Curious Generalized Fibonacci Numbers

: A generalization of the well-known Fibonacci sequence is the k − Fibonacci sequence whose ﬁrst k terms are 0, . . . ,0,1 and each term afterwards is the sum of the preceding k terms. In this paper, we ﬁnd all k -Fibonacci numbers that are curious numbers (i.e., numbers whose base ten representation have the form a · · · ab · · · ba · · · a ) . This work continues and extends the prior result of Trojovský, who found all Fibonacci numbers with a prescribed block of digits, and the result of Alahmadi et al., who searched for k -Fibonacci numbers, which are concatenation of two repdigits.

However, many papers have been written on Diophantine equations involving repdigits and terms of certain linear recurrence sequences. It should be mentioned that Luca [3] in 2000 showed that 55 and 11 are the largest repdigits in the Fibonacci and Lucas sequences, respectively. Since then, this result has been generalized and extended in various directions. For example, Erduvan and Keskin [4] found all repdigits expressible as products of two Fibonacci or Lucas numbers. Faye and Luca [5] looked for repdigits in the usual Pell sequence; using some elementary methods, they concluded that there are no Pell numbers larger than 10 that are repdigits. We also mention the work of Normenyo, Luca and Togbé [6] who found all repdigits expressible as sums of three Pell numbers. Shortly afterwards, they extended their work to four Pell numbers [7].
The Fibonacci sequence has been generalized in many ways. Here, we consider, for an integer k ≥ 2, the k−Fibonacci sequence F (k) = (F n the nth k-Fibonacci number. The Fibonacci numbers are obtained for k = 2, while when, for example, k = 3, the resulting sequence is a tribonacci sequence. Diophantine problems involving k−Fibonacci numbers and repdigits were recently an active research field in number theory. For example, a conjecture (proposed by Marques [8]) about repdigits in k−Fibonacci sequences was proved by Bravo-Luca [9]. Alahmadi et al. [10] generalized recently the results mentioned above by showing that only repdigits with at least two digits as a product of consecutive k-Fibonacci numbers occur only for (k, ) = (2, 1), (3,1), extending the works [11,12], which dealt with the particular cases of Fibonacci and Tribonacci numbers. See [13] for a problem involving repdigits in generalized pell sequences. In addition, Bravo-Luca [14] found all repdigits, which are sums of two k−Fibonacci numbers (see [15] for a product version). Problems concerning powers of two and coincidences in generalized Fibonacci numbers can be found in [16,17]. Finally, Alahmadi et al. [18] determined all k−Fibonacci numbers that are concatenations of two repdigits, while Trojovský [19] found all Fibonacci numbers with a prescribed block of digits.
In this paper, we determine all curious numbers, which are k-Fibonacci numbers, i.e., which continues and extends the works in [18,19]. More precisely, we solve the Diophantine equation as follows: in positive integers n, k, m, , a and b with k ≥ 2, a, b ∈ {0, 1, . . . , 9} and a = b. Before presenting our main theorem, it is important to mention that in Equation (1), we assumed , m ≥ 1 and a = b since otherwise, the problem reduces to finding all k−Fibonacci numbers that are repdigits or concatenations of two repdigits; these problems have been already solved in [9,18] (see also [20]). In addition, note that when a = 0, our problem also reduces to determining all k−Fibonacci numbers that are concatenations of two repdigits. Thus, throughout this paper we also assume that a ≥ 1. Our result is the following. Theorem 1. The only curious generalized Fibonacci number is F As an immediate consequence of Theorem 1, we have the following corollary.

Corollary 1.
There are no curious numbers that are powers of two.

Preliminary Results
In this section, we present some basic properties of the k−Fibonacci sequence and give some important estimations needed for the sequel. Additionally, we present a lower bound for a nonzero linear form in logarithms of algebraic numbers and state two reduction lemmas, which will be the key tool used in this paper to reduce the upper bounds. All these facts will be used in the proof of Theorem 1.

On k-Fibonacci Numbers
The first direct observation about the k−Fibonacci sequence is that the first k + 1 non-zero terms in F (k) are powers of two, namely, the following: while the next term is F In fact, the inequality (see [16]) n < 2 n−2 holds for all n ≥ k + 2.
Next, F (k) is a linear recurrence sequence of the following characteristic polynomial: The classic study of linear recurrence sequences (see [21]) is based on knowledge of the roots of their characteristic polynomials. While studying the roots of Ψ k (z), it is usual to work with the shifted polynomial as follows: Except for the extra root at z = 1, ψ k (z) has the same roots as Ψ k (x). By Descartes' rule of signs, the polynomial Ψ k (z) has exactly one positive real root, for example, z = α(k). Since Ψ k (1) = 1 − k and Ψ k (2) = 1, it follows that α(k) ∈ (1, 2). In fact, it is known that 2(1 − 2 −k ) < α(k) < 2 (see [22] (Lemma 2.3) or [23] (Lemma 3.6)). Thus, α(k) approaches 2 as k tends to infinity. To simplify the notation, we shall omit the dependence on k of α.
Miles [24] showed that the roots of Ψ k (z) are distinct, and the remaining k − 1 roots of Ψ k (z) different from α lie inside the unit disk. He showed this by reducing the equation Ψ k (z) = 0 to a form where Rouch's theorem could be applied. This fact was reproved by Miller [25] by an elementary argument. In particular, α is a Pisot number of degree k, and Ψ k (z) is an irreducible polynomial over Q[z].
We consider for k ≥ 2, the function f k (z) := (z − 1)/(2 + (k + 1)(z − 2)). With this notation, Dresden and Du proved in [26] the following: which for all n ≥ 1 and k ≥ 2. From (4), we can write the following: Furthermore, we obtain the following: which hold for all k ≥ 2, where α =: α 1 , α 2 , . . . , α k are all the zeros of Ψ k (z). So, by computing norms from Q(α) to Q, we see that the number f k (α) is not an algebraic integer. Proofs for this fact and inequalities (6) can be found in [27]. Additionally, it was proved in [9] the following: n ≤ α n−1 holds for all n ≥ 1 and k ≥ 2.
We finish this subsection with the following estimate, due to Bravo, Gómez and Luca [28], which will be one of the key points in addressing the large values of k (see also [29]). Lemma 1. Let k ≥ 2 and suppose that r < 2 k/2 . Then, the following holds:

Linear Forms in Logarithms
We need to use a Baker-type lower bound for a nonzero linear form in logarithms of algebraic numbers. We begin by recalling some basic notions from algebraic number theory.
Let η be an algebraic number of degree d over Q with minimal primitive polynomial over the integers m(z) : , where the leading coefficient a 0 is positive and the η (i) 's are the conjugates of η. The logarithmic height of η is given by the following: In order to illustrate this, we can use the facts that the minimal primitive polynomial of α is Ψ k (z), Q(α) = Q( f k (α)) and that | f k (α (i) )| ≤ 1 for all i = 1, . . . , k and k ≥ 2 (see (6)), to prove the following: See [27] for further details of the proof of (8). In addition, if η = p/q is a rational number with gcd(p, q) = 1 and q > 0, then h(η) = log max{|p|, q}. Finally, the following properties of h(·) are used in the next sections: As a consequence of the above properties, one can easily deduce the next lemma.
Lemma 1. Let k ≥ 2 and s = 0 be the integers and suppose that |s| ≤ 10 ε for some integer ε ≥ 1. Then, we have the following: Furthermore, if ε = 1, then we have the following: Our main tool is the following lower bound for a non-zero linear form in logarithms of algebraic numbers, due to Matveev [30].
Theorem 1 (Matveev's theorem). Let K be a number field of degree D over Q, γ 1 , . . . , γ t be positive real numbers of K, and b 1 , . . . , b t rational integers. Put Λ : Then, assuming that Λ = 0, we have the following:

Reduction Tools
To lower the bounds arising from applying Theorem 1, we will use a result from the theory of continued fractions. The following lemma is a slight variation of a result, due to Dujella and Pethő [31]. We shall use the version given by Bravo, Gómez and Luca (see [27] [Lemma 1]).

Lemma 2.
Let τ be an irrational number, and let A, B, µ be real numbers with A > 0 and B > 1. Assume that M is a positive integer. Let p/q be a convergent of the continued fraction of τ such that q > 6M and put := µq − M τq , where · denotes the distance from the nearest integer. If > 0, then there is no solution of the inequality in positive integers u, v and w with u ≤ M and w ≥ log(Aq/ )/ log B.
The above lemma cannot be applied when µ is an integer linear combination of 1 and τ since then, < 0. In this case, we use the following nice property of continued fractions (see Theorem 8.2.4 and top of page 263 in [32]).
. If x, y ∈ Z with x > 0, then we have the following: We finish with the following simple facts concerning the exponential function. We list it as a lemma for further reference, and its proof can be found in [33].

Lemma 4.
For any non-zero real number x, we have the following:

Proof Theorem 1
Assume throughout that (n, k, a, b, , m) is a solution of Equation (1). First, we note that n ≤ 3 is impossible since F (k) n must have at least 3 digits in its decimal representation. Thus, we assume n ≥ 4. We now want to establish a relationship between the variables of (1). For this purpose, we use Equations (1), (3) and (7) to obtain the following: giving the following: 2 + m < (n − 2) log 2 log 10 + 1 and n − 2 < (2 + m) log 10 log α .
Thus, from now on, we suppose that n ≥ k + 2.

Bounding n in Terms of k
In this subsection, we want to find an upper bound for n in terms of k. To do this, we use (5) and rewrite (1) in two different forms, namely, the following: where we have put X := a · 10 − (a − b). For future calculations, it will be important to note the following: We now take absolute value in relations given by (13); doing some straightforward calculations, we obtain the following: 9 f k (α)α n−1 − a · 10 2 +m < 11 · 10 +m , 9 f k (α)α n−1 − 10 +m X < 11 · 10 .
Dividing both sides of each one of the above inequalities (15) by a · 10 2 +m and 10 +m X, respectively, and rearranging some terms, we obtain the following: At this point, we claim that the left-hand sides of (16) and (17) are not zero. Indeed, if these were zero, we would then obtain the following, respectively: a · 10 2 +m = 9 f k (α)α n−1 and 10 +m X = 9 f k (α)α n−1 , Conjugating with an automorphism σ of the Galois group of Ψ k (x) over Q such that σ(α) = α i for some i > 1, taking absolute values and using the fact that |9 f k (α i )α n−1 i | < 9, we obtain the following, respectively: a · 10 2 +m < 9 and 10 +m X < 9, However, these lead to a contradiction since the following are true: a · 10 2 +m ≥ 10 3 and 10 +m X ≥ 10 2 .

An Inequality for in Terms of k
In order to apply Matveev's theorem on (16) with the parameters γ 1 , γ 2 and γ 3,1 , we take A 1 , A 2 as mentioned before and A 3 := 6k log k (by (18)) to obtain the following: where we use that 1 + log k < 3 log k and 1 + log n < 2 log n hold for all k ≥ 2 and n ≥ 4, respectively. Comparing (16) and (20) and performing the respective calculations, we obtain the following: < 4 × 10 12 k 4 log 2 k log n.

An Inequality for m in Terms of k
In light of (19) and (21), we deduce the following: h(γ 3,2 ) < 10 13 k 4 log 2 k log n.
This allows us to choose now A 3 := 10 13 k 5 log 2 k log n. We then apply Matveev's theorem on (17) with the parameters γ 1 , γ 2 and γ 3,2 to obtain the following: Using now (17) and (22), we have the following:

An Inequality for n in Terms of k
We finally use (21) and (23) combined with (10) to assert the following: n log 2 n < 1.2 × 10 26 k 8 log 3 k.
In order to upper bound n polinomially in terms of k, we use an analytical argument of Guzmán and Luca [34] who proved that if m ≥ 1, T > (4m 2 ) m and T > x/ log m x, then x < 2 m T log m T. In our case, we take T := 1.2 × 10 26 k 8 log 3 k and m := 2 to obtain from (24) the following lemma. Lemma 5. If (n, k, a, b, , m) is a solution of Equation (1) with n ≥ k + 2, then the following holds: 2 + m < n < 5 × 10 30 k 8 log 5 k.

The Case of Large k
Suppose that k > 430. Note that for such values of k, we have the following: Then, by Lemma 5, we obtain that the inequality n < 2 k/2 is satisfied when k > 430, and therefore, we are in the hypothesis of Lemma 1. Applying the above lemma and Equation (1), we obtain the following: · 10 2 +m · 2 −(n−2) − 1 < 3 · 10 +m 2 n−2 + 1 2 k/2 < 30 10 where we have used that 10 +m /2 n−2 < 10/10 (see (9)). Consequently, we have the following: a 9 · 10 2 +m · 2 −(n−2) − 1 < 30 where θ := (log 10)/(log 2) and λ := min{k/2, θ }. Again, in order to use the result of Matveev, we take t := 3 and the following: We begin by noticing that the three numbers γ 1 , γ 2 , γ 3 are positive rational numbers, so we can take K := Q for which D := 1. To see why the left-hand side of (25) is not zero, note that, otherwise, we would obtain a · 10 2 +m = 9 · 2 n−2 , which is impossible since its left-hand side is divisible by 5 while its right-hand side is not.
Clearly, we can take A 1 := log 9, A 2 := log 10 and A 3 := log 2. Here, we can also take B := n. Then, Matveev's theorem together with a straightforward calculation gives the following: a 9 · 10 2 +m · 2 −(n−2) − 1 > exp −1.1 × 10 12 log n , where we use 1 + log n < 2 log n holds for all n ≥ 4. Comparing (25) and (26), taking logarithms and then performing the respective calculations, we arrive at the following: Note that, if λ = k/2, then k < 3.6 × 10 12 log n. Since log n < 73 log k holds for all k > 430 by Lemma 5, we obtain k < 2.7 × 10 14 log k, giving k < 10 16 . For the case when λ = θ , we have < 5.5 × 10 11 log n. Here, proceeding as in (25), we obtain the following: The same argument used before also shows that the left-hand side of (27) is not zero. With a view toward applying Matveev's theorem, we take the same parameters as in the previous application, except by γ 1 and b 2 , which, in this case, are given by X/9 and + m, respectively. As before, K := Q, D := 1, A 2 := log 10, A 3 := log 2 and B := n. Moreover, by (14), we have the following: h(γ 1 ) = log X ≤ ( + 1) log 10 < 1.3 × 10 12 log n.

Reducing the Bound on k.
We now want to reduce our bound on k by using Lemma 2. Let the following hold: Γ 1 := log(a/9) + (2 + m) log 10 − (n − 2) log 2.
Applying now Lemma 2 to inequality (28) for each a ∈ {1, 2, . . . , 8} we find, with the help of Mathematica, that λ ≤ 960. For the case a = 9, we cannot use Lemma 2 because the corresponding value of is always negative. However, one can see that if a = 9, then the resulting inequality from (28) has the following shape: with γ := θ being an irrational number and x := 2 + m, y := n − 2 ∈ Z. In order to apply Lemma 3 on the left-hand side of (29), we define [a 0 , a 1 , a 2 , a 3 , .
In any case, by the same arguments used to get (29), we obtain 2 k/2 < 1.7 × 10 290 , which implies that k ≤ 1928. Thus, k ≤ 1955 holds for any choice of λ. Then, by Lemma 5, 2 + m < 2.7 × 10 61 := M. With this new choice of M, Lemma 2 applied to inequality (28) implies that λ ≤ 222 (including the case a = 9). If λ = k/2, then k ≤ 444, while if λ = θ , we have that ≤ 66. We finally apply Lemma 2 with M := 2.7 × 10 61 to inequality (30) for all a, b ∈ {0, 1, . . . , 9} with a = b, a ≥ 1 and 1 ≤ ≤ 66, except in the special cases mentioned above. With the help of Mathematica, we find that k ≤ 457. The same upper bound for k holds in the special cases. So, k ≤ 457 holds for any choice of λ. Finally, taking M := 8.2 × 10 55 and repeating the previous procedure, we obtain k ≤ 422, which is a contradiction. Hence, the Equation (1) has no solutions for k > 430.

The Case of Small k
Suppose now that k ∈ [2,430]. Note that for each of these values of k, Lemma 5 gives us absolute upper bounds for n. However, these upper bounds are so large and will be reduced by using Lemma 2 once again. To do this, we put the following: log α − (2 + m) log 10 + log((9 f k (α))/a) (n − 1) log α − ( + m) log 10 + log((9 f k (α))/X) .
Note that τ 1 clearly is an irrational number because α and 10 are multiplicatively independent. Next, we shall apply Lemma 2 to (31) and (32). For this purpose, we put also M k := 5 × 10 30 k 8 log 5 k, which is an upper bound on n − 1 by Lemma 5.
Finally, we use Mathematica to display the values of F