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Repdigits As Sums of Two Generalized Lucas Numbers

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Repdigits As Sums of Two Generalized Lucas Numbers Turkish Journal of Mathematics Turk J Math (2021) 45: 1166 – 1179 http://journals.tubitak.gov.tr/math/ © TÜBİTAK Research Article doi:10.3906/mat-2011-59 Repdigits as sums of two generalized Lucas numbers Sai Gopal RAYAGURU1;∗, Jhon Jairo BRAVO2 1Department of Mathematics, National Institute of Technology Rourkela, Odisha, India 2Department of Mathematics, University of Cauca, Popayán, Colombia Received: 18.11.2020 • Accepted/Published Online: 20.03.2021 • Final Version: 20.05.2021 Abstract: A generalization of the well–known Lucas sequence is the k -Lucas sequence with some fixed integer k ≥ 2. The first k terms of this sequence are 0;:::; 0; 2; 1, and each term afterwards is the sum of the preceding k terms. In this paper, we determine all repdigits, which are expressible as sums of two k -Lucas numbers. This work generalizes a prior result of Şiar and Keskin who dealt with the above problem for the particular case of Lucas numbers and a result of Bravo and Luca who searched for repdigits that are k -Lucas numbers. Key words: Generalized Lucas number, repdigit, linear form in logarithms, reduction method 1. Introduction For an integer k ≥ 2, let the k -generalized Fibonacci sequence (or simply, the k -Fibonacci sequence) F (k) := (k) (Fn )n≥2−k be defined by the linear recurrence sequence of order k (k) (k) (k) ··· (k) ≥ Fn = Fn−1 + Fn−2 + + Fn−k for all n 2; (k) (k) ··· (k) (k) with initial conditions F−(k−2) = F−(k−3) = = F0 = 0 and F1 = 1. The above sequence is one among the several generalizations of the Fibonacci sequence (Fn)n≥0 . For example, when k = 2, this coincides with the Fibonacci sequence, while, when k = 3, this sequence is also known as the Tribonacci sequence. Let us now consider the k -generalized Lucas sequence (or simply, the k -Lucas sequence) L(k) := (k) (Ln )n≥2−k whose terms satisfy the recurrence relation of order k (k) (k) (k) ··· (k) ≥ Ln = Ln−1 + Ln−2 + + Ln−k for all n 2; (k) (k) ··· (k) (k) (k) (k) with initial conditions L−(k−2) = L−(k−3) = = L−1 = 0, L0 = 2 and L1 = 1. The expression Ln th denotes the n term of the k -Lucas sequence. The usual Lucas sequence (Ln)n≥0 is obtained when k = 2. Repdigits are natural numbers having a single digit in their decimal expansion, i.e., numbers of the form 10` − 1 a for some ` ≥ 1 where a 2 f1; 2; ··· ; 9g: 9 ∗Correspondence: [email protected] 2010 AMS Mathematics Subject Classification: 11B39, 11J86 1166 This work is licensed under a Creative Commons Attribution 4.0 International License. RAYAGURU and BRAVO/Turk J Math During the past decade, there has been a flurry of activity regarding finding all members of certain classical recurrence sequences which are repdigits. For example, repdigits in Fibonacci, Lucas, Pell, Pell–Lucas, balancing and Lucas-balancing numbers have been explored in [7, 9, 14]. A similar study has been carried out by replacing Fibonacci, Lucas, balancing and Lucas-balancing numbers by their respective consecutive products (see [8, 10, 14]). In [15], Rayaguru and Panda extended their previous work [14] by exploring the existence of repdigits that are products of balancing and Lucas-balancing numbers with their indices in arithmetic progressions. Repdigits in k -Fibonacci and k -Lucas sequences have been studied in [2, 3]. An investigation for the repdigits that are sums of two k -Fibonacci numbers has been done in [4]. In a recent paper [17], Şiar and Keskin searched for the repdigits that are sums of two Lucas numbers. As a generalization to the work of Şiar and Keskin [17] and to the work of Bravo and Luca [3], we address a similar problem with the k -Lucas sequence L(k) i.e., we determine all the solutions of the Diophantine equation 10` − 1 L(k) + L(k) = a (1.1) n m 9 in nonnegative integers n; m; k; a and ` with k ≥ 2, n ≥ m, 1 ≤ a ≤ 9 and ` ≥ 2. Our result is the following theorem. Theorem 1.1 All solutions of the Diophantine equation (1.1) in nonnegative integers n; m; k; a, and ` with k ≥ 2, n ≥ m, 1 ≤ a ≤ 9, and ` ≥ 2 are (4) L4 + L3 = 11 L9 + L1 = 77 2L5 = 44 (3) (3) 2L5 = 22 L12 + L5 = 333 L7 + L0 = 66 (3) (3) (3) (3) L6 + L3 = 22 L4 + L1 = 11 L7 + L6 = 99 (3) (3) (k) (k) 8 ≥ L7 + L3 = 33 L5 + L2 = 22 L7 + L2 = 99; k 7 Our proof combines linear forms in logarithms, reduction techniques, and some estimates from [3] to deal with large values of k . In this paper, we follow the approach and the presentation described in [4]. 2. Auxiliary results We start with some properties of L(k) . First, it is known that the characteristic polynomial of the sequence L(k) , namely k k−1 Ψk(x) = x − x − · · · − x − 1; is irreducible over Q[x] and has just one zero α := α(k) outside the unit circle. To simplify the notation, we shall omit the dependence on k of α. The other zeros of Ψ (x) are strictly inside the unit circle. Furthermore, k α 2 2(1 − 2−k); 2 (see, for example, [12, 13, 19]). We now consider for an integer s ≥ 2, the function x − 1 f (x) = for x > 2(1 − 2−s): s 2 + (s + 1)(x − 2) With the above notation, we have the following properties of L(k) . 1167 RAYAGURU and BRAVO/Turk J Math Lemma 2.1 [3, Lemma 2] Let k ≥ 2 be an integer. Then n−1 (k) n (a) α ≤ Ln ≤ 2α for all n ≥ 1; (b) L(k) satisfies the following Binet - like formula: Xk (k) − n−1 Ln = (2αi 1)fk(αi)αi ; i=1 where α = α1; : : : ; αk are roots of Ψk(x); (k) n−1 (c) jLn − (2α − 1)fk(α)α j < 3=2 holds for all n ≥ 2 − k ; (k) n−2 (d) If 2 ≤ n ≤ k , then Ln = 3 · 2 . (k) Note that property (a) of Lemma 2.1 also holds for n = 0. Moreover, using the identity Ln = (k) − (k) ≥ 2Ln−1 Ln−k−1 , which holds for all n 3, one can prove that (k) · n−2 ≥ Ln < 3 2 holds for all n k + 1: To solve Diophantine equations involving repdigits and terms of linear recurrence sequences, many authors have used lower bounds for linear forms in logarithms of algebraic numbers. These bounds play an important role while solving such Diophantine equations. Before presenting the result that will be used here, it is worth recalling some important aspects from algebraic number theory. Let η be an algebraic number of degree d having the minimal polynomial Xd Yd d−j (j) f(x) = ajx = a0 (x − η ) 2 Z[x]; j=0 j=1 (j) where the aj ’s are relatively prime integers with a0 > 0 and the η ’s are conjugates of η . The logarithmic height of η is given by ! 1 Xd h(η) = log a + maxf0; logjη(j)jg : d 0 j=1 In particular, if η = a=b is a rational number with gcd(a; b) = 1 and b > 1, then h(η) = log(maxfjaj; bg). The following properties [18, Property 3.3] of the logarithmic height will be used with or without further reference as and when needed. h(η ± γ) ≤ h(η) + h(γ) + log2; h(ηγ±1) ≤ h(η) + h(γ); h(ηs) = jsj h(η); s 2 Z: A modified version of a result of Matveev [11] appears in [5]. Let L be a real algebraic number field of degree D. Let γ1; γ2; : : : ; γt be positive real algebraic numbers in L and b1; b2; : : : ; bt be rational integers. Put Yt fj j j jg bi − B = max b1 ;:::; bt and Λ = γi 1: i=1 1168 RAYAGURU and BRAVO/Turk J Math Theorem 2.2 [5, Theorem 9.4] If Λ =6 0, then t+3 4:5 2 jΛj > exp(−1:4 × 30 × t × D (1 + logD)(1 + logB)A1 ··· At); where Ai ≥ maxfDh(γi); jlogγij; 0:16g for all i = 1; : : : ; t: Using the above theorem and properties of the logarithmic height, we will obtain upper bounds for the index of k -Lucas numbers. We will also use the following estimate which can be derived from [3] and will be the key point in addressing the large values of k . Lemma 2.3 Let k ≥ 12 and r be positive integers such that r − 1 < 2k=2 . Then 4 (2α − 1)f (α)αr−1 = 3 · 2r−2(1 + ζ(r; k)) where jζ(r; k)j < : k 2k=2 Proof It follows from [3, p. 150] that δ (2α − 1)f (α)αr−1 = 3 · 2r−2 + 3 · 2r−1η + + ηδ; k 2 where η and δ are real numbers such that 2k 2r+2 jηj < and jδj < : 2k 2k=2 Thus r−1 r−2 (2α − 1)fk(α)α = 3 · 2 (1 + ζ(r; k)); where δ ηδ jζ(r; k)j = 2η + + 3 · 2r−1 3 · 2r−2 4k 3 11k < + + 2k 2k=2 23k=2 4 < for all k ≥ 12: 2k=2 2 Lemma 2.4 For k ≥ 2, let α be the dominant root of the characteristic polynomial Ψk(x) of the k -Lucas sequence, and consider the function fk(x) defined in (2). Then h(fk(α)) < log(k + 1) + log4 < 3logk; where h(·) represents the logarithmic height function. Moreover, if r > 1 is an integer satisfying r − 1 < 2k=2 , then δ (2α − 1)f (α)αr−1 = 3 · 2r−2 + + 3 · 2r−1η + ηδ; k 2 1169 RAYAGURU and BRAVO/Turk J Math where δ and η are real numbers such that 2r+2 2k jδj < and jηj < : 2k=2 2k In [6], Dujella and Pethö gave a version of the reduction method based on the Baker–Davenport lemma.
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