Arab. J. Math. (2021) 10:469–480 https://doi.org/10.1007/s40065-021-00317-1 Arabian Journal of Mathematics Salah Eddine Rihane · Alain Togbé Repdigits as products of consecutive Padovan or Perrin numbers Received: 22 April 2020 / Accepted: 8 February 2021 / Published online: 28 March 2021 © The Author(s) 2021 Abstract A repdigit is a positive integer that has only one distinct digit in its decimal expansion, i.e., a number ( m − )/ ≥ ≤ ≤ of the form a 10 1 9, for some m 1and1 a 9. Let (Pn)n≥0 and (En)n≥0 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. Mathematics Subject Classification 11B39 · 11J86 1 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 (Pn)n≥0 be the Padovan sequence satisfying the recurrence relation Pn+3 = Pn+1 + Pn with initial conditions P0 = 0andP1 = P2 = 1. Let (En)n≥0 be the Perrin sequence following the same recursive pattern as the Padovan sequence, but with initial conditions E0 = 2, E1 = 0, and E2 = 1. Pn and En 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 [1–9,11,12]. 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. Theorem 1.1 The Diophantine equation 10m − 1 Pn ···Pn+(− ) = a , (1.1) 1 9 S. E. Rihane Department of Mathematics, Institute of Science and Technology, University Center of Mila, Mila, Algeria E-mail: [email protected] A. Togbé (B) Department of Mathematics, Statistics, and Computer Science, Purdue University Northwest, 1401 S, U.S. 421, Westville, IN 46391, USA E-mail: [email protected] 123 470 Arab. J. Math. (2021) 10:469–480 has no solution in positive integers n,,m, a, with 1 ≤ a ≤ 9 and m ≥ 2. Theorem 1.2 The only solution of the Diophantine equation: 10m − 1 En ···En+(− ) = a , (1.2) 1 9 in positive integers n,,m, a, with 1 ≤ a ≤ 9 and m ≥ 2 is (n,,m, a) = (11, 1, 2, 2), i.e., E11 = 22. Theorem 1.3 The Diophantine equation 10m − 1 En ···En+(k− ) Pn+k ···Pn+k+(− ) = a (1.3) 1 1 9 has no solution in positive integers n, k,,m, a, with 1 ≤ a ≤ 9 and m ≥ 2. Theorem 1.4 The Diophantine equation 10m − 1 Pn ···Pn+(k− ) En+k ···En+k+(− ) = a (1.4) 1 1 9 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,and1.4. 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. 2 The tools We start by recalling some useful properties of Padovan and Perrin sequences. The characteristic equation of 3 {Pn}n≥0 and {En}n≥0 is z − 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: s s s Ps = cαα + cβ β + cγ γ , for all s ≥ 0 (2.1) and s s s Es = α + β + γ , for all s ≥ 0, (2.2) where 1 + α 1 + β 1 + γ cα = , cβ = , cγ = = cβ . −α2 + 3α + 1 −β2 + 3β + 1 −γ 2 + 3γ + 1 It is easy to see that α ∈ (1.32, 1.33) , |β|=|γ |∈(0.86, 0.87) , cα ∈ (0.72, 0.73) and |cβ |=|cγ |∈ (0.24, 0.25). By the facts that β = α−1/2eiθ and γ = α−1/2e−iθ ,forsomeθ ∈ (0, 2π), we can show that: s 1 P = cαα + e , with |e | < , for all s ≥ 1 (2.3) s s s αs/2 and 2 E = αs + e , with |e | < , for all s ≥ 1. (2.4) s s s αs/2 Further, we have: s−2 s−1 α ≤ Ps ≤ α , for all s ≥ 4 (2.5) 123 Arab. J. Math. (2021) 10:469–480 471 and s−2 s+1 α ≤ Es ≤ α , for all s ≥ 2. (2.6) For a prime number p and a non-zero integer r,thep-adic order υp(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: ⎧ ≡ , , , ( ), ⎨⎪ 0 if n 0 1 2 5 mod 7 υ2(n + 4) + 1 if n ≡ 3 (mod 7), υ2 (Pn) = ⎩⎪ υ2((n + 3)(n + 17)) + 1 if n ≡ 4 (mod 7), υ2((n + 1)(n + 8)) + 1 if n ≡ 6 (mod 7). Lemma 2.2 For n ≥ 1, we have: ⎧ ⎪ 0 if n ≡ 0, 3, 5, 6 (mod 7), ⎪ ⎨ 1 if n ≡ 2 (mod 14), υ2 (En) = 2 if n ≡ 9 (mod 14), ⎪ υ ( − ) + ≡ ( ), ⎩⎪ 2 n 1 1 if n 1 mod 7 1 if n ≡ 4 (mod 7). The next tools are related to the transcendental approach to solve Diophantine equations. For any non-zero γ Q Z d − γ ( j) algebraic number of degree d over , whose minimal polynomial over is a j=1 X , we denote by: ⎛ ⎞ d 1 ( ) h(γ ) = ⎝log |a|+ log max 1, γ j ⎠ d j=1 the usual absolute logarithmic height of γ . To prove our main results, we use lower bounds for linear forms in logarithms to bound the index n appearing in Eqs. (1.1), (1.2), (1.3), and (1.4). We need the following result of Bugeaud, Mignotte, and Siksek [2], which is a modified version of the result of Matveev [10]. Lemma 2.3 Let γ1,...,γs be real algebraic numbers and let b1,...,bs be non-zero rational integer numbers. Let D be the degree of the number field Q(γ1,...,γs) over Q and let A j be a positive real number satisfying: A j = max{Dh(γ ), | log γ |, 0.16} for j = 1,...,s. Assume that: B ≥ max{|b1|,...,|bs|}. γ b1 ···γ bs = If 1 s 1, then: |γ b1 ···γ bs − |≥ (− ( , )( + ) ··· ), 1 s 1 exp C s D 1 log B A1 As where C(s, D) := 1.4 · 30s+3 · s4.5 · D2(1 + log D). 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]). Let ϑ1,ϑ2,β ∈ R be given, and let x1, x2 ∈ Z be unknowns. Let: = β + x1ϑ1 + x2ϑ2. (2.7) Let c, δ be positive constants. Set X = max{|x1|, |x2|}.LetX0, Y be positive numbers. Assume that: | | < c · exp(−δ · Y ), (2.8) Y ≤ X ≤ X0. (2.9) 123 472 Arab. J. Math. (2021) 10:469–480 When β = 0in(2.7), we get = x1ϑ1 + x2ϑ2. Put ϑ =−ϑ1/ϑ2. We assume that x1 and x2 are coprime. Let the continued fraction expansion of ϑ be given by [a0, a1, a2,...], and let the kth convergent of ϑ be pk/qk for k = 0, 1, 2,.... We may assume without loss of generality that |ϑ1| < |ϑ2| and that x1 > 0. We have the following results. Lemma 2.4 [14, Lemma 3.2] Let: A = max ak+1, 0≤k≤Y0 where √ log( 5X0 + 1) Y0 =−1 + √ . 1+ 5 log 2 If (2.8) and (2.9) hold for x1,x2 and β = 0, then: 1 c(A + 2)X Y < log 0 . (2.10) δ |ϑ2| When β = 0in(2.7), put ϑ =−ϑ1/ϑ2 and ψ = β/ϑ2. Then, we have = ψ − x1ϑ + x2. ϑ2 Let p/q be a convergent of ϑ with q > X0. For a real number x,welet x =min{|x − n|, n ∈ Z} be the distance from x to the nearest integer. We have the following result. Lemma 2.5 [14, Lemma 3.3] Suppose that: 2X qψ > 0 . q Then, the solutions of (2.8) and (2.9) satisfy: 1 q2c Y < log .