Repdigits as Product of Terms of k-Bonacci Sequences

For any integer k≥2, the sequence of the k-generalized Fibonacci numbers (or k-bonacci numbers) is defined by the k initial values F−(k−2)(k)=⋯=F0(k)=0 and F1(k)=1 and such that each term afterwards is the sum of the k preceding ones. In this paper, we search for repdigits (i.e., a number whose decimal expansion is of the form aa…a, with a∈[1,9]) in the sequence (Fn(k)Fn(k+m))n, for m∈[1,9]. This result generalizes a recent work of Bednařík and Trojovská (the case in which (k,m)=(2,1)). Our main tools are the transcendental method (for Diophantine equations) together with the theory of continued fractions (reduction method).

Abstract: For any integer k ≥ 2, the sequence of the k-generalized Fibonacci numbers (or k-bonacci numbers) is defined by the k initial values F (k) −(k−2) = · · · = F (k) 0 = 0 and F (k) 1 = 1 and such that each term afterwards is the sum of the k preceding ones. In this paper, we search for repdigits (i.e., a number whose decimal expansion is of the form aa . . . a, with a ∈ [1,9]) in the sequence (F (k) n F (k+m) n

Introduction
We start by recalling that the Fibonacci sequence (F n ) n is defined by the recurrence with initial values F 0 = 0 and F 1 = 1 (see, e.g., [1][2][3]). This sequence admits many generalizations and one of the most known is its higher order version. The Fibonacci sequence is a binary (or second order) recurrence and then, for any integer k ≥ 2, the sequence of the k-generalized Fibonacci numbers (or k-bonacci numbers) is defined by the kth order recurrence Clearly, for k = 2, we have the Fibonacci numbers and for k = 3, we have Tribonacci numbers (which is one of the most well-studied generalizations of Fibonacci numbers).
On the other hand, a repdigit (short for "repeated digit") is a number of the form where ≥ 1 and a ∈ [1,9] (here, as usual, for integers x < y, we denote [x, y] = {x, x + 1, . . . , y}), that is, a number with only one distinct digit (in this case a) in its decimal expansion. We point out that many authors have been interested in solving Diophantine equations involving repdigits (their sums, products, concatenations, etc.) and some special forms of linear recurrences (like their product, sums, etc.). For some works in this direction, we refer the reader to  and the references therein.
Luca [34], in 2000, and Marques [35], in 2012, proved that the largest repdigits in the Fibonacci and Tribonacci sequence are F 10 = 55 and T 8 = 44, respectively. Recently, Bednařík and Trojovská [36] and Trojovský [37] found all repdigits of the form F n T n and F n + T n , respectively.
The aim of this paper is to continue this program and generalize the main result of [36]. More precisely, we search for repdigits which are the product of the nth k-bonacci number by the nth (k + m)-bonacci number, for m ∈ [1,9]. Our main result is the following: does not have a solution in positive integers n, a, k, , with k ≥ 2, > 1 and a, m ∈ [1,9].
The main tools in the proof is the transcendental method (lower bounds for linear logarithm of real algebraic numbers) together with the theory of continued fractions (reduction method). It is important to stress that the method can be implemented for any given range of values for m. However, we chose m ∈ [1,9] in order to avoid too much time of computation (by using Mathematica software).

Auxiliary Results
The results of this section can be found in the classical literature about this kind of Diophantine equation (see, for example, [20] and the references therein). For this reason, we shall present here these tools as succinctly as possible.
The first useful result is due to Dresden and Du [38] (Theorem 1) who proved that with |E n,k | < 1/2, where α is the dominant root of the polynomial ψ k (x) := x k − ∑ k−1 j=0 x j . Moreover, we have the notation g(x, y) := (x − 1)/(2 + (y + 1)(x − 2)). Furthermore, Bravo and Luca [39] showed that Another very useful ingredient is the following result à la Baker: Lemma 1. Let γ 1 , . . . , γ t ∈ R be algebraic numbers and let b 1 , . . . , b t be nonzero integer numbers. Let D be the degree of the number field extension Q(γ 1 , . . . , γ t )/Q and let A j ∈ be any constant such that In addition, choose a constant B for which This result is a version of a Matveev theorem [40] due to Bugeaud et al. (see its proof in [41]). In the previous statement, h(γ) denotes the logarithmic height of an -degree algebraic number γ. This function satisfies the following properties (the proof of these facts can be found in [42]): Finally, the last tool was proved by Dujella and Pethő [43] (Lemma 5(a)): Lemma 3. Let M be a positive integer and let p/q be a convergent of the continued fraction expansion of the irrational number γ such that q > 6M. Let (A, B) ∈ R >0 × R >1 and define := µq −M γq , where µ is a given real number. If > 0, then there is no solution to the Diophantine inequality in positive integers m, n and k satisfying In the previous statement, we used the notation x = min{|x − n| : n ∈ Z}. Now, we are ready to deal with the proof of the theorem.
Thus, all conditions of Lemma 3 are satisfied for A = 104 and B = 1.4 and hence there is no solution to inequality (19) for and k satisfying − 1 < M and k ≥ log(A q 454 / ) log B .
By repeating this process again for the new M = 3.8 · 10 49 (we use q 120 ), we obtain k ≤ 365 and n < 5.7 · 10 493 . To conclude, we apply one more time Lemma 3 for the new choice of M := 3.8 · 10 49 (for q 107 ) and hence k ≤ 322. This contradicts our assumption that k ≥ 323. In conclusion, there is no solution to the Diophantine Equation (3) for k ≥ 323.

Other Similar Equations: The Elementary Method
It is important to notice that an elementary method does not provide a reasonable approach to deal with the Equation (1). The possible reasons can be because the product (in the left-hand side) possesses only two terms, n can be much larger than k + m as well as the very limited knowledge about arithmetic properties of repdigits (from that equation we infer only that must be a composite number).
However, we shall provide here some similar Diophantine equations which can be solved by using basic tools.
The first one is when the order is larger than the index. More precisely Proposition 1. The Diophantine equation does not have a solution in positive integers n, a, k, , with k ≥ 2, > 1 and a ∈ [1,9].
The proof is similar to the previous one by using that either n ≤ k + 1 or k ≤ n + 1.
In the previous propositions, we used the fact that F (k) n is a power of 2, when n ≤ k + 1. However, we can use another approach (since F (k) n is never a power of two, for n > k + 1, see [39]) for equations related to the product of "many" consecutive k-bonacci numbers. More precisely: does not have a solution in positive integers n, a, k, , with k ≥ 2, > 1 and a ∈ [1,9].
For proving this, we notice that the recurrence of (F (k) n ) n is a (k + 1)-periodic sequence (mod 2). Thus, the sequence (F (k) n ) n contains infinitely many even numbers (for example, F is a multiple of 16 and so it can not be a repdigit.

Conclusions
For any integer k ≥ 2, the sequence of the k-generalized Fibonacci numbers (or kbonacci numbers) is defined by the k initial values F n F (k+m) n = a(10 − 1)/9 for positive integers k, n, , m and a, with k ≥ 2, ≥ 2 and a, m ∈ [1,9]. In particular, the only repdigits, which can be written as a product of nth terms of two generalized Fibonacci sequences with consecutive orders, has only one digit. Our approach to proving this fact is to combine the Baker's theory (on lower bounds for linear forms in the logarithms) with a reduction method from the theory of continued fractions (due to Dujella and Pethő). In the concluding section, we present some similar problems which can be solved by using only elementary tools.
Author Contributions: P.T. conceived the presented idea on the conceptualization, methodology, and investigation. Writing-review and editing and preparation of program procedures in Mathematica were done by P.C. All authors have read and agreed to the published version of the manuscript.