Repdigits in k generalized Pell sequence − Zafer S¸iara and Refik Keskinb aBing¨ol University, Mathematics Department, Bing¨ol/TURKEY bSakarya University, Mathematics Department, Sakarya/TURKEY [email protected], [email protected], September 29, 2020 Abstract (k) Let k ≥ 2 and let (Pn )n≥2−k be k-generalized Pell sequence defined by (k) (k) (k) (k) Pn = 2Pn−1 + Pn−2 + ... + Pn−k for n ≥ 2 with initial conditions (k) (k) (k) (k) (k) P−(k−2) = P−(k−3) = · · · = P−1 = P0 = 0, P1 = 1. In this paper, we deal with the Diophantine equation 10m − 1 P (k) = d n 9 in positive integers n,m,d with m ≥ 2 and 1 ≤ d ≤ 9. We show that (k) repdigits in the sequence Pn , which have at least two digits, n≥2−k (3) (4) are the numbers P5 = 33 and P6 = 88. Keywords: Repdigit, Fibonacci and Lucas numbers, Exponential Diophan- tine equations, Linear forms in logarithms; Baker’s method AMS Subject Classification(2010): 11B39, 11D61, 11J86, arXiv:2009.13387v1 [math.NT] 24 Sep 2020 1 Introduction (k) Let k 2 be an integer. Let the linear recurrence sequence Gn of ≥ n 2 k order k define by ≥ − (k) (k) (k) (k) Gn = rGn 1 + Gn 2 + ... + Gn k (1) − − − 1 (k) (k) (k) for n 2 with the initial conditions G (k 2) = G (k 3) = = G 1 = 0, − (k) ≥ (k) − − − − ··· G0 = a, and G1 = b. For (a,b,r) = (0, 1, 1) and (a,b,r) = (2, 1, 1), the se- (k) (k) quence Gn is called k generalized Fibonacci sequence Fn n 2 k − n 2 k ≥ − ≥ − (k) and k generalized Lucas sequence Ln (see [2, 3]). For (a,b,r) = − n 2 k ≥ − (k) (0, 1, 2) and (a,b,r)=(2, 2, 2), the sequence Gn is called k generalized n 2 k − ≥ − (k) (k) Pell sequence Pn and k generalized Pell-Lucas sequence Qn , n 2 k − n 2 k respectively (see [10]). ≥ The− terms of these sequences are called k generalized ≥ − Fibonacci numbers, k generalized Lucas numbers, k generalized Pell− numbers and k generalized Pell-Lucas− numbers, respectively.− When k = 2, we have − Fibonacci, Lucas, Pell and Pell-Lucas sequences, (Fn)n 0 , (Ln)n 0 , (Pn)n 0 , ≥ ≥ ≥ and (Qn)n 0, respectively. A repdigit≥ is a positive integer whose digits are all equal. Recently, some mathematicians have investigated the repdigits in the above sequences for k =2 or general k. In [11], Luca determined that the largest repdigits in the sequences (2) (2) (2) (2) Fn and Ln are F10 = 55 and L5 = 11. In [9], the authors have n 0 n 0 ≥ ≥ (2) (2) found all repdigits in the sequences Pn and Qn . Here, they n 0 n 0 ≥ (2) ≥ (2) showed that the largest repdigits in these sequences are P3 = 5 and Q2 =6. (3) In [12], Marques proved that the largest repdigits in the sequence Fn n 1 (3) ≥− are F8 = 44. Besides, for general case of k, in [5], Bravo and Luca handled the Diophantine equation 10m 1 F (k) = d − (2) n 9 and showed that this equation has the solutions (n,k,d,m) = (10, 2, 5, 2), (8, 3, 4, 2) in positive integers n,m,k,d with k 2, 1 d 9 and m 2. The same au- ≥ ≤ ≤ (k≥) thors, in [3], considered the equation (2) for the sequence Ln , and they n 2 k have given the solutions of this equation by (n,k,d,m)=(5 , 2, 1≥, 2)−, (5, 4, 2, 2). In this paper, we will deal with the Diophantine equation 10m 1 P (k) = d − (3) n 9 in positive integers n,m,d with m 2 and 1 d 9. We will show that the (k) ≥ ≤ ≤ repdigits in the sequence Pn , which have at least two digits, are the n 2 k (3) (4) ≥ − numbers P5 = 33 and P6 = 88. 2 2 Preliminaries (k) It can be seen that the characteristic polynomial of the sequence Pn n 2 k is ≥ − k k 1 Ψ (x)= x 2x − x 1. (4) k − −···− − We know from Lemma 1 in given [17] that this polynomial has exactly one positive real root located between 2 and 3. We denote the roots of the polynomial in (4) by α1, α2,...,αk. Particuarly, let α = α1 denote positive real root of Ψk(x). The positive real root α = α(k) is called dominant root of Ψk(x). The other roots are strictly inside the unit circle. In [4], the Binet- like formula for k generalized Pell numbers are given by − k (α 1) P (k) = j − αn. (5) n α2 1+ k(α2 3α + 1) j j=1 j j j X − − It was also showed in [4] the contribution of the roots inside the unit circle to the formula (4) is very small, namely that the approximatiıon 1 P (k) g (α)αn < (6) n − k 2 holds for all n 2 k, where ≥ − z 1 g (z)= − . (7) k (k + 1)z2 3kz + k 1 − − The proof of the following inequality is given in [14]. (αj 1) 2 −2 2 (8) αj 1+ k(αj 3αj + 1) ≤ − − for k 2, where αj ’s for j =1, 2,...,k are the roots of the polynomial in (4). Throughout≥ this paper, α denotes the positive real root of the polynomial (k) given in (4). The following relation between α and Pn given in [4] is valid for all n 1. ≥ n 2 (k) n 1 α − P α − . (9) ≤ n ≤ Also, Kılı¸c[10] proved that (k) Pn = F2n 1 (10) − for all 1 n k +1. ≤ ≤ Lemma 1 ([4], Lemma 3.2)Let k,l 2 be integers. Then (a) If k > l, then α(k) > α(l),≥where α(k) and α(l) are the values of α relative to k and l, respectively. 2 k 2 (b) ϕ (1 ϕ− ) <α<ϕ , where ϕ is golden ratio. 2− 1 (c) gk(ϕ )= ϕ+2 . (d) 0.276 <gk(α) < 0.5. 3 For solving the equation (3), we use linear forms in logarithms and Baker’s theory. For this, we will give some notions, theorem, and lemmas related to linear forms in logarithms and Baker’s theory. Let η be an algebraic number of degree d with minimal polynomial d d d 1 (a) a x + a x − + + a = a X η Z[x], 0 1 ··· d 0 − ∈ i=1 Y (a) where the ai’s are integers with gcd(a0,...,an) = 1 and a0 > 0 and η ’s are conjugates of η. Then 1 d h(η)= log a + log max η(a) , 1 (11) d 0 | | i=1 ! X n o is called the logarithmic height of η. In particularly, if η = a/b is a rational number with gcd(a,b) = 1 and b 1, then h(η) = log (max a ,b ) . We give some properties of the≥ logarithmic height whose{| proofs| } can be found in [7]: h(η γ) h(η)+ h(γ) + log 2, (12) ± ≤ 1 h(ηγ± ) h(η)+ h(γ), (13) ≤ h(ηm)= m h(η). (14) | | In [14], using the above properties of the logarithmic height, the authors have proved the inequality h(g (α)) 4 log k for k 3, (15) k ≤ ≥ which will be used in the main theorem, where gk(α) is as defined in (7). Now we give a theorem deduced from Corollary 2.3 of Matveev [13] and provides a large upper bound for the subscript n in the equation (3) (also see Theorem 9.4 in [6]). Theorem 2 Assume that γ1,γ2,...,γt are positive real algebraic numbers in a real algebraic number field K of degree D, b1,b2,...,bt are rational integers, and Λ := γb1 γbt 1 1 ··· t − is not zero. Then Λ > exp 1.4 30t+3 t4.5 D2(1 + log D)(1 + log B)A A A , | | − · · · 1 2 ··· t where B max b ,..., b , ≥ {| 1| | t|} and A max Dh(γ ), log γ , 0.16 for all i =1, . , t. i ≥ { i | i| } 4 The following lemma was proved by Dujella and Peth˝o[8] and is a variation of a lemma of Baker and Davenport [1]. This lemma will be used to reduce the upper bound for the subscript n in the equation (3). For any real number x, we let x = min x n : n Z be the distance from x to the nearest integer. || || {| − | ∈ } Lemma 3 Let M be a positive integer, let p/q be a convergent of the continued fraction of the irrational number γ such that q > 6M, and let A,B,µ be some real numbers with A > 0 and B > 1. Let ǫ := µq M γq . If ǫ > 0, then there exists no solution to the inequality || || − || || w 0 < uγ v + µ < AB− , | − | in positive integers u, v, and w with log(Aq/ǫ) u M and w . ≤ ≥ log B The following lemma can be found in [16]. Lemma 4 Let a, x R. If 0 <a< 1 and x < a, then ∈ | | log(1 a) log(1 + x) < − − x | | a · | | and a x < ex 1 . | | 1 e a · | − | − − 3 Main Theorem Theorem 5 The only solution of Diophantine equation (3) in positive integers (n,m) with 1 d 9 are given by (n,k,d,m)=(5, 3, 3, 2), (6, 4, 8, 2). ≤ ≤ m Proof. (k) 10 1 Assume that Pn = d 9− with n 1, m, k 2 and 1 d 9. If m 10 1 (k)≥ ≥ ≤ ≤ 1 n k +1, then we have d 9− = Pn = F2n 1 by (10).