 2 ( log k + 2 log (log k)), sice ( log k + 2 log (log k)) < 48 log k for all k 2, the Sice 5 c < (5) < k 4 (log k) 3. < α 1 ad α 2 < 2.3 c + 1 < < 3.4 (c + 1) + 2 < 5 c+1, the the iequality holds. I the sequel, assume that k [2, 372]. I order to apply Lemma 2.2, let t := ( 1) log α + log g (α, k) c log 5. By the equatio (4), we ca write that e t 1 < c. Sice e t 1 > 0, the t > 0. Together with the equatio (5), ( 1) log α + log g (α, k) c log 5 < c < 3 (1.6) c < 3 (1.6) < 6.37 (1.14) holds. Dividig both sides of the above iequality by log 5, we get ( log α ) log g (α, k) log α + c < 6.37 (6) log 5 log 5 log 5 (1.14) < 3.96 (1.14). With γ := log α log 5, log g (α, k) log α µ :=, log 5 A := 3.96, B := 1.14,

6 284 N. IRMAK ad M. ALP the iequality (6) yields 0 < γ c + µ < A B. It is obvious that γ is a irratioal umber. Take M := k 4 (log k) 3. We use the Lemma 2.2 for each case k [2, 372]. Mathematica programme reveals that maximum value of log (Aq/ε) / log B is 1941, Hece, we deduce that possible solutios of the diophatie equatio (1) are i the rage k [2, 372] ad [4, 1941]. I order to decrease the upper boud for, we use the iequality ( log 3 F (k) 5 log 5 δ ) ) log 5 (F (k) δ for 0 δ 5. Sice we assume that max{a, b} c, this ( iequality must hold. For k [2, 372] ad [4, 1941], the iequality log 3 F (k) 5 log 5 δ ) ( ) log 5 F (k) δ yields that 30. We go through the solutios of the diophatie equatio. We fid them as i Theorem 1.1. From ow o, assume that k 373. Uder this coditio, the iequality (7) < k 4 (log k) 3 < 2 k 4 holds. Usig the same argumets i [2] with Lemma 2.3, we have g (α, k) α 1 = δ η + ηδ, where δ < 2 ad η < 2k. The 2 k/2 2 k 2 a + 3 b + 5 c + δ 2 2 = 2 a + 3 b + 5 c g (α, k) α 1 + δ η + ηδ < k/2, where we used the facts 1/2 1 < 1/2 k/2, 4k/2 k < 1/2 k/2 ad 8k/2 3k/2 < 1/2 k/2 for k > 363. If we divide both sides by 2 2, the we get 2 a + 3 b + 5 c < 5 2. k/2 This iequality gives that 1 The facts 2 2a < 5 c < So, 5c 2 2 < a + 3 b k/ δ 2 2. < 2 2 ad b < 5 c < a + 3 b k/ δ 12 < k. 1 5c 2 2 < k. < 2 2 yield that

7 k-generalized FIBONACCI NUMBERS 285 Now we apply the Lemma 2.1 agai. We take t := 2, D := 1, γ 1 := 5, γ 2 := 2, b 1 := c, b 2 := ( 2), B :=,A 1 := log 5 ad A 2 := log 2. The the Lemma 2.2 yields that e (1+log ) log 5 log 2 < k. Takig both sides with logarithm fuctio, we have 3k 10 log 2 log 12 < log 5 log log, where we use the fact 1 + log < 3 2 log. Sice ( log < log k 4 (log k) 3) < 11 log k for k 373, the 3k 10 log 2 log 7 < log k. Whe we solve the above iequality, we get 2 k < ad 4 < I order to reduce the upper boud of, we use Lemma 2.2 agai. Let τ := c log 5 ( 2) log 2. Sice 5 c < 2 2, the c log 5 ( 2) log 2 < 0 yields that τ < 0. Thus, we obtai τ < e τ 1 = e τ e τ 1 < k, where we use that e τ < 2 sice e τ 1 < 1 2. Sice 2 k, the k log 2 c log 5 ( 2) log 2 c log 5 < 24 (2 0.3) k. After dividig both sides by log 5, we get k log 2 log 5 c < 15 ( 2 0.3) k. Let γ := log 2 log 5, [a 0, a 1, a 2,...] = [0, 2, 3, 9,...] be the cotiued fractio of γ ad p k /q k deote kth coverget. The we have q 142 > Furthermore, a M := max {a i ; i = 0, 1,..., 142}. The we fid a M = a 137 = Usig the properties of the cotiued fractio, we get that 1 ( 2) γ c > (a M + 2) ( 2). It yields that The, by the iequality (7), ( 2) < ( 2) γ c < 9 (2 0.3) k k < ( 2) < k 4 (log k) 3 holds. This iequality implies that k 361. This case is already treated.

8 286 N. IRMAK ad M. ALP Therefore, Theorem 1.1 is completed. Refereces 1. Bugeaud Y., Migotte M. ad Siksek S., Classical ad modular approaches to expoetial Diophatie equatio. I. Fiboacci ad Lucas perfect powers, A. of Math. (2) 163(3) (2006), Bravo J. J. ad Luca F., Powers of two i geeralized Fiboacci sequece, Rev. Colombiaa Math. 46 (2012), Bravo J. J. ad Luca F., O a cojecture about repdigits i k-geeralized Fiboacci sequeces, Publ. Math. Debrece 82(3-4) (2013), Dresde G. P. ad Du Z., A simplified Biet formula for K-geeralized Fiboacci umbers, J. Iteger Seq. 17(4) (2014), Article , 9 pp. 5. Dujella A. ad Pethö A., A Geeralizatio of a Theorem of Baker ad Daveport, Quart. J. Math. Oxford 49(3) (1998), Irmak N. ad Szalay L., Triboacci umbers close to the sum 2 a + 3 b + 5 c, Math. Scad. 118 (1), (2016), Luca F. ad Szalay L., Fiboacci umbers of the form p a ± p b + 1, Fiboacci Quart., 45 (2007), Marques D. ad Togbe A., Fiboacci ad Lucas umbers of the form 2 a + 3 b + 5 c, Proc. Japa Acad., Ser. A 89(2013), Marques M., Geeralized Fiboacci umbers of the form 2 a + 3 b + 5 c, Bull. Braz. Math. Soc. New Series 45(3) (2014), Matveev E. M., A Explicit Lower Boud for a Homogeeous Ratioal Liear Form i the Logarithms of Algebraic Numbers, Izv. Math. 64(6) (2000), N. Irmak, Correspodig author, Ömer Halisdemir Uiversity, Art ad Sciece Faculty, Mathematics Departmet, 51240, Niğde, Turkey, irmak@igde.edu.tr M. Alp, Ömer Halisdemir Uiversity, Art ad Sciece Faculty, Mathematics Departmet, 51240, Niğde, Turkey, muratalp@igde.edu.tr