A Brief Study on the k-Dimensional Repunit Sequence

Abstract

In this paper, we aim to introduce and investigate the bidimensional, tridimensional and k-dimensional extension of Repunit numbers, with a particular focus on their recurrence relations, key properties, and various sum identities.

1. Introduction

The Repunit numbers ${\left\{R_n\right\}}_n$ form a numerical sequence given by $\{0,1,11,111,\dots \}$, where each term is defined by the recursive relation $R_{n+1}=10R_n+1$ for all non-negative integer n, with the initial term $R_0=0$. This sequence, identified as $A002275$ in the OEIS [1], please consult the cited sources; see [2,3,4,5] for further details. Was further studied by Santos and Costa [6], who demonstrated that it also satisfies a Horadam-like recurrence relation:

$$R_{n+1}=11R_n-10R_{n-1},\mathrm{for}\mathrm{all}n\ge 1,$$

(1)

with the initial terms $R_0=0$ and $R_1=1$. The first observation of the correlation between the Horadam recurrence (Fibonacci type) and the Repunit numbers was documented in reference [7].

Recall that the characteristic equation for the Repunit sequence is $r^2=11r-10$. This equation corresponds to the characteristic equation of a Horadam-type sequence. Solving it for r yields the roots of the characteristic equation. According to [6], the roots are $r_1=10$ and $r_2=1$. With these roots, the Binet formula offers a direct method to compute the n-th Repunit number without the need for iteration through the sequence:

$$R_n={\displaystyle \frac{{10}^n-1}{9}}.$$

(2)

In Costa et al. [8], the Repunit sequence ${\left\{R_n\right\}}_n$ was extended to include negative indexes. For $n\ge 1$, the Repunit number with a negative index is defined as follows:

$$R_{-n}=-{\displaystyle \frac{R_n}{{10}^n}}.$$

In this paper, we introduce and analyze k-dimensional Repunit sequences, with a particular focus on the bidimensional Repunit sequence. This sequence is studied in more detail in order to establish a theoretical framework that supports its extension to tridimensional and k-dimensional Repunit sequences. By making use of Binet’s formula for the ordinary or classical Repunit numbers, we derive and present several identities that are associated with these bidimensional sequences.

There is a substantial literature devoted to the bidimensional, tridimensional, and n-dimensional versions of various numerical sequences. For example, in [9] a study examines bidimensional and tridimensional identities for Fibonacci numbers in their complex form, while [10] introduces and investigates Gaussian Fibonacci numbers along with their bidimensional recurrent relations. In [11], the authors provide an overview of the Leonardo sequence and discuss the bidimensional recurrent relations derived from its one-dimensional model. Similarly, in [12,13], the bidimensional and tridimensional recurrent relations of the Narayana sequence defined from its one-dimensional recursive model are examined. Recently, in [14,15], the authors have presented the bidimensional extensions of balancing, Lucas-balancing, cobalancing and Lucas-cobalancing numbers, along with an exploration of the properties of these new bidimensional sequences. These and other related works have served as inspiration for studying new bidimensional versions of two additional numerical sequences, building on their unidimensional models.

The article is structured as follows: the next section presents the requisite background on ordinary Repunit numbers, which constitutes the foundation for the subsequent analysis. In Section 3, we introduce the bidimensional versions of this sequence and investigate some of their properties. Section 3.1 presents some of the identities satisfied by the sequence, while Section 3.2 focuses on the partial sums of its terms. In Section 4, defines the tridimensional Repunit sequence and presents results related to the bidimensional Repunit sequence. Finally, the definition and properties are extended to those of the k-dimensional Repunit sequence.

2. Background and Preliminaries Results

In this section, we revisit several important results related to the unidimensional version of Repunit sequence. These foundational results will play a crucial role in establishing new findings for the k-dimensional forms of the same sequence.

Now, we present some auxiliary results about the difference and the sum of two Repunit terms. The first is quite simple and immediately verifiable, and can be deducted from the [16].

Lemma 1.

For all non-negative integers n and k, with $n\ge k$ it follows that

$$\left(a\right)R_n-R_k=R_{n-k}·{10}^k;$$

$$\left(b\right)R_n+R_k=R_{n-k}·{10}^k+2·R_k.$$

Specifically, when $n=k+1$, one has

$$(a.\mathit{1})R_{k+1}-R_k={10}^k,$$

$$(b.\mathit{1})R_{k+1}+R_k={10}^k+2·R_k,$$

where ${\left\{R_n\right\}}_n$ is the Repunit sequence.

A consequence of Lemma 1 is that the difference $R_{2n+1}-R_{2n}$ is a perfect square, the same result can be deduced from [17].

Lemma 2.

Let n be any non-negative integer, then

$$R_{2n+1}-R_{2n}={10}^{2n},$$

where ${\left\{R_n\right\}}_n$ is the Repunit sequence.

The next result can be found in [18].

Lemma 3.

Let n be any non-negative integer. We have

$$R_{n+1}-10R_{n-1}={10}^n+1,$$

where ${\left\{R_n\right\}}_n$ is the Repunit sequence.

Furthermore, we present an auxiliary result that will be useful in more advanced stages of the investigation.

Proposition 1.

Let n be any non-negative integer. We have

$$R_{2n}-2R_n=9R_n^2,$$

where ${\left\{R_n\right\}}_n$ is the Repunit sequence.

Proof. 

A straightforward calculation, and making use of Equation (2), shows that

$$\begin{array}{ccc}\hfill R_{2n}& =& {\displaystyle \frac{{10}^{2n}-1}{9}}={\displaystyle \frac{({10}^n-1)({10}^n+1)}{9}}={\displaystyle \frac{({10}^n-1)[({10}^n-1)+2]}{9}}\hfill \\ & =& {\displaystyle \frac{({10}^n-1)({10}^n-1)}{9}}+2{\displaystyle \frac{{10}^n-1}{9}}=9R_n^2+2R_n,\hfill \end{array}$$

which proves the result. □

Next, consider the sequence of partial sums ${\sum }_{k=0}^n{R}_k=R_0+R_1+R_2+\cdots +R_n$ for n, where ${\left\{R_n\right\}}_n$ denotes the Repunit sequence. Regarding these sums, we have the following results:

Lemma 4

([6], Proposition 6.1). Let ${\left\{R_n\right\}}_n$ be the Repunit sequence, then

$$\begin{array}{ccc}\hfill \left(a\right)& {\sum }_{k=0}^n{R}_k& ={\displaystyle \frac{10R_n-n}{9}},\hfill \\ \hfill \left(b\right)& {\sum }_{k=0}^n{R}_{2k}& ={\displaystyle \frac{{10}^2{R}_{2n}-n{R}_2}{99}},\hfill \\ \hfill \left(c\right)& {\sum }_{k=0}^n{R}_{2k+1}& ={\displaystyle \frac{R_{2n+3}-(n+1)R_2}{99}}.\hfill \end{array}$$

3. Bidimensional Repunit Sequence

As noted by Harman [19], the numbers denoted by $(n,m)$ represent Gaussian integers of the form $(n,m)=n+mi$, where n and m are integers. These and other authors also discuss the extension of one-dimensional identities of the Fibonacci sequence to two, three or higher dimensions by means of n-dimensional recursive relations, see [9,10,11,20,21,22,23,24], among others.

To explore the extension of the Repunit sequence into a complex or bidimensional framework, this article investigates into the recurrence relations governing the bidimensional case of this sequence, providing a comprehensive analysis of its structure and properties.

In this section, we define bidimensional Repunit sequence and provide some properties of this new sequence of numbers. We start considering the following definition,

Definition 1.

For integers $m\ge 0$ and $n\ge 0$ the bidimensional Repunit numbers ${\left\{R_{(m,n)}\right\}}_{(m,n)}$ are defined recursively by

$$R_{(m,n)}=:\left\{\begin{array}{cc}11R_{(m-1,n)}-10R_{(m-2,n)},& for all n and m\ge 2\\ 11R_{(m,n-1)}-10R_{(m,n-2)},& for all m and n\ge 2\end{array}\right.,$$

(3)

with initial terms $R_{(0,0)}=0$, $R_{(1,0)}=1$, $R_{(0,1)}=i$, $R_{(1,1)}=1+i$ and where $i^2=-1$ is the imaginary unit.

The Definition 1 is correct in the sense that $R_{(m,n)}$ does not depend on the path we use for calculation. For instance, to find $R_{(2,2)}$ we see that

$$\left\{\begin{array}{cc}11R_{(1,2)}-10R_{(0,2)},& \mathrm{path} (a);\\ 11R_{(2,1)}-10R_{(2,0)},& \mathrm{path} (b).\end{array}\right.$$

• Consider path (a): we have

  $$\begin{array}{ccc}\hfill R_{(1,2)}& =& 11R_{(1,1)}-10R_{(1,0)}\hfill \\ & =& 11(1+i)-10·1\hfill \\ & =& 11+11i-10=1+11i,\hfill \end{array}$$

  and

  $$\begin{array}{ccc}\hfill R_{(0,2)}& =& 11R_{(0,1)}-10R_{(0,0)}\hfill \\ & =& 11·i-10·0=11i,\hfill \end{array}$$

  hence,

  $$\begin{array}{ccc}\hfill R_{(2,2)}& =& 11R_{(1,2)}-10R_{(0,2)}\hfill \\ & =& 11(1+11i)-10\left(11i\right)\hfill \\ & =& 11+121i-110i=11+11i.\hfill \end{array}$$
• Consider path (b): we obtain

  $$\begin{array}{ccc}\hfill R_{(2,1)}& =& 11R_{(1,1)}-10R_{(0,1)}\hfill \\ & =& 11(1+i)-10·i\hfill \\ & =& 11+i,\hfill \end{array}$$

  and

  $$\begin{array}{ccc}\hfill R_{(2,0)}& =& 11R_{(1,0)}-10R_{(0,0)}\hfill \\ & =& 11·1-10·0=11,\hfill \end{array}$$

  hence,

  $$\begin{array}{ccc}\hfill R_{(2,2)}& =& 11R_{(2,1)}-10R_{(2,0)}\hfill \\ & =& 11(11+i)-10\left(11\right)\hfill \\ & =& 121+11i-110=11+11i,\hfill \end{array}$$

  which coincides with the value found for $R_{(2,2)}$ where we used the path (a).

For illustrative example, consider two branches of the bidimensional Repunit sequence:

$${\left\{R_{(m,0)}\right\}}_m=\{0,1,11,111,1111,11111,\dots \},$$

and

$${\left\{R_{(0,n)}\right\}}_n=\{0,i,11i,111i,1111i,1111i,11111i,\dots \}.$$

3.1. Some Properties

In the following, a brief overview of the properties of bidimensional Repunit numbers is provided.

Proposition 2.

Let m and n denote arbitrary non-negative integers. The following properties hold:

$$\begin{array}{ccc}\hfill R_{(m,0)}& =& R_m;\hfill \end{array}$$

(4)

$$\begin{array}{ccc}\hfill R_{(0,n)}& =& R_{n}i;\hfill \end{array}$$

(5)

$$\begin{array}{ccc}\hfill R_{(m,1)}& =& R_m+i;\hfill \end{array}$$

(6)

$$\begin{array}{ccc}\hfill R_{(1,n)}& =& 1+R_{n}i;\hfill \end{array}$$

(7)

where ${\left\{R_{(m,n)}\right\}}_{(m,n)}$ is the bidimensional Repunit sequence, ${\left\{R_n\right\}}_n$ is the Repunit sequence, and $i^2=-1$.

Proof. 

For Equation (4), the proof is carried out by induction on m. For $m=0$ and $m=1$, we have: $R_{(0,0)}=0=R_0$ and $R_{(1,0)}=1=R_1.$ Thus, the property is true for $m=0$ and $m=1$. Assume that the property is true for all values less than or equal to m, that is, $R_{(k,0)}=R_k$ for $k\le m$. We need to prove that the property holds for $m+1$. By (3), we have:

$$R_{(m+1,0)}=11R_{(m,0)}-10R_{(m-1,0)}.$$

Using the inductive hypothesis, we have:

$$R_{(m+1,0)}=11R_m-10R_{m-1}.$$

But by Equation (1), we conclude

$$R_{(m+1,0)}=R_{m+1}.$$

Therefore, by the principle of mathematical induction, property from Equation (4) is true for all non-negative integers m.

In a similar way, the proof of Equation (5) can be done by induction on n.

Again, in order to prove the Equation (6), we proceed by induction on m. For $m=0$ and $m=1$, we have: $R_{(0,1)}=i=0+i=R_0+i$ and $R_{(1,1)}=1+i=R_1+i.$ Thus, the property is true for $m=0$ and $m=1$. Assume that the property is true for all values less than or equal to m, that is, $R_{(k,1)}=R_k+i$ for $k\le m$. We need to prove that the property holds for $m+1$. By (3), we have:

$$R_{(m+1,1)}=11R_{(m,1)}-10R_{(m-1,1)}.$$

Using the inductive hypothesis, we have:

$$\begin{array}{ccc}\hfill R_{(m+1,1)}& =& 11(R_m+i)-10(R_{m-1}+i)\hfill \\ & =& (11R_m-10R_{m-1})+(11-10)i\hfill \\ & =& (11R_m-10R_{m-1})+i.\hfill \end{array}$$

But by Equation (1), we conclude

$$R_{(m+1,1)}=R_{m+1}+i.$$

Therefore, by the principle of mathematical induction, property from Equation (6) is true for all non-negative integers m.

In a similar way, the proof of Equation (7) can be done by induction on n. □

The following result establishes a connection between the bidimensional Repunit and the sequence of classical Repunits and provides the Gaussian Repunit sequence studied in [25].

Theorem 1.

For non-negative integers n and m, the bidimensional Repunit numbers are described as follows:

$$R_{(m,n)}=R_m+R_{n}i,$$

where ${\left\{R_{(m,n)}\right\}}_{(m,n)}$ is the bidimensional Repunit sequence, ${\left\{R_n\right\}}_n$ is the Repunit sequence, and $i^2=-1$.

Proof. 

We start by applying induction to n, with m held constant. Let us consider $n=0$. Note that by initial conditions $R_{(0,0)}=0=0+0·i=R_0+R_{0}i$ and $R_{(1,0)}=1=1+0·i=R_1+R_{0}i$. For $m\ge 2$ held constant, we have

$$\begin{array}{ccc}\hfill R_{(m,0)}& =& R_m=R_m+0·i\hfill \\ & =& R_m+R_{0}i,\hfill \end{array}$$

which is true by Equation (4), and since $R_0=0$. Suppose the theorem is true for all $k\le n$. Let us prove for $n+1$. By (3) we have

$$R_{(m,n+1)}=11R_{(m,n)}-10R_{(m,n-1)}.$$

Now, by the induction hypothesis, we get

$$\begin{array}{ccc}\hfill R_{(m,n+1)}& =& 11R_{(m,n)}-10R_{(m,n-1)}\hfill \\ & =& 11(R_m+R_{n}i)-10(R_m+R_{n-1}i)\hfill \\ & =& R_m+(11R_n-10R_{n-1})i,\hfill \end{array}$$

following Equation (1) that

$$\begin{array}{c}\hfill R_{(m,n+1)}=R_m+R_{n+1}i,\end{array}$$

which verifies the result.

Let us now fix n and perform the induction on m. For $m=0$ note that $R_{(0,1)}=i=0+1·i=R_0+R_{1}i$ and given the initial terms of ${\left\{R_n\right\}}_n$, we have

$$\begin{array}{ccc}\hfill R_{(0,n)}& =& R_{n}i=0+R_{n}i\hfill \\ & =& R_0+R_{n}i,\hfill \end{array}$$

which is true by Equation (5). Suppose the theorem is true for $k\le m$. Let us prove for $m+1$. By (3), we have

$$R_{(m+1,n)}=11R_{(m,n)}-10R_{(m-1,n)}.$$

Now, by the induction hypothesis, we get

$$\begin{array}{ccc}\hfill R_{(m+1,n)}& =& 11R_{(m,n)}-10R_{(m-1,n)}\hfill \\ & =& 11(R_m+R_{n}i)-10(R_{m-1}+R_{n}i)\hfill \\ & =& (11R_m-10R_{m-1})+R_{n}i,\hfill \end{array}$$

following Equation (1) that

$$\begin{array}{c}\hfill R_{(m+1,n)}=R_{m+1}+R_{n}i,\end{array}$$

as we wanted to show. □

Then, according to the previous result and Binet’s formula, Equation (2), we can obtain the Binet formula for bidimensional Repunit sequence, as follows.

Corollary 1.

[Binet’s formula] For non-negative integers n and m, we have

$$R_{(m,n)}={\displaystyle \frac{{10}^m-1}{9}}+{\displaystyle \frac{{10}^n-1}{9}}i,$$

or equivalently

$$R_{(m,n)}={\displaystyle \frac{{10}^m+{10}^{n}i-(1+i)}{9}},$$

where ${\left\{R_{(m,n)}\right\}}_{(m,n)}$ is the bidimensional Repunit sequence, and $i^2=-1$.

The above corollary is a special case of Proposition 4 in [25].

From Theorem 1 follows the next two results.

Proposition 3.

Let us consider non-negative integers $m,n$ and k, with $m\ge k$ and $n\ge k$. Let ${\left\{R_{(m,n)}\right\}}_{(m,n)}$ be the bidimensional Repunit sequence, then the identities are verified:

$$\left(a\right)R_{(m,n)}-R_{(k,n)}=R_{m-k}{10}^k,$$

$$\left(b\right)R_{(m,n)}-R_{(m,k)}=R_{n-k}{10}^{k}i,$$

where ${\left\{R_m\right\}}_m$ is the Repunit sequence, and $i^2=-1$.

Proof. 

(a) According to Theorem 1 we have

$$\begin{array}{ccc}& & R_{(m,n)}-R_{(k,n)}\hfill \\ & =& R_m+R_{n}i-(R_k+R_{n}i)\hfill \\ & =& R_m-R_k.\hfill \end{array}$$

By Lemma 1, item (a), we get

$$\begin{array}{c}\hfill R_{(m,n)}-R_{(k,n)}=R_{m-k}{10}^k,\end{array}$$

and this completes the proof.

(b) The proof of this is similar to that of item (a). □

A direct consequence of the Proposition 3 is the next result.

Corollary 2.

Let us consider non-negative integers m and n. Let ${\left\{R_{(m,n)}\right\}}_{(m,n)}$ be the bidimensional Repunit sequence, then the identities are verified:

$$\left(a\right)R_{(m+1,n)}-R_{(m,n)}={10}^m,$$

$$\left(b\right)R_{(m,n+1)}-R_{(m,n)}={10}^{n}i,$$

$$\left(c\right)R_{(2m+1,n)}-R_{(2m,n)}={10}^{2m},$$

$$\left(d\right)R_{(m,2n+1)}-R_{(m,2n)}={10}^{2n}i,$$

where $i^2=-1$.

Proposition 4.

For non-negative integers m and n and for the bidimensional Repunit sequence ${\left\{R_{(m,n)}\right\}}_{(m,n)}$, the identities are verified:

$$\left(a\right)R_{(m+1,n)}-10R_{(m-1,n)}=({10}^m+1)-9R_{n}i,$$

$$\left(b\right)R_{(m,n+1)}-10R_{(m,n-1)}=-9R_m+({10}^n+1)i,$$

where ${\left\{R_m\right\}}_m$ is the Repunit sequence, and $i^2=-1$.

Proof. 

(a) According to Theorem 1 we have

$$\begin{array}{ccc}\hfill R_{(m+1,n)}-10R_{(m-1,n)}& =& (R_{m+1}+R_{n}i)-10\left[R_{m-1}+R_{n}i\right]\hfill \\ & =& (R_{m+1}-10R_{m-1})-9R_{n}i.\hfill \end{array}$$

By Lemma 3 we get

$$\begin{array}{c}\hfill R_{(m+1,n)}-10R_{(m-1,n)}=({10}^m+1)-9R_{n}i,\end{array}$$

and this completes the proof.

(b) The proof follows a similar approach to the one used in item (a). □

Combining the Proposition 1 and Theorem 1 we have:

Proposition 5.

For non-negative integers m and n and for the bidimensional Repunit sequence ${\left\{R_{(m,n)}\right\}}_{(m,n)}$, the identities are verified:

$$\left(a\right)R_{(2m,n)}-2R_{(m,n)}=9R_m^2-R_{n}i,$$

$$\left(b\right)R_{(m,2n)}-2R_{(m,n)}=-R_m+9R_n^{2}i,$$

where ${\left\{R_n\right\}}_n$ is the Repunit sequence, and $i^2=-1$.

3.2. Some Sum Formulas

In this section, we provide results concerning the partial sums of terms of the bidimensional Repunit sequence ${\left\{R_{(m,n)}\right\}}_{(m,n)}$. We begin by examining the sequence of partial sums, given by

$$\sum _{m=0}^k{R}_{(m,n)}=R_{(0,n)}+R_{(1,n)}+R_{(2,n)}+\cdots +R_{(k,n)}$$

for n, where ${\left\{R_{(m,n)}\right\}}_{(m,n)}$ represents the bidimensional Repunit sequence.

We will present two results involving the partial sum and alternating partial sum of the terms of the bidimensional Repunit sequence ${\left\{R_{(m,n)}\right\}}_{(m,n)}$.

Proposition 6.

For non-negative integers m and n, let ${\left\{R_{(m,n)}\right\}}_{(m,n)}$ be the bidimensional Repunit sequence and k a non-negative integer. Then:

$$\begin{array}{ccc}\hfill \left(a\right)& {\sum }_{m=0}^k{R}_{(m,n)}& ={\displaystyle \frac{10R_k-k}{9}}+(k+1)R_{n}i,\hfill \\ \hfill \left(b\right)& {\sum }_{m=0}^k{R}_{(2m,n)}& ={\displaystyle \frac{{10}^2{R}_{2k}-k{R}_2}{99}}+(k+1)R_{n}i,\hfill \\ \hfill \left(c\right)& {\sum }_{m=0}^k{R}_{(2m+1,n)}& ={\displaystyle \frac{{10}^2{R}_{2k+3}-(k+1)R_2}{99}}+(k+1)R_{n}i,\hfill \end{array}$$

where ${\left\{R_n\right\}}_n$ is the Repunit sequence.

Proof. 

(a) By Lemma 4, item (a) and Theorem 1 we have,

$$\begin{array}{ccc}\hfill \sum _{m=0}^k{R}_{(m,n)}& =& \sum _{m=0}^k(R_m+R_{n}i)=\sum _{m=0}^k{R}_m+\sum _{m=0}^k{R}_{n}i\hfill \\ & =& {\displaystyle \frac{10R_k-k}{9}}+(k+1)R_{n}i;\hfill \end{array}$$

which verifies the result.

(b) Again, by Lemma 4, item (b) and Theorem 1 we have,

$$\begin{array}{ccc}\hfill \sum _{m=0}^k{R}_{(2m,n)}& =& \sum _{m=0}^k(R_{2m}+R_{n}i)=\sum _{m=0}^k{R}_{2m}+\sum _{m=0}^k{R}_{n}i\hfill \\ & =& {\displaystyle \frac{{10}^2{R}_{2k}-k{R}_2}{99}}+(k+1)R_{n}i;\hfill \end{array}$$

which proves the result.

(c) Similarly, by Lemma 4, item (c) and Theorem 1 we have:

$$\begin{array}{ccc}\hfill \sum _{m=0}^k{R}_{(2m+1,n)}& =& \sum _{m=0}^k(R_{2m+1}+R_{n}i)\hfill \\ & =& {\displaystyle \frac{{10}^2{R}_{2k+3}-(k+1)R_2}{99}}+(k+1)R_{n}i\hfill \end{array}$$

and this completes the proof. □

Consider now the sequence of alternating partial sums given by

$$\sum _{m=0}^k{(-1)}^m{R}_{(m,n)}=R_{(0,n)}-R_{(1,n)}+R_{(2,n)}-R_{(3,n)}+\cdots +{(-1)}^k{R}_{(k,n)}$$

for n, where ${\left\{R_{(m,n)}\right\}}_{(m,n)}$ denotes the bidimensional Repunit sequence.

Combining items (b) and (c) of the Proposition 6 we have

Proposition 7.

For non-negative integers m, n and k, let ${\left\{R_{(m,n)}\right\}}_{(m,n)}$ be the bidimensional Repunit sequence. Then:

$$\begin{array}{cc}\hfill \left(a\right)\sum _{m=0}^{2k+1}{(-1)}^m{R}_{(m,n)}& ={\displaystyle \frac{{10}^2(R_{2k}-R_{2k+3})+R_2}{99}},\\ \hfill \left(b\right)\sum _{m=0}^{2k}{(-1)}^m{R}_{(m,n)}& ={\displaystyle \frac{{10}^2(R_{2k}-R_{2k+1})}{99}},\end{array}$$

where ${\left\{R_n\right\}}_n$ is the Repunit sequence.

Now, we examine the sequence of partial sums, given by

$$\sum _{n=1}^k{R}_{(m,n)}=R_{(m,0)}+R_{(m,1)}+\cdots +R_{(m,k)}$$

for $m\ge 0$, where ${\left\{R_{(m,n)}\right\}}_{(m,n)}$ represents the bidimensional Repunit sequence.

In a similar way to Proposition 6, we have

Proposition 8.

For non-negative integers m, n and k, let ${\left\{R_{(m,n)}\right\}}_{(m,n)}$ be the bidimensional Repunit sequence. Then:

$$\begin{array}{ccc}\hfill \left(a\right)& {\sum }_{n=0}^k{R}_{(m,n)}& =(k+1)R_m+{\displaystyle \frac{10R_k-k}{9}}i,\hfill \\ \hfill \left(b\right)& {\sum }_{n=0}^k{R}_{(m,2n)}& =(k+1)R_m+{\displaystyle \frac{{10}^2{R}_{2k}-k{R}_2}{99}}i,\hfill \\ \hfill \left(c\right)& {\sum }_{n=0}^k{R}_{(m,2n-1)}& =(k+1)R_m+{\displaystyle \frac{{10}^2{R}_{2k+3}-(k+1)R_2}{99}}i,\hfill \end{array}$$

where ${\left\{R_n\right\}}_n$ is the Repunit sequence.

Consider now the sequence of alternating partial sums given by

$$\sum _{n=0}^k{(-1)}^n{R}_{(m,n)}=R_{(m,0)}-R_{(m,1)}+R_{(m,2)}-R_{(m,3)}+\cdots +{(-1)}^k{R}_{(m,k)}$$

for $m\ge 0$, where ${\left\{R_{(m,n)}\right\}}_{(m,n)}$ denotes the bidimensional Repunit sequence.

Combining items (b) and (c) of the Proposition 8 we have

Proposition 9.

For non-negative integers m, n and k, let ${\left\{R_{(m,n)}\right\}}_{(m,n)}$ be the bidimensional Repunit sequence. Then:

$$\begin{array}{cc}\hfill \left(a\right)\sum _{n=0}^{2k+1}{(-1)}^k{R}_{(m,n)}& ={\displaystyle \frac{{10}^2(R_{2k}-R_{2k+3})+R_2}{99}}i,\\ \hfill \left(b\right)\sum _{n=0}^{2k}{(-1)}^k{R}_{(m,n)}& ={\displaystyle \frac{{10}^2(R_{2k}-R_{2k+1})}{99}}i,\end{array}$$

where ${\left\{R_n\right\}}_n$ is the Repunit sequence.

4. k-Dimensional Repunit Sequence

Initially, in this Section we define a tridimensional Repunit sequence and we provide an enumeration of its properties. The following definition is offered as a point of departure:

Definition 2.

For integers $n_1\ge 0$, $n_2\ge 0$ and $n_3\ge 0$ the tridimensional Repunit sequence ${\left\{R_{(n_1,n_2,n_3)}\right\}}_{(n_1,n_2,n_3)}$ is defined recursively by

$$R_{(n_1,n_2,n_3)}=:\left\{\begin{array}{cc}11R_{(n_1-1,n_2,n_3)}-10R_{(n_1-2,n_2,n_3)},& for all n_2,n_3 and n_1\ge 2\\ 11R_{(n_1,n_2-1,n_3)}-10R_{(n_1,n_2-2,n_3)},& for all n_1,n_3 and n_2\ge 2\\ 11R_{(n_1,n_2,n_3-1)}-10R_{(n_1,n_2,n_3-2)},& for all n_1,n_2 and n_3\ge 2\end{array}\right.,$$

with initial terms $R_{(0,0,0)}=0$, $R_{(1,0,0)}=1$, $R_{(0,1,0)}=i$, $R_{(0,0,1)}=j$, $R_{(1,1,0)}=1+i$, $R_{(1,0,1)}=1+j$, $R_{(0,1,1)}=i+j$, $R_{(1,1,1)}=1+i+j$, where $i^2=-1$ and $j^2=-1$ are imaginary units.

The Definition 2 is valid because $R_{(n_1,n_2,n_3)}$ is independent of the path chosen for its computation. For example, we will describe 3 different paths to obtain $R_{(2,2,2)}$. So, to calculate $R_{(2,2,2)}$, we observe that:

$$R_{(2,2,2)}=\left\{\begin{array}{cc}11R_{(1,2,2)}-10R_{(0,2,2)},& \mathrm{path} (a);\\ 11R_{(2,1,2)}-10R_{(2,0,2)},& \mathrm{path} (b);\\ 11R_{(2,2,1)}-10R_{(2,2,0)},& \mathrm{path} (c).\end{array}\right.$$

To begin, we perform and outline some preliminary calculations:

$$\begin{array}{ccc}\hfill R_{(0,0,2)}=& 11R_{(0,0,1)}-10R_{(0,0,0)}=11·j-10·0& =11j;\hfill \\ \hfill R_{(0,2,0)}=& 11R_{(0,1,0)}-10R_{(0,0,0)}=11·i-10·0& =11i;\hfill \\ \hfill R_{(2,0,0)}=& 11R_{(1,0,0)}-10R_{(0,0,0)}=11·1-10·0& =11;\hfill \\ \hfill R_{(0,1,2)}=& 11R_{(0,1,1)}-10R_{(0,1,0)}=11·(i+j)-10·i& =i+11j;\hfill \\ \hfill R_{(1,0,2)}=& 11R_{(1,0,1)}-10R_{(1,0,0)}=11·(1+j)-10·1& =1+11j;\hfill \\ \hfill R_{(0,2,1)}=& 11R_{(0,1,1)}-10R_{(0,0,1)}=11·(i+j)-10·j& =11i+j;\hfill \\ \hfill R_{(1,2,0)}=& 11R_{(1,1,0)}-10R_{(1,0,0)}=11·(1+i)-10·1& =1+11i;\hfill \\ \hfill R_{(2,0,1)}=& 11R_{(1,0,1)}-10R_{(0,0,1)}=11·(1+j)-10·j& =11+j;\hfill \\ \hfill R_{(2,1,0)}=& 11R_{(1,1,0)}-10R_{(0,1,0)}=11·(1+i)-10·i& =11+i;\hfill \\ \hfill R_{(1,1,2)}=& 11R_{(1,1,1)}-10R_{(1,1,0)}=11·(1+i+j)-10·(1+i)& =1+i+11j;\hfill \\ \hfill R_{(1,2,1)}=& 11R_{(1,1,1)}-10R_{(1,0,1)}=11·(1+i+j)-10·(1+j)& =1+11i+j;\hfill \\ \hfill R_{(2,1,1)}=& 11R_{(1,1,1)}-10R_{(0,1,1)}=11·(1+i+j)-10·(i+j)& =11+i+j.\hfill \end{array}$$

• Consider path (a): we have

  $$\begin{array}{ccc}\hfill R_{(1,2,2)}& =& 11R_{(1,2,1)}-10R_{(1,2,0)}\hfill \\ & =& 11(1+11i+j)-10·(1+11i)\hfill \\ & =& 1+11i+11j,\hfill \end{array}$$

  and

  $$\begin{array}{ccc}\hfill R_{(0,2,2)}& =& 11R_{(0,2,1)}-10R_{(0,2,0)}\hfill \\ & =& 11·(11i+j)-10·11i=11i+11j,\hfill \end{array}$$

  hence,

  $$\begin{array}{ccc}\hfill R_{(2,2,2)}& =& 11R_{(1,2,2)}-10R_{(0,2,2)}\hfill \\ & =& 11(1+11i+11j)-10(11i+11j)\hfill \\ & =& 11+11i+11j.\hfill \end{array}$$
• Consider path (b): we obtain

  $$\begin{array}{ccc}\hfill R_{(2,1,2)}& =& 11R_{(1,1,2)}-10R_{(0,1,2)}\hfill \\ & =& 11(1+i+11j)-10·(i+11j)\hfill \\ & =& 11+i+11j,\hfill \end{array}$$

  and

  $$\begin{array}{ccc}\hfill R_{(2,0,2)}& =& 11R_{(1,0,2)}-10R_{(0,0,2)}\hfill \\ & =& 11·(1+11j)-10·\left(11j\right)\hfill \\ & =& 11+11j,\hfill \end{array}$$

  hence,

  $$\begin{array}{ccc}\hfill R_{(2,2,2)}& =& 11R_{(2,1,2)}-10R_{(2,0,2)}\hfill \\ & =& 11(11+i+11j)-10(11+11j)\hfill \\ & =& 11+11i+11j,\hfill \end{array}$$
• Consider path (c): we have the following

  $$\begin{array}{ccc}\hfill R_{(2,2,1)}& =& 11R_{(2,1,1)}-10R_{(2,0,1)}\hfill \\ & =& 11(11+i+j)-10·(11+j)\hfill \\ & =& 11+11i+j,\hfill \end{array}$$

  and

  $$\begin{array}{ccc}\hfill R_{(2,2,0)}& =& 11R_{(2,1,0)}-10R_{(2,0,0)}\hfill \\ & =& 11·(11+i)-10·\left(11\right)\hfill \\ & =& 11+11i,\hfill \end{array}$$

  hence,

  $$\begin{array}{ccc}\hfill R_{(2,2,2)}& =& 11R_{(2,2,1)}-10R_{(2,2,0)}\hfill \\ & =& 11(11+11i+j)-10(11+11i)\hfill \\ & =& 11+11i+11j,\hfill \end{array}$$

  this result matches the value obtained for $R_{(2,2,2)}$ when calculated using either path (a) or path (b).

We will illustrate, as an example, one branch of the tridimensional Repunit sequence:

$${\left\{R_{(0,0,n)}\right\}}_n=\{0,1j,11j,111j,1111j,11111j,\dots \}$$

In a similar way to what we have done in Section 3, we will have similar results in this section, but here we will only present the statement.

Proposition 10.

Let $n_1$, $n_2$ and $n_3$ denote arbitrary non-negative integers. The following properties hold:

$$\begin{array}{ccc}\hfill R_{(n_1,0,0)}& =& R_{n_1};\hfill \\ \hfill R_{(0,n_2,0)}& =& R_{n_2}i;\hfill \\ \hfill R_{(0,0,n_3)}& =& R_{n_3}j;\hfill \\ \hfill R_{(n_1,1,0)}& =& R_{n_1}+i;\hfill \\ \hfill R_{(n_1,0,1)}& =& R_{n_1}+j;\hfill \\ \hfill R_{(n_1,1,1)}& =& R_{n_1}+i+j;\hfill \\ \hfill R_{(1,n_2,0)}& =& 1+R_{n_2}i;\hfill \\ \hfill R_{(1,n_2,1)}& =& 1+R_{n_2}i+j;\hfill \\ \hfill R_{(1,0,n_3)}& =& 1+R_{n_3}j;\hfill \\ \hfill R_{(1,1,n_3)}& =& 1+i+R_{n_3}j;\hfill \end{array}$$

where ${\left\{R_{(n_1,n_2,n_3)}\right\}}_{(n_1,n_2,n_3)}$ is the tridimensional Repunit sequence, ${\left\{R_n\right\}}_n$ is the Repunit sequence, and $i^2=j^2=-1$.

The following result links tridimensional Repunit to the classical Repunit sequence.

Proposition 11.

For non-negative integers $n_1$, $n_2$ and $n_3$, the tridimensional Repunit numbers are described as follows:

$$R_{(n_1,n_2,n_3)}=R_{n_1}+R_{n_2}i+R_{n_3}j,$$

where ${\left\{R_{(n_1,n_2,n_3)}\right\}}_{(n_1,n_2,n_3)}$ is the tridimensional Repunit sequence, ${\left\{R_n\right\}}_n$ is the Repunit sequence, and $i^2=j^2=-1$.

Thus, using the previous result and Binet’s formula from Equation (2), we derive the Binet formula for tridimensional Repunit sequence, as follows.

Corollary 3.

[Binet’s formula] For non-negative integers $n_1$, $n_2$ and $n_2$, we have

$$R_{(n_1,n_2,n_3)}={\displaystyle \frac{{10}^{n_1}-1}{9}}+{\displaystyle \frac{{10}^{n_2}-1}{9}}i+{\displaystyle \frac{{10}^{n_3}-1}{9}}j,$$

or equivalently

$$R_{(n_1,n_2,n_3)}={\displaystyle \frac{{10}^{n_1}+{10}^{n_2}i+{10}^{n_3}j-(1+i+j)}{9}},$$

where ${\left\{R_{(n_1,n_2,n_3)}\right\}}_{(n_1,n_2,n_3)}$ is the tridimensional Repunit sequence, and $i^2=j^2=-1$.

The next results are an extension of the results of the tridimensional Repunit, so we will only present the statement with respect to the third component.

Proposition 12.

For all non-negative integers $n_1$, $n_2$, $n_3$ and k, with $n_3\ge k$. Let ${\left\{R_{(n_1,n_2,n_3)}\right\}}_{(n_1,n_2,n_3)}$ be the tridimensional Repunit sequence, them the following identities are verified:

$$\left(a\right)R_{(n_1,n_2,n_3)}-R_{(n_1,n_2,k)}=R_{n_3-k}{10}^{k}j,$$

$$\left(b\right)R_{(n_1,n_2,n_3+1)}-R_{(n_1,n_2,n_3-1)}=-9R_{n_1}-9R_{n_2}i+({10}^{n_3}+1)j,$$

$$\left(c\right)R_{(n_1,n_2,2n_3)}-2R_{(n_1,n_2,n_3)}=-R_{n_1}-R_{n_2}i+9R_{n_3}^{2}j,$$

where ${\left\{R_m\right\}}_m$ is the Repunit sequence, and $i^2=j^2=-1$.

Now let us extend the Definition 1 and so we introduce the k-dimensional Repunit sequence for all integers $k\ge 2$,

Definition 3.

For integers $n_1\ge 0,n_2\ge 0,\dots ,n_k\ge 0$ and $k\ge 2$. The k-dimensional Repunit sequence ${\left\{R_{(n_1,n_2,\dots ,n_k)}\right\}}_{(n_1,n_2,\dots ,n_k)}$ is defined recursively by

$$\left\{\begin{array}{ccc}{R}_{(n_1+1,n_2,\dots ,n_k)}& =& 11R_{(n_1,n_2,\dots ,n_k)}-10R_{(n_1-1,n_2,\dots ,n_k)}\\ R_{(n_1,n_2+1,\dots ,n_k)}& =& 11R_{(n_1,n_2,\dots ,n_k)}-10R_{(n_1,n_2-1,\dots ,n_k)}\\ ⋮& & ⋮\\ R_{(n_1,n_2,\dots ,n_{k-1},n_k+1)}& =& 11R_{(n_1,n_2,\dots ,n_{k-1},n_k)}-10R_{(n_1,n_2,\dots ,n_{k-1},n_k-1)}\end{array}\right.,$$

with initial terms $R_{(0,0,0,0,\dots ,0)}=0$, $R_{(1,0,0,0\dots ,0)}=1$, $R_{(0,1,0,0,\dots ,0)}=i_2$, $R_{(0,0,1,0,\dots ,0)}=i_3$, $R_{(1,1,0,0,\dots ,0}=1+i_2$, $R_{(1,0,1,0,\dots ,0)}=1+i_3$, $R_{(0,1,1,0,\dots ,0)}=i_2+i_3$, $R_{(1,1,1,0,\dots ,0)}=1+i_2+i_3,\dots$, $R_{(1,1,1,1,\dots ,1)}=1+i_2+i_3+\dots +i_k$, where $i_2^2=\cdots =i_k^2=-1$ are imaginary units.

Combining the Definition 3 and the results established in Proposition 11 and Corollary 3, we state the following key result:

Proposition 13.

For any k-tuple $(n_1,n_2,n_3,\dots ,n_k)$ of non-negative integers, the general term of the set ${\left\{R_{(n_1,n_2,n_3,\dots ,n_k)}\right\}}_{(n_1,n_2,n_3,\dots ,n_k)}$ is given by:

$$R_{(n_1,n_2,n_3,\dots ,n_k)}=R_{n_1}+R_{n_2}{i}_2+\cdots +R_{n_k}{i}_k,$$

or equivalently:

$$R_{(n_1,n_2,n_3,\dots ,n_k)}={\displaystyle \frac{{10}^{n_1}+{10}^{n_2}{i}_2+\cdots +{10}^{n_k}{i}_k-(1+i_2+\cdots +i_k)}{9}},$$

where ${\left\{R_{(n_1,n_2,\dots ,n_k)}\right\}}_{(n_1,n_2,n_3,\dots ,n_k)}$ is the k-dimensional Repunit sequence, ${\left\{R_n\right\}}_{n_j},(j=1,2,\dots ,k)$ is the Repunit sequence, and $i_2^2=\cdots =i_k^2=-1$.

According the reference [8] the results are also valid for negative indexes. Specifically, we have:

$$R_{(-n_1,-n_2,-n_3,\dots ,-n_k)}=R_{-n_1}+R_{-n_2}{i}_2+\cdots +R_{-n_k}{i}_k.$$

Thus, all the properties verified in this research can also be extended to the case where the indexes are negative.

5. Conclusions

The unidimensional model of the Repunit sequence permitted the introduction of imaginary units, facilitating the development of bidimensional and tridimensional recurrences and leading the way to generalizations to k-dimensional forms. This complexification of the classical or unidimensional Repunit sequence was achieved by extending the recurrence formula and the initial terms, which resulted in its generalization to a k-dimensional structure. This work presents a comprehensive analysis of the complexification process of the Repunit sequence and its relationship with the Horadam sequence. The complexification was achieved by incorporating the imaginary units and extending it to an algebraic representation in higher dimensions. Consequently, the bidimensional recurrence relations of the Repunit sequence were established, along with several identities in their complex forms. These contributions offer valuable insights into the structural complexity of these numbers through a rigorous mathematical analysis of their recurrence equation. These results highlight the versatility of Repunit numbers and their generalizations, providing a foundation for further exploration of their properties in both theoretical and applied contexts.