Tribonacci numbers that are concatenations of two repdigits

Let \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$ (T_{n})_{n\ge 0} $$\end{document}(Tn)n≥0 be the sequence of Tribonacci numbers defined by \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$ T_0=0 $$\end{document}T0=0, \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$ T_1=T_2=1$$\end{document}T1=T2=1, and \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$ T_{n+3}= T_{n+2}+T_{n+1} +T_n$$\end{document}Tn+3=Tn+2+Tn+1+Tn for all \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$ n\ge 0 $$\end{document}n≥0. In this note, we use of lower bounds for linear forms in logarithms of algebraic numbers and the Baker-Davenport reduction procedure to find all Tribonacci numbers that are concatenations of two repdigits.


Introduction
A repdigit is a positive integer R that has only one distinct digit when written in its decimal expansion. That is, R is of the form for some positive integers d, with ≥ 1 and 0 ≤ d ≤ 9. The sequence of repdigits is sequence A010785 on the On-Line Encyclopedia of Integer Sequences (OEIS) [8].
The sequence of Tribonacci numbers is sequence A000073 on the OEIS. The first few terms of this sequence are given by

Main result
In this paper, we study the problem of finding all Tribonacci numbers that are concatenations of two repdigits. More precisely, we completely solve the Diophantine equation T n = d 1 · · · d 1 1 times d 2 · · · d 2 2 times = d 1 10 q − 1 9 · 10 2 + d 2 10 2 − 1 9 , (2.1) in non-negative integers (n, d 1 , d 2 , 1 , 2 ) with n ≥ 0, 1 ≥ 2 ≥ 1, and d 1 , Our main result is the following. Our method of proof involves the application of Baker's theory for linear forms in logarithms of algebraic numbers, and the Baker-Davenport reduction procedure. Computations are done with the help of a computer program in Mathematica. Let (F n ) n≥0 be the sequence of Fibonacci numbers given by F 0 = 0, F 1 = 1, and F n+2 = F n+1 + F n for all n ≥ 0, and (B n ) n≥0 be the sequence of balancing numbers given by B 0 = 0, B 1 = 1, and B n+2 = 6B n+1 − B n for all n ≥ 0. This paper is inspired by the results of Alahmadi et al. [1], in which they show that the only Fibonacci numbers that are concatenations of two repdigits are F n ∈ {13, 21, 34, 55, 89, 144, 233, 377}, and Rayaguru and Panda [10], who showed that B n ∈ {35} is the only balancing number that can be written as a concatenation of two repdigits. Other related interesting results in this direction include: the result of Bravo and Luca [3], the result of Trojovský [11], the result of Qu and Zeng [9], and the result of Boussayoud et al. [2].

The Tribonacci sequence
Here, we recall some important properties of the Tribonacci sequence {T n } n≥0 . The characteristic equation and Further, the Binet formula for the general terms of the Tribonacci sequence is given by T n = aα n + bβ n + cγ n for all n ≥ 0, and the minimal polynomial of a over the integers is given by has zeros a, b, c with max{|a|, |b|, |c|} < 1. Numerically, the following estimates hold: From (3.1), (3.2), and (3.5), it is easy to see that the contribution the complex conjugate roots β and γ , to the right-hand side of (3.3), is very small. In particular, setting e(n) := T n − aα n = bβ n + cγ n then |e(n)| < 1 α n/2 , (3.6) holds for all n ≥ 1. The proof of the last inequality in (3.6) follows from the fact that |β| = |γ | = α − 1 2 and |b| = |c| < 0.36 (by 3.5). That is, for any n ≥ 1, Furthermore, by induction, one can prove that α n−2 ≤ T n ≤ α n−1 holds for all n ≥ 1. (3.7) Let K := Q(α, β) be the splitting field of the polynomial over Q. Then, [K, Q] = 6. Furthermore, [Q(α) : Q] = 3. The Galois group of K over Q is given by Thus, we identify the automorphisms of G with the permutations of the zeros of the polynomial . For example, the permutation (αγ ) corresponds to the automorphism σ : α → γ, γ → α, β → β.

Linear forms in logarithms
Let η be an algebraic number of degree d with minimal primitive polynomial over the integers where the leading coefficient a 0 is positive and the η (i) 's are the conjugates of η. Then the logarithmic height of η is given by In particular, if η = p/q is a rational number with gcd( p, q) = 1 and q > 0, then h(η) = log max{| p|, q}. The following are some of the properties of the logarithmic height function h(·), which will be used in the next section of this paper without reference: We recall the result of Bugeaud et al. ([4], Theorem 9.4), which is a modified version of the result of Matveev [7], which is one of our main tools in this paper. Theorem 3.1 Let η 1 , · · · , η t be positive real algebraic numbers in a real algebraic number field K ⊂ R of degree D, b 1 , · · · , b t be nonzero integers, and assume that

Reduction procedure
During the calculations, we get upper bounds on our variables which are too large, thus we need to reduce them. To do so, we use some result from the theory of continued fractions. For a nonhomogeneous linear form in two integer variables, we use a slight variation of a result due to Dujella and Pethő ( [5], Lemma 5a). For a real number X , we write X := min{|X − n| : n ∈ Z} for the distance from X to the nearest integer. The following Lemma is also useful. It is due to Gúzman Sánchez and Luca ([6], Lemma 7).

The small ranges
With the help of Mathematica, we checked all the solutions to the Diophantine equation (2.1) in the ranges 0 ≤ d 2 < d 1 ≤ 9 and 1 ≤ 2 ≤ 1 ≤ n ≤ 200 and found only the solutions stated in Theorem 2.1. From now on we assume that n > 200.

The initial bound on n
We rewrite (2.1) as Thus, We prove the following lemma, which gives a relation on the size of n versus 1 + 2 .
Thus, dividing both sides by d 1 · 10 1 + 2 we get that Next, we apply Theorem 3.1 on (4.5). First, we need to check that 1 = 0. If it were, then we would get that aα n = d 1 9 · 10 1 + 2 .
Now, we apply the automorphism σ of the Galois group G on both sides and take absolute values as follows. By Lemma 4.1, we have that 1 + 2 < n. Therefore, we can take B := n. Observe that Comparing the above inequality with (4.4) gives leading to 1 log 10 < 1.96 · 10 14 (1 + log n). (4.6) Step 2 By (3.6), we rewrite (4.1) as from which we deduce that Thus, dividing both sides by 9aα n we get that (4.7) Next, we apply Theorem 3.1 on (4.8). First, we need to check that 2 = 0. If not, then we would get that As before, we apply the automorphism σ of the Galois group G on both sides and take absolute values as follows.
We have just proved the following lemma.

Reducing the bounds
The bounds given in Lemma 4.2 are too large to carry out meaningful computation. Thus, we need to reduce them. To do so, we apply Lemma 3.1 as follows. First, we return to (4.4) and put The inequality (4.4) can be rewritten as Assume that 1 ≥ 2, then the right-hand side in the above inequality is at most 7/25 < 1/2. The inequality |e z − 1| < w for real values of z and w implies that z < 2w. Thus, which implies that Dividing through by log α gives For fixed 0 ≤ d 2 < d 1 ≤ 9 and 1 ≤ 1 ≤ 35, we return to (4.7) and put 2 := 2 log 10 − n log α + log From the inequality (4.7), we have that Since n > 200, the right-hand side of the above inequality is less than 1/2. Thus, the above inequality implies that | 1 | < 4 α n , which leads to 2 log 10 − n log α + log Dividing through by log α gives, 2 log 10 log α − n + log (d 1 · 10 1 − (d 1 − d 2 ))/9a log α < 4 α n log α .
We take the same τ and its convergent p/q = p 62 /q 62 as before. We choose 2 < 10 32 := M.
Thus, we have that n ≤ 143, contradicting the assumption that n > 200. Hence, Theorem 2.1 is proved.
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