S S symmetry Article On Repdigits as Sums of Fibonacci and Tribonacci Numbers Pavel Trojovský Department of Mathematics, Faculty of Science, University of Hradec Králové, 500 03 Hradec Králové, Czech Republic; [email protected]; Tel.: +42-049-333-2860 Received: 17 September 2020; Accepted: 21 October 2020; Published: 26 October 2020 Abstract: In this paper, we use Baker’s theory for nonzero linear forms in logarithms of algebraic numbers and a Baker-Davenport reduction procedure to find all repdigits (i.e., numbers with only one distinct digit in its decimal expansion, thus they can be seen as the easiest case of palindromic numbers, which are a “symmetrical” type of numbers) that can be written in the form Fn + Tn, for some n ≥ 1, where (Fn)n≥0 and (Tn)n≥0 are the sequences of Fibonacci and Tribonacci numbers, respectively. Keywords: Diophantine equations; repdigits; Fibonacci; Tribonacci; Baker’s theory MSC: 11B39; 11J86 1. Introduction A palindromic number is a number that has the same form when written forwards or backwards, i.e., of the form c1c2c3 ... c3c2c1 (thus it can be said that they are “symmetrical” with respect to an axis of symmetry). The first 19th palindromic numbers are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, 44, 55, 66, 77, 88, 99 and clearly they are a repdigits type. A number n is called repdigit if it has only one repeated digit in its decimal expansion. More precisely, n has the form ! 10` − 1 n = a , 9 for some ` ≥ 1 and a 2 [1, 9] (as usual, we set [a, b] = fa, a + 1, ... , bg, for integers a < b). An old open problem consists in proving the existence of infinitely many prime repunit numbers (sequence A002275 in OEIS [1]), where the `th repunit is defined as 10` − 1 R` = , 9 (it is an easy exercise that if R` is prime, then so is `). There are many articles that address Diophantine equations concerning Fibonacci and Lucas numbers (see, e.g., [2–13]). In the last years, many authors have worked on Diophantine problems related to repdigits (e.g., their sums, concatenations) and linear recurrences (e.g., their product, sums). For more about this subject, we refer the reader to [14–24] and references therein. Remark 1. We remark that the definition of repdigit is not restricted to decimal expansion. In fact, a repdigit in base g ≥ 2, has the form Symmetry 2020, 12, 1774; doi:10.3390/sym12111774 www.mdpi.com/journal/symmetry Symmetry 2020, 12, 1774 2 of 7 ! g` − 1 a , g − 1 for some ` ≥ 1 and a 2 [1, g − 1]. In this work, we shall study two well-known recurrence sequences. The first one is the omnipresent sequence (Fn)n. These numbers are defined by the second order linear recurrence Fn+2 = Fn+1 + Fn, for all n ≥ 0 with initial values F0 = 0 and F1 = 1 (see, e.g., [25]). The sequence of Tribonacci numbers (Tn)n (generalizes the Fibonacci sequence) is defined by the third-order recurrence Tn+3 = Tn+2 + Tn+1 + Tn, for all n ≥ 0 which begins with T0 = 0 and T1 = T2 = 1 (see, e.g., [26,27]). We remark that Luca showed that F10 = 55 is the largest repdigit in the Fibonacci sequence, while Marques [28] proved that the largest repdigit in the Tribonacci sequence is T8 = 44. In this paper, we continue this program by searching for repdigits, which are the sum of a Fibonacci and a Tribonacci number (both with the same index). More specifically, our main result is the following: Theorem 1. The only solutions of the Diophantine equation ! 10` − 1 F + T = a , (1) n n 9 in positive integers (n, `, a), with a 2 [1, 9], are (n, `, a) 2 f(1, 1, 2), (2, 1, 2), (3, 1, 4), (4, 1, 7)g. 2. Auxiliary Results First, we recall a very useful non-recursive formula for the nth Fibonacci numbers. The Binet’s formula is: fn − (−f)−n Fn = p , (2) 5 p where f = (1 + 5)/2. By a simple inductive argument, we can obtain that: n−2 n−1 f ≤ Fn ≤ f , for all n ≥ 1. (3) Also, we can write fn Fn = p + n, (4) 5 p p n where jnj < 1/ 5 (actually, one has the asymptotic formula Fn = (f / 5)(1 + o(1))). For the Tribonacci sequence, in 1982, Spickerman [29] found the following “Binet-like” formula: an bn gn Tn = + + , for all n ≥ 1, (5) −a2 + 4a − 1 −b2 + 4b − 1 −g2 + 4g − 1 p 3 2 1/3 where a, b, g arep the roots of polynomial x − x − x − 1. Numerically, if w1 := (19 + 3 33) and 1/3 w2 := (19 − 3 33) , then Symmetry 2020, 12, 1774 3 of 7 p 1 1 a = 3 (1 + w1 + w2), b = 6 (2 − (w1 + w2) + i 3(w1 − w2)), and g = b. Another very helpful formula provided by Spickermann is the following: a T = an , n (a − b)(a − g) where, as usual, bxe is the nearest integer to x. In particular, it holds that 0 n Tn = a a + h, (6) where jhj < 1/2 and a0 := a/(a − b)(a − g). Moreover, again by an inductive argument, we can deduce that n−2 n−1 a ≤ Tn ≤ a , for all n ≥ 1. (7) The main approach to attack Theorem1 is the Baker’s theory about lower bounds for linear forms in logarithms. The next result is due to Matveev [30] according to Bugeaud, Mignotte and Siksek [9]: Lemma 1. Let a1, a2, a3 2 R be algebraic numbers and let b1, b2, b3 be nonzero integer numbers. Define L = b1 log a1 + b2 log a2 + b3 log a3. Let D = [Q(a1, a2, a3) : Q] (degree of field extension) and let A1, A2, A3 be real numbers such that Aj ≥ maxfDh(aj), j log ajj, 0.16g, for j 2 f1, 2, 3g. Take B ≥ maxf1, maxfjbjjAj/A1; 1 ≤ j ≤ 3gg. If L 6= 0, then 2 log jLj ≥ −C1D A1 A2 A3 log(1.5eDB log(eD)), where 4 5.5 2 C1 = 6750000 · e (20.2 + log(3 D log(eD))). In the previous statement, the logarithmic height of a t-degree algebraic number a is defined by ! 1 t h(a) = log jaj + ∑ log maxf1, ja(j)jg , t j=1 (j) where a is the leading coefficient of the minimal polynomial of a (over Z), (a )1≤j≤t are the algebraic conjugates of a. Some helpful properties of h(x) are in the following lemma (see Property 3.3 of [31]): Lemma 2. Let x and y be algebraic numbers. Then h(xy) ≤ h(x) + h(y); h(x + y) ≤ h(x) + h(y) + log 2; h(ar) = jrj · h(a), for all r 2 Q. Our last ingredient is a reduction method provided by Dujella and Peth˝o[32], which is itself a variation of the result of Baker and Davenport [33]. For x 2 R, set kxk = minfjx − nj : n 2 Zg = jx − bxej for the distance from x to the nearest integer. Symmetry 2020, 12, 1774 4 of 7 Lemma 3. For a positive integer M, let p/q be a convergent of the continued fraction of g 62 Q, such that q > 6M, and let m, A and B be real numbers, with A > 0 and B > 1. If the number e = kmqk − Mkgqk is positive, then there is no solution to the Diophantine inequality 0 < mg − n + m < A · B−m in integers m, n > 0 with log(Aq/e) ≤ m < M. log B See Lemma 5 of [32]. Now, we are in a position to prove our main theorem. 3. The Proof of Theorem1 3.1. Finding an Upper Bound for n and ` By using (4) and (6) in (1), we have ! fn 10` − 1 p + n + a0an + h = a . 5 9 We can rewrite the previous equality as ` 0 n 10 n a a − a < 2.4f , (8) 9 p where we used that jnj ≤ 1/ 5 and jhj < 1/2. After dividing by a0an, we obtain a 8 1 − a−n10` < , (9) 9a0 (1.13)n where we used that a/f > 1.13 and a0 > 0.3. Let us define L = ` log 10 − n log a + log qa, 0 where qa := a/9a , with a 2 [1, 9]. It follows from (9) that 8 jeL − 1j < . (10) (1.13)n ` n Now, we claim that L is nonzero. Indeed, on the contrary, we would have 10 qa = a and so a · 10` = 9a0an. However, the minimal polynomial of a0, namely 44x3 − 2x − 1, has all its roots inside the unit circle and also jbj = jgj < 1. Thus, we can conjugate the relation a · 10` = 9a0an by the Galois automorphism a ! b in order to obtain a · 10` = 9(b0)bn. By applying absolute values in the previous expression, we get 10` ≤ jaj · 10` = 9jb0jjbjn < 9 which contradicts the fact that ` ≥ 1. When L > 0, then L < eL − 1 < 8 · (1.13)−n, while for L < 0, we can use that 1 − e−jLj = jeL − 1j < 8 · (1.13)−n to infer that 8 · (1.13)−n jLj < ejLj − 1 < < 8 · (1.13)−n+1. 1 − 8 · (1.13)−n Symmetry 2020, 12, 1774 5 of 7 Hence, we have jLj < 8 × (1.13)−n+1. Therefore log jLj < −(n − 1) log(1.13) + log 8. (11) Now, we can apply Lemma1 for the choice of a1 := 10, a2 := a, a3 = qa, b1 = `, b2 = −n, b3 = 1, 0 0 10 where qa := a/(9a ).