Perrin numbers which are concatenations of two distinct repdigits

. We determine all Perrin numbers that are concatenations of two repdigits.

The Padovan numbers and Perrin numbers share many similar properties. In particular, they have the same recurrence relation, the difference being that the Padovan numbers are initialized via Pad(0) = 0 and Pad(1) = Pad(2) = 1. This means that the two sequences also have the same characteristic equation.
Despite the similarities, the two sequences also have some stark differences. For instance, the Perrin numbers satisfy the remarkable divisibility property that if n is prime, then n divides P n . One can easily confirm that this does not hold for the Padovan numbers.
We ignore the d 1 = d 2 case for the time being, since it has been covered within a more general context in an upcoming paper, where we study the reverse question of repdigits which are sums of Perrin numbers. In any case, the only such Perrin number which is a solution of the above Diophantine equation is P 11 = 22.

Preliminary results.
In this section we collect some facts about Perrin numbers and other preliminary lemmas that are crucial to our main argument.

Some properties of the Perrin numbers.
Recall that the characteristic equation of the Perrin sequence is given by φ(x) := x 3 − x − 1 = 0, with zeros α, β and γ = β given by: For all n ≥ 0, Binet's formula for the Perrin sequence tells us that the n th Perrin number is given by Numerically, we have the following estimates for the quantities {α, β, γ}: 1.32 <α < 1.33, It follows that the difference between the right hand side of equation (2) and α n becomes quite small as n increases. More specifically, let e(n) := P n − α n = β n + γ n . Then, |e(n)| < 3 α n/2 for all n ≥ 1.
Lemma 2 follows from a simple inductive argument, and the fact that α 3 = α + 1, from the characteristic polynomial φ. We therefore identify the automorphisms of G with the permutation group of the zeroes of φ. We highlight the permutation (αβ), corresponding to the automorphism σ : α → β, β → α, γ → γ, which we use later to obtain a contradiction on the size of the absolute value of a certain bound.

Linear forms in logarithms.
We use the standard procedure of obtaining certain estimates for linear forms in (nonzero) logarithms associated with the sequence of interest. Whilst the upper bounds are obtained via a manipulation of Binet's formula for the given sequence, the lower bounds require the use of the celebrated Baker's theorem on linear forms in logarithms. We begin this section by defining the (logarithmic) Weil height of an algebraic number.
Let η be an algebraic number of degree d with minimal polynomial where the leading coefficient a 0 is positive and the α j 's are the conjugates of α. The logarithmic height of η is given by Note that, if η = p q ∈ Q is a rational number in reduced form with q > 0, then the above definition reduces to h(η) = log max{|p|, q}. We list some well known properties of the height function below, which we shall subsequently use without reference: The version of Baker's theorem most appropriate for our applications is the one proved by Bugeaud, Mignotte and Siksek in ([1], Theorem 9.4, pp. 989), we quote it below.

Reduction procedure.
The bounds on the variables obtained via Baker's theorem are usually too large for computational purposes. In order to get further refinements, we use the Baker-Davenport reduction procedure. The variant we apply here is the one due to Dujella and Pethő ([4], Lemma 5a). For a real number r, we denote by r the quantity min{|r − n| : n ∈ Z}, the distance from r to the nearest integer.
Lemma 4 (Dujella, Pethő). Let κ 0, A, B and µ be real numbers such that A > 0 and B > 1. Let M > 1 be a positive integer and suppose that p q is a convergent of the continued fraction expansion of κ with q > 6M. Let If ε > 0, then there is no solution of the inequality We remark that Lemma 4 cannot be applied when µ = 0 (since then ε < 0). For this case, we use the following well known technical result from Diophantine approximation, known as Legender's criterion.
Lemma 5 (Legendre). Let κ be real number and x, y integers such that Then x/y = p k /q k is a convergent of κ. Furthermore, let M and N be a nonnegative integers such that q N > M. Then putting a(M) := max{a i : i = 0, 1, 2, . . . , N}, the inequality holds for all pairs (x, y) of positive integers with 0 < y < M.
3. Proof of the main result.

The low range.
We used a computer program to check all the solutions of the Diophantine equation (1) for the parameters d 1 d 2 ∈ {0, . . . , 9}, d 1 > 0 and 1 ≤ l, m, ≤ n ≤ 500. We only found the solutions listed in Theorem 3. Henceforth, we assume n > 500.

The initial bound on n.
We note that (1) can be rewritten as The next lemma relates the sizes of n and l + m.
Taking the logarithm on both sides, we get n log α < (l + m) log 10 + 3 log α.
We examine (4) in two different steps as follows.
Step 1. From equations (2) and (4), we have that Therefore, We thus have that where we used the fact that n > 500. Dividing both sides by d 1 × 10 l+m , we get We let We would like to compare the upper bound on |Γ 1 | with the lower bound we deduce from Theorem 3. Note that Γ 1 0, since this would imply that α n = 10 l+m ×d 1 9 . If this is the case, then applying the automorphism σ on both sides of the preceeding equation and taking absolute values, we have that which is false. We thus have that Γ 1 0.
Step 2. We rewrite equation (4) as That is Hence, Dividing throughout by 9α n , we have that We let α −n × 10 m − 1.
As before, we have that Γ 2 0 because this would imply that which in turn implies that which is false. Before applying Theorem 3, we define the following parameters: In order to determine what A 1 will be, we need to find the find the maximum of the quantities h(η 1 ) and | log η 1 |.
With the notation of Lemma 6, we let r = 2, L = n and H = 1.10 × 10 44 and notice that this data meets the conditions of the lemma. Applying the lemma, we have that n < 2 2 × 1.1 × 10 44 × (log 1.1 × 10 44 ) 2 .

The reduction procedure.
We note that the bounds from Lemma 8 are too large for computational purposes. However, with the help of Lemma 4, they can be considerably sharpened. The rest of this section is dedicated towards this goal. We proceed as in [3]. Using equation (6), we define the quantity Λ 1 as Λ 1 = − log(Γ 1 + 1) = (l + m) log 10 − n log α − log 9 d 1 .
Equation (5) can thus be rewritten as If l ≥ 2, then the above inequality is bounded above by 1 3 . Recall that if x and y are real numbers such that |e x − 1| < y, then x < 2y. We therefore conclude that |Λ 1 | < 92 10 l . Equivalently, (l + m) log 10 − n log α − log 9 d 1 < 92 10 l .
The continued fraction expansion of τ is given by as the first one for which the denominator q = q 106 > 3.6 × 10 48 = 6M. Maintaining the notation of Lemma 4, we computed M τq and obtained M τq < 0.0393724. The smallest (positive) value of µq we obtained satisfies µq > 0.0752711, corresponding to d 1 = 3. We thus choose = 0.0358987 < µq − M τq . We deduce that l ≤ log(332q/ ) log 10 < 53.
In the case l < 2, we have that l < 2 < 53. It follows that l ≤ 53 holds in all cases.
We conclude that n ≤ 454.