Repdigits as products of consecutive Padovan or Perrin numbers

A repdigit is a positive integer that has only one distinct digit in its decimal expansion, i.e., a number of the form a(10m-1)/9\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$a(10^m-1)/9$$\end{document}, for some m≥1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$m\ge 1$$\end{document} and 1≤a≤9\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$1 \le a \le 9$$\end{document}. Let Pnn≥0\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\left( P_n\right) _{n\ge 0}$$\end{document} and Enn≥0\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\left( E_n\right) _{n\ge 0}$$\end{document} be the sequence of Padovan and Perrin numbers, respectively. This paper deals with repdigits that can be written as the products of consecutive Padovan or/and Perrin numbers.


Introduction
A positive integer is called a repdigit if it has only one distinct digit in its decimal expansion. The sequence of numbers with repeated digits is included in Sloane's On-Line Encyclopedia of Integer Sequences (OEIS) [13] as the sequence A010785.
Let (P n ) n≥0 be the Padovan sequence satisfying the recurrence relation P n+3 = P n+1 + P n with initial conditions P 0 = 0 and P 1 = P 2 = 1. Let (E n ) n≥0 be the Perrin sequence following the same recursive pattern as the Padovan sequence, but with initial conditions E 0 = 2, E 1 = 0, and E 2 = 1. P n and E n are called nth Padovan number and nth Perrin number, respectively. The Padovan and Perrin sequences are included in the OEIS [13] as the sequences A000931 and A001608, respectively.
Finding some specific properties of sequences is of big interest since the famous result of Bugeaud, Mignotte, and Siksek [2]. One can also see [1][2][3][4][5][6][7][8][9]11,12]. Marques and the second author [9] studied repdigits as products of consecutive Fibonacci numbers. Irmak and the second author [5] studied repdigits as products of consecutive Lucas numbers. Rayaguru and Panda [11] studied repdigits as products of consecutive Balancing and Lucas-Balancing numbers. It is natural to ask what will happen if we consider Padovan and Perrin numbers. This is the aim of this paper. Therefore, in this paper, we investigate repdigits which can be written as the product of consecutive Padovan or/and Perrin numbers. More precisely, we prove the following results. has no solution in positive integers n, , m, a, with 1 ≤ a ≤ 9 and m ≥ 2.

Theorem 1.3 The Diophantine equation
has no solution in positive integers n, k, , m, a, with 1 ≤ a ≤ 9 and m ≥ 2.

Theorem 1.4 The Diophantine equation
has no solution in positive integers n, k, , m, a, with 1 ≤ a ≤ 9 and m ≥ 2.
Here is the outline of this paper. In Sect. 2, we will recall the results that will be used to prove Theorems 1.1, 1.2, 1.3, and 1.4. In Sect. 3, first, we will use Baker's method and 2-adic valuation of Padovan numbers to obtain a bound for n that is too high to completely solve Eq. (1.1). We will then need to apply twice the reduction method of de Weger to find a very low bound for n, which enables to run a program to find the small solutions of Eq. (1.1). We will use the same method in the next sections to prove the remaining theorems.
Computations are done with the help of a computer program in Maple.

The tools
We start by recalling some useful properties of Padovan and Perrin sequences. The characteristic equation of {P n } n≥0 and {E n } n≥0 is z 3 − z − 1 = 0 and has one real root α and two complex roots β and γ = β. The Binet formulae for the Padovan and Perrin numbers are respectively: and It is easy to see that α ∈ (1.32, 1.33) , |β| = |γ | ∈ (0.86, 0.87) , c α ∈ (0.72, 0.73) and |c β | = |c γ | ∈ (0.24, 0.25). By the facts that β = α −1/2 e iθ and γ = α −1/2 e −iθ , for some θ ∈ (0, 2π), we can show that: For a prime number p and a non-zero integer r , the p-adic order υ p (r ) is the exponent of the highest power of a prime p which divides r . The following two results, due to Irmak [4], characterize the 2-adic order of Padovan and Perrin numbers, respectively. Lemma 2.1 For n ≥ 1, we have: if n ≡ 6 (mod 7).

Lemma 2.2
For n ≥ 1, we have: The next tools are related to the transcendental approach to solve Diophantine equations. For any non-zero algebraic number γ of degree d over Q, whose minimal polynomial over Z is a d j=1 X − γ ( j) , we denote by: If γ b 1 1 · · · γ b s s = 1, then: After getting the upper bound of n, which is generally too large, the next step is to reduce it. For this reduction purpose, we present a variant of the reduction method of Baker and Davenport due to de Weger [14]).

Lemma 2.5 [14, Lemma 3.3]
Suppose that: Then, the solutions of (2.8) and (2.9) satisfy: We conclude this section by recalling the following lemma that we need in the sequel:
Dividing through by c α α n+ ( −1)/2 and taking the absolute value, we deduce that: where:: To find a lower bound for 1 , we take the parameters s := 3: Conjugating the above relation by the Galois automorphism σ := (αβ), and then taking absolute values on both sides of the resulting equality, we obtain: This is a contradiction. Thus, 1 = 0. Next, we give estimates to A i for i = 1, 2, 3. By the properties of the absolute logarithmic height, we have: Now, we need to estimate h(c α ). For that, the minimal polynomial of c α is 23X 3 − 23x 2 + 6X − 1. Therefore, h(c α ) = 1 3 log 23, and thus: h(γ 1 ) ≤ 2 log 9 + 2 log 23.

Proof of Theorem 1.2
In this section, we will use the same method for the proof of Theorem 1.1 to completely prove Theorem 1.2. However, for the sake of completeness, we will give some details.

Absolute bounds on the variables
First of all, we give the number of factors in the Diophantine equation (1.2).
With the help of Maple, we get n < 7.1 × 10 15 .

Proofs of Theorems 1.3 and 1.4
We will use the same method as above to only show Theorem 1.3 as the proof of Theorem 1.4 is similar.