Fibonacci Numbers with a Prescribed Block of Digits

Abstract

In this paper, we prove that $F_{22}=17711$ is the largest Fibonacci number whose decimal expansion is of the form $ab\dots bc\dots c$. The proof uses lower bounds for linear forms in three logarithms of algebraic numbers and some tools from Diophantine approximation.

1. Introduction

Let ${\left(F_n\right)}_{n\ge 0}$ be the Fibonacci sequence given by second-order recurrence $F_{n+2}=F_{n+1}+F_n$, for $n\ge 0$, with initial conditions $F_0=0$ and $F_1=1$. A few terms of this sequence are

$$0,1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765,10946,17711,\dots .$$

In the last decades, many results on Diophantine properties of Fibonacci numbers have been proved with the use of refined tools in number theory. For instance, Bugeaud, Mignotte, and Siksek [1] settled the problem of Fibonacci perfect power numbers (i.e., the equation $F_n=y^t$, for $t>1$) by combing the powerful Baker’s theory with the Modular method (used by Wiles in the proof of the Fermat Last Theorem). See a generalization of their result in [2].

We remark that digital problems involving Fibonacci numbers have received much attention in the literature. A first result in this direction was proved, in 2000, by F. Luca [3] who showed that the largest Fibonacci number with only one distinct digit is $F_{10}=55$. After this, many authors worked on repdigits (i.e., integers having only one distinct digit in its decimal expansion) as expressions related to sum, product of terms of binary recurrences (see [4,5,6,7,8,9,10,11,12,13] and references therein). However, the related problem of finding all Fibonacci numbers with only two distinct digits remains open.

The aim of this paper is to continue this program. In fact, our main result searches for all Fibonacci numbers of the form $ab\dots bc\dots c$, which provides a generalization for Luca’s result ($a=b=c$). More precisely, our main result is the following:

Theorem 1.

Let $a\in [1,9]$ and $b,c\in [0,9]$. The largest solution of the Diophantine equation

$$F_n=a\underset{m}{\underbrace{b\dots b}}\underset{\ell }{\underbrace{c\dots c}}=a·{10}^{m+\ell }+b·\left(\frac{{10}^m-1}{9}\right){10}^{\ell }+c·\left(\frac{{10}^{\ell }-1}{9}\right),$$

(1)

in positive integers $m,n$, and ℓ, with $2\le \ell \le m$, is $n=22$. Explicitly, $F_{22}=17711$.

Our proof combines two deep techniques in number theory, namely, the Baker’s theory on linear forms in logarithms and some tools from Diophantine approximation (a reduction method due to Baker and Davenport).

2. Auxiliary Results

In this section, we shall present some results which will be useful in the proofs.

Let $\alpha =(1+\sqrt{5})/2$ and $\beta =-1/\alpha$. The Binet’s formula asserts that $F_n=({\alpha }^n-{\beta }^n)/\sqrt{5}$. From this formula, it is possible to deduce that the estimates

$${\alpha }^{n-2}\le F_n\le {\alpha }^{n-1},$$

(2)

hold for all $n\ge 1$. In addition, this Binet’s formula allows us to manipulate our Diophantine equation to obtain upper bounds for some linear forms in three logarithms. Thus, in order to obtain lower bounds for these forms, we shall use the celebrated Baker’s theory. Among these lower bounds, we decided to use one which was proved in ([1], Theorem 9.4).

Lemma 1.

Let ${\gamma }_1,\dots ,{\gamma }_t$ be real algebraic numbers and let $b_1,\dots ,b_t$ be nonzero rational integer numbers. Let D be the degree of the number field $\mathbb{Q}({\gamma }_1,\dots ,{\gamma }_t)$ over $\mathbb{Q}$ and let $A_j$ be a positive real number satisfying

$$A_j\ge max\{Dh\left({\gamma }_j\right),|log{\gamma }_j|,0.16\},$$

where $j=1,\dots ,t$. Assume that

$$B\ge max\left\{\right|b_1|,\dots ,|b_t\left|\right\}.$$

If ${\gamma }_1^{b_1}\cdots {\gamma }_t^{b_t}\ne 1$, then

$$|{\gamma }_1^{b_1}\cdots {\gamma }_t^{b_t}-1|\ge exp(-1.4·{30}^{t+3}·t^{4.5}·D^2(1+logD)(1+logB)A_1\cdots A_t).$$

Here, the logarithmic height of an n-degree algebraic number $\alpha$ is defined as

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

where a is the leading coefficient of the minimal primitive polynomial of $\alpha$ (over $\mathbb{Z}$) and ${\left({\alpha }^{\left(j\right)}\right)}_{1\le j\le n}$ are the (algebraic) conjugates of $\alpha$.

With these lower and upper bounds, we shall obtain an upper bound for n which is, in general, very large and then the next step is to reduce it. For that, we shall use a reduction method which is originated from Diophantine approximation. Here, we shall use a result due to Dujella and Pethö [14] (which is a variant of a famous method due to Baker–Davenport). For a real number x we use $\Vert x\Vert =min\left\{\right|x-n|:n\in \mathbb{Z}\}$ for the distance from x to the nearest integer (the so-called Nint function).

Lemma 2.

Let $M>0$ be an integer and let $\gamma ,\mu$ be real numbers, such that $\gamma \notin \mathbb{Q}$. Let $p/q$ be a convergent of the continued fraction expansion of γ such that $q>6M$ and $ϵ:=\Vert \mu q\Vert -M\Vert \gamma q\Vert$ > 0. Then there is no solution to the Diophantine inequality

$$0<m\gamma -n+\mu <A·B^{-m}$$

in positive integers $m,n$ with

$$\frac{log(Aq/ϵ)}{logB}\le m<M.$$

After presetting these tools, we can now prove our main result.

3. The Proof of The Theorem

3.1. Finding a Bound on N

By the Binet’s formula and the identity in Equation (1), we have

$${\displaystyle \frac{{\alpha }^n}{\sqrt{5}}}-\left(a+{\displaystyle \frac{b}{9}}\right){10}^{m+\ell }=\left({\displaystyle \frac{c-b}{9}}\right){10}^{\ell }+{\displaystyle \frac{{\beta }^n}{\sqrt{5}}}-{\displaystyle \frac{c}{9}}.$$

(3)

Thus,

$$\left|{\displaystyle \frac{{\alpha }^n}{\sqrt{5}}}-\left(a+{\displaystyle \frac{b}{9}}\right){10}^{m+\ell }\right|<2·{10}^{\ell }.$$

On dividing through by $(a+b/9){10}^{m+\ell }$, we obtain

$$\left|{\left(a+{\displaystyle \frac{b}{9}}\right)}^{-1}{\displaystyle \frac{{\alpha }^n}{\sqrt{5}}}{10}^{-(m+\ell )}-1\right|<{\displaystyle \frac{2}{{10}^m}},$$

(4)

where we used the fact that $a+b/9\ge 1$. Now, we are in a position to apply Lemma 1, but first we must prove that $|{\left(\sqrt{5}(a+b/9)\right)}^{-1}{10}^{-(m+\ell )}{\alpha }^n-1|\ne 0$. Indeed, in the contrary case, we would get that ${\alpha }^{2n}\in \mathbb{Q}$ which is an absurd. Thus, let us take

$${\gamma }_1:=\sqrt{5}(a+b/9),{\gamma }_2:=\alpha ,{\gamma }_3:=10,b_1:=-1,b_2:=n,b_3:=-(m+\ell ).$$

Note that $\mathbb{Q}({\gamma }_1,{\gamma }_2,{\gamma }_3)=\mathbb{Q}\left(\sqrt{5}\right)$ and then $D=2$. The conjugates of ${\gamma }_1,{\gamma }_2$, and ${\gamma }_3$ are ${\gamma }_1^{\prime }=-{\gamma }_1,{\gamma }_2^{\prime }=\beta ,{\gamma }_3^{\prime }={\gamma }_3$, respectively. Surely, ${\gamma }_2$ and ${\gamma }_3$ are algebraic integers, while the minimal polynomial of ${\gamma }_1$ is $(X-{\gamma }_1)(X-{\gamma }_1^{\prime })=X^2-5{(a+b/9)}^2$ which is a divisor of $81X^2-5{(9a+b)}^2$. Therefore,

$$h\left({\gamma }_1\right)\le {\displaystyle \frac{1}{2}}\left(log81+2log\left(10\sqrt{5}\right)\right)<5.4.$$

(A more relaxed upper bound for $h\left({\gamma }_1\right)$ could be found by using the well-known property that $h(x+yz)\le h\left(x\right)+h\left(y\right)+h\left(z\right)+log2$). In addition, $h\left({\gamma }_2\right)=(log\alpha )/2<0.75$ and $h\left({\gamma }_3\right)=log10<2.31$. Let us take $A_1:=10.9,A_2=1.5$, and $A_3=4.7$. Of course, we can assume that $n>14$. Thus,

$$\left|{\left(a+{\displaystyle \frac{b}{9}}\right)}^{-1}{\displaystyle \frac{{\alpha }^n}{\sqrt{5}}}{10}^{-(m+\ell )}-1\right|>exp\left(-7.6·{10}^{13}(1+logB)\right),$$

(5)

where $B\ge max\{n,m+\ell \}$

By combining the estimates in Equations (4) and (5), we get

$$m<3.4·{10}^{13}(1+logB).$$

Now, we have that $F_n$ has $m+\ell +1$ digits and so

$$m+\ell +1=⌊{\displaystyle \frac{log{F}_n}{log10}}⌋+1.$$

(6)

Since $m\ge \ell$ and $\lfloor x\rfloor +1>x$, we have

$$2m+1>{\displaystyle \frac{log{F}_n}{log10}}\ge (n-2){\displaystyle \frac{log\alpha }{log10}}>0.2(n-2)$$

so $10m+6\ge n$. Thus, we can take $B:=10m+6$, which yields to

$$m<3.4·{10}^{13}(1+log(10m+6))$$

and therefore $m\le 1.3·{10}^{15}$.

3.2. Reducing the Bound

Now, let us write $\Lambda :=nlog\alpha -(m+\ell )log10+log\left({(a+b/9)}^{-1}{\left(\sqrt{5}\right)}^{-1}\right)$. We know that $e^x-1>x$, for all $x\in {\mathbb{R}}^{+}$. By supposing that $\Lambda >0$ (the other case is completely similar, where we used the fact that $|e^x-1|=1-e^{-\left|x\right|}$, if $x<0$), we can rewrite Equation (4) as

$$0<nlog\alpha -(m+\ell )log10+log\left({(a+b/9)}^{-1}{\left(\sqrt{5}\right)}^{-1}\right)<1.8·{10}^{-m}.$$

Since $m>\left(0.1\right)n-0.7$ (because $2m+1>0.2(n-2)$), we can divide the previous inequality by $log10$, to obtain

$$0<n\gamma -(m+\ell )+\mu <4·{\left(1.2\right)}^{-n},$$

(7)

where $\gamma :=log\alpha /log10$ and $\mu :=log\left({(a+b/9)}^{-1}{\left(\sqrt{5}\right)}^{-1}\right)/log10$.

Clearly $\gamma$ is an irrational number (because ${\alpha }^k$ is irrational for any non-zero integer k). Let us denote $p_n/q_n$ as the nth convergent of its continued fraction.

In order to reduce our bound on m, we shall use Lemma 2. Now, since $n\le 10m+6<1.4·{10}^{16}$, we choose $M=1.4·{10}^{16}$. Thus,

$${\displaystyle \frac{p_{38}}{q_{38}}}={\displaystyle \frac{1426134855866370784}{6824015306170795931}},$$

then $q_{38}\ge 6824015306170795931>8.4·{10}^{16}=6M$. Furthermore, we have $M\Vert q_{38}\gamma \Vert <0.0013$. On the other hand, by computing $\Vert q_{38}\mu \Vert$, for $a\in [1,9]$ and $b\in [0,9]$, we have that the minimal value of this expression is obtained when $a=8,b=2$ and is $>0.0028$. Hence,

$$ϵ=\Vert q_{38}\mu \Vert -M\Vert q_{38}\gamma \Vert >0.0015.$$

We notice the all the hypotheses of the Lemma 2 are fulfilled, where $A=4$ and $B=1.2$, so, by that lemma, there is no solution of the inequality in Equation (7) (and then for the Diophantine Equation (1)) for n in the range

$$[⌊log(A{q}_{38}/ϵ)/logB⌋+1,M]=[281,1.4·{10}^{16}].$$

Since $n<M$, we get $n\le 280$. By using Equation (6), we deduce that

$$m+\ell \le (n-1){\displaystyle \frac{log\alpha }{log10}}-1<58.4$$

and so $m+\ell \le 58$. Since $m+\ell +1\ge 5$, it is seen that $F_n$ has at least 5 digits yielding $n\ge 21$. A simple search in the list of the Fibonacci numbers $F_n$ in the range $n\in [21,280]$ (see Table A1 in Appendix A), returns only $F_{22}=17711$ with the required properties. This completes the proof. □

4. Conclusions

In this paper we have been interested in finding all Fibonacci numbers which are special concatenation of digits. In particular, we show that $F_{22}=17711$ is the largest Fibonacci number whose decimal expansion is of the form $ab\dots bc\dots c$, where $a,b$, and c are decimal digits. Our approach to the proof is based on the combination of lower bounds for linear forms in logarithms (due to Baker) with reduction methods (due to Dujella–Pethö).