Lucas Numbers Which Are Concatenations of Two Repdigits

Abstract

In this paper, we find all Lucas numbers written in the form $\overline{c\cdots cd\cdots d}$, where $\overline{c\cdots cd\cdots d}$ is the concatenation of two repdigits in base 10 with $c,d\in \{0,1,\cdots ,9\}$, $c\ne d$ and $c>0$.

1. Introduction

Linear form in logarithms has many important applications in solving Diophantine equations [1,2,3,4]. In 2002, by applying linear form in logarithms, A. Dujella and B. Jadrijević [1] showed that the solutions to quartic Thue equations $x^4-4c{x}^{3}y+(6c+2)x^2{y}^2+4cx{y}^3+y^4=1$ are only $(x,y)=(\pm 1,0)$ and $(0,\pm 1)$ for an integer $c\ge 3$. Suppose that ${\left\{F_n\right\}}_{n\ge 0}$ is the Fibonacci sequence given by $F_{n+2}=F_{n+1}+F_n$, with initial values $F_0=0$ and $F_1=1$ and let ${\left\{L_n\right\}}_{n\ge 0}$ be the Lucas sequence defined by $L_{n+2}=L_{n+1}+L_n$, where $L_0=2$ and $L_1=1$. In 2011, F. Luca and R. Oyono [2] concluded that there is no solution $(m,n,s)$ to the Diophantine equation $F_m^s+F_{m+1}^s=F_n$ for integers $m\ge 2,n\ge 1,s\ge 3$ by applying linear form in logarithms. There are many papers in the literature which solve Diophantine equations related to Fibonacci numbers and Lucas numbers [3,4,5,6,7,8,9,10,11,12,13,14]. In 2013, D. Marques and A. Togbé [3] found all solutions $(n,a,b,c)$ to the Diophantine equation $F_n=2^a+3^b+5^c$ and $L_n=2^a+3^b+5^c$ for integers $n,a,b,c$ with $0\le max\{a,b\}\le c$. In 2019, B. D. Bitim [4] investigated the solutions $(n,m,a)$ to the Diophantine equation $L_n-L_m=2·3^a$ for nonnegative integers $n,m,a$ with $n>m$. Let p be a prime number and $max\{a,b\}\ge 2$, in 2009, F. Luca and P. Stǎnicǎ [5] concluded that there are only finitely many positive integer solutions $(n,p,a,b)$ to the Diophantine equation $F_n=p^a\pm p^b$.

Assume that $q\ge 2$ is an integer. A positive number $n\in \mathbb{N}$ is called a base $q-$ repdigit if $n=c\frac{q^t-1}{q-1}$, for some $t\ge 1$ and $c\in \{1,2,\dots ,q-1\}$. When $q=10$, n is simply called a repdigit. We use ${\overline{B_1\cdots B_t}}_{\left(q\right)}$ to express an integer’s base $-q$ representation which is the concatenation of the base $-q$ representations of positive integers $B_1,\dots ,B_t$. We ignore writing q if $q=10$. Then we can denote the repdigit n by $n=\overline{\underset{ttimes}{\underbrace{c\cdots c}}}$ and the concatenation of two repdigits in base 10 is $\overline{\underset{stimes}{\underbrace{c\cdots c}}\underset{ttimes}{\underbrace{d\cdots d}}},$ where $c,d\in \{0,1,\cdots ,9\}$, $c\ne d,c>0,s\ge 1$ and $t\ge 1$. There are many papers in the literature on investigating Diophantine equations related to repdigits [8,9,11,12,13,14,15,16,17,18,19,20,21]. In 2000, Luca [15] proved that if $F_n=a\frac{{10}^m-1}{9}$ and $L_n=a\frac{{10}^m-1}{9}$ for some $a\in \{0,1,\cdots ,9\}$ and $m\ge 1$, then $n=0,1,2,3,4,5,6,10$ and $n=0,1,2,3,4,5$ respectively. In 2012, all repdigits in base 10 expressible as sums of three Fibonacci numbers were found in [16]. In 2018, all repdigits in base 10 which are sums of four Fibonacci or Lucas numbers were determined in [17]. In 2019, all solutions to the Diophantine equation $F_n=\overline{\underset{mtimes}{\underbrace{a\cdots a}}\underset{ltimes}{\underbrace{b\cdots b}}}$ were found in [18], where $a,b\in \{0,1,\cdots ,9\}$ and $a>0$. For the research of concatenations of two repdigits in balancing numbers, Padovan numbers and Tribonacci numbers, please refer to the literature [19,20,21] respectively.

In this paper, we find all Lucas numbers which are concatenations of two repdigits. More precisely, we have the following result.

Theorem 1.

If

$$L_n=\overline{\underset{stimes}{\underbrace{c\cdots c}}\underset{ttimes}{\underbrace{d\cdots d}}},$$

(1)

with $c,d\in \{0,1,\cdots ,9\}$, $c\ne d,c>0,s\ge 1$ and $t\ge 1$, then

$$(n,L_n)\in \{(6,18),(7,29),(8,47),(9,76),(11,199),(12,322)\}.$$

2. Preliminaries

Firstly, the Binet’s formula for Lucas sequence is

$$L_n={\alpha }^n+{\beta }^n,n\ge 0,$$

where $\alpha =\frac{1+\sqrt{5}}{2}$ and $\beta =\frac{1-\sqrt{5}}{2}$. For all positive integers n, we have the following inequality

$${\alpha }^{n-1}\le L_n\le {\alpha }^{n+1}.$$

(2)

Secondly, we recall the definition and properties for logarithmic height of an algebraic number. Let $\eta$ be an algebraic number of degree m and suppose that the minimal primitive polynomial of $\eta$ is $f\left(X\right):=a_0{\prod }_{i=1}^m(X-{\eta }^{\left(i\right)})\in \mathbb{Z}\left[X\right]$ with $a_0>0$. We give the logarithmic height of $\eta$ by

$$h\left(\eta \right):={\displaystyle \frac{1}{m}}\left(log{a}_0+\sum _{i=1}^{m}logmax\left\{\right|{\eta }^{\left(i\right)}|,1\}\right).$$

In this paper, for any two integers a and b, we denote the greatest common divisor of a and b by $gcd(a,b)$. Specifically, $h\left(\eta \right)=logmax\left\{\right|p|,q\}$ when $\eta =\frac{p}{q}\in \mathbb{Q}$ with $gcd(p,q)=1$ and $q>0$. We have the following properties of the logarithmic height $h(·)$:

$$h(\eta \pm \gamma )\le h\left(\eta \right)+h\left(\gamma \right)+log2,$$

$$h\left(\eta {\gamma }^{\pm 1}\right)\le h\left(\eta \right)+h\left(\gamma \right),$$

$$h\left({\eta }^k\right)=\left|k\right|h\left(\eta \right)(k\in \mathbb{Z}).$$

We need the following lemma to prove our theorem.

Lemma 1.

(see [22]) Let $d_{\mathbb{L}}$ be the degree of an algebraic number field $\mathbb{L}$ over $\mathbb{Q}$ and $\mathbb{L}\subseteq \mathbb{R}$. Let ${\gamma }_1,{\gamma }_2,\cdots ,{\gamma }_l\in \mathbb{L}$ be non-zero elements and let $b_1,\cdots ,b_l$ be rational integers. If $\Gamma :={\gamma }_1^{b_1}\cdots {\gamma }_l^{b_l}-1\ne 0$, then

$$|\Gamma |\ge \exp (-1.4·{30}^{l+3}{l}^{4.5}{d}_{\mathbb{L}}^2(1+\log d_{\mathbb{L}})(1+\log B)A_1\cdots A_l),$$

where $A_j$ are real numbers such that

$$A_j\ge \max \{d_{\mathbb{L}}h\left({\gamma }_j\right),|\log {\gamma }_j|,0.16\}$$

for $j=1,\cdots ,l$ and $B\ge \max \left\{\right|b_1|,\cdots ,|b_l|,3\}$.

Thirdly, we need the following Lemma 2 and Lemma 3 to reduce some large upper bounds on the variables in the course of our calculations.

Lemma 2.

(see [23]) Let M be a positive integer and let $\frac{p}{q}$ be a convergent of the continued fraction of the irrational number α such that $q>6M$, and let $A,B,\tau$ be some real numbers with $A>0$ and $B>1$. Let $ϵ:=\parallel \tau q\parallel -M\parallel \alpha q\parallel$, where $\parallel ·\parallel$ denotes the distance from the nearest integer. If $ϵ>0$, then there exists no solution to the inequality

$$0<|u\alpha -v+\tau |<A{B}^{-\omega }$$

in positive integers $u,v$, and ω with $u\le M$ and $w\ge \frac{\log (Aq/ϵ)}{\log B}$.

Lemma 3.

(see [24]) Let τ be an irrational number, M be a positive integer and $\frac{p_k}{q_k}(k=0,1,2,\dots )$ be all the convergents of the continued fraction $[a_0,a_1,\dots ]$ of τ. Let N be such that $q_N>M$. Then putting $a_M:=\max \{a_i:i=0,1,\dots ,N\}$, the inequality

$$|m\tau -n|>\frac{1}{(a_M+2)m}$$

holds for all pairs $(n,m)$ of integers with $0<m<M$.

3. Proof of Theorem 1

3.1. Bounding n

According to (1), we get

$$\begin{array}{cc}\hfill L_n& =\overline{\underset{stimes}{\underbrace{c\cdots c}}\underset{ttimes}{\underbrace{d\cdots d}}}\hfill \\ \hfill & =\overline{\underset{stimes}{\underbrace{c\cdots c}}}·{10}^t+\overline{\underset{ttimes}{\underbrace{d\cdots d}}}\hfill \\ \hfill & =\frac{1}{9}(c{10}^{s+t}-(c-d){10}^t-d).\hfill \end{array}$$

(3)

Suppose that $n>1000$. From inequality (2), we can get ${\alpha }^{n-1}\le L_n<{10}^{s+t}$ and ${10}^{s+t-1}\le L_n\le {\alpha }^{n+1}$, which implies that

$$(s+t)\log 10-\log 10-\log \alpha \le n\log \alpha <(s+t)\log 10+\log \alpha .$$

(4)

Thus, we can get

$$4.78(s+t)-5.8<n<4.79(s+t)+1.$$

(5)

From (5), we get $s+t>\frac{n-1}{4.79}>208$ and $n>s+t$. According to (3) and Binet’s formulae for Lucas sequences, we get

$$|9{\alpha }^n-c{10}^{s+t}|=|-9{\beta }^n-((c-d){10}^t+d)|\le 9{\alpha }^{-n}+9·{10}^t+9<27·{10}^t,$$

(6)

which implies that

$$\left|\frac{9}{c}{\alpha }^n{10}^{-s-t}-1\right|<\frac{27}{{10}^s}.$$

(7)

Let ${\Gamma }_1$:$=\frac{9}{c}{\alpha }^n{10}^{-s-t}-1$, then ${\Gamma }_1\ne 0$. If ${\Gamma }_1=0$, then ${\alpha }^n=\frac{{10}^{s+t}c}{9}\in \mathbb{Q}$, thus we have $\frac{{10}^{s+t}c}{9}=\frac{{(1+\sqrt{5})}^n}{2^n}=\frac{f+g\sqrt{5}}{2^n}$, where $f,g\in \mathbb{Z},f>0,g>0$, this implies that $\sqrt{5}=\frac{\frac{{10}^{s+t}c{2}^n}{9}-f}{g}\in \mathbb{Q}$, which is impossible. According to Lemma 1, we take $l=3$, ${\gamma }_1=\frac{9}{c},{\gamma }_2=\alpha ,{\gamma }_3=10$ and $b_1=1,b_2=n,b_3=-s-t.$ Thus, we have $\mathbb{L}=\mathbb{Q}\left(\alpha \right),d_{\mathbb{L}}=[\mathbb{L}:\mathbb{Q}]=2.$ Note that $h\left({\gamma }_1\right)=h\left(\frac{9}{c}\right)\le \log 9,h\left({\gamma }_2\right)=\frac{1}{2}\log \alpha ,h\left({\gamma }_3\right)=\log 10$. Thus, we can take $A_1=2\log 9,$$A_2=0.5,$$A_3=4.8$. Note that $B=\max \left\{\right|b_1|,|b_2|,|b_3|,3\}=\max \{1,n,s+t,3\}=n$. Hence, we get

$$\mid {\Gamma }_1\mid >\exp (-C_1(1+\log n)),$$

(8)

where $C_1=1.025\times {10}^{13}$. Thus from (7) and (8), we can get

$$s\log 10<C_1(1+\log n)+\log 27.$$

(9)

We rewrite Equation (3), then we get

$$\left|{\alpha }^n-\frac{c{10}^s-(c-d)}{9}·{10}^t\right|=\left|{\beta }^n+\frac{d}{9}\right|\le {\alpha }^{-n}+1<2.$$

(10)

It follows that

$$\begin{array}{cc}\hfill \left|\frac{c{10}^s-(c-d)}{9}·{\alpha }^{-n}·{10}^t-1\right|& <\frac{2}{{\alpha }^n}.\hfill \end{array}$$

(11)

Let ${\Gamma }_2:=\frac{c{10}^s-(c-d)}{9}·{\alpha }^{-n}·{10}^t-1$, then ${\Gamma }_2\ne 0$. If ${\Gamma }_2=0$, then ${\alpha }^n=\frac{c{10}^s-(c-d)}{9}·{10}^t\in \mathbb{Q}$, which is false. According to Lemma 1, we take $l=3$,${\gamma }_1=\frac{c{10}^s-(c-d)}{9},{\gamma }_2=\alpha ,{\gamma }_3=10$ and $b_1=1,b_2=-n,b_3=t.$ Thus, we have $\mathbb{L}=\mathbb{Q}\left(\alpha \right),d_{\mathbb{L}}=[\mathbb{L}:\mathbb{Q}]=2.$ From (9), we can get

$$\begin{array}{cc}\hfill h\left({\gamma }_1\right)& \le h(c{10}^s-(c-d))+h\left(9\right)\hfill \\ \hfill & \le 3\log 9+s\log 10+\log 2\hfill \\ \hfill & \le C_1(1+\log n)+\log 27+3\log 9+\log 2\hfill \\ \hfill & \le 1.03·{10}^{13}·(1+\log n),\hfill \end{array}$$

(12)

and we have $h\left({\gamma }_2\right)=\frac{1}{2}\log \alpha ,h\left({\gamma }_3\right)=\log 10$. Thus, we can take $A_1=2.06·{10}^{13}·(1+\log n),$$A_2=0.5,$$A_3=4.8$. Note that $B=\max \left\{\right|b_1|,|b_2|,|b_3|,3\}=\max \{1,n,t,3\}=n$. Hence, we get

$$\mid {\Gamma }_2\mid >\exp (-C_2{(1+\log n)}^2),$$

(13)

where $C_2=4.8\times {10}^{25}$. Thus from (11) and (13), we can get

$$n\log \alpha <C_2{(1+\log n)}^2+\log 2,$$

(14)

this implies that $n<4.8\times {10}^{29}$. Hence we can conclude that

$$s+t<\frac{n+5.8}{4.78}<1.01·{10}^{29}.$$

To sum up, we have the lemma as follows.

Lemma 4.

If $(n,c,d,s,t)$ is a solution in non-negative integers of Equation (1), with $c,d\in \{0,1,\cdots ,9\}$, $c\ne d$ and $c>0$, then

$$s+t<n<4.8·{10}^{29},s+t<1.01·{10}^{29}.$$

3.2. Reducing the Bound on n

We use the Lemmas 2 and 3 to reduce the bound for n. Let

$${\mathsf{\Lambda }}_1:=(s+t)\log 10-n\log \alpha -\log \frac{9}{c}.$$

From (7), we conclude that

$$\left|e^{-{\mathsf{\Lambda }}_1}-1\right|<\frac{27}{{10}^s}.$$

(15)

If $s\ge 2$, then $|e^{-{\mathsf{\Lambda }}_1}-1|<\frac{27}{{10}^s}<\frac{1}{2}$, which implies that $\frac{1}{2}<e^{-{\mathsf{\Lambda }}_1}<\frac{3}{2}$. If ${\mathsf{\Lambda }}_1>0$, then $0<{\mathsf{\Lambda }}_1<e^{{\mathsf{\Lambda }}_1}-1=e^{{\mathsf{\Lambda }}_1}(1-e^{-{\mathsf{\Lambda }}_1})<\frac{54}{{10}^s}$. If ${\mathsf{\Lambda }}_1<0$, then $0<|{\mathsf{\Lambda }}_1|<e^{|{\mathsf{\Lambda }}_1|}-1=e^{-{\mathsf{\Lambda }}_1}-1<\frac{27}{{10}^s}$. In any case, it is always holds true $0<|{\mathsf{\Lambda }}_1|<\frac{54}{{10}^s}$, which implies

$$0<\left|(s+t)\frac{\log 10}{\log \alpha }-n-\frac{\log \frac{9}{c}}{\log \alpha }\right|<\frac{\frac{54}{\log \alpha }}{{10}^s}.$$

(16)

The continued fraction of $\frac{\log 10}{\log \alpha }$ is $[a_0,a_1,a_2,a_3,a_4,\cdots ]=[4,1,3,1,1,1,6,\cdots ]$, and let $\frac{p_k}{q_k}$ be its kth convergent. Note that $s+t<1.01·{10}^{29}$ by Lemma 4. It is easy to see that $\frac{\log 10}{\log \alpha }$ is irrational. In fact, if $\frac{\log 10}{\log \alpha }=\frac{p}{q}(p,q\in \mathbb{Z}$ and $p>0,q>0,\gcd (p,q)=1)$, then ${\alpha }^p={10}^q\in \mathbb{Q}$, which is an absurdity. For all $c\in \{1,\cdots ,8\}$, according to (16) and Lemma 2, we take $M=1.01·{10}^{29}$ and $q_{60}>6M$, hence we get the minimum value of $ϵ$ is $0.061483\dots$ and $s<34$. If $c=9$, from (16), we get

$$0<\left|(s+t)\frac{\log 10}{\log \alpha }-n\right|<\frac{\frac{54}{\log \alpha }}{{10}^s}.$$

(17)

According to Lemma 3, we take $M=1.01·{10}^{29}$ and $q_{60}>M$, hence we get $a_M:=\max \{a_i:i=0,1,\dots ,60\}=106$ and we have

$$\left|(s+t)\frac{\log 10}{\log \alpha }-n\right|>\frac{1}{(a_M+2)(s+t)}>\frac{1}{108·1.01·{10}^{29}}.$$

(18)

Thus, from (17) and (18), we get

$$\frac{1}{108·1.01·{10}^{29}}<\frac{\frac{54}{\log \alpha }}{{10}^s},$$

this leads to $s<34$. So we always have $s<34$.

Let

$${\mathsf{\Lambda }}_2:=t\log 10-n\log \alpha +\log \frac{c{10}^s-(c-d)}{9}.$$

From (11) and $n>1000$, we conclude that

$$\left|e^{{\mathsf{\Lambda }}_2}-1\right|<\frac{2}{{\alpha }^n}<\frac{1}{2},$$

(19)

which implies that $\frac{1}{2}<e^{{\mathsf{\Lambda }}_2}<\frac{3}{2}$. If ${\mathsf{\Lambda }}_2>0$, then $0<{\mathsf{\Lambda }}_2<e^{{\mathsf{\Lambda }}_2}-1<\frac{2}{{\alpha }^n}$. If ${\mathsf{\Lambda }}_2<0$, then $0<|{\mathsf{\Lambda }}_2|<e^{-{\mathsf{\Lambda }}_2}-1=e^{-{\mathsf{\Lambda }}_2}(1-e^{{\mathsf{\Lambda }}_2})<\frac{4}{{\alpha }^n}$. In any case, since $0<|{\mathsf{\Lambda }}_2|<\frac{4}{{\alpha }^n}$, thus we have

$$0<\left|t\frac{\log 10}{\log \alpha }-n+\frac{\log \frac{c{10}^s-(c-d)}{9}}{\log \alpha }\right|<\frac{\frac{4}{\log \alpha }}{{\alpha }^n},$$

(20)

where $s\le 33,c\in \{1,\cdots ,9\}$ and $d\in \{0,1,\cdots ,9\}$. For inequality (20), we consider the following two cases: if $(s,c,d)\ne (1,1,0)$, according to (20) and Lemma 2, we take $M=1.01\times {10}^{29}$ and $q_{60}>6M$, hence we obtain 25 negative values of $ϵ$, the minimum value in the values of positive $ϵ$ is $0.00004477\dots$ and $n<171$. For the values of $(s,c,d)$ corresponding to the 25 negative values of $ϵ$, we take $q_{63}>6M$, according to (20) and Lemma 2, we get the minimum value in the values of $ϵ$ is $0.005613\dots$ and $n<168$. If $(s,c,d)=(1,1,0)$, from (20), we get

$$0<\left|t\frac{\log 10}{\log \alpha }-n\right|<\frac{\frac{4}{\log \alpha }}{{\alpha }^n}.$$

(21)

According to Lemma 3, we take $M=1.01·{10}^{29}$ and $q_{60}>M$, hence we get $a_M:=\max \{a_i:i=0,1,\dots ,60\}=106$ and we have

$$\left|t\frac{\log 10}{\log \alpha }-n\right|>\frac{1}{(a_M+2)t}>\frac{1}{108·1.01·{10}^{29}}.$$

(22)

Thus, from (21) and (22), we get

$$\frac{1}{108·1.01·{10}^{29}}<\frac{\frac{4}{\log \alpha }}{{\alpha }^n},$$

which leads to $n<153$. In summary, we have $n<171$. This contradicts the assumption $n>1000$. Finally, we search for the solutions to (1) in the range $n\le 1000$ by applying a program written in Mathematica and we obtain the solutions $(n,L_n)\in \{(6,18),(7,29),(8,47),(9,76),(11,199),(12,322)\}$. We complete the proof.

4. Conclusions and Future Research

For a fixed integer $k\ge 2$, let ${\left\{F_n^{\left(k\right)}\right\}}_{n\ge 2-k}$ be the $k-$ generalized Fibonacci sequence defined by $F_n^{\left(k\right)}=F_{n-1}^{\left(k\right)}+F_{n-2}^{\left(k\right)}+\cdots +F_{n-k}^{\left(k\right)}$ with the initial values $F_{-(k-2)}^{\left(k\right)}=F_{-(k-3)}^{\left(k\right)}=\cdots =F_0^{\left(k\right)}=0$, $F_1^{\left(k\right)}=1$ and ${\left\{L_n^{\left(k\right)}\right\}}_{n\ge 2-k}$ be the $k-$ generalized Lucas sequence given by $L_n^{\left(k\right)}=L_{n-1}^{\left(k\right)}+L_{n-2}^{\left(k\right)}+\cdots +L_{n-k}^{\left(k\right)}$ with the initial values $L_{-(k-2)}^{\left(k\right)}=L_{-(k-3)}^{\left(k\right)}=\cdots =L_{-1}^{\left(k\right)}=0$, $L_0^{\left(k\right)}=2,L_1^{\left(k\right)}=1$. Suppose that $c,d\in \{0,1,\cdots ,9\}$, $c\ne d,c>0,s\ge 1$ and $t\ge 1,$ our aim is to solve the two Diophantine equations

$$F_n^{\left(k\right)}=\overline{\underset{stimes}{\underbrace{c\cdots c}}\underset{ttimes}{\underbrace{d\cdots d}}}$$

(23)

and

$$L_n^{\left(k\right)}=\overline{\underset{stimes}{\underbrace{c\cdots c}}\underset{ttimes}{\underbrace{d\cdots d}}}.$$

(24)

For $k=2$ and $k=3$, the Diophantine Equation (23) has been solved in [18] and [21], respectively. In this paper, we solve the Diophantine Equation (24) for the case of $k=2$. Our future research work is to solve the Diophantine Equations (23) and (24) completely for the case of $k\ge 3$. In addition, for the main Mathematica programs used in this paper, readers can refer to Appendix A.