k−Generalized Fibonacci numbers as Concatenation of Three Repdigits

1. Introduction

One characteristic of a palindromic number is that it doesn’t alter when its digits are switched. A repdigit is a particular kind of palindromic number that is made up of a single digit that is repeated several times in base 10. The mathematical expression for a repdigit is $a\left({10}^m-1\right)/9$ for some $m\ge 1$ and $1\le a\le 9$. Interestingly, in the trivial situation of a repdigit, $m=1$, the outcome is just the digit itself.

The Fibonacci sequence $\left(F_n\right)$ is defined recursively by

$$F_n=F_{n-1}+F_{n-2},\mathrm{for}n\ge 2$$

with initial terms 0 and 1, and each subsequent term is the sum of the two preceding terms. An explicit formula for the n-th Fibonacci number, known as the Binet formula, is given by

$$F_n={\displaystyle \frac{{\alpha }^n-{\beta }^n}{\sqrt{5}}},$$

where

$$\alpha ={\displaystyle \frac{1+\sqrt{5}}{2}}\mathrm{and}\beta ={\displaystyle \frac{1-\sqrt{5}}{2}}$$

are the roots of the characteristic equation $x^2-x-1=0.$

The k-generalized Fibonacci sequence or simply the k-Fibonacci sequence denoted as ${\left(F_n^{\left(k\right)}\right)}_{n\ge 2-k}$, for an integer $k\ge 2$. The first k terms are equal to 0, and subsequent term is 1. The sum of the preceding k phrases determines each next term in the series. More specifically, for every $n\ge 2$, the recurrence relation for the k-generalized Fibonacci sequence is

$$F_n^{\left(k\right)}=F_{n-1}^{\left(k\right)}+F_{n-2}^{\left(k\right)}+\dots +F_{n-k}^{\left(k\right)}.$$

The initial values are given by $F_{-(k-2)}^{\left(k\right)}=F_{-(k-3)}^{\left(k\right)}=\dots =F_0^{\left(k\right)}=0$ and $F_1^{\left(k\right)}=1$. For instance, when $k=2$, the sequence corresponds to the standard Fibonacci numbers. When $k=3$, the sequence is known as the Tribonacci numbers, and for $k=4$, it is referred to as the Tetranacci numbers. This pattern continues as k increases.

In recent years, diophantine equations with repdigits and terms from binary recurrence sequences have attracted a lot of interest. Specifically, research on Fibonacci numbers and how they relate to repdigits has produced a number of significant findings. According to Luca [12], the Fibonacci sequence’s biggest repdigit is 55. A hypothesis by Marques [13] was later confirmed by Bravo and Luca [3], who showed that no repdigit with two or more digits exists in any k-Fibonacci sequence for $k>3$. Fibonacci numbers, which may be represented as the concatenation of two repdigits, were studied by Alahmadi et al. [1]. This study was later expanded to include k-Fibonacci numbers in [2]. In 2023, Erduvan and Keskin [10] found all Fibonacci numbers which are concatenation of three repdigits. In this paper, we investigate the representation of k-Fibonacci numbers as concatenation of three repdigits. In 2023, Erduvan and Keskin [10] found all Fibonacci numbers which are concatenation of three repdigits. In this paper, we investigate the representation of k-Fibonacci numbers as concatenation of three repdigits. Specifically, we focus on expressing k-Fibonacci numbers in the form

$$F_n^{\left(k\right)}=\overline{\underset{m_1\mathrm{times}}{\underbrace{a\dots a}}\underset{m_2\mathrm{times}}{\underbrace{b\dots b}}\underset{m_3\mathrm{times}}{\underbrace{c\dots c}}},$$

(1.1)

extending the work of Erduvan and Keskin [10]. More specifically, we present the following result.

Theorem 1.1.

For $k\ge 2$ and $n\ge k+2$, the Diophantine equation

$$\begin{array}{cc}\hfill F_n^{\left(k\right)}& =\overline{\underset{m_1\mathrm{times}}{\underbrace{a\dots a}}\underset{m_2\mathrm{times}}{\underbrace{b\dots b}}\underset{m_3\mathrm{times}}{\underbrace{c\dots c}}}=\underset{m_1\mathrm{times}}{\underbrace{\overline{a\dots a}}}\times {10}^{m_2+m_3}+\underset{m_2\mathrm{times}}{\underbrace{\overline{b\dots b}}}\times {10}^{m_3}+\underset{m_3\mathrm{times}}{\underbrace{\overline{c\dots c}}}\hfill \\ \hfill & ={\displaystyle \frac{1}{9}}\left(a{10}^{m_1+m_2+m_3}-(a-b){10}^{m_2+m_3}-(b-c){10}^{m_3}-c\right),\hfill \end{array}$$

(1.2)

has exactly 28 positive integer solutions $(n,k,m_1,m_2,m_3,a,b,c)$ with $1\le d_1\le 9$, $0\le d_2,d_3\le 9$, and $m_1,m_2,m_3\ge 1$.

$$\begin{array}{cccc}{F}_{15}^{\left(2\right)}=610\hfill & F_{16}^{\left(2\right)}=987\hfill & F_{22}^{\left(2\right)}=17711\hfill & F_{10}^{\left(3\right)}=149\hfill \\ F_{11}^{\left(3\right)}=274\hfill & F_{12}^{\left(3\right)}=504\hfill & F_{13}^{\left(3\right)}=927\hfill & F_9^{\left(4\right)}=108\hfill \\ F_{10}^{\left(4\right)}=208\hfill & F_{11}^{\left(4\right)}=401\hfill & F_{15}^{\left(4\right)}=5536\hfill & F_9^{\left(5\right)}=120\hfill \\ F_{10}^{\left(5\right)}=236\hfill & F_{11}^{\left(5\right)}=464\hfill & F_{12}^{\left(5\right)}=912\hfill & F_9^{\left(6\right)}=125\hfill \\ F_{10}^{\left(6\right)}=248\hfill & F_{11}^{\left(6\right)}=492\hfill & F_{12}^{\left(6\right)}=976\hfill & F_9^{\left(7\right)}=127\hfill \\ F_{10}^{\left(7\right)}=253\hfill & F_{11}^{\left(7\right)}=504\hfill & F_{12}^{\left(7\right)}=1004\hfill & F_{20}^{\left(7\right)}=24888\hfill \\ F_{11}^{\left(8\right)}=509\hfill & F_{15}^{\left(9\right)}=8144\hfill & F_{14}^{\left(10\right)}=4088\hfill & F_{18}^{\left(15\right)}=65533\hfill \end{array}$$

Furthermore, for $n<k+2,F_n^{\left(k\right)}$ is a power of 2, and the only solutions toEquation (2)are

$$F_9^{\left(k\right)}=128,(fork\ge 8),F_{10}^{\left(k\right)}=256,(fork\ge 9),\mathrm{and}{F}_{11}^{\left(k\right)}=512,(fork\ge 10).$$

2. Auxiliary Results

Matveev’s finding on lower bounds for nonzero linear forms of logarithms of algebraic numbers will be used frequently to solve the Diophantine equations. These bounds are crucial for effectively resolving these equations. We will start by going over the main ideas and significant findings from algebraic number theory.

Consider $\delta$ is an an algebraic number with minimal polynomial

$$g\left(X\right)=r_0(X-{\delta }^{\left(1\right)})···(X-{\delta }^{\left(k\right)})\in \mathbb{Z}\left[X\right].$$

The conjugates of $\delta$ are ${\delta }^{\left(i\right)}$’s and $r_0>0$. Consequently, the $absolute$ $logarithmic$ $height$ of $\delta$ is given by

$$h\left(\delta \right)={\displaystyle \frac{1}{k}}(log{r}_0+\sum _{j=1}^{k}max\{0,log|{\delta }^{\left(j\right)}\left|\right\}).$$

If $\delta ={\displaystyle \frac{r}{s}}$, $(s\ne 0)$ is a rational number with $gcd(r,s)=1,$ then $h\left(\delta \right)=log(max\{\left|r\right|,s\left\}\right).$

The following are some properties of the absolute logarithmic height, with their proofs available in [4] [Theorem B5]. Let $\rho$ and $\delta$ be two algebraic numbers, then

$$\begin{array}{cc}\hfill & \left(i\right)h(\rho \pm \delta )\le h\left(\rho \right)+h\left(\delta \right)+log2,\hfill \\ \hfill & \left(ii\right)h\left(\rho {\delta }^{\pm 1}\right)\le h\left(\rho \right)+h\left(\delta \right),\hfill \\ \hfill & \left(iii\right)h\left({\rho }^k\right)=\left|k\right|h\left(\rho \right).\hfill \end{array}$$

Expanding on the previous notations, we present a theorem that improves upon a result by Matveev [14], as further developed by Bugeaud et al. [5]. This theorem establishes a precise upper bound for the variables in Equation (2.1).

Theorem 2.1.

[14]. Let ${\gamma }_1,\dots ,{\gamma }_l\in \mathbb{L}$ be positive real numbers in an algebraic number field $\mathbb{L}$ of degree $d_{\mathbb{L}}$ and let $e_1,\dots ,e_l$ be nonzero integers. Consider $\mathsf{\Gamma }=\prod _{i=1}^l{\gamma }_i^{e_i}-1$ . If $\mathsf{\Gamma }\ne 0$, then

$$log\left|\mathsf{\Gamma }\right|>-1.4·{30}^{l+3}·l^4.5·d_{\mathbb{L}}^2(1+log{d}_{\mathbb{L}}^2)(1+log\left(D\right))A_1{A}_2···A_l,$$

where $D\ge max\left\{\right|e_1|,\dots ,|e_l\left|\right\}$ and $A_1,\dots ,A_l$ are positive integers such that $A_j\ge h^{\prime }\left({\gamma }_j\right)=max\{d_{\mathbb{L}}h\left({\gamma }_j\right),|log{\gamma }_j|,0.16\},$ for $j=1,\dots ,l$.

Another approach employed in our proofs is the Baker-Davenport reduction method, introduced by Dujella and Petho [9]. This method will be utilized to refine the upper bounds on the variables involved.

Lemma 2.1.

[9]Assume that M is the upper bound of u and $q>6M$ represents the n-th convergent of the continued fraction corresponding to an irrational number τ. Consider some real numbers $A,B,\mu$ where $A>0$ and $B>1$. Using $∥.∥$ to indicate the distance to the nearest integer, define $ϵ=∥\mu q∥-M∥\tau q∥$. Then, as long as $ϵ>0$ with

$$w\ge {\displaystyle \frac{log(Aq/ϵ)}{logB}},$$

the inequality $0<|u\tau -v+\mu |<A{B}^{-w}$ has no solution, where $u,v,w\in {\mathbb{Z}}^{+}.$

The previously stated lemma is not applicable for $\mu =0,$ since $ϵ<0.$ Rather, in these cases, we use the following well-known continuing fraction characteristic.

Lemma 2.2.

[15]. Let $r/s=p_i/q_i$ represent the convergences of the continued fraction expansion $[a_0,a_1,\dots ]$ of the irrational number τ. Define

$$a_M=max\{a_i\mid 0\le i\le N+1\},$$

where $M,N\in \mathbb{N}$ such that $q_N\le M<q_{N+1}$. Then, the inequality

$$\left|\tau -{\displaystyle \frac{r}{s}}\right|>{\displaystyle \frac{1}{(a_M+2)r^2}}$$

holds for all $r<M$ and any pair of positive integers $(r,s)$ with $s>0$.

Lemma 2.3.

[11] Assume that $z\ge 1$ and $T>0$ make $T>{\left(4z^2\right)}^z$ and $T>P/{(logP)}^z$. It follows that $P<2^{z}T{(logT)}^z$.

2.3.1. k-Generalized FIBONACCI Numbers

The initial $k+1$ nonzero terms of $F_n^{\left(k\right)}$ can be explicitly identified as powers of 2, specifically:

$$F_1^{\left(k\right)}=1,F_2^{\left(k\right)}=1,F_3^{\left(k\right)}=2,F_4^{\left(k\right)}=4,\dots ,F_{k+1}^{\left(k\right)}=2^{k-1}.$$

The subsequent term is given by $F_{k+2}^{\left(k\right)}=2^k-1.$

The characteristic polynomial of the k–generalized Fibonacci sequence, denoted by $F^{\left(k\right)}={\left\{F_n^{\left(k\right)}\right\}}_{n\ge -(k-2)}$, is defined as

$${\Psi }_k\left(x\right)=x^k-x^{k-1}-\dots -x-1.$$

This polynomial has an unique real root greater than 1, $\alpha =\alpha \left(k\right)$, which is located inside the range $(2(1-2^{-k}),2)$ and is irreducible over $\mathbb{Q}\left[x\right]$ (see [8]). Often known as the dominant root of $F^{\left(k\right)},$ the root $\alpha$ will now be simply represented as $\alpha$, with its reliance on k removed for convenience. The roots of ${\Psi }_k\left(x\right)$ are denoted by ${\alpha }^{\left(1\right)},{\alpha }^{\left(2\right)},\dots ,{\alpha }^{\left(k\right)}$, where we adopt the convention

$$\alpha ={\alpha }^{\left(1\right)}.$$

The function $f_k\left(z\right)$ for every integer $k\ge 2$ is defined as

$$f_k\left(z\right)={\displaystyle \frac{z-1}{2}}+(k+1)(z-2),\mathrm{for}z\in \mathbb{C}.$$

(2.1)

Using these notations, Dresden and Du introduced in [8] a "Binet-like" formula for the terms of $F^{\left(k\right)}$:

$$F_n^{\left(k\right)}=\sum _{i=1}^k{f}_k\left({\alpha }^{\left(i\right)}\right){\left({\alpha }^{\left(i\right)}\right)}^{n-1}.$$

(2.2)

They also showed that the influence of the roots within the unit circle in (2.2) is minimal. Their approximation, which holds for all $n\ge -(k-2)$, is as follows:

$$|F_n^{\left(k\right)}-f_k\left(\alpha \right){\alpha }^{n-1}|<{\displaystyle \frac{1}{2}}.$$

(2.3)

Moreover, Bravo and Luca concluded in [3] that $F_n^{\left(k\right)}$ meets the inequality

$${\alpha }^{n-2}\le F_n^{\left(k\right)}\le {\alpha }^{n-1},\mathrm{for}n\ge 1\mathrm{and}k\ge 2.$$

(2.4)

Lemma 2.4.

[3] Let $k\ge 2$, and let α represent the dominant root of the sequence ${\left\{F_n^{\left(k\right)}\right\}}_{n\ge -(k-2)}$, with $f_k\left(z\right)$ defined as in (3). Then

(i)

The inequalities

$${\displaystyle \frac{1}{2}}<f_k\left(\alpha \right)<{\displaystyle \frac{3}{4}},\mathrm{and}\left|f_k\left({\alpha }^{\left(i\right)}\right)\right|<1,2\le i\le k,$$

are satisfied.

(ii)

The logarithmic height of $f_k\left(\alpha \right)$ is bounded above by

$$h\left(f_k\left(\alpha \right)\right)<3logk.$$

Lemma 2.5.

[6] For $1\le n<2^{k/2}$ and $k\ge 10$, we have

$$F_n^{\left(k\right)}=2^{n-2}(1+\zeta ),\mathrm{where}\left|\zeta \right|<{\displaystyle \frac{5}{2^{k/2}}}.$$

(2.5)

Lemma 2.6.

All solutions of Equation (1.2)satisfy

$$(m_1+m_2+m_3)log10-2\le nlog\alpha <(m_1+m_2+m_3)log10+2.$$

Proof. The result follows directly from the fact that ${\alpha }^{n-2}<F_n^{\left(k\right)}<{\alpha }^{n-1}$. The estimate in (2.4) can be used to determine that

$${\alpha }^{n-2}<F_n^{\left(k\right)}<{10}^{m_1+m_2+m_3}.$$

After rearranging and applying the logarithm to both sides, we get

$$nlog\alpha <(m_1+m_2+m_3)log10+2log\alpha .$$

Under the condition $log\alpha <1$ for $k\ge 2$, this expression further simplifies to $nlog\alpha <(m_1+m_2+m_3)log10+2.$ For the lower bound, the inequality ${10}^{m_1+m_2+m_3-1}\le F_n^{\left(k\right)}<{\alpha }^{n-1}$ is employed. We obtain

$$(m_1+m_2+m_3-1)log10\le (n-1)log\alpha ,$$

by taking logarithms on both sides and its rearrangement leads to

$$nlog\alpha \ge (m_1+m_2+m_3)log10-2.$$

The ultimate boundaries are determined by combining these findings:

$$(m_1+m_2+m_3)log10-2\le nlog\alpha <(m_1+m_2+m_3)log10+2.$$

Lemma 2.7.

There are no powers of two in $F_n^{\left(k\right)}$ with more than three digits which are concatenation of three repdigits.

Proof. We begin with the expression

$$2^n=\overline{\underset{m_1\mathrm{times}}{\underbrace{a\dots a}}\underset{m_2\mathrm{times}}{\underbrace{b\dots b}}\underset{m_3\mathrm{times}}{\underbrace{c\dots c}}}\mathrm{where}a,b,c\in \{0,1,\dots ,9\},a\ne 0.$$

Substituting the mathematical expression of repdigits, this expression can be reformulated as

$$9\times 2^n=a\left({10}^{m_1}-1\right){10}^{m_2+m_3}+b\left({10}^{m_2}-1\right){10}^{m_3}+c({10}^{m_3}-1)$$

If $c=0,$ then $5\mid 9\times 2^n$, a contradiction. Consequently, c is a member of the set $\{1,\dots ,9\}$. Specifically, ${\nu }_2\left(c\right)\le 3$, where ${\nu }_2\left(z\right)$ denotes the exponent of 2 in the factorization of an integer z. Analyzing the $2-$adic evaluation on both sides of the equation

$$9\times 2^n-a\left({10}^{m_1}-1\right){10}^{m_2+m_3}-b\left({10}^{m_2}-1\right){10}^{m_3}-c{10}^{m_3}=c,$$

it is deduced that $m_3\le 3$ for $n\ge 4$. Next, consider the expression:

$$9\times 2^n-a{10}^{m_1+m_2+m_3}-(b-a){10}^{m_2+m_3}=(c-b){10}^{m_3}-c.$$

The term $(c-b){10}^{m_3}-c\ne 0$, (otherwise $5\mid 9\times 2^n$), and it’s absolute value is at most $9\times {10}^3+9<2^{14}$. For $n\ge 14$, we examine the $2-$adic valuations of both sides of the inequality

$$9\times 2^n-a{10}^{m_1+m_2+m_3}-(b-a){10}^{m_2+m_3}=(c-b){10}^{m_3}-c$$

and deduce that $m_2+m_3\le 13$. Consequently, the absolute value of $(b-a){10}^{m_2+m_3}+(c-b){10}^{m_3}-c$ is at most $9\times {10}^{13}+9\times {10}^3+9<2^{47}.$ For $n\ge 47$, further comparison of the $2-$adic valuations in the inequality

$$9\times 2^n-a{10}^{m_1+m_2+m_3}=(b-a){10}^{m_2+m_3}+(c-b){10}^{m_3}-c$$

yields $m_1+m_2+m_3\le 46$. Under these constraints, it follows that

$$2^n<{10}^{46}+{10}^{13}+{10}^3+1<2^{153}.$$

Therefore, $n<153$, and a numerical verification concludes the proof.

3. The Proof of Theorem 1.1

We begin by assuming that the Diophantine Equation (2) holds. The repdigit cases $a=b=c$, $a=b=0$, $b=c=0$, and $a=0,b=c$ are excluded from consideration, as they have been thoroughly addressed in [3]. The scenario $n\le 1000$ is then examined. It is sufficient to assume that $k\le 1000$ since, if $k>1000$, then $n<k$, therefore $F_n^{\left(k\right)}$ is a power of 2. In this case, Lemma 2.7 establishes that Equation (2)) has no solutions with $m_1+m_2+m_3\ge 4$. Furthermore, $F_n^{\left(k\right)}\le 2^{n-1}\le 2^{999}$, implying that $F_n^{\left(k\right)}$ has at most 300 digits. The list of all $F_n^{\left(k\right)}$ for $2\le k,n\le 1000$ and the list of all numbers that are concatenation of three repdigits with a total of at most 300 digits were created in order to address this situation. These two lists were combined to provide the solutions given in Theorem 1.1. We assume that $n>1000$ from this point on.

3.1. An Upper Bound on n in Terms of k

As explained below, we will now analyse Equation (1.2) in three different scenarios.

Case 1: From equations (1.2), (2.2), and (2.3),

$$9f_k\left(\alpha \right){\alpha }^{n-1}-a{10}^{m_1+m_2+m_3}=9{\zeta }_n-\left((a-b){10}^{m_2+m_3}-(b-c){10}^{m_3}-c\right),\left|{\zeta }_n\right|<1/2$$

is obtained.

Taking the absolute value on both sides of the equation yields

$$\begin{array}{cc}\hfill \left|9f_k\left(\alpha \right){\alpha }^{n-1}-a{10}^{m_1+m_2+m_3}\right|& =\left|9{\zeta }_n-\left((a-b){10}^{m_2+m_3}-(b-c){10}^{m_3}-c\right)\right|\hfill \\ \hfill & \le 9/2+\left(9·{10}^{m_2+m_3}+9·{10}^{m_3}+9\right)\hfill \\ \hfill & <9/2+{10}^{m_3}(9·{10}^{m_2}+9+0.9)\hfill \\ \hfill & <9/2+{10}^{m_3+m_2}(9+0.9+0.09)\hfill \\ \hfill & <10.035\times {10}^{m_2+m_3}.\hfill \end{array}$$

(3.1)

Dividing both sides of (3.1) by $a{10}^{m_1+m_2+m_3}$ results in

$$\left|\left({\displaystyle \frac{9f_k\left(\alpha \right)}{a}}\right){\alpha }^{n-1}{10}^{-m_1-m_2-m_3}-1\right|<{\displaystyle \frac{10.035}{{10}^{m_1}}}$$

(3.2)

Define

$${\mathsf{\Gamma }}_1=\left({\displaystyle \frac{9f_k\left(\alpha \right)}{a}}\right){\alpha }^{n-1}{10}^{-m_1-m_2-m_3}-1.$$

Notably, ${\mathsf{\Gamma }}_1\ne 0,$ as the vanishing of $\mathsf{\Gamma }$ would imply that

$${\displaystyle \frac{9f_k\left(\alpha \right)}{a}}={\displaystyle \frac{{10}^{m_1+m_2+m_3}}{{\alpha }^{n-1}}}.$$

For some $i\in \{2,\dots ,k\}$, applying a non-trivial Galois automorphism $\sigma$ of $\mathbb{Q}\left[\alpha \right]$ that maps $\alpha$ to ${\alpha }^{\left(i\right)}$ results in

$${\displaystyle \frac{9f_k\left({\alpha }^{\left(i\right)}\right)}{a}}={\displaystyle \frac{{10}^{m_1+m_2+m_3}}{{\left({\alpha }^{\left(i\right)}\right)}^{n-1}}}.$$

Using the estimate from Lemma 2.4 (i) and the fact that $\left|{\alpha }^{\left(i\right)}\right|<1$ for $i=2,\dots ,k$, and taking absolute values on both sides, we arrive at

$$9>{\displaystyle \frac{\left|9f_k\left({\alpha }^{\left(i\right)}\right)\right|}{a}}={\displaystyle \frac{{10}^{m_1+m_2+m_3}}{{\left|{\alpha }^{\left(i\right)}\right|}^{n-1}}}>{10}^{m_1+m_2+m_3},$$

which leads to a contradiction.

Adopting the notation from Theorem 2.1, we choose $t=3$ and define the following parameters:

$${\gamma }_1={\displaystyle \frac{9f_k\left(\alpha \right)}{a}},{\gamma }_2=\alpha ,{\gamma }_3=10,e_1=1,e_2=n-1,e_3=-m_1-m_2-m_3.$$

As ${10}^{m_1+m_2+m_3-1}<F_n^{\left(k\right)}<{\alpha }^{n-1}$, it follows that $m_1+m_2+m_3\le n$. So, we set $D=n$. Additionally, since $\mathbb{L}=\mathbb{Q}\left({\eta }_1,{\eta }_2,{\eta }_3\right)=\mathbb{Q}\left(\alpha \right)$, we can figure out that $d_{\mathbb{L}}=k$. $h\left({\gamma }_2\right)=(log\alpha )/k\le 1/k$ and $h\left({\gamma }_3\right)=log10$ are obtained by using the properties of absolute logarithmic height. Furthermore, using Lemma 2.4 (ii), we establish that

$$h\left({\gamma }_1\right)\le h(9/a)+h\left(f_k\left(\alpha \right)\right)\le log9+3logk\le 7logk\mathrm{for}\mathrm{all}k\ge 2.$$

Accordingly, we define

$$A_1=7klogk,A_2=1,A_3=klog10.$$

A lower bound for $log|{\mathsf{\Gamma }}_1|$ can be determined by using Theorem 2.1 as follows:

$$\begin{array}{cc}\hfill log\left|{\mathsf{\Gamma }}_1\right|& >-1.4·{30}^6{3}^{4.5}{k}^2(1+logk)(1+logn)(7klogk)(klog10)\hfill \\ & >-5.8\times {10}^{12}{k}^4{(logk)}^2(1+logn).\hfill \end{array}$$

(3.3)

In this derivation, we use the relation $1+logk\le 2.5logk$ valid for all $k\ge 2$. A comparison of Equation (3.3) with Equation (3.2) leads to

$$m_{1}log10-log10.035<5.8·{10}^{12}{k}^4{(logk)}^2(1+logn),$$

which simplifies to

$$m_{1}log10<5.9·{10}^{12}{k}^4{(logk)}^2(1+logn).$$

(3.4)

Case 2: Proceeding with the second rearrangement of Equation (1.2) as

$$f_k\left(\alpha \right){\alpha }^{n-1}-\left({\displaystyle \frac{a{10}^{m_1}-(a-b)}{9}}\right){10}^{m_2+m_3}={\zeta }_n-{\displaystyle \frac{(b-c){10}^{m_3}}{9}}-{\displaystyle \frac{c}{9}},$$

(3.5)

we get

$$\left|9f_k\left(\alpha \right){\alpha }^{n-1}-\left(a{10}^{m_1}-(a-b)\right){10}^{m_2+m_3}\right|=\left|9{\zeta }_n-(b-c){10}^{m_3}-c\right|<10.35·{10}^{m_3}.$$

(3.6)

Dividing both sides of (3.6) by $\left(a{10}^{m_1}-\left(a-b\right)\right){10}^{m_2+m_3}$, we arrive at

$$\left|1-\left({\displaystyle \frac{9f\left(\alpha \right)}{a{10}^{m_1}-\left(a-b\right)}}\right){\alpha }^{n-1}{10}^{-m_2-m_3}\right|<{\displaystyle \frac{1.16}{{10}^{m_2}}}.$$

(3.7)

Let us introduce the following parameters:

$$({\gamma }_1,{\gamma }_2,{\gamma }_3)=({\displaystyle \frac{9f\left(\alpha \right)}{a{10}^{m_1}-\left(a-b\right)}},\alpha ,10)$$

and

$$(e_1,e_2,e_3)=(1,n-1,-m_2-m_3).$$

With these definitions, we proceed to apply Theorem 2.1. In this scenario, $d_{\mathbb{L}}=k$ since $({\gamma }_1,{\gamma }_2$,${\gamma }_3)\in \mathbb{L}=\mathbb{Q}\left(\alpha \right)$. We define

$${\mathsf{\Gamma }}_2=1-\left({\displaystyle \frac{9f\left(\alpha \right)}{a{10}^{m_1}-\left(a-b\right)}}\right){\alpha }^{n-1}{10}^{-m_2-m_3}.$$

Following the same reasoning used previously for ${\mathsf{\Gamma }}_1$, we conclude that ${\mathsf{\Gamma }}_2\ne 0$. Using properties of the absolute logarithmic height, we derive

$$\begin{array}{cc}\hfill h\left({\gamma }_1\right)& =h\left({\displaystyle \frac{9f\left(\alpha \right)}{a{10}^{m_1}-\left(a-b\right)}}\right)\hfill \\ & <3log9+log2+5.9·{10}^{12}·k^4{(logk)}^2(1+logn)+3logk\hfill \\ & <6·{10}^{12}·k^4{(logk)}^2(1+logn).\hfill \\ \hfill h\left({\gamma }_2\right)& =h\left(\alpha \right)={\displaystyle \frac{log\alpha }{k}}\hfill \\ \hfill h\left({\gamma }_3\right)& =h\left(10\right)=log10<2.31.\hfill \end{array}$$

So, we can assign $A_1=6·{10}^{12}·k^5{(logk)}^2(1+logn),A_2=1$, and $A_3=klog10$. Since $m_2+m_3<n$ and $D\ge max\left\{\left|1\right|,|n-1|,\left|-m_2-m_3\right|\right\}$, we set $D=n$.

By using Theorem 2.1 and inequality (3.7), we arrive at

$$\left|{\mathsf{\Gamma }}_2\right|>exp\left(-C·(1+logn)·(1+logk)·klog10·A_1·A_2·A_3\right),$$

where $C=1.4·{30}^6·3^{4.5}·k^2$. Thus, $m_{2}log10<6.1·{10}^{24}·k^8{(1+logn)}^2{(logk)}^3$ is obtained. We may approximate $6.1{(1+logn)}^2$ by $7{(logn)}^2$ for $n>{10}^{10}$ and by substituting this estimate, we obtain

$$m_{2}log10<7·{10}^{24}·k^8{(logn)}^2{(logk)}^3.$$

(3.8)

Case 3: The third rearrangement of Equation (1.2) yields

$$9f\left(\alpha \right){\alpha }^{n-1}-\left(a{10}^{m_1+m_2}-\left(a-b\right){10}^{m_2}-\left(b-c\right)\right){10}^{m_3}=-({\displaystyle \frac{9{\zeta }_n}{2}}+c).$$

(3.9)

We get the following by taking the absolute values of Equation (3.9)’s both sides:

$$\left|9f\left(\alpha \right){\alpha }^{n-1}-\left(a{10}^{m_1+m_2}-\left(a-b\right){10}^{m_2}-\left(b-c\right)\right){10}^{m_3}\right|\le 9{\zeta }_n+c<{\displaystyle \frac{9}{2}}+9<13.5$$

This brings us to the key inequality:

$$\left|9f\left(\alpha \right){\alpha }^{n-1}-\left(a{10}^{m_1+m_2}-\left(a-b\right){10}^{m_2}-\left(b-c\right)\right){10}^{m_3}\right|<13.5.$$

(3.10)

Upon dividing both sides of (3.10) by $9f\left(\alpha \right){\alpha }^{n-1}$, we obtain

$$\left|1-\left({\displaystyle \frac{\left(a{10}^{m_1+m_2}-\left(a-b\right){10}^{m_2}-\left(b-c\right)\right)}{9f\left(\alpha \right)}}\right){\alpha }^{-(n-1)}{10}^{m_3}\right|\le {\displaystyle \frac{3}{{\alpha }^{n-1}}}.$$

(3.11)

Define

$${\mathsf{\Gamma }}_3=1-\left({\displaystyle \frac{\left(a{10}^{m_1+m_2}-\left(a-b\right){10}^{m_2}-\left(b-c\right)\right)}{9f\left(\alpha \right)}}\right){\alpha }^{-(n-1)}{10}^{m_3}.$$

If ${\mathsf{\Gamma }}_3=0$, then

$${\displaystyle \frac{{\alpha }^{n-1}}{{10}^{m_3}}}=\left({\displaystyle \frac{\left(a{10}^{m_1+m_2}-\left(a-b\right){10}^{m_2}-\left(b-c\right)\right)}{9f\left(\alpha \right)}}\right).$$

(3.12)

So, ${\mathsf{\Gamma }}_3\ne 0$. By applying an automorphism of $\mathbb{L}$ that maps $\alpha$ to ${\alpha }^{\left(i\right)}$, where $i\ge 2$, and subsequently implementing the absolute value operation, we derive

$${\displaystyle \frac{|{\alpha }^{\left(i\right)}{|}^{n-1}}{{10}^{m_3}}}=\left({\displaystyle \frac{\left(a{10}^{m_1+m_2}-\left(a-b\right){10}^{m_2}-\left(b-c\right)\right)}{\left|9f\right({\alpha }^{\left(i\right)}\left)\right|}}\right).$$

(3.13)

Taking the ratio of Equations (3.12) and (3.13), and applying Lemma 2.4 (i), leads to

$$1>|f_k\left({\alpha }^{\left(i\right)}\right)\left|\right|{\alpha }^{\left(i\right)}{|}^{n-1}=f_k\left(\alpha \right){\alpha }^{n-1}>{\displaystyle \frac{1}{2}}{\left({\displaystyle \frac{1+\sqrt{5}}{2}}\right)}^{n-1},$$

a contradiction since $n>1000.$ Next, we apply Theorem 2.1 with

$$\begin{array}{cc}\hfill ({\gamma }_1,e_1)& =\left({\displaystyle \frac{\left(a{10}^{m_1+m_2}-\left(a-b\right){10}^{m_2}-\left(b-c\right)\right)}{9f\left(\alpha \right)}},1\right),\hfill \\ \hfill ({\gamma }_2,e_2)& =(\alpha ,-(n-1\left)\right),\mathrm{and}\hfill \\ \hfill ({\gamma }_3,e_3)& =(10,m_3).\hfill \end{array}$$

In the field $mathbbL=mathbbQ\left(alpha\right)$, the parameters $gamm{a}_1$, $gamm{a}_2$, and $gamm{a}_3$ are all positive real values, suggesting that $d_{m}athbbL=k$. Using the absolute logarithmic height’s characteristics, we get

$$\begin{array}{cc}\hfill h\left({\gamma }_1\right)& =h\left({\displaystyle \frac{\left(a{10}^{m_1+m_2}-\left(a-b\right){10}^{m_2}-\left(b-c\right)\right)}{9f\left(\alpha \right)}}\right)\hfill \\ & \le h\left(9f\left(\alpha \right)\right)+h\left(a{10}^{m_1+m_2}\right)+h\left(\left(a-b\right){10}^{m_2}\right)+h\left(\left(b-c\right)+2log2\right.\hfill \\ & <10.9+\left(m_1+m_2\right)log10+m_{2}log10\hfill \\ \hfill h\left({\gamma }_2\right)& ={\displaystyle \frac{log\alpha }{k}}\hfill \\ \hfill h\left({\gamma }_3\right)& =log10<2.31.\hfill \end{array}$$

We derive $h\left({\gamma }_1\right)<12.3·{10}^{24}·k^8{(logk)}^3{(1+logn)}^2$ using (3.4) and (3.8). Consequently, we set $A_1=12.3·{10}^{24}·k^9{(logk)}^3{(1+logn)}^2,A_2=1$, and $A_3=klog10$ instead. This leads us to choose $D=n$ since $D\ge max\left\{\left|1\right|,|-(n-1)|,\left|m_3\right|\right\}$. Therefore, using Theorem 2.1, we get

$$3·{\alpha }^{-(n-1)}>\left|{\mathsf{\Gamma }}_3\right|>exp\left(-4.1·{10}^{36}·k^{12}·(1+logk){(logk)}^3·{(1+logn)}^3\right)$$

or

$$(n-1)log\alpha -log3<4.1·{10}^{36}·k^{12}·(1+logk){(logk)}^3·{(1+logn)}^3.$$

This implies $n<8.9·{10}^{36}·k^{12}·(1+logk){(logk)}^3·{(1+logn)}^3.$ As we have assumed $n>{10}^{10}$, so we have $8.9{(1+logn)}^3\le 11{(logn)}^3$. this implies

$$n<11·{10}^{36}·k^{12}·(1+logk){(logk)}^3·{(logn)}^3.$$

(3.14)

Now by applying Lemma 2.3, with $(1+logk)\le 2.5logk$, we have

$${\displaystyle \frac{n}{{(logn)}^3}}<2.8\times {10}^{37}{k}^{12}{(logk)}^4=T.$$

We therefore obtain

$$\begin{array}{cc}& <2^3·T(logT)\hfill \\ & <2.3·{10}^{38}{k}^{12}{(logk)}^4·{\left(log\left(2.8\right)+37log10+12logk+4(loglogk)\right)}^3\hfill \\ & <2.3·{10}^{38}{k}^{12}{(logk)}^4·{\left(134.3logk\right)}^3\hfill \\ & <5.8·{10}^{44}·k^{12}{(logk)}^7.\hfill \end{array}$$

Lemma 2.6 and the fact that $\alpha <2$ imply that

$$(m_1+m_2+m_3)log10<nlog\alpha +2.$$

This results in

$$m_1+m_2+m_3<{\displaystyle \frac{1}{log10}}\left((5.8·{10}^{44}·k^{12}{(logk)}^7(log2))+2\right)<2\times {10}^{44}{k}^{12}{(logk)}^7.$$

The conclusions derived up to this point are summarized in the following lemma.

Lemma 3.2.

The solutions to Equation (1.2) are constrained by the inequalities

$$m_1+m_2+m_3<2\times {10}^{44}{k}^{12}{(logk)}^7\mathrm{and}n<5.8\times {10}^{44}{k}^{12}{(logk)}^7.$$

3.3. An Upper Bound on n in the Case of Large k

In order to apply Lemma 2.5, $n<2^{k/2}$ must be ensured. By applying Lemma 3.2, this condition is satisfied if

$$5.8\times {10}^{44}{k}^{12}{(logk)}^7<2^{k/2}$$

and for $k\ge 554$, this inequality is valid. Assuming $k\ge 554$, we may write:

$$F_n^{\left(k\right)}=2^{n-2}(1+\zeta ),\left|\zeta \right|<{\displaystyle \frac{5}{2^{k/2}}}.$$

(3.15)

In case of k, we have $\left|\zeta \right|<1/2$, which leads to $2^{n-2}\in \left[(2/3)F_n^{\left(k\right)}/2,2F_n^{\left(k\right)}\right].$ This yields $2^{n-2}>(2/3)F_n^{\left(k\right)}\ge (2/3){10}^{m_1+m_2+m_3-1}$. Substituting Equation (3.15) into Equation (1.2) produces

$$2^{n-2}(1+\zeta )={\displaystyle \frac{1}{9}}\left(a{10}^{m_1+m_2+m_3}-(a-b){10}^{m_2+m_3}-b\right).$$

This can be rewritten as

$$\left|2^{n-2}-(a/9){10}^{m_1+m_2+m_3}\right|\le 2^{n-2}\left|\zeta \right|+{\displaystyle \frac{|a-b|{10}^{m_2+m_3}+|b-c|{10}^{m_3}+c}{9}}\le {\displaystyle \frac{5\times 2^{n-2}}{2^{k/2}}}+9.99\times {10}^{m_2+m_3}.$$

Hence,

$$\begin{array}{cc}\hfill \left|(a/9){10}^{m_1+m_2+m_3}{2}^{-(n-2)}-1\right|& \le {\displaystyle \frac{5}{2^{k/2}}}+{\displaystyle \frac{9.99\times {10}^{m_2+m_3}}{2^{n-2}}}\hfill \\ \hfill & \le {\displaystyle \frac{5}{2^{k/2}}}+{\displaystyle \frac{9.99\times 1.5}{{10}^{m_1-1}}}\hfill \\ \hfill & \le 155max\left\{{\displaystyle \frac{1}{2^{k/2}}},{\displaystyle \frac{1}{{10}^{m_1}}}\right\}.\hfill \end{array}$$

(3.16)

Assign

$${\mathsf{\Gamma }}_4=(a/9){10}^{m_1+m_2+m_3}{2}^{-(n-2)}-1.$$

It is clear that ${\mathsf{\Gamma }}_4\ne 0$, because if it were zero, the equation ${10}^{m_1+m_2+m_3}a=9·2^{n-2}$ would arise. This is not possible as $5\mid {10}^{m_1+m_2+m_3}a$, but $5\nmid 9·2^{n-2}$. Here, we have set the parameters

$$\left({\gamma }_1,{\gamma }_2,{\gamma }_3,e_1,e_2,e_3,t\right)=(a/9,10,2,1,m_1+m_2+m_3,-(n-2),3).$$

It can be observed that $\mathbb{L}({\gamma }_1,{\gamma }_2,{\gamma }_3)=\mathbb{Q}$, so $d_{\mathbb{L}}=1$. $A_1=log9\ge h\left({\gamma }_1\right),A_2=log10,\mathrm{and}{A}_3=log2$. Since $m_1+m_2+m_3\le n$, we have $D=n\ge max\{1,m_1+m_2+m_3,|-(n-2)\left|\right\}$. By virtue of Theorem 2.1, we derive

$$log\left|{\mathsf{\Gamma }}_4\right|>-6\times {10}^{11}(1+logn),$$

which simplifies to

$$min\{(k/2)log2,m_{1}log10\}<6\times {10}^{11}(1+logn)+log155<{10}^{12}logn.$$

This inequality is valid for $n>1000$. Assume that

$$(k/2)log2=min\{(k/2)log2,m_{1}log10\}.$$

This assumption results in the inequality

$$k<{\displaystyle \frac{2}{log2}}\left({10}^{12}logn\right)<3\times {10}^{12}logn.$$

Substituting this result into the bound of n from Lemma 3.2, and taking into account that $n\ge k$ (because otherwise, $F_n^{\left(k\right)}$ would be a power of 2, which contradicts Lemma 2.7), we obtain

$$\begin{array}{cc}\hfill n& <5.8\times {10}^{44}{k}^{12}{(logk)}^7\hfill \\ \hfill & <5.8\times {10}^{44}{(3\times {10}^{12})}^{12}{(logn)}^{19}\hfill \\ \hfill & <3.1\times {10}^{194}{(logn)}^{19}.\hfill \end{array}$$

Applying Lemma 2.3, we get

$$n<2^{19}T{(logT)}^{19}<4\times {10}^{250},$$

(3.17)

where $T=3.1\times {10}^{194}$. Assume

$$min\{(k/2)log2,m_{1}log10\}=m_{1}log10.$$

As a result, we have

$$m_{1}log10<{10}^{12}logn.$$

(3.18)

Equation (12) can be rewritten as

$$2^{n-2}(1+\zeta )-\left({\displaystyle \frac{a{10}^{m_1}-(a-b)}{9}}\right){10}^{m_2+m_3}={\displaystyle \frac{(c-b){10}^{m_3}}{9}}-{\displaystyle \frac{c}{9}}.$$

Rearranging this equation, we get

$$\left|2^{n-2}-\left({\displaystyle \frac{a{10}^{m_1}-(a-b)}{9}}\right){10}^{m_2+m_3}\right|=\left|-2^{n-2}\zeta -{\displaystyle \frac{(b-c){10}^{m_3}}{9}}-{\displaystyle \frac{c}{9}}\right|.$$

(3.19)

Dividing by $2^{n-2}$ and implementing Lemma 2.5, we get

$$\begin{array}{cc}\hfill \left|\left({\displaystyle \frac{a{10}^{m_1}-(a-b)}{9}}\right){10}^{m_2+m_3}·2^{-(n-2)}-1\right|& \le \left|\zeta \right|+{\displaystyle \frac{|b-c|{10}^{m_3}}{9·2^{n-2}}}+{\displaystyle \frac{c}{9·2^{n-2}}}\hfill \\ \hfill & \le 22max\left\{{\displaystyle \frac{1}{2^{k/2}}},{\displaystyle \frac{1}{{10}^{m_1+m_2}}}\right\}.\hfill \end{array}$$

(3.20)

Let

$${\mathsf{\Gamma }}_5=\left({\displaystyle \frac{a{10}^{m_1}-(a-b)}{9}}\right){10}^{m_2+m_3}·2^{-(n-2)}-1.$$

Here also ${\mathsf{\Gamma }}_5\ne 0$. If ${\mathsf{\Gamma }}_5=0$, then the equation becomes

$$2^{n-2}\zeta ={\displaystyle \frac{(c-b){10}^{m_3}}{9}}-{\displaystyle \frac{c}{9}}.$$

This leads to

$$\begin{array}{cc}\hfill F_n^{\left(k\right)}& =2^{n-2}+2^{n-2}\zeta \hfill \\ & =2^{n-2}+{\displaystyle \frac{(c-b){10}^{m_3}}{9}}-{\displaystyle \frac{c}{9}}.\hfill \end{array}$$

Since $F_n^{\left(k\right)}$ is an integer, it follows that $b,c\in \{0,9\}$ and $b\ne c$.

The case $(b,c)=(9,0)$ leads to

$$\begin{array}{cc}\hfill 2^{n-2}-{10}^{m_3}=F_n^{\left(k\right)}& =a\left({\displaystyle \frac{{10}^{m_1}-1}{9}}\right){10}^{m_2+m_3}+9\left({\displaystyle \frac{{10}^{m_2}-1}{9}}\right){10}^{m_3}\hfill \\ & =a\left({\displaystyle \frac{{10}^{m_1}-1}{9}}\right){10}^{m_2+m_3}+{10}^{m_2+m_3}-{10}^{m_3}.\hfill \end{array}$$

This implies ${10}^{m_2+m_3}\mid 2^{n-2}$, which is a contradiction.

If $(b,c)=(0,9)$, then

$$\begin{array}{cc}\hfill 2^{n-2}+{10}^{m_3}-1=F_n^{\left(k\right)}& =a\left({\displaystyle \frac{{10}^{m_1}-1}{9}}\right){10}^{m_2+m_3}+9\times \left({\displaystyle \frac{{10}^{m_3}-1}{9}}\right)\hfill \\ & =a\left({\displaystyle \frac{{10}^{m_1}-1}{9}}\right){10}^{m_2+m_3}+{10}^{m_3}-1,\hfill \end{array}$$

and so, ${10}^{m_2+m_3}\mid 2^{n-2}$, a contradiction. Therefore, ${\mathsf{\Gamma }}_5\ne 0$. Now we can proceed to apply Theorem 2.1. Here we have

$$\left({\gamma }_1,{\gamma }_2,{\gamma }_3,e_1,e_2,e_3,t\right)=\left({\displaystyle \frac{a{10}^{m_1}-(a-b)}{9}},10,2,1,m_2,-(n-2),3\right)$$

Now,

$$\begin{array}{cc}\hfill h\left({\gamma }_1\right)& \le log9+log\left(a{10}^{m_1}\right)+log(a-b)+log2\hfill \\ & \le 3log9+log2+{10}^{12}logn\hfill \\ & <1.1\times {10}^{12}logn.\hfill \end{array}$$

Furthermore, since $d_{\mathbb{L}}=1$, so we assign $A_1=1.1\times {10}^{12}logn,A_2=h\left({\gamma }_2\right)=log10$ and $A_3=h\left({\gamma }_3\right)=log2$. By implementing Theorem 2.1, we get

$$log\left|{\mathsf{\Gamma }}_5\right|>-2.6\times {10}^{23}(1+logn)logn>-5\times {10}^{23}{(logn)}^2.$$

This bound is compared to equation (3.20), and since the right-hand side of (3.20) has a minimum occurs at $k/2$, we get

$$(k/2)log2-log22<5\times {10}^{23}{(logn)}^2,$$

which leads to

$$k<{\displaystyle \frac{2}{log2}}\left(5\times {10}^{23}{(logn)}^2+log22\right)<1.5\times {10}^{24}{(logn)}^2.$$

Using Lemma 3.2 we derive

$$\begin{array}{cc}\hfill n& <5.8\times {10}^{44}{k}^{12}{(logk)}^7\hfill \\ \hfill & <5.8\times {10}^{44}{(1.5\times {10}^{24}{(logn)}^2)}^{12}{(logn)}^7\hfill \\ \hfill & <7.6\times {10}^{334}{(logn)}^{31},\hfill \end{array}$$

as $k<n$. Application of Lemma 2.3 yields

$$n<2^{31}·T{(logT)}^{31}<5.2\times {10}^{433},$$

(3.21)

where $T=7.6\times {10}^{334}.$ Assume that

$$min\{(k/2)log2,(m_1+m_2)log10\}=(m_1+m_2)log10.$$

In this scenario, we conclude

$$(m_1+m_2)log10<5·{10}^{23}{(logn)}^2.$$

(3.22)

Next, by rewriting Equation (3.9), we obtain

$$2^{n-2}-{\displaystyle \frac{\left(a{10}^{m_1+m_2}-\left(a-b\right){10}^{m_2}-\left(b-c\right)\right)}{9}}{10}^{m_3}=-2^{n-2}\zeta -{\displaystyle \frac{c}{9}}.$$

(3.23)

Taking the absolute values on both sides and dividing through by $2^{n-2}$, Equation (3.23) becomes

$$\begin{array}{cc}\hfill \left|1-{\displaystyle \frac{\left(a{10}^{m_1+m_2}-\left(a-b\right){10}^{m_2}-\left(b-c\right)\right)}{9}}·2^{-(n-2)}·{10}^{m_3}\right|& \le \left|\zeta \right|+\left|{\displaystyle \frac{c}{9·2^{n-2}}}\right|\hfill \\ \hfill & \le 6max\left\{{\displaystyle \frac{1}{2^{k/2}}},{\displaystyle \frac{1}{2^{n-2}}}\right\}.\hfill \end{array}$$

(3.24)

On the right-hand side, $k/2$ is the least power of 2. Assume the reverse, that the minimal exponent exceeds $k/2$. In this case, $k/2\ge n-2$, resulting in $n<k$. Yet, when $n<k$, we have $F_n^{\left(k\right)}=2^{n-2}$, and by Lemma 2.7, this cannot be a concatenation of two repdigits because $n>1000$. Therefore, the minimum exponent of 2 must be $k/2$. Let

$${\mathsf{\Gamma }}_6=1-{\displaystyle \frac{\left(a{10}^{m_1+m_2}-\left(a-b\right){10}^{m_2}-\left(b-c\right)\right)}{9}}·2^{-(n-2)}·{10}^{m_3}.$$

Here ${\mathsf{\Gamma }}_6\ne 0$. In fact, if ${\mathsf{\Gamma }}_6=0$, we then get $2^{n-2}\zeta =-c/9$. Consequently, $F_n^{\left(k\right)}=2^{n-2}+2^{n-2}\zeta =2^{n-2}-c/9$ and since this expression is an integer, it follows that c is either 0 or 9. The case $c=0$ leads to $F_n^{\left(k\right)}=2^{n-2}$ which we have seen that it is impossible. Thus, if $c=9$ and $F_n^{\left(k\right)}=2^{n-2}-1$, then

$$\begin{array}{cc}\hfill 2^{n-2}-1=F_n^{\left(k\right)}& =a\left({\displaystyle \frac{{10}^{m_1}-1}{9}}\right){10}^{m_2+m_3}+b\left({\displaystyle \frac{{10}^{m_2}-1}{9}}\right){10}^{m_3}+9\times \left({\displaystyle \frac{{10}^{m_3}-1}{9}}\right)\hfill \\ & =a\left({\displaystyle \frac{{10}^{m_1}-1}{9}}\right){10}^{m_2+m_3}+b\left({\displaystyle \frac{{10}^{m_2}-1}{9}}\right){10}^{m_3}+{10}^{m_3}-1,\hfill \end{array}$$

which implies ${10}^{m_3}\mid 2^{n-2}$, a contradiction.

So, ${\mathsf{\Gamma }}_6\ne 0$. So, Here we have

$$\left({\gamma }_1,{\gamma }_2,{\gamma }_3,e_1,e_2,e_3\right)=\left({\displaystyle \frac{\left(a{10}^{m_1+m_2}-\left(a-b\right){10}^{m_2}-\left(b-c\right)\right)}{9}},2,10,1,-(n-2),m_3\right).$$

An analogous calculation to the one carried out for ${\mathsf{\Gamma }}_3$ in Subsection 3.1 reveals that

$$\begin{array}{cc}\hfill h\left({\gamma }_1\right)& \le log9+log\left(a{10}^{m_1+m_2}\right)+log\left((a-b){10}^{m_2}\right)+log(b-c)+log2\hfill \\ & \le 4log9+2log2+(m_1+m_2)log10+m_{1}log10\hfill \\ & \le 4log9+2log2+5·{10}^{23}{(logn)}^2+{10}^{12}logn<1.2\times {10}^{25}logn,\hfill \end{array}$$

as ${(logn)}^2<23.5logn$ for $n>{10}^{10}$. Further, $d_{\mathbb{L}}=1$, so we assign $A_1=1.2\times {10}^{25}logn,A_2=h\left({\gamma }_2\right)=log10$ and $A_3=h\left({\gamma }_3\right)=log2$. By virtue of Theorem 2.1,

$$log\left|{\mathsf{\Gamma }}_6\right|>-2.8\times {10}^{36}(1+logn)logn>-5.1\times {10}^{36}{(logn)}^2.$$

By analyzing the derived bound in relation to Equation (3.24) and noting that the minimum value on the right-hand side of (3.24) occurs at $k/2$, we obtain

$$(k/2)log2-log6<5.1\times {10}^{36}{(logn)}^2,$$

which can be further simplified to

$$k<{\displaystyle \frac{2}{log2}}\left(5.1\times {10}^{37}{(logn)}^2+log6\right)<1.5\times {10}^{37}{(logn)}^2.$$

Using Lemma 3.2 and the condition $k<n$, we may deduce:

$$\begin{array}{cc}\hfill n& <5.8\times {10}^{44}{k}^{12}{(logk)}^7\hfill \\ \hfill & <5.8\times {10}^{44}{(1.5\times {10}^{37}{(logn)}^2)}^{12}{(logn)}^7\hfill \\ \hfill & <7.6\times {10}^{490}{(logn)}^{31}.\hfill \end{array}$$

Now, by applying Lemma 2.3, we get

$$n<2^{31}T{(logT)}^{31}<7.3\times {10}^{594},\mathrm{where}T=7.6\times {10}^{490}.$$

(3.25)

After comparing the estimates in (3.25) and (3.21), we may conclude that the bound in (3.25) is valid, with specific confirmation for $k\ge 554$. The lemma 3.2 provides an instantaneous constraint on n for $k<554$, with the condition $n<9.4\times {10}^{471}<7.3\times {10}^{594}$. Consequently, in all scenarios, the estimate in (3.25) remains valid. We formalize this as a lemma.

Lemma 3.4.

If $F_n^{\left(k\right)}$ is a concatenation of three repdigits, then

$$1000<n<7.3\times {10}^{594}.$$

3.5. Reducing the Upper Bound

We begin by considering the estimate provided in (3.16). Let

$${\Lambda }_4=(m_1+m_2+m_3)log10-(n-2)log2+log(a/9).$$

It is worth noting that ${\mathsf{\Gamma }}_4=e^{{\Lambda }_4}-1$ and as ${\mathsf{\Gamma }}_4\ne 0$, so ${\Lambda }_4\ne 0$. Assuming $m_1\ge 2$, it can be observed that the right-hand side of Equation (3.16) is bounded above by $1/2$. The inequality $|e^x-1|<y$ guarantees that x is smaller than $2y$ for real values $x,y$. With $w=min\{(k/2)log2,m_{1}log10\}$, we obtain

$$\left|{\Lambda }_4\right|<{\displaystyle \frac{310}{e^w}},$$

and this leads to

$$|(m_1+m_2+m_3)log10-(n-2)log2+log(a/9)|<{\displaystyle \frac{310}{e^w}}.$$

We obtain

$$|(m_1+m_2+m_3){\displaystyle \frac{log10}{log2}}-(n-2)+{\displaystyle \frac{log(a/9)}{log2}}|<{\displaystyle \frac{(310/log2)}{e^w}}$$

(3.26)

by dividing each side of the aforementioned inequality by $log2$. In order to implement Lemma 2.1, let us define

$$u=m_1+m_2+m_3,v=n-2,\tau ={\displaystyle \frac{log10}{log2}},\mu ={\displaystyle \frac{log(a/9)}{log2}},A=448,\mathrm{and}B=e.$$

We can set $M=7.3·{10}^{594}$ as an upper bound of $m_1+m_2+m_3$. The 1187-th convergent of $\tau$, denoted as $q_{1187}$, has a denominator that exceeds $6M$. So, for $1\le d\le 9$, a quick computation with $Mathematica$ yields the inequality $0<ϵ=∥\mu q_{1187}∥-M∥\tau q_{1187}∥=0.305954$. Applying Lemma 2.1 to the inequality (3.26), we obtain

$$w<log\left(A{q}_{1187}/\epsilon \right)<1380.$$

Since $(k/2)log2>1380$ for $k>4000$, we may conclude that $w=m_{1}log10$. Consequently, $m_1<1380/log10<599$. For $a\ne 9$, this is true since $\mu =0$ for $a=9$. In this instance, we have

$$\left|\tau -{\displaystyle \frac{n-2}{m_1+m_2+m_3}}\right|<{\displaystyle \frac{A}{(m_1+m_2+m_3)B^w}}.$$

Assuming $B^w>2AM$, the right-hand side of the above inequality is strictly less than $1/\left(2{(m_1+m_2+m_3)}^2\right)$. Consequently, the ratio $(n-2)/(m_1+m_2+m_3)$ corresponds to a convergent of $\tau$. Therefore, it follows that $(n-2)/(m_1+m_2+m_3)=p_j/q_j$ for some $j\le 1187$, where $p_j=(n-2)/d$ and $q_j=(m_1+m_2+m_3)/d$, with $d=gcd(n-2,m_1+m_2+m_3)$. According to Lemma 2.2, when $a_M=max\left\{a_i:0\le i\le 1187\right\}=5393$, the left-hand side of the inequality is bigger than $1/\left((a_M+2)q_j^2\right)$. As a result, we get

$$B^w\le 5395A(m_1+m_2+m_3)<5395AM,$$

which implies

$$w<log\left(5395AM\right)<1384.$$

In both circumstances, that is, if $a\in \{1,\dots ,8\}$ or if $a=9$, we thus conclude that $m_1<601$ given $k>4000$. Next, consider

$${\Lambda }_5=(m_2+m_3)log10-(n-2)log2+log\left({\displaystyle \frac{a{10}^{m_1}-(a-b)}{9}}\right).$$

Observe that ${\mathsf{\Gamma }}_5=e^{{\Lambda }_5}-1$. Since we have shown that ${\mathsf{\Gamma }}_5\ne 0$, it follows that ${\Lambda }_5\ne 0$. Referring to inequality (3.20), the right-hand side is smaller than $1/2$. Thus,

$$\left|{\Lambda }_5\right|<{\displaystyle \frac{44}{e^w}},$$

(3.27)

which leads to

$$\left|(m_2+m_3){\displaystyle \frac{log10}{log2}}-(n-2)-{\displaystyle \frac{log\left(\left(a{10}^{m_1}-(a-b)\right)/9\right)}{log2}}\right|<{\displaystyle \frac{64}{e^w}}.$$

(3.28)

As per notations of Lemma 2.1, here we have

$$u=m_2+m_3,v=n-2,\mu ={\displaystyle \frac{log\left(\left(a{10}^{m_1}-(a-b)\right)/9\right)}{log2}},\tau ={\displaystyle \frac{log10}{log2}},A=64,\mathrm{and}B=e.$$

Taking the same $M=7.3·{10}^{594}$, we get that $q_{1187}>6M$. A computer calculation for $a,b\in \{0,1,\dots ,9\}$, with the conditions $a\ne 0$, $b\ne a$, and $m_1\in [1,601]$ indicates that $0<ϵ=∥\mu q_{1187}∥-M∥\tau q_{1187}∥=0.30966404$, implying that

$$w<{\displaystyle \frac{log\left(A{q}_{1187}{\epsilon }^{-1}\right)}{logB}}<1378.$$

Consequently, $k<4000$ for $w=(k/2)log2<1378$, which results in a contradiction. The implication is that $(m_1+m_2)<1378/log10<598$ since $w=(m_1+m_2)log10$. Every $\mu$ is covered by this, except for the nine triples$(a,b,m_1)$ :

$$\begin{array}{cc}\hfill & (1,0,1),(1,9,1),(2,0,1),(3,9,1),(4,0,1)\hfill \\ \hfill & (4,9,1),(5,0,1),(7,9,1),(8,0,1).\hfill \end{array}$$

(3.29)

For these triples, it can be readily verified that

$$\mu +2=\left\{\begin{array}{cc}2,\hfill & \mathrm{if}(a,b,m_1)=(1,0,1);\hfill \\ 3,\hfill & \mathrm{if}(a,b,m_1)=(1,9,1),(2,0,1);\hfill \\ 4,\hfill & \mathrm{if}(a,b,m_1)=(3,9,1),(4,0,1);\hfill \\ 5,\hfill & \mathrm{if}(a,b,m_1)=(7,9,1),(8,0,1).\hfill \\ 1+{\displaystyle \frac{log10}{log2}},\hfill & \mathrm{if}(a,b,m_1)=(4,9,1),(5,0,1).\hfill \end{array}\right.$$

Under these circumstances, inequality (3.28) becomes

$${\Lambda }_5=|u\left({\displaystyle \frac{log10}{log2}}\right)-(n-i)|<64·B^{-w},\mathrm{for}i=2,3,4,5$$

and

$${\Lambda }_5=|(u+1)\left({\displaystyle \frac{log10}{log2}}\right)-(n-1)|<64·B^{-w},\mathrm{for}i=1+{\displaystyle \frac{log10}{log2}}.$$

Both of these inequalities lead us to the conclusion that $B^w/(64\times 5395)<7.3·{10}^{594}$ using the continuous fraction expansion of $log10/log2$. The original assumption that $k>4000$ is contradicted by this finding, which suggests that $k<4000$. Consequently, it must hold that $w=(m_1+m_2)log10$. Therefore, we obtain the bound $(m_1+m_2)<1382/log10<600$. In both scenarios—whether $a\in 1,\dots ,8$ or $a=9$ —we conclude that $(m_1+m_2)<600$ under the condition $k>4000$.

Next, define

$${\Lambda }_6=m_{3}log10-(n-2)log2+log\left({\displaystyle \frac{a{10}^{m_1+m_2}-(a-b){10}^{m_2}-(b-c)}{9}}\right).$$

It follows that ${\mathsf{\Gamma }}_6=e^{{\Lambda }_6}-1$. Since we have established that ${\mathsf{\Gamma }}_6\ne 0$, it necessarily follows that ${\Lambda }_6\ne 0$. Turning our attention to inequality (3.24), we observe that its right-hand side is less than $1/2$. This leads us to the result

$$|{\Lambda }_6|<{\displaystyle \frac{12}{2^{k/2}}},$$

which indicates that

$$\left|m_{3}log10-(n-2)log2+log\left({\displaystyle \frac{a{10}^{m_1+m_2}-(a-b){10}^{m_2}-(b-c)}{9}}\right)\right|<{\displaystyle \frac{12}{2^{k/2}}}.$$

(3.30)

By dividing both sides of this inequality by $log2$, we obtain

$$\left|\left(m_3\right){\displaystyle \frac{log10}{log2}}-(n-2)-{\displaystyle \frac{log\left({\displaystyle \frac{a{10}^{m_1+m_2}-(a-b){10}^{m_2}-(b-c)}{9}}\right)}{log2}}\right|<{\displaystyle \frac{18}{2^{k/2}}}.$$

In order to implement Lemma 2.1, let us assign

$$\begin{array}{cc}\hfill & u=m_3,v=n-2,\tau ={\displaystyle \frac{log10}{log2}},\mu ={\displaystyle \frac{log\left({\displaystyle \frac{a{10}^{m_1+m_2}-(a-b){10}^{m_2}-(b-c)}{9}}\right)}{log2}},\hfill \\ & A=18,\mathrm{and}B=2.\hfill \end{array}$$

We again set $M=7.3·{10}^{594}$ and utilize the same continued fraction approximation $p_{1187}/q_{1187}$. A computer calculation for $m_1\in [1,602]$, and $m_1+m_2<600$ showed that $0<ϵ=∥\mu q_{1187}∥-M∥\tau q_{1187}∥=0.2998877$ and

$$k/2<{\displaystyle \frac{log\left(A{q}_{1187}{\epsilon }^{-1}\right)}{logB}}<1987,$$

implies $k<4000$, a contradiction except for the three triples $(a,b,c)$:

$$(1,5,9),(3,1,9),(6,3,9).$$

(3.31)

For these triples, we have

$$\mu +2=\left\{\begin{array}{cc}6,\hfill & \mathrm{if}(a,b,c)=(1,5,9);\hfill \\ 7,\hfill & \mathrm{if}(a,b,c)=(3,1,9);\hfill \\ 8,\hfill & \mathrm{if}(a,b,c)=(6,3,9).\hfill \end{array}\right.$$

In these cases, inequality (3.30) turns into

$${\Lambda }_6=|m_3\left({\displaystyle \frac{log10}{log2}}\right)-(n-i)|<18·B^{-w},\mathrm{for}i=6,7,8.$$

Thus, by using Lemma 2.2, we have $B^w/(18\times 5395)<7.3·{10}^{594}$, which gives

$$w<{\displaystyle \frac{log\left(5395AM\right)}{log2}}<1992.$$

We have $k<3984$, which violates our expectation of $k>4000$. Thus, we demonstrated that if $k\ge 554$, then $k\le 4000$. Next, suppose $k>600$. Using Lemma 3.2, we may conclude that $n<2·{10}^{83}.$

Using the same technique for ${\mathsf{\Gamma }}_4$ and $M=2·{10}^{83}$, we discover that $q_{172}>6M$. Consequently, this implies $ϵ=0.385393$. Implementation of Lemma 2.1 results $w<201$. If $k>600$, then the inequality $(k/2)log2<201$ leads to $k<580$, which results in a contradiction. Thus, we must have $w=m_{1}log10<201$, implying that $m_1<87$.

For the case $a=9$, the previous argument gives $B^w<5395AM$, which implies $w<206$. Since $k>600$, this implies that $m_1<90$. Thus, $m_1<90$ applies to all a values. Moving on to ${\mathsf{\Gamma }}_5$, we apply the same value for M with $m_1<90$ to give $ϵ=0.025490933.$ This yields $w<202$, implying $k<582$. This is again a contradiction since $k>600$. So, we must have $w=(m_1+m_2)log10<202$, which implies $(m_1+m_2)<87$.

If $(a,b,m_1)$ is one of the 9 triples shown at (3.29), we get $B^w<5395AM$. So, now $w=(k/2)log2<204$, implies that $k<588,$ which is again a contradiction. We must have $w=(m_1+m_2)log10<88$, implying that $m_1+m_2<88$ regardless of $\mu$ values. Next, we’ll look at ${\mathsf{\Gamma }}_6$. Using $m_1<90$ in ${\mathsf{\Gamma }}_6$ to calculate M yields $ϵ=0.0017344.$ This indicates that $w<294$, giving $k<588$, resulting in a contradiction because $k>600$. If $(a,b,c)$ is one of the three triples at (3.31), then $B^w<5395AM$. Hence, we have the inequality $w=(k/2)<293$, which implies $k<586$, leading to a contradiction because $k>600$.

So, from now on, assume that $k\le 600.$ Now, by Lemma 3.2, we get the same $M=2·{10}^{83}$ for $m_1+m_2+m_3$.

So, we have by (3.2),

$${\Lambda }_1=\left|(m_1+m_2+m_3)log10-(n-1)log\alpha +log{\displaystyle \frac{9f_k\left(\alpha \right)}{a}}\right|<{\displaystyle \frac{10.035}{{10}^{m_1}}}.$$

(3.32)

Assuming $m_1\ge 2$, the right–hand side of the above equation is less than 1/2. As the inequality $|e^z-1|<y$ for real values of z and y leads to the conclusion that $z<2y$, hence, we have ${\Lambda }_1<20.07/{10}^{m_1}$. Dividing Equation (3.32) by $log\alpha$, we get

$$\left|{\displaystyle \frac{(m_1+m_2+m_3)log10}{log\alpha }}-(n-1)+{\displaystyle \frac{log\left({\displaystyle \frac{9f_k\left(\alpha \right)}{a}}\right)}{log\alpha }}\right|<{\displaystyle \frac{42}{{10}^{m_1}}}.$$

By applying Lemma 2.1, we have $q_{172}>6M$ and we get $ϵ=0.362058$, which implies $m_1<86$. Next, consider the expression in (3.7) with $m_1<86$.

$${\Lambda }_2=\left|(m_2+m_3)log10-(n-1)log\alpha +log\left({\displaystyle \frac{9f_k\left(\alpha \right)}{a{10}^{m_1}-(a-b)}}\right)\right|<{\displaystyle \frac{1.16}{{10}^{m_2}}}.$$

(3.33)

Assuming $m_2\ge 1$, the right-hand side in the above inequality is limited to a maximum of 1/2. Therefore, by the similar argument done for ${\Lambda }_1$, we can assert $|{\Lambda }_2|<2.32/{10}^{m_1}.$ Dividing Equation (3.33) by $log\alpha$, we get

$$\left|{\displaystyle \frac{(m_2+m_3)log10}{log\alpha }}-(n-1)+{\displaystyle \frac{log\left({\displaystyle \frac{9f_k\left(\alpha \right)}{a{10}^m-(a-b)}}\right)}{log\alpha }}\right|<{\displaystyle \frac{5}{{10}^{m_2}}}.$$

Now by applying Lemma2.1, we get $ϵ=0.021281$, this implies $m_2<86.$

Next move to (3.11). Here we have

$${\Lambda }_3=\left|m_{3}log10-(n-1)log\alpha +log\left({\displaystyle \frac{a{10}^{m_1+m_2}-(a-b){10}^{m_2}-(b-c)}{9f_k\left(\alpha \right)}}\right)\right|<{\displaystyle \frac{3}{{\alpha }^{n-1}}}.$$

(3.34)

Since $n>1000$, the right–hand side of (3.34) is less than 1/2. So, we have

$$|{\Lambda }_3|<{\displaystyle \frac{6}{{\alpha }^{n-1}}}.$$

Dividing both sides of (3.34) by $log\alpha$, we get

$$\left|{\displaystyle \frac{m_{3}log10}{log\alpha }}-(n-1)+{\displaystyle \frac{log\left({\displaystyle \frac{a{10}^{m_1+m_2}-(a-b){10}^{m_2}-(b-c)}{a{10}^m-(a-b)}}\right)}{9f_k\left(\alpha \right)}}\right|<{\displaystyle \frac{13}{{10}^{n-1}}}.$$

Finally, by implementing Lemma 2.1, we get $ϵ=0.399699$, this implies $n-1<412.$

So, we have $n<413$, which contradicts our assumption that $n>1000.$