Repdigits as Product of Fibonacci and Tribonacci Numbers

Abstract

In this paper, we study the problem of the explicit intersection of two sequences. More specifically, we find all repdigits (i.e., numbers with only one repeated digit in its decimal expansion) which can be written as the product of a Fibonacci by a Tribonacci number (both with the same indexes). To work on this problem, our approach is to combine lower bounds from the Baker’s theory with reduction methods (based on the theory of continued fractions) due to Dujella and Pethö.

1. Introduction

Before starting with the main problem of this paper, we recall some nomenclature and symbols for the convenience of the reader:

The Fibonacci sequence ${\left(F_n\right)}_n$ is defined by the recurrence

$$F_{n+1}=F_n+F_{n-1},$$

(1)

with initial values $F_0=0$ and $F_1=1$.

The Tribonacci numbers t n ${\left(T_n\right)}_n$ are defined by the third-order recurrence

$$T_{n+1}=T_n+T_{n-1}+T_{n-2},$$

(2)

with initial values $T_0=0$ and $T_1=T_2=1$.

A repdigit (short for “repeated digit”) is also a number of the form

$$a\left({\displaystyle \frac{{10}^{\ell }-1}{9}}\right),$$

(3)

where $\ell \ge 1$ and $a\in [1,9]$ (here, for integers $a<b$, we set $[a,b]=\{a,a+1,\cdots ,b\}$), that is, a number with only one distinct digit in its decimal expansion.

The main subject of this work is the Diophantine equations. It is almost unnecessary to stress that these objects play an important role in the Number Theory—for example, the equations $x^2+y^2=z^2$ (Pythagoras equation), $x^2-D{y}^2=c$ (Pell equation), and $x^n+y^n=z^n$ (Fermat equation) intrigued several mathematicians at different times. It is also important to notice that their studies contributed strongly to the advance of mathematics. There are many articles that address Diophantine equations concerning Fibonacci and Lucas numbers (see, e.g., [1,2,3,4,5,6,7,8]). The linear forms in logarithms, which were probably firstly used for solving Diophantine equations in Dujella and Jadrijević [9], have proved to be a very effective tool for finding solutions to all these equations.

Recently, many authors have been interested in solving Diophantine equations involving repdigits (their sums, products concatenations, etc.) and some special types of linear recurrences (usually their product, sums, etc.), where we refer the reader to [10,11,12,13,14,15,16,17,18] and references therein.

We point out that Luca [19] and Marques [20] proved that the largest repdigits in the Fibonacci and Tribonacci sequence are $F_{10}=55$ and $T_8=44$, respectively.

The aim of this paper is to continue the study of Diophantine problems involving recurrence sequences and repdigits. More precisely, we search for repdigits which are the product of Fibonacci and Tribonacci numbers with the same index. Our main result is the following:

Theorem 1.

The only solutions of the Diophantine equation

$$F_n{T}_n=a\left({\displaystyle \frac{{10}^{\ell }-1}{9}}\right),$$

(4)

in positive integers $n,a$ and ℓ, with $a\in [1,9]$, are

$$(n,\ell ,a)\in \left\{\right(1,1,1),(2,1,1),(3,1,4\left)\right\}.$$

(5)

2. Auxiliary Results

We recall a well-known non-recursive formula for generating Fibonacci numbers. Binet’s formula asserts that:

$$F_n={\displaystyle \frac{{\varphi }^n-{(-\varphi )}^{-n}}{\sqrt{5}}},$$

(6)

where $\varphi =(1+\sqrt{5})/2$ (the golden number). With this formula, we can deduce that:

$${\varphi }^{n-2}\le F_n\le {\varphi }^{n-1},\mathrm{for}\mathrm{all}n\ge 1.$$

(7)

It is also possible to infer that

$$F_n={\displaystyle \frac{{\varphi }^n}{\sqrt{5}}}+\nu ,$$

(8)

where $\left|\nu \right|\le 1/\sqrt{5}$.

Spickerman [21] (in 1982) found the following “Binet-like” formula for Tribonacci numbers (see also [22] (pp. 527–536) for some properties of this sequence):

$$T_n={\displaystyle \frac{{\alpha }^n}{-{\alpha }^2+4\alpha -1}}+{\displaystyle \frac{{\beta }^n}{-{\beta }^2+4\beta -1}}+{\displaystyle \frac{{\gamma }^n}{-{\gamma }^2+4\gamma -1}},\mathrm{for}\mathrm{all}n\ge 1,$$

(9)

where $\alpha ,\beta ,\gamma$ are the roots of $x^3-x^2-x-1=0$. More precisely, we have

$$\begin{array}{lll}\hfill \alpha & =& \frac{1}{3}+\frac{1}{3}{(19-3\sqrt{33})}^{1/3}+\frac{1}{3}{(19-3\sqrt{33})}^{1/3},\hfill \end{array}$$

(10)

$$\begin{array}{lll}\hfill \beta & =& \frac{1}{3}-\frac{1}{6}(1+i\sqrt{3}){(19-3\sqrt{33})}^{1/3}-\frac{1}{6}(1-i\sqrt{3}){(19+3\sqrt{33})}^{1/3},\hfill \end{array}$$

(11)

$$\begin{array}{lll}\hfill \gamma & =& \frac{1}{3}-\frac{1}{6}(1-i\sqrt{3}){(19-3\sqrt{33})}^{1/3}-\frac{1}{6}(1+i\sqrt{3}){(19+3\sqrt{33})}^{1/3}.\hfill \end{array}$$

(12)

Another very useful formula provided by Spickermann is

$$T_n=⌊{\displaystyle \frac{\alpha }{(\alpha -\beta )(\alpha -\gamma )}}{\alpha }^n⌉,$$

(13)

where, as usual, $\lfloor x\rceil$ denotes the nearest integer to x (the so-called Nint function). In particular, the formula

$$T_n={\alpha }^{\prime }{\alpha }^n+\eta ,$$

(14)

holds, where $\left|\eta \right|<1/2$ and ${\alpha }^{\prime }:=\alpha /(\alpha -\beta )(\alpha -\gamma )$. Moreover, since ${\alpha }^{-2}<{\alpha }^{\prime }=0.33622\cdots <\alpha$, the previous identity implies that

$${\alpha }^{n-3}<T_n<{\alpha }^{n+2},\mathrm{for}\mathrm{all}n\ge 1.$$

(15)

To prove Theorem 1, we will use Baker’s theory. Specifically, we shall use a lower bound for a linear form in three logarithms:

Lemma 1.

Let ${\alpha }_1,{\alpha }_2,{\alpha }_3\in \mathbb{R}$ be algebraic numbers and let $b_1,b_2,b_3$ be non-zero integer numbers. Define

$${\Lambda }=b_{1}log{\alpha }_1+b_{2}log{\alpha }_2+b_{3}log{\alpha }_3.$$

(16)

Let $D=[\mathbb{Q}({\alpha }_1,{\alpha }_2,{\alpha }_3):\mathbb{Q}]$ (degree of field extension) and let $A_1,A_2,A_3$ be real numbers, such that

$$A_j\ge max\{D\mathrm{h}\left({\alpha }_j\right),|log{\alpha }_j|,0.16\},forj\in \{1,2,3\}.$$

(17)

Take

$$B\ge max\{1,max\{|b_j|A_j/A_1;1\le j\le 3\left\}\right\}.$$

(18)

If ${\Lambda }\ne 0$, then

$$log\left|{\Lambda }\right|\ge -C_1{D}^2{A}_1{A}_2{A}_{3}log(1.5eDBlog\left(eD\right)),$$

(19)

where

$$C_1=6750000·e^4(20.2+log(3^{5.5}{D}^{2}log\left(eD\right))).$$

(20)

The proof of this result can be found in [23].

In the previous statement, the logarithmic height of a t-degree algebraic number $\alpha$ is defined by

$$\mathrm{h}\left(\alpha \right)={\displaystyle \frac{1}{t}}\left({\displaystyle log\left|a\right|+\sum _{j=1}^{t}logmax\{1,|{\alpha }^{\left(j\right)}\left|\right\}}\right),$$

(21)

where a is the leading coefficient of the minimal polynomial of $\alpha$ (over $\mathbb{Z}$), and ${\left({\alpha }^{\left(j\right)}\right)}_{1\le j\le t}$ are the algebraic conjugates of $\alpha$. The next lemma provides some useful properties of this function (we refer to [24] for the proof of the following facts):

Lemma 2.

We have

i. 

$\mathrm{h}\left(xy\right)\le \mathrm{h}\left(x\right)+\mathrm{h}\left(y\right)$;

ii. 

$\mathrm{h}\left(xy\right)\le \mathrm{h}\left(x\right)+\mathrm{h}\left(y\right)+log2$;

iii. 

$\mathrm{h}\left({\alpha }^r\right)=\left|r\right|·\mathrm{h}\left(\alpha \right)$, for all $r\in \mathbb{Q}$.

Our last tool is a reduction method provided by a variant of the well-known Baker-Davenport lemma, proved by Dujella and Pethő. For $x\in \mathbb{R}$, set $\parallel x\parallel =min\left\{\right|x-n|:n\in \mathbb{Z}\}=|x-\lfloor x\rceil |$ for the distance from x to the nearest integer. We refer the reader to Lemma 5 in [25] for the proof of the following lemma.

Lemma 3.

For a positive integer M, let $p/q$ be a convergent of the continued fraction of $\gamma \notin \mathbb{Q}$, such that $q>6M$, and let $\mu ,A$, and B be real numbers, with $A>0$ and $B>1$. If the number $ϵ=\parallel \mu q\parallel -M\parallel \gamma q\parallel$ is positive, then there is no solution to the Diophantine inequality

$$0<m\gamma -n+\mu <A·B^{-m}$$

(22)

in integers $m,n>0$ with

$${\displaystyle \frac{log(Aq/ϵ)}{logB}}\le m<M.$$

(23)

Now, we are ready to prove the main theorem.

3. The Proof of Theorem 1

3.1. Finding an Upper Bound for n and ℓ

By using Equations (8) and (14) in Equation (4), we have

$$\left({\displaystyle \frac{{\varphi }^n}{\sqrt{5}}}+\nu \right)\left({\alpha }^{\prime }{\alpha }^n+\eta \right)=a\left({\displaystyle \frac{{10}^{\ell }-1}{9}}\right).$$

(24)

After some manipulations, we arrive at

$$\left|{\displaystyle \frac{{\alpha }^{\prime }{\left(\alpha \varphi \right)}^n}{\sqrt{5}}}-a{\displaystyle \frac{{10}^{\ell }}{9}}\right|<1.6{\alpha }^n,$$

(25)

where we used that $\left|\nu \right|\le 1/\sqrt{5},\left|\eta \right|<1/2$ and $\varphi <1.7<\alpha$. On dividing through by ${\alpha }^{\prime }{\left(\alpha \varphi \right)}^n/\sqrt{5}$, we get

$$\left|1-{\displaystyle \frac{a\sqrt{5}}{9{\alpha }^{\prime }}}{\displaystyle \frac{{10}^{\ell }}{9}}\right|<{\displaystyle \frac{12}{{\varphi }^n}},$$

(26)

where we used that ${\alpha }^{\prime }>0.3$. Define

$${\Lambda }=\ell log10-nlog\left(\varphi \alpha \right)+log{\theta }_a,$$

(27)

where ${\theta }_a:=a\sqrt{5}/9{\alpha }^{\prime }$ (for $a\in [1,9]$). Then, Equation (26) can be rewritten as

$$|e^{{\Lambda }}-1|<{\displaystyle \frac{12}{{\varphi }^n}}.$$

(28)

First, we claim that ${\Lambda }\ne 0$. To obtain a contradiction, suppose that ${\Lambda }=0$, and thus, ${10}^{\ell }{\theta }_a={\left(\varphi \alpha \right)}^n$, and so ${\varphi }^{2n}\in \mathbb{Q}\left(\alpha \right)$. Since $\left[\mathbb{Q}\right(\alpha ):\mathbb{Q}]=3$, then ${\varphi }^{2n}$ is either a rational or a 3-degree algebraic number. However, $\varphi$ is a quadratic algebraic number, and since $\mathbb{Q}\left({\varphi }^n\right)\subseteq \mathbb{Q}\left(\varphi \right)$, then the degree of ${\varphi }^{2n}$ is either 1 or 2. So, we conclude that ${\varphi }^{2n}\in \mathbb{Q}$, which is an absurd, since (by the Binomial theorem) ${\varphi }^{2n}=A_n+B_n\sqrt{5}$, for some positive rational numbers $A_n$ and $B_n$. Therefore, we have that ${\Lambda }\ne 0$.

If ${\Lambda }>0$, then ${\Lambda }<e^{{\Lambda }}-1<12{\varphi }^{-n}$ (see Equation (28)). If ${\Lambda }<0$, then $1-e^{-\left|{\Lambda }\right|}=|e^{{\Lambda }}-1|<12{\varphi }^{-n}$. Thus, we get

$$\left|{\Lambda }\right|<e^{\left|{\Lambda }\right|}-1<{\displaystyle \frac{{\varphi }^{-n}}{1-{\varphi }^{-n}}}<{\varphi }^{-n+1}.$$

(29)

Hence, we have $\left|{\Lambda }\right|<12{\varphi }^{-n+1}$. Therefore,

$$log\left|{\Lambda }\right|<-(n-1)log\varphi +log12.$$

(30)

Now, we are in the position to apply Lemma 1. For that, take

$${\alpha }_1=10,{\alpha }_2=\varphi \alpha ,{\alpha }_3={\theta }_a,b_1=\ell ,b_2=-n,b_3=1.$$

(31)

Since $\mathbb{Q}({\alpha }_1,{\alpha }_2,{\alpha }_3)=\mathbb{Q}(\alpha ,\varphi )$, then $D\le 6$, and so $C_1<1.4·{10}^{10}$.

By using the properties of the logarithm height, we deduce that $\mathrm{h}\left({\alpha }_1\right)=\mathrm{h}\left(10\right)=log10$ and $\mathrm{h}\left({\alpha }_2\right)\le \mathrm{h}\left(\varphi \right)+\mathrm{h}\left(\alpha \right)=(log\varphi )/2+(log\alpha )/3<5(log2)/6$. Additionally,

$$\mathrm{h}\left({\alpha }_3\right)=\mathrm{h}\left({\theta }_a\right)\le \mathrm{h}\left(a\right)+\mathrm{h}\left(\sqrt{5}\right)+\mathrm{h}\left(9\right)+\mathrm{h}\left({\alpha }^{\prime }\right)<8.6,$$

(32)

where we used the definition of ${\alpha }^{\prime }$ together with the fact that $\mathrm{h}\left(\alpha \right)=\mathrm{h}\left(\beta \right)=\mathrm{h}\left(\gamma \right)$ (since they are algebraic conjugates). Thus, we can take

$$A_1=14,A_2=4\mathrm{and}{A}_3=52.$$

(33)

If $\ell \ge 4$, we have

$$max\{1,max\{|b_j|A_j/A_1;1\le j\le 3\left\}\right\}=max\{\ell ,2n/7\}.$$

(34)

Now, we combine the bounds in Equations (7) and (15) to derive

• ${\varphi }^{n-2}{\alpha }^{n-3}<F_n{T}_n=a({10}^{\ell }-1)/9<{10}^{\ell }⟹n<4.8\ell +2$
• ${\varphi }^{n-1}{\alpha }^{n+2}>F_n{T}_n=a({10}^{\ell }-1)/9>{10}^{\ell -1}⟹4.7\ell -6<n$.

In conclusion, we have

$$4.7\ell -6<n<4.8\ell +2.$$

(35)

However, $4.7\ell -6\ge \ell$ for $\ell >1$, and so, we can choose $B:=n$. Hence, by Lemma 1, we get

$$log\left|{\Lambda }\right|>-1.5\times {10}^{15}log\left(69n\right).$$

(36)

By combining the estimates Equations (30) and (36), we obtain

$$n<2.9\times {10}^{15}log\left(69n\right).$$

(37)

From this inequality, we deduce that $n<1.3\times {10}^{17}$, and by the estimate $4.7\ell -6<n$, we infer that $\ell <3\times {10}^{16}$.

3.2. Reducing the Bound

Now, we need to reduce the upper bound for n and ℓ. For that, we may suppose, with no loss of generality, that ${\Lambda }>0$ (the other case is simply a mimic, considering that $0<{{\Lambda }}^{\prime }=-{\Lambda }$).

Since $0<{\Lambda }<12{\varphi }^{-n+1}$, we obtain

$$0<\ell log10-nlog\left(\varphi \alpha \right)+log{\theta }_a<1536·{12}^{-\ell }.$$

(38)

On dividing through by $log\left(\varphi \alpha \right)$, we have

$$0<\ell {\gamma }^{\prime }-n+{\mu }_a<1448·{12}^{-\ell },$$

(39)

with ${\gamma }^{\prime }:=log10/log\left(\varphi \alpha \right)$ and ${\mu }_a:=log{\theta }_a/log\left(\varphi \alpha \right)$.

Clearly, ${\gamma }^{\prime }$ is an irrational number (because $\alpha$ and $\varphi$ are multiplicatively independent). Therefore, we shall denote $p_n/q_n$ as the n-th convergent of the (infinite) continued fraction of ${\gamma }^{\prime }$.

To reduce our bound on ℓ, we shall apply Lemma 3. For that, we choose $M=3\times {10}^{16}$, and so

$${\displaystyle \frac{p_{40}}{q_{40}}}={\displaystyle \frac{1014377782556875091}{480446934735032799}}$$

(40)

is enough of an approximant of ${\gamma }^{\prime }$ to fulfill the hypotheses of that lemma. Indeed, $q_{40}=480,446,934,735,032,799>$ $6M$. Additionally, by defining

$${ϵ}_a:=\parallel {\mu }_a{q}_{40}\parallel -M\parallel {\gamma }^{\prime }{q}_{40}\parallel$$

(41)

for $a\in [1,9]$, we have that ${min}_{a\in [1,9]}{ϵ}_a={ϵ}_9=0.0089973\cdots$ (here, we used Mathematica software (Wolfram Mathematica version 12, Wolfram Research of Champaign, Illinois, USA), see Appendix A).

Hence, the hypotheses of Lemma 3 are fulfilled for the choice of $A=1448$ and $B=12$. Thus, there is no solution of the Diophantine inequality in Equation (39) for ℓ belonging to the range

$$\left[⌊{\displaystyle \frac{log(A{q}_{40}/{ϵ}_a)}{logB}}⌋+1,M\right]=[22,3\times {10}^{16}].$$

(42)

Since $\ell <M$, then $\ell \le 21$ and so $n\le 103$. Thus, we prepare a Mathematica routine which shows that the solutions of $F_n{T}_n=a({10}^{\ell }-1)/9$, in the range $n\in [1,103],\ell \in [1,22]$ and $a\in [1,9]$, are

$$(n,\ell ,a)\in \left\{\right(1,1,1),(2,1,1),(3,1,4\left)\right\}.$$

(43)

This completes the proof. □

4. Conclusions

In this paper, we solved the Diophantine equation $F_n{T}_n=a({10}^{\ell }-1)/9$, where ${\left(F_n\right)}_n$ and ${\left(T_n\right)}_n$ are the Fibonacci and Tribonacci sequences, respectively, in positive integers $n,\ell$ and a, with $a\in [1,9]$. In other words, we found all repdigits (i.e., positive integers with only one distinct digit in its decimal expansion) which can be written as a product of a Fibonacci number and a Tribonacci number (both with the same index). In particular, we proved that the only repdigits with the desired property are the trivial ones, that is, those with only one digit ($\ell =1$). To prove this result, we combined the theory of lower bounds for linear forms in the logarithm of algebraic numbers (from Baker’s theory) with reduction methods from Diophantine approximation (based on the theory of continued fractions) due to Dujella and Pethö.