Repdigits as products of consecutive Padovan or Perrin numbers

Abstract

A repdigit is a positive integer that has only one distinct digit in its decimal expansion, i.e., a number of the form $a(10^m-1)/9$, for some $m\ge 1$ and $1 \le a \le 9$. Let $\left( P_n\right) _{n\ge 0}$ and $\left( E_n\right) _{n\ge 0}$ 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.

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 $(P_n)_{n\ge 0}$ 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 $(E_n)_{n\ge 0}$ 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.

Theorem 1.1

The Diophantine equation

$$\begin{aligned} P_{n}\cdots P_{n+\left( \ell -1\right) }=a\left( \frac{10^{m}-1}{9}\right) , \end{aligned}$$

(1.1)

has no solution in positive integers $n,\ell ,m,a,$ with $1\le a\le 9$ and $m\ge 2$.

Theorem 1.2

The only solution of the Diophantine equation:

$$\begin{aligned} E_{n}\cdots E_{n+\left( \ell -1\right) }=a\left( \frac{10^{m}-1}{9}\right) , \end{aligned}$$

(1.2)

in positive integers $n,\ell ,m,a,$ with $1\le a\le 9$ and $m\ge 2$ is $\left( n,\ell ,m,a\right) =\left( 11,1,2,2\right)$, i.e., $E_{11} = 22$.

Theorem 1.3

The Diophantine equation

$$\begin{aligned} E_{n}\cdots E_{n+\left( k-1\right) }P_{n+k} \cdots P_{n+k+\left( \ell -1\right) }=a\left( \frac{10^{m}-1}{9}\right) \end{aligned}$$

(1.3)

has no solution in positive integers $n,k,\ell ,m,a,$ with $1\le a\le 9$ and $m\ge 2$.

Theorem 1.4

The Diophantine equation

$$\begin{aligned} P_{n}\cdots P_{n+\left( k-1\right) }E_{n+k} \cdots E_{n+k+\left( \ell -1\right) }=a\left( \frac{10^{m}-1}{9}\right) \end{aligned}$$

(1.4)

has no solution in positive integers $n,k,\ell ,m,a,$ with $1\le a\le 9$ and $m\ge 2$.

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.

2 The tools

We start by recalling some useful properties of Padovan and Perrin sequences. The characteristic equation of $\{P_n\}_{n\ge 0}$ and $\{E_n\}_{n\ge 0}$ is $z^{3}-z-1=0$ and has one real root $\alpha$ and two complex roots $\beta$ and $\gamma =\overline{\beta }$. The Binet formulae for the Padovan and Perrin numbers are respectively:

$$\begin{aligned} P_{s}=c_{\alpha }\alpha ^{s}+c_{\beta }\beta ^{s}+c_{\gamma }\gamma ^{s}, \quad \text {for all} \quad s\ge 0 \end{aligned}$$

(2.1)

and

$$\begin{aligned} E_{s}=\alpha ^{s}+\beta ^{s}+\gamma ^{s}, \quad \text {for all} \quad s\ge 0, \end{aligned}$$

(2.2)

where

$$\begin{aligned} c_{\alpha }=\frac{1+\alpha }{-\alpha ^2+3\alpha +1},\quad c_{\beta }=\frac{1+\beta }{-\beta ^2+3\beta +1}, \quad c_{\gamma }=\frac{1+\gamma }{-\gamma ^2+3\gamma +1}=\overline{c_{\beta }}. \end{aligned}$$

It is easy to see that $\alpha \in \left( 1.32, 1.33\right) , |\beta |=|\gamma | \in \left( 0.86, 0.87\right) , c_{\alpha } \in \left( 0.72, 0.73\right)$ and $|c_{\beta }|=|c_{\gamma }| \in \left( 0.24, 0.25\right)$.

By the facts that $\beta =\alpha ^{-1/2}e^{i\theta }$ and $\gamma =\alpha ^{-1/2}e^{-i\theta }$, for some $\theta \in (0,2\pi )$, we can show that:

$$\begin{aligned} P_{s} = c_{\alpha }\alpha ^{s}+e_s, \qquad \mathrm{with} \quad |e_s|<\frac{1}{\alpha ^{s/2}} ,\quad \text {for all} ~~ s\ge 1 \end{aligned}$$

(2.3)

and

$$\begin{aligned} E_{s} = \alpha ^{s}+e'_s, \qquad \mathrm{with} \quad |e'_s|<\frac{2}{\alpha ^{s/2}}, \quad \text {for all} ~~ s\ge 1. \end{aligned}$$

(2.4)

Further, we have:

$$\begin{aligned} \alpha ^{s-2}\le P_{s}\le \alpha ^{s-1}, \quad \text {for all} \quad s \ge 4 \end{aligned}$$

(2.5)

and

$$\begin{aligned} \alpha ^{s-2}\le E_{s}\le \alpha ^{s+1}, \quad \text {for all} \quad s \ge 2. \end{aligned}$$

(2.6)

For a prime number p and a non-zero integer r, the p-adic order $\upsilon _{p}(r)$ 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 $n\ge 1$, we have:

$$\begin{aligned} \upsilon _{2}\left( P_{n}\right) = \left\{ \begin{array}{lll} 0 &{}\quad \text{ if } n\equiv 0,1,2, 5 &{} \pmod {7},\\ \upsilon _{2}(n+4)+1 &{}\quad \text{ if } n\equiv 3 &{} \pmod {7},\\ \upsilon _{2}((n+3)(n+17))+1 &{}\quad \text{ if } n\equiv 4 &{} \pmod {7},\\ \upsilon _{2}((n+1)(n+8))+1 &{}\quad \text{ if } n\equiv 6 &{} \pmod {7}. \end{array}\right. \end{aligned}$$

Lemma 2.2

For $n\ge 1$, we have:

$$\begin{aligned} \upsilon _{2}\left( E_{n}\right) = \left\{ \begin{array}{lll} 0 &{} \quad \text{ if } n\equiv 0,3,5,6 &{} \pmod {7},\\ 1 &{} \quad \text{ if } n\equiv 2 &{} \pmod {14},\\ 2 &{} \quad \text{ if } n\equiv 9 &{} \pmod {14},\\ \upsilon _{2}(n-1)+1 &{} \quad \text{ if } n\equiv 1 &{} \pmod {7},\\ 1 &{} \quad \text{ if } n\equiv 4 &{} \pmod {7}.\\ \end{array}\right. \end{aligned}$$

The next tools are related to the transcendental approach to solve Diophantine equations. For any non-zero algebraic number $\gamma$ of degree d over $\mathbb Q$, whose minimal polynomial over $\mathbb Z$ is $a\prod _{j=1}^d \left( X-\gamma ^{(j)} \right)$, we denote by:

$$\begin{aligned} h(\gamma ) = \frac{1}{d} \left( \log |a| + \sum _{j=1}^d \log \max \left( 1, \left| \gamma ^{(j)}\right|\right) \right) \end{aligned}$$

the usual absolute logarithmic height of $\gamma$.

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 $\gamma _1,\ldots ,\gamma _s$ be real algebraic numbers and let $b_1,\ldots ,b_s$ be non-zero rational integer numbers. Let D be the degree of the number field $\mathbb Q(\gamma _1,\ldots ,\gamma _s)$ over $\mathbb Q$ and let $A_j$ be a positive real number satisfying:

$$\begin{aligned} A_j=\max \{Dh(\gamma ),|\log \gamma |,0.16\}\quad \text {for}\; j=1,\ldots ,s. \end{aligned}$$

Assume that:

$$\begin{aligned} B\ge \max \{|b_1|,\ldots ,|b_s|\}. \end{aligned}$$

If $\gamma _1^{b_1}\cdots \gamma _s^{b_s}\ne 1$, then:

$$\begin{aligned} |\gamma _1^{b_1}\cdots \gamma _s^{b_s}-1|\ge \exp (-C(s,D)(1+\log B) A_1\cdots A_s), \end{aligned}$$

where $C(s,D):=1.4\cdot 30^{s+3}\cdot s^{4.5}\cdot D^2(1+\log D).$

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 $\vartheta _1,\vartheta _2,\beta \in {\mathbb {R}}$ be given, and let $x_1,x_2\in \mathbb {Z}$ be unknowns. Let:

$$\begin{aligned} \Lambda =\beta +x_1\vartheta _1 +x_2\vartheta _2. \end{aligned}$$

(2.7)

Let c, $\delta$ be positive constants. Set $X=\max \{|x_1|,|x_2|\}$. Let $X_0, Y$ be positive numbers. Assume that:

$$\begin{aligned}&|\Lambda |<c\cdot \exp (-\delta \cdot Y), \end{aligned}$$

(2.8)

$$\begin{aligned}&Y\le X\le X_0. \end{aligned}$$

(2.9)

When $\beta =0$ in (2.7), we get

$$\begin{aligned}\Lambda =x_1\vartheta _1 +x_2\vartheta _2.\end{aligned}$$

Put $\vartheta =-{\vartheta _1}/{\vartheta _2}$. We assume that $x_1$ and $x_2$ are coprime. Let the continued fraction expansion of $\vartheta$ be given by

$$\begin{aligned}\,[a_0,a_1,a_2,\ldots ],\end{aligned}$$

and let the kth convergent of $\vartheta$ be ${p_k}/{q_k}$ for $k=0,1,2,\ldots$. We may assume without loss of generality that $|\vartheta _1|<|\vartheta _2|$ and that $x_1>0$. We have the following results.

Lemma 2.4

[14, Lemma 3.2] Let:

$$\begin{aligned} A=\max _{0\le k\le Y_0} a_{k+1}, \end{aligned}$$

where

$$\begin{aligned} Y_0=-1+\dfrac{\log (\sqrt{5}X_0+1)}{\log \left( \frac{1+\sqrt{5}}{2}\right) }. \end{aligned}$$

If (2.8) and (2.9) hold for $x_1$, $x_2$ and $\beta =0$, then:

$$\begin{aligned} Y<\frac{1}{\delta }\log \left( \frac{c(A+2)X_0}{|\vartheta _2|}\right) . \end{aligned}$$

(2.10)

When $\beta \ne 0$ in (2.7), put $\vartheta =-{\vartheta _1}/{\vartheta _2}$ and $\psi ={\beta }/{\vartheta _2}$. Then, we have

$$\begin{aligned}\frac{\Lambda }{\vartheta _2}=\psi -x_1\vartheta +x_2.\end{aligned}$$

Let p/q be a convergent of $\vartheta$ with $q>X_0$. For a real number x, we let $\Vert x\Vert =\min \{|x-n|, n\in {\mathbb Z}\}$ be the distance from x to the nearest integer. We have the following result.

Lemma 2.5

[14, Lemma 3.3] Suppose that:

$$\begin{aligned}\parallel q \psi \parallel >\frac{ 2X_0}{q}.\end{aligned}$$

Then, the solutions of (2.8) and (2.9) satisfy:

$$\begin{aligned} Y<\frac{1}{\delta }\log \left( \frac{q^2c}{|\vartheta _2|X_0} \right) .\end{aligned}$$

We conclude this section by recalling the following lemma that we need in the sequel:

Lemma 2.6

[14, Lemma 2.2, page 31] Let $a, x \in {\mathbb {R}}$ and $0< a < 1$. If $|x| < a$, then:

$$\begin{aligned} \left| \log (1+x)\right| < \dfrac{-\log (1-a)}{a} |x| \end{aligned}$$

and

$$\begin{aligned} |x| < \dfrac{a}{1-e^{-a}} \left| e^x-1\right| . \end{aligned}$$

3 Proof of Theorem 1.1

In this section, we will use Baker’s method and the p-adic valuation to completely prove Theorem 1.1.

3.1 Absolute bounds on the variables

We start by giving the number of factors $\ell$ in the Diophantine equation (1.1).

Lemma 3.1

If Diophantine equation (1.1) has solutions, then $\ell \le 6$.

Proof

Note that, for all $1\le a\le 9$, we have:

$$\begin{aligned} \upsilon _{2}\left( a\left( \frac{10^{m}-1}{9}\right) \right) = \upsilon _{2}(a) \le 3. \end{aligned}$$

Therefore, if $\upsilon _2 (P_{n}P_{n+1}\cdots P_{n+(\ell -1)})\ge 4$, then Diophantine equation (1.1) has no solution.

Let $x\in \left\{ 0,1,2,\ldots ,6\right\}$, such that $n\equiv x\pmod {7}$. Suppose that $x=2$, and hence, $n+1 \equiv 3 \pmod 7$, $n+2 \equiv 4 \pmod 7$, and $n+4 \equiv 6 \pmod 7$. Therefore, by Lemma 2.1, we get:

$$\begin{aligned} \upsilon _2(P_nP_{n+1}\cdots P_{n+4})= 3\upsilon _2(n+5)+\upsilon _2(n+12)+\upsilon _2(n+19)+3. \end{aligned}$$

Clearly, if n is odd, then $n+5$ and $n+19$ are even; otherwise, $n+12$ is odd. Thus, we get $\upsilon _2(P_nP_{n+1}\cdots P_{n+4})\ge 4$. Therefore, Diophantine equation (1.1) has no solution if $\ell \ge 5$ in this case.

The other cases can be treated using a similar method. As a conclusion, we get Table 1. Thus, we deduce that $\ell \le 6$.

Table 1 2-adic order of product of consecutive Padovan numbers

$\square$

Now, we give an upper bound for n and m.

Lemma 3.2

If $(n,\ell ,m,a)$ is a positive integer solution of (1.1) with $n\ge 15$, $m\ge 2$, $1\le a\le 9$, and $1\le \ell \le 6$, then:

$$\begin{aligned} m\le \ell n +\ell (\ell -3)/2\quad \text {and} \quad n< 1.8\times 10^{16}. \end{aligned}$$

Proof

By (1.1) and (2.5), we have:

$$\begin{aligned} 10^{m-1}<a\left( \dfrac{10^m-1}{9}\right) =P_nP_{n+1}\cdots P_{n+(\ell -1)}< \alpha ^{\ell n +\frac{\ell (\ell -3)}{2}}. \end{aligned}$$

Thus, we get:

$$\begin{aligned} m\le \ell n +\ell (\ell -3)/2. \end{aligned}$$

(3.1)

Now, by (2.3), we obtain:

$$\begin{aligned} P_{n} \cdots P_{n+(\ell -1)}= & {} (c_{\alpha }\alpha ^{n}+e_n)\cdots (c_{\alpha }\alpha ^{n+(\ell -1)}+e_{n+\ell -1})\\= & {} c_{\alpha }^{\ell }\alpha ^{\ell n+\ell (\ell -1)/2} + r_1(c_{\alpha },\alpha ,n,\ell ),\nonumber \end{aligned}$$

(3.2)

where $r_1(c_{\alpha },\alpha ,n,\ell )$ involves the part of the expansion of the previous line that contains the product of powers of $c_{\alpha },~\alpha$ and the errors $e_i$, for $i=n,\ldots n+(\ell -1)$. Moreover, $r_1(c_{\alpha },\alpha ,n,\ell )$ is the sum of 63 terms with maximum absolute value $c_{\alpha }^{\ell -1}\alpha ^{(\ell -1)n +\ell (\ell -1)/2}\alpha ^{-n/2}$.

The equality (3.2) enables us to express (1.1) into the form:

$$\begin{aligned} \frac{a}{9}10^{m}-c_{\alpha }^{\ell }\alpha ^{\ell n+\ell (\ell -1)/2}=\frac{a}{9}+r_1(c_{\alpha },\alpha ,n,\ell ). \end{aligned}$$

Dividing through by $c_{\alpha }^{\ell }\alpha ^{\ell n+\ell (\ell -1)/2}$ and taking the absolute value, we deduce that:

$$\begin{aligned} \left| \Gamma _1\right|&\le \left( \frac{a}{9}+|r_1(c_{\alpha },\alpha ,n,\ell )|\right) \cdot c_{\alpha }^{-\ell }\alpha ^{-(\ell n+\ell (\ell -1)/2)}\\&< (1+63c_{\alpha }^{\ell -1}\alpha ^{(\ell -1)n+\ell (\ell -1)/2}\alpha ^{-n/2})\cdot c_{\alpha }^{-\ell }\alpha ^{-(\ell n+\ell (\ell -1)/2)}\nonumber \\&\le 64c_{\alpha }^{-1}\alpha ^{-3n/2}<89\alpha ^{-3n/2},\nonumber \end{aligned}$$

(3.3)

where::

$$\begin{aligned} \Gamma _1=\frac{a}{9c_{\alpha }^{\ell }}\alpha ^{-(\ell n+\ell (\ell -1)/2)}10^{m}-1. \end{aligned}$$

(3.4)

To find a lower bound for $\Gamma _1$, we take the parameters $s:=3$:

$$\begin{aligned} (\gamma _1, b_1):=((a/9)c_{\alpha }^{-\ell },1),\; (\gamma _{2},b_2):=(\alpha ,-(\ell n+\ell (\ell -1)/2))\; \mathrm{and} \; (\gamma _{3},b_3):=(10,m), \end{aligned}$$

in Lemma 2.3. For our choices, we have $\gamma _1,\gamma _2,\gamma _3\in \mathbb {Q}(\alpha )$, with degree $D:=3$. To apply Lemma 2.3 it is necessary to show that $\Gamma _1\ne 0$. If we assume the contrary, thus we get:

$$\begin{aligned} a\cdot 10^{m}/9=c_{\alpha }^{\ell }\alpha ^{\ell n+\ell (\ell -1)/2}. \end{aligned}$$

Conjugating the above relation by the Galois automorphism $\sigma :=( \alpha \beta )$, and then taking absolute values on both sides of the resulting equality, we obtain:

$$\begin{aligned} 1<a\cdot 10^{m}/9=|c_{\beta }|^{\ell }|\beta |^{\ell n+\ell (\ell -1)/2}<1. \end{aligned}$$

This is a contradiction. Thus, $\Gamma _1\ne 0$. Next, we give estimates to $A_i$ for $i=1,2,3$. By the properties of the absolute logarithmic height, we have:

$$\begin{aligned}h(\gamma _{1})\le h(d)+h(9)+\ell h(c_{\alpha })\le 2\log 9+\ell h(c_{\alpha }). \end{aligned}$$

Now, we need to estimate $h(c_{\alpha })$. For that, the minimal polynomial of $c_{\alpha }$ is $23X^{3}-23x^2+6X-1$. Therefore, $h(c_{\alpha })=\frac{1}{3}\log 23$, and thus:

$$\begin{aligned} h(\gamma _{1})\le 2\log 9+2\log 23. \end{aligned}$$

On the other hand, $h(\gamma _{2})=\frac{1}{3}\log \alpha$ and $h(\gamma _{3})=\log 10$. Therefore, we take $A_{1}:=32$, $A_{2}:=0.3$ and $A_{3}:=7$. Finally, by (3.1) and the fact that $\ell \le 6$, we take $B:=6 n +15$. Applying Lemma 2.3, we get a lower bound for $|\Gamma _1|$, which by comparing it to (3.3) leads to:

$$\begin{aligned} \frac{3n}{2}\log \alpha -\log 89<1.82\times 10^{14}(1+\log (6n+15)).\\ \end{aligned}$$

Hence, we get:

$$\begin{aligned} n<4.4\times 10^{14}(1+\log (6n+15)). \end{aligned}$$

Therefore, we obtain $n<1.8\times 10^{16}$. $\square$

3.2 Reducing n

To lower the bound of n, we will use Lemma 2.5.

Let:

$$\begin{aligned} \Lambda _1 :=m\log 10-(\ell n+\ell (\ell -1)/2)\log \alpha + \log (a/9c_{\alpha }^{\ell }).\end{aligned}$$

Therefore, (3.3) can be rewritten as $|e^{\Lambda _1}-1| < 89/\alpha ^{3n/2}$. Furthermore, if $n\ge 15$ $|\Gamma _1|<0.16$. Therefore, by applying Lemma 2.6, we deduce that:

$$\begin{aligned} |\Lambda _1|<-\dfrac{\log (1-0.16)}{0.16} |\Gamma _1|< 97 \exp \left( -0.42 n\right) . \end{aligned}$$

(3.5)

Put:

$$\begin{aligned} \vartheta _1:=-\log \alpha ,\quad \vartheta _2:=\log 10,\quad \psi := \log \left( \dfrac{a}{9c_{\alpha }}\right) ,\quad c:=97,\quad \delta :=0.42.\end{aligned}$$

Furthermore, as $\max \{m,\ell n+\ell (\ell -1)/2\}<1.1\times 10^{17}$, then we take $X_0=1.1\times 10^{17}$. Using Maple, one can see that:

$$\begin{aligned} q_{43}=76200291125177096225 \end{aligned}$$

satisfies the conditions of Lemma 2.5 for all $1\le a\le 9$ and $1\le \ell \le 6$. Therefore, Lemma 2.5 implies that if the Diophantine equation (1.1) has solutions, then:

$$\begin{aligned} n\le \dfrac{1}{0.42} \times \log \left( \dfrac{76200291125177096225^2 \times 97}{\log 10 \times 1.1\times 10^{17}}\right) < 134. \end{aligned}$$

Now, we reduce again this new bound of n. In this application of Lemma 2.5, we take $X_0=813$ and see that $q_{10}=869219$ satisfies the conditions of Lemma 2.5. Thus, we obtain:

$$\begin{aligned} n\le \dfrac{1}{0.42} \times \log \left( \dfrac{869219^2 \times 97}{\log 10 \times 813}\right) <59. \end{aligned}$$

Hence, it remains to check Eq. (1.1) for $1\le n \le 58$, $1\le \ell \le 6$, $2\le m \le 363$ and $1\le a \le 9$. A quick inspection using Maple reveals that Diophantine equation (1.1) has no solutions. This completes the proof of Theorem 1.1.

4 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.

4.1 Absolute bounds on the variables

First of all, we give the number of factors in the Diophantine equation (1.2).

Lemma 4.1

The Diophantine equation (1.2) has a solution if $\ell \le 7$:

Proof

Let $x\in \left\{ 0,1,2,\ldots ,13\right\}$, such that $n\equiv x\pmod {14}$. Assume that $x=8$, and hence, $n \equiv 1 \pmod 7$, $n+1 \equiv 9 \pmod {14}$, and $n+3 \equiv 4 \pmod 7$. Thus, Lemma 2.2 gives:

$$\begin{aligned} \upsilon _2(E_nE_{n+1}\cdots E_{n+3})= \upsilon _2(n-1)+4=4, \end{aligned}$$

because $n\equiv 8 \pmod {14}$ leads to $\upsilon _2(n-1)=0$. We give the other results in Table 2.

Table 2 2-adic order of product of consecutive Perrin numbers

As $\upsilon _{2}\left( a\left( \frac{10^{m}-1}{9}\right) \right) = \upsilon _{2}(a) \le 3$, for all $1\le a\le 9$, then it follows from Table 2 that $\ell \le 7$. $\square$

Now, we will show the following lemma.

Lemma 4.2

If $(n,\ell ,m,a)$ is a positive integer solution of (1.2) with $n\ge 20$, $m\ge 2$, $1\le a\le 9$, and $1\le \ell \le 10$, then:

$$\begin{aligned} m\le \ell n+\ell (\ell +1)/2\quad \text {and}\quad n < 7.1\times 10^{15}. \end{aligned}$$

Proof

First, we assume that $n\ge 20$. Combining (1.2) and (2.6), we obtain:

$$\begin{aligned} 10^{m-1}<\dfrac{a(10^m-1)}{9}=E_nE_{n+1}\cdots E_{n+(\ell -1)}<\alpha ^{\ell n+\frac{\ell (\ell +1)}{2}}. \end{aligned}$$

Thus, we have:

$$\begin{aligned} m\le \ell n +\ell (\ell +1)/2. \end{aligned}$$

(4.1)

Now, by (2.4), we get:

$$\begin{aligned} E_{n} \cdots E_{n+(\ell -1)}= & {} (\alpha ^{n}+e_n)\cdots (\alpha ^{n+(\ell -1)}+e_{n+\ell -1})\\= & {} \alpha ^{\ell n+\ell (\ell -1)/2} + r_2(\alpha ,n,\ell ),\nonumber \end{aligned}$$

(4.2)

where $r_2(\alpha ,n,\ell )$ involves the part of the expansion of the previous line that contains the product of powers of $~\alpha$ and the errors $e'_i$, for $i=n,\ldots n+(\ell -1)$. Moreover, $r_2(\alpha ,n,\ell )$ is the sum of 127 terms with maximum absolute value $2^{\ell }\alpha ^{(\ell -1)n +\ell (\ell -1)/2}\alpha ^{-n/2}$.

The equality (4.2) enables us to express (1.2) as:

$$\begin{aligned} \frac{a}{9}10^{m}-\alpha ^{\ell n+\ell (\ell -1)/2}=\frac{a}{9}+r_2(\alpha ,n,\ell ). \end{aligned}$$

Multiplying both sides by $\alpha ^{-(\ell n+\ell (\ell -1)/2)}$ and taking the absolute value, we conclude that:

$$\begin{aligned} \left| \Gamma _2\right|&\le \left( \frac{a}{9}+|r_2(\alpha ,n,\ell )|\right) \cdot \alpha ^{-(\ell n+\ell (\ell -1)/2)}\\&< (1+127\cdot 2^{\ell }\alpha ^{(\ell -1)n+\ell (\ell -1)/2}\alpha ^{-n/2})\cdot \alpha ^{-(\ell n+\ell (\ell -1)/2)}\nonumber \\&\le 128\cdot 2^{\ell }\alpha ^{-3n/2}<16384\alpha ^{-3n/2},\nonumber \end{aligned}$$

(4.3)

where:

$$\begin{aligned} \Gamma _2=\frac{a}{9}\alpha ^{-(\ell n+\ell (\ell -1)/2)}10^{m}-1. \end{aligned}$$

(4.4)

Now, we use Lemma 2.3 to find a lower bound for $\Gamma _2$, with the parameters $s:=3$:

$$\begin{aligned} (\gamma _{1}, b_1):=((a/9),1), ~~ (\gamma _{2},b_2):=(\alpha ,-(\ell n+\ell (\ell -1)/2)),\quad \mathrm{and} \quad (\gamma _{3},b_3):=(10,m). \end{aligned}$$

The number field containing $\gamma _{1},\gamma _{2},\gamma _{3}$ is $\mathbb {Q}(\alpha )$, which degree is $D:=3$.

We next justify that $\Gamma _2\ne 0$. Indeed, if this were zero, we would then get:

$$\begin{aligned} \alpha ^{\ell n+\ell (\ell -1)/2}=a\cdot 10^{m}/9. \end{aligned}$$

Conjugating the above relation by the Galois automorphism $\sigma :=( \alpha \beta )$, and then taking absolute values on both sides of the resulting equality, we obtain:

$$\begin{aligned} |\beta |^{\ell n+\ell (\ell -1)/2}=a\cdot 10^{m}/9, \end{aligned}$$

which is not possible, because $|\beta |^{\ell n+\ell (\ell -1)/2}<1$ and $a\cdot 10^{m}/9>10$. Thus, $\Gamma _2\ne 0$. Next, $h(\gamma _{1})\le h(d)+h(9)\le 2\log 9$, $h(\gamma _{2})=\frac{1}{3}\log \alpha$ and $h(\gamma _{3})=\log 10$. Thus, we take $A_{1}:=13.2$, $A_{2}:=0.3$ and $A_{3}:=7$. According to (4.1) and the fact that $\ell \le 10$, we take $B:=7 n +28$. Applying Lemma 2.3, we get a lower bound for $|\Gamma _2|$, and taking into account inequality (4.3), we obtain:

$$\begin{aligned} \exp \left( -7.5\times 10^{13}(1+\log (7 n+28))\right) <\frac{16384}{\alpha ^{3n/2}}. \end{aligned}$$

Taking the logarithm of both sides of the above inequality, we obtain:

$$\begin{aligned} n<1.78\times 10^{14}(1+\log (7n+28)). \end{aligned}$$

With the help of Maple, we get $n<7.1\times 10^{15}$. $\square$

4.2 Reducing n

To lower the bound of n, we will use Lemmas  2.4 and 2.5. We will proceed with three successive reductions.

Let:

$$\begin{aligned} \Lambda _2 :=m\log 10-(\ell n+\ell (\ell -1)/2)\log \alpha + \log (a/9). \end{aligned}$$

Therefore, (4.3) can be rewritten as $|e^{\Lambda _2}-1| < 16384\alpha ^{-3n/2}$. Furthermore, if $n\ge 30$, then $|\Gamma _1|<0.06$. Therefore, Lemma 2.6 leads to:

$$\begin{aligned} |\Lambda _2|<-\dfrac{\log (1-0.06)}{0.06} |\Gamma _2|< 16897 \exp \left( -0.42 n\right) . \end{aligned}$$

If $a\ne 9$, then we take:

$$\begin{aligned} \vartheta _1:=-\log \alpha ,\quad \vartheta _2:=\log 10,\quad \psi := \log \left( \dfrac{a}{9}\right) ,\quad c:=16897,\quad \delta :=0.42\end{aligned}$$

in Lemma 2.5. Furthermore, as $\max \{m,\ell n+\ell (\ell -1)/2\}<5\times 10^{16}$, we can take $X_0=5\times 10^{16}$. One can use Maple to see that:

$$\begin{aligned} q_{43}=76200291125177096225 \end{aligned}$$

satisfies the conditions of Lemma 2.5, for all $1\le a\le 8$ and $1\le \ell \le 10$. Therefore, from Lemma 2.5, we deduce that if Diophantine equation (1.2) has solutions, then:

$$\begin{aligned} n\le \dfrac{1}{0.42} \times \log \left( \dfrac{76200291125177096225^2 \times 16897}{\log 10 \times 5\times 10^{16}}\right) < 148. \end{aligned}$$

Now, we again reduce this new bound of n. In this application of Lemma 2.5, we take $X_0=1057$ and see that $q_{9}=139228$ satisfies the conditions of Lemma 2.5. Thus, we obtain:

$$\begin{aligned} n\le \dfrac{1}{0.42} \times \log \left( \dfrac{139228^2 \times 16897}{\log 10 \times 1057}\right) <62. \end{aligned}$$

If $a=9$, then $\Lambda _2$ becomes:

$$\begin{aligned} \Lambda _2=m\log 10-(\ell n+\ell (\ell -1)/2)\log \alpha , \end{aligned}$$

so we apply Lemma 2.4 in this case by choosing

$$\begin{aligned} c:=16897,\quad \delta :=0.42, \quad X_0=5\times 10^{16},\quad Y_0=80.5763\ldots . \end{aligned}$$

Maple gives us $\max \nolimits _{0\le k\le 81}a_{k+1}=49$, and thus, by Lemma 2.4, we have:

$$\begin{aligned} n\le \dfrac{1}{0.42}\times \log \left( \dfrac{16897\times 51 \times 5\times 10^{16}}{\log 10}\right) <123. \end{aligned}$$

Now, we proceed to the third reduction of the bound of n. To apply Lemma 2.4, we take $X_0=882$ and find that $\max \nolimits _{0\le k\le 15}a_{k+1}=8$. Therefore, we get:

$$\begin{aligned} n\le \dfrac{1}{0.42}\times \log \left( \dfrac{16897\times 10 \times 882}{\log 10}\right) <43. \end{aligned}$$

Therefore, $n\le 62$ holds in all cases. Hence, it remains to check Eq. (1.2) for $1\le n \le 62$, $1\le \ell \le 10$, $2\le m \le 462$, and $1\le a \le 9$. By a fast computation with Maple in these ranges, we conclude that the quadruple $(n,\ell ,m,d)=(11,1,2,2)$ is the only solution of Diophantine equation (1.2). This completes the proof of Theorem 1.2.

5 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.

5.1 Absolute bounds on the variables

By Lemmas 3.1 and 4.1, we claim that $k\le 7$ and $\ell \le 6$. Indeed, one has:

$$\begin{aligned} \upsilon _2 (E_n\cdots E_{n+(k-1)} P_{n+k}\cdots P_{n+k+(\ell -1)})>3, \end{aligned}$$

and then, Diophantine equation (1.3) has no solution.

Now, we will prove the following lemma.

Lemma 5.1

If $(n,k,\ell ,m,a)$ is a positive integer solution of (1.3) with $n\ge 30$, $m\ge 2$, $1\le a\le 9$, $1\le k\le 7$, and $1\le \ell \le 6$, then:

$$\begin{aligned} m\le (k+\ell ) n+\frac{k(k+1)}{2}+\frac{\ell (2k+\ell -3)}{2} \quad \text {and}\quad n < 1.88\times 10^{17}. \end{aligned}$$

Proof

First, assume that $n\ge 30$. Combining (1.3), (2.5), and (2.6), we get:

$$\begin{aligned} \begin{array}{lcl} 10^{m-1}<\dfrac{a(10^m-1)}{9}&{}=&{}E_nE_{n+1}\cdots E_{n+(k-1)}P_{n+k}\cdots P_{n+k+\left( \ell -1\right) }\\ &{}<&{}\alpha ^{(k+\ell ) n+\frac{k(k+1)}{2}+\frac{\ell (2k+\ell -3)}{2}}. \end{array} \end{aligned}$$

Taking the logarithm of both sides of the above inequality leads to:

$$\begin{aligned} m\le (k+\ell ) n+\frac{k(k+1)}{2}+\frac{\ell (2k+\ell -3)}{2}. \end{aligned}$$

(5.1)

Now, by (2.3) and (2.4), we get:

$$\begin{aligned}&E_nE_{n+1}\cdots E_{n+(k-1)}P_{n+k}\cdots P_{n+k+\left( \ell -1\right) }\\&\quad = \alpha ^{(k+\ell ) n+k(k-1)/2+\ell (2k+\ell -1)/2} + r_3(c_{\alpha },\alpha ,n,k,\ell ),\nonumber \end{aligned}$$

(5.2)

where $r_3(c_{\alpha },\alpha ,n,k,\ell )$ involves the part of the expansion of the previous line that contains the product of powers of $c_{\alpha }~\alpha$ and the errors $e_{k+j}$ and $e'_i$, for $i=n,\ldots , n+(k-1)$ and $j=n,\ldots , n+(\ell -1)$. Moreover, $r_3(c_{\alpha },\alpha ,n,k,\ell )$ is the sum of 8191 terms with maximum absolute value $2^kc_{\alpha }^{\ell }\alpha ^{(k+\ell -1) n+k(k-1)/2+\ell (2k+\ell -1)/2}\alpha ^{-n/2}$.

The equality (5.2) enables us to express (1.3) as:

$$\begin{aligned} \frac{a}{9}10^{m}-c_{\alpha }^{\ell }\alpha ^{(k+\ell ) n+k(k-1)/2+\ell (2k+\ell -1)/2}=\frac{a}{9}+r_3(c_{\alpha },\alpha ,n,k,\ell ). \end{aligned}$$

Multiplying through by $c_{\alpha }^{-\ell }\alpha ^{-((k+\ell ) n+k(k-1)/2+\ell (2k+\ell -1)/2)}$ and taking the absolute value, we deduce that:

$$\begin{aligned} \left| \Gamma _3\right|&\le \left( \frac{a}{9}+|r_3(c_{\alpha },\alpha ,n,k,\ell )|\right) \cdot \alpha ^{-((k+\ell ) n+k(k-1)/2+\ell (2k+\ell -1)/2)}\\&< (1+8191\cdot 2^k c_{\alpha }^{\ell }\alpha ^{(k+\ell -1) n+k(k-1)/2+\ell (2k+\ell -1)/2}\alpha ^{-n/2})\cdot \alpha ^{-((k+\ell ) n+k(k-1)/2+\ell (2k+\ell -1)/2)}\nonumber \\&\le 8192\cdot 2^kc_{\alpha }^{-1}\alpha ^{-3n/2}<1452072\alpha ^{-3n/2},\nonumber \end{aligned}$$

(5.3)

where:

$$\begin{aligned} \Gamma _3=\frac{a}{9c_{\alpha }^{\ell }}\alpha ^{(k+\ell ) n+k(k-1)/2+\ell (2k+\ell -1)/2}10^{m}-1. \end{aligned}$$

(5.4)

Now, we use Lemma 2.3 to find a lower bound for $\Gamma _3$, with the parameters $s:=3$ and:

$$\begin{aligned} (\gamma _{1}, b_1):=((d/9c_{\alpha }^{\ell }),1), ~~ (\gamma _{2},b_2):=(\alpha ,-((k+\ell ) n+k(k-1)/2+\ell (2k+\ell -1)/2)), ~~ (\gamma _{3},b_3):=(10,m). \end{aligned}$$

The algebraic numbers $\gamma _{1},\gamma _{2},\gamma _{3}$ belong to the number field $\mathbb {Q}(\alpha )$, which degree is $D:=3$. We claim that $\Gamma _3\ne 0$. Otherwise, we get:

$$\begin{aligned} c_{\alpha }^{\ell }\alpha ^{(k+\ell ) n+k(k-1)/2+\ell (2k+\ell -1)/2}=a\cdot 10^{m}/9. \end{aligned}$$

Conjugating the above relation by the automorphism $\sigma :=( \alpha \beta )$, and then taking absolute values on both sides of the resulting equality, we obtain:

$$\begin{aligned} |c_{\beta }^{\ell }||\beta |^{(k+\ell ) n+k(k-1)/2+\ell (2k+\ell -1)/2}=a\cdot 10^{m}/9, \end{aligned}$$

which is not possible, because $|c_{\beta }^{\ell }||\beta |^{(k+\ell ) n+k(k-1)/2+\ell (2k+\ell -1)/2}<1$ and $a\cdot 10^{m}/9>10$. Thus, $\Gamma _3\ne 0$. Next, $h(\gamma _{1})\le h(d)+h(9)\le 2\log 9$, $h(\gamma _{2})=\frac{1}{3}\log \alpha$ and $h(\gamma _{3})=\log 10$. Thus, we can take $A_{1}:=13.2$, $A_{2}:=0.3$ and $A_{3}:=7$. According to (5.1) and the facts $k\le 7$ and $\ell \le 6$, we take $B:=13n+85$. Applying Lemma 2.3 we get a lower bound for $|\Gamma _3|$, which, by comparing it to (5.3), leads to:

$$\begin{aligned} \exp \left( -1.82\times 10^{14}(1+\log (13n+85)\right) <\frac{1452072}{\alpha ^{3n/2}}. \end{aligned}$$

Hence, we get:

$$\begin{aligned} n<4.32\times 10^{14}(1+\log (13n+85)). \end{aligned}$$

Therefore, we obtain $n<1.88\times 10^{17}$. $\square$

5.2 Reducing n

To lower the bound of n, we will use Lemma 2.5.

Let

$$\begin{aligned} \Lambda _3 :=m\log 10-((k+\ell ) n+k(k-1)/2+\ell (2k+\ell -1)/2)\log \alpha + \log (a/9c_{\alpha }^{\ell }).\end{aligned}$$

Therefore, (5.3) can be rewritten as $|e^{\Lambda _3}-1| < 1452072\alpha ^{-3n/2}$. Furthermore, if $n\ge 40$ then $|\Gamma _3|<0.07$. Therefore, by applying Lemma 2.6, we deduce that:

$$\begin{aligned} |\Lambda _3|<-\dfrac{\log (1-0.07)}{0.07} |\Gamma _3|< 1505399 \exp \left( -0.42 n\right) . \end{aligned}$$

Put:

$$\begin{aligned} \vartheta _1:=-\log \alpha ,\quad \vartheta _2:=\log 10,\quad \psi := \log \left( \dfrac{a}{9c_{\alpha }}\right) ,\quad c:=1505399,\quad \delta :=0.42.\end{aligned}$$

Furthermore, as $\max \{m,(k+\ell ) n+k(k-1)/2+\ell (2k+\ell -1)/2\}<2.45\times 10^{18}$, then we take $X_0=2.45\times 10^{18}$. We use Maple to find that:

$$\begin{aligned} q_{45}=20674124943023524548605 \end{aligned}$$

satisfies the conditions of Lemma 2.5 for all $1\le a\le 9$, $1\le k\le 10$ and $1\le \ell \le 6$. Therefore, by Lemma 2.5, if Diophantine equation (1.3) has solutions, then:

$$\begin{aligned} n\le \dfrac{1}{0.42} \times \log \left( \dfrac{20674124943023524548605^2 \times 1505399}{\log 10 \times 2.45\times 10^{18}}\right) < 176. \end{aligned}$$

Now, we reduce one more time this new bound of n. In this application of Lemma 2.5, we take $X_0=2360$ and observe that $q_{10}=869219$ is a good candidate to verify the conditions of Lemma 2.5. Thus, we obtain:

$$\begin{aligned} n\le \dfrac{1}{0.42} \times \log \left( \dfrac{869219^2 \times 1505399}{\log 10 \times 2360}\right) <79. \end{aligned}$$

Hence, it remains to check Eq. (1.1) for $1\le n \le 78$, $1\le k\le 7$, $1\le \ell \le 6$, $2\le m \le 1216$, and $1\le a \le 9$. For this, we use a simple routine written in Maple which (in a few minutes) does not return any solution of Diophantine equation (1.3) in these ranges. This completes the proof of Theorem 1.3.