On Repdigits as Sums of Fibonacci and Tribonacci Numbers

: 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 F n + T n , for some n ≥ 1, where ( F n ) n ≥ 0 and ( T n ) n ≥ 0 are the sequences of Fibonacci and Tribonacci numbers, respectively.


Introduction
A palindromic number is a number that has the same form when written forwards or backwards, i.e., of the form c 1 c 2 c 3 . . . c 3 c 2 c 1 (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 n = a 10 − 1 9 , for some ≥ 1 and a ∈ [1,9] (as usual, we set [a, b] = {a, a + 1, . . . , b}, 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 (it is an easy exercise that if R is prime, then so is ).

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 for some ≥ 1 and a ∈ [1, g − 1].
In this work, we shall study two well-known recurrence sequences. The first one is the omnipresent sequence (F n ) n . These numbers are defined by the second order linear recurrence F n+2 = F n+1 + F n , for all n ≥ 0 with initial values F 0 = 0 and F 1 = 1 (see, e.g., [25]). The sequence of Tribonacci numbers (T n ) n (generalizes the Fibonacci sequence) is defined by the third-order recurrence T n+3 = T n+2 + T n+1 + T n , for all n ≥ 0 which begins with T 0 = 0 and T 1 = T 2 = 1 (see, e.g., [26,27]). We remark that Luca showed that F 10 = 55 is the largest repdigit in the Fibonacci sequence, while Marques [28] proved that the largest repdigit in the Tribonacci sequence is T 8 = 44.
In the previous statement, the logarithmic height of a t-degree algebraic number α is defined by where a is the leading coefficient of the minimal polynomial of α (over Z), (α (j) ) 1≤j≤t are the algebraic conjugates of α. 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
Our last ingredient is a reduction method provided by Dujella and Pethő [32], which is itself a variation of the result of Baker and Davenport [33]. For x ∈ R, set x = min{|x − n| : n ∈ Z} = |x − x | for the distance from x to the nearest integer.

Lemma 3.
For a positive integer M, let p/q be a convergent of the continued fraction of γ ∈ Q, such that q > 6M, and let µ, A and B be real numbers, with A > 0 and B > 1. If the number = µq − M γq is positive, then there is no solution to the Diophantine inequality See Lemma 5 of [32]. Now, we are in a position to prove our main theorem.

Reducing the Bound
The next step is to use some reduction method in order to reduce the bounds for n and . For that, let us suppose, without loss of generality, that Λ > 0 (the other case can be handled in the same way by observing that 0 < Λ = −Λ).
This finishes the proof.

Conclusions
In this paper, we solve the Diophantine equation F n + T n = a(10 − 1)/9, where (F n ) n and (T n ) n are the Fibonacci and Tribonacci sequences, respectively. In other words, we found all repdigits (i.e., positive integers with only one distinct digit in its decimal expansion), which can be written as the sum of a Fibonacci number and a Tribonacci number with the same index. In particular, we proved that the only repdigits with the desired property are the trivial ones, i.e., with only one digit ( = 1). To prove that, we use Baker's theory and a reduction method due to Dujella and Pethő.