On the Incomplete Edouard and Incomplete Edouard–Lucas Numbers

Abstract

This study introduces two new sequences: the incomplete Edouard and the incomplete Edouard–Lucas numbers. In addition, we establish some of the properties, identities, and recurrence relations of these sequences. The relations of these new sequences with balancing, Lucas-balancing, incomplete balancing, and incomplete Lucas-balancing numbers are explored, and some new identities are provided. Moreover, the functions for generating the incomplete Edouard and incomplete Edouard–Lucas numbers are stated.

1. Introduction

Sequences of integers are interesting because of their applications in several areas. Therefore, numerous sequences have been introduced and studied. Two of them, which are the central theme of this article, were introduced by Soykan in the article [1] and are called the sequence of Edouard numbers ${\left\{E_n\right\}}_{n\ge 0}$ and sequence of Edouard–Lucas numbers ${\left\{K_n\right\}}_{n\ge 0}.$ In [1], the author established the Binet formulas, generating functions, and some of the identities of the these recently published sequences of numbers.

The sequence of Edouard numbers ${\left\{E_n\right\}}_{n\ge 0}$ and the sequence of Edouard–Lucas numbers ${\left\{K_n\right\}}_{n\ge 0}$ are given by the third-order recurrence relations and initial conditions

$$E_n=7E_{n-1}-7E_{n-2}+E_{n-3},E_0=0,E_1=1,E_2=7,$$

(1)

$$K_n=7K_{n-1}-7K_{n-2}+K_{n-3},K_0=3,K_1=7,K_2=35,$$

(2)

respectively. Relations (1) and (2) can be rewritten as non-homogeneous relations, as given below:

$$E_n=6E_{n-1}-E_{n-2}+1,E_0=0,E_1=1,$$

$$K_n=6K_{n-1}-K_{n-2}-4,K_0=3,K_1=7.$$

These sequences are new to the literature and the motivation for their study is the interesting fact that they have a closed connection with balancing numbers. The concept of balancing numbers was first introduced by Behera and Panda [2] in the year 1999 in connection with a Diophantine equation. It consists of finding a natural number n such that $1+2+3\cdots +(n+1)=(n+1)+(n+2)+\cdots +(n+r)$ for some other natural number $r.$ The sequence of balancing numbers is denoted by ${\left\{B_n\right\}}_{n\ge 1}$. In [2], Behera and Panda first obtained the recurrence relation

$$B_{n+1}=6B_n-B_{n-1},$$

for $n\ge 2$ and the initial conditions $B_0=1$ and $B_1=6.$ They developed the Binet formula, generating function, and several identities by solving this recurrence relation as a second-order linear homogeneous difference equation. Since b is a balancing number if and only if the number $8b^2+1$ is a perfect square, then the definition of a related number is as given in [3], namely, the Lucas-balancing numbers.

We use ${\left\{C_n\right\}}_{n\ge 0}$ to denote the sequence of Lucas-balancing numbers corresponding to the balancing numbers ${\left\{B_n\right\}}_{n\ge 0}.$ The n-th Lucas-balancing number is defined by

$$C_n=\sqrt{8{\left(B_n\right)}^2+1}.$$

It is interesting to note that the sequence of Lucas-balancing numbers satisfies the same recurrence relation as that of balancing numbers, namely,

$$C_{n+1}=6C_n-C_{n-1},$$

for $n\ge 2$ and initial conditions $C_0=1$ and $C_1=3$ (see more in [2,3,4,5,6,7,8,9,10,11,12]).

Soykan, in [1], provided the relations

$$4E_n=B_{n+1}-B_n-1,$$

(3)

$$K_n=2C_n+1,$$

(4)

where $E_n$ and $K_n$ denote, respectively, the Edouard and Edouard–Lucas numbers of order n. Moreover, in [12] (Theorem 2.1), the binomial expression was established for $B_n$ and $C_n,$ which is given by

$$B_n=\sum _{j=0}^{⌊(n-1)/2⌋}{(-1)}^j\left(\begin{array}{c}n-(j+1)\\ j\end{array}\right)6^{n-(2j+1)},$$

(5)

$$C_n=3\sum _{j=0}^{⌊n/2⌋}{(-1)}^j{\displaystyle \frac{n}{n-j}}\left(\begin{array}{c}n-j\\ j\end{array}\right)6^{n-(2j+1)},$$

(6)

where $⌊·⌋$ denotes the floor function.

Therefore, since there is a connection between the Edouard, Edouard–Lucas, and balancing numbers, it is possible to derive a binomial expression for $E_n$ and $K_n.$ In addition, using these binomial expressions, we can establish the incomplete Edouard and incomplete Edouard–Lucas numbers, which is a goal of this article.

Incomplete numbers are a widely studied subject and consist of the numerical sequence defined by the truncated sums of the binomial expression of a given numerical sequence. This class of sequences of numbers has revealed many interesting results. For example, the class of incomplete Fibonacci and Lucas numbers and their generalization as bivariate sequences and polynomials were explored in [13,14,15,16,17,18,19,20,21]. The incomplete k-Pell, k-Pell–Lucas, and modified k-Pell numbers were introduced in [22]. In addition, the incomplete Jacobsthal and Jacobsthal–Lucas numbers were investigated in [23]. However, the most interesting study for this article is the study of incomplete balancing and Lucas-balancing numbers, which were introduced and studied in [12]. The incomplete balancing numbers $B_n\left(k\right)$ and the incomplete Lucas-balancing numbers $C_n\left(k\right)$ are defined by

$$B_n\left(k\right)=\sum _{j=0}^k{(-1)}^j\left(\begin{array}{c}n-(j+1)\\ j\end{array}\right)6^{n-(2j+1)},0\le k\le ⌊{\displaystyle \frac{n-1}{2}}⌋,$$

(7)

and

$$C_n\left(k\right)=3\sum _{j=0}^k{(-1)}^j{\displaystyle \frac{n}{n-j}}\left(\begin{array}{c}n-j\\ j\end{array}\right)6^{n-(2j+1)},0\le k\le ⌊{\displaystyle \frac{n}{2}}⌋,$$

(8)

for any natural number $n.$

Next, we will introduce some identities that will be useful in this paper. In [12], the following identities were verified:

$$B_n\left(⌊{\displaystyle \frac{n-1}{2}}⌋-1\right)=\left\{\begin{array}{cc}{B}_n\pm 3n,& \mathrm{if}n\mathrm{is}\mathrm{even},\\ B_n\pm 1,& \mathrm{if}n\mathrm{is}\mathrm{odd},\end{array}\right.$$

(9)

for $n\ge 3,$

$$\sum _{k=0}^{⌊{\displaystyle \frac{n-1}{2}}⌋}{B}_n\left(k\right)=\left\{\begin{array}{cc}(3n{C}_n-B_n)/12,\hfill & \mathrm{if}n\mathrm{is}\mathrm{even}\hfill \\ (3nCn+7B_n)/16,\hfill & \mathrm{if}n\mathrm{is}\mathrm{odd},\hfill \end{array}\right.$$

(10)

$$B_{n+2}(k+1)=6B_{n+1}(k+1)-B_n\left(k\right),$$

(11)

for $0\le k<\lfloor {\displaystyle \frac{n-2}{2}}\rfloor .$

In this paper, motivated by the connection between the balancing, Lucas–balancing, Edouard, and Edouard–Lucas numbers, we will introduce the incomplete Edouard and incomplete Edouard–Lucas numbers. We will provide some of the properties of these new sequences of numbers and establish identities and the generating function for each sequence.

This article is organized as follows: In Section 2, we will define the incomplete Edouard and incomplete Edouard–Lucas numbers and provide some of the properties of these sequences of numbers. Section 3 is devoted to establishing some identities involving these sequences of numbers. In Section 4, their generating functions are provided and, finally, conclusions are stated.

2. The Incomplete Edouard and Incomplete Edouard–Lucas Numbers

This section will introduce the incomplete Edouard and incomplete Edouard–Lucas numbers. In addition, some properties of these numbers will be established. First, we will establish the binomial expressions of the Edouard and Edouard–Lucas numbers. Consider the following lemma:

Lemma 1.

For all $n>0$ and $0\le j<⌊{\displaystyle \frac{n}{2}}⌋$, we know that

$$6\left({\displaystyle \genfrac{}{}{0pt}{}{n-j}{j}}\right)-\left({\displaystyle \genfrac{}{}{0pt}{}{n-j-1}{j}}\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{n-j-1}{j}}\right)\left({\displaystyle \frac{5n-4j}{n-2j}}\right).$$

According Equation (5), we obtain

$$\begin{array}{ccc}& & B_{n+1}-B_n\hfill \\ & =& \sum _{j=0}^{⌊n/2⌋}{(-1)}^j\left(\begin{array}{c}(n+1)-(j+1)\\ j\end{array}\right)6^{(n+1)-(2j+1)}\hfill \\ & & -\sum _{j=0}^{⌊(n-1)/2⌋}{(-1)}^j\left(\begin{array}{c}n-(j+1)\\ j\end{array}\right)6^{n-(2j+1)}\hfill \\ & =& \sum _{j=0}^{⌊n/2⌋}{(-1)}^j\left(\begin{array}{c}n-j\\ j\end{array}\right)6^{n-2j}+\sum _{j=0}^{⌊(n-1)/2⌋}{(-1)}^{j+1}\left(\begin{array}{c}n-(j+1)\\ j\end{array}\right)6^{n-(2j+1)}.\hfill \end{array}$$

Then, by using Lemma 1 and considering odd $n,$ we have

$$\begin{array}{c}\hfill B_{n+1}-B_n=\sum _{j=0}^{⌊n/2⌋}{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{n-j-1}{j}}\right)\left({\displaystyle \frac{5n-4j}{n-2j}}\right)6^{n-(2j+1)}.\end{array}$$

Otherwise, if n is even, we find that

$$\begin{array}{c}\hfill B_{n+1}-B_n=\sum _{j=0}^{⌊(n-1)/2⌋}{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{n-j-1}{j}}\right)\left({\displaystyle \frac{5n-4j}{n-2j}}\right)6^{n-(2j+1)}+{(-1)}^{n/2}.\end{array}$$

Therefore, since Equation (3) holds, we provide the binomial expression for Edouard numbers as follows.

Proposition 1.

For all $n>0,$ the following identity holds:

$$4E_n=\left\{\begin{array}{cc}\sum _{j=0}^{⌊n/2⌋}{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{n-(j+1)}{j}}\right)\left({\displaystyle \frac{5n-4j}{n-2j}}\right)6^{n-(2j+1)}-1,& foroddn\\ \sum _{j=0}^{⌊(n-1)/2⌋}{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{n-j-1}{j}}\right)\left({\displaystyle \frac{5n-4j}{n-2j}}\right)6^{n-(2j+1)}+{(-1)}^{n/2}-1,& forevenn\end{array}\right..$$

Similarly, using Equation (4), we obtain the binomial expression for Edouard–Lucas numbers, which is presented in the following proposition.

Proposition 2.

For all $n>0,$ the following identity holds:

$$\begin{array}{ccc}\hfill K_n& =& 1+\sum _{j=0}^{⌊n/2⌋}{(-1)}^j{\displaystyle \frac{n}{n-j}}\left(\begin{array}{c}n-j\\ j\end{array}\right)6^{n-2j}.\hfill \end{array}$$

Propositions 1 and 2 give us the binomial expressions for incomplete Edouard and incomplete Edouard–Lucas numbers. Moreover, the binomial expressions allow us to define the incomplete Edouard and incomplete Edouard–Lucas numbers, as demonstrated below.

Definition 1.

For any positive integer n, the incomplete Edouard numbers $E_n\left(k\right)$ are defined by

$$4E_n\left(k\right)=\sum _{j=0}^k{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{n-(j+1)}{j}}\right)\left({\displaystyle \frac{5n-4j}{n-2j}}\right)6^{n-(2j+1)}-1,$$

(12)

for $0\le k<⌊{\displaystyle \frac{n-1}{2}}⌋$ and $E_n\left(⌊{\displaystyle \frac{n-1}{2}}⌋\right)=E_n$.

Definition 2.

For any positive integer $n,$ the incomplete Edouard–Lucas numbers $K_n\left(k\right)$ are defined by

$$K_n\left(k\right)=\sum _{j=0}^k{(-1)}^j\left({\displaystyle \frac{n}{n-j}}\right)\left(\begin{array}{c}n-j\\ j\end{array}\right)6^{n-2j}+1,0\le k\le ⌊{\displaystyle \frac{n}{2}}⌋.$$

(13)

Table 1. Incomplete Edouard numbers.

n/k	0	1	2	3	4	5
1	4
2	28
3	179	168
4	1080	984
5	6479	5723	5740
6	38,878	33,262	33,460
7	233,279	193,103	195,047	195,024
8	1,399,680	1,119,744	1,137,024	1,136,688
9	8,398,079	6,485,183	6,629,039	6,625,079	6,625,108
10	50,388,478	37,511,422	38,654,494	38,613,454	38,613,964
11	302,330,879	216,670,463	225,441,791	225,051,695	225,058,715	225,058,680

and

Table 2. Incomplete Edouard–Lucas numbers.

n/k	0	1	2	3	4	5
1	7
2	37	35
3	217	199
4	1297	1153	1155
5	7777	6697	6727
6	46,657	38,881	39,205	39,203
7	279,937	225,505	228,529	228,487
8	1,679,617	1,306,369	1,332,289	1,331,713	1,331,715
9	10,077,697	7,558,273	7,768,225	7,761,745	7,761,799
10	60,466,177	43,670,017	45,302,977	45,238,177	45,239,077	45,239,075

show us the values of $4E_n\left(k\right)$ and $K_n\left(k\right)$ for $n=1,\dots ,11$ and $k=0,\dots ,5.$ Next, we will provide some of the properties of these new sequences of numbers, some of which can be verified directly by looking at Table 1 and Table 2. The following proposition is derived directly from Definitions 1 and 2.

Proposition 3.

For all $n\ge 1$, we know that

$$E_n\left(⌊{\displaystyle \frac{n-1}{2}}⌋\right)=E_n,$$

$$K_n\left(⌊{\displaystyle \frac{n}{2}}⌋\right)=K_n.$$

(14)

$$4E_n\left(0\right)=5·6^{n-1}-1,$$

$$K_n\left(0\right)=6^n+1.$$

The result below gives us the value of $K_n(⌊{\displaystyle \frac{n}{2}}⌋-1)$ for odd n and even n in terms of the n-th Edouard–Lucas number.

Proposition 4.

For all integers $n\ge 2,$ the following identity holds:

$$K_n(⌊{\displaystyle \frac{n}{2}}⌋-1)=\left\{\begin{array}{cc}{K}_n-2{(-1)}^{{\displaystyle \frac{n}{2}}},& forevenn,\\ K_n+{(-1)}^{\lfloor {\displaystyle \frac{n}{2}}\rfloor }(12\lfloor {\displaystyle \frac{n}{2}}\rfloor +6),& foroddn.\end{array}\right.$$

Proof. 

Suppose n is even, i.e, $n=2t.$ We know that $⌊{\displaystyle \frac{n}{2}}⌋=⌊{\displaystyle \frac{2t}{2}}⌋=t.$ The term $⌊{\displaystyle \frac{n}{2}}⌋$ in Equation (13) is

$${(-1)}^t\left(\begin{array}{c}2t-t\\ t\end{array}\right)\left({\displaystyle \frac{2t}{2t-t}}\right)6^{2t-2t}={(-1)}^{t}1·2·1={(-1)}^{t}2.$$

Therefore, by using Equation (14), we obtain

$$\begin{array}{ccc}\hfill K_n\left(⌊{\displaystyle \frac{n}{2}}⌋-1\right)& =& K_n\left(⌊{\displaystyle \frac{n}{2}}⌋\right)-{(-1)}^t\left(\begin{array}{c}2t-t\\ t\end{array}\right)\left({\displaystyle \frac{2t}{2t-t}}\right)6^{2t-2t}\hfill \\ & =& K_n-2{(-1)}^{{\displaystyle \frac{n}{2}}}.\hfill \end{array}$$

Otherwise, suppose $n=2t+1.$ We know that $⌊{\displaystyle \frac{n}{2}}⌋=⌊{\displaystyle \frac{2t+1}{2}}⌋=t.$ In this case, the term $⌊{\displaystyle \frac{n}{2}}⌋$ in Equation (13) is

$$\begin{array}{ccc}& & {(-1)}^t\left(\begin{array}{c}2t+1-t\\ t\end{array}\right)\left({\displaystyle \frac{2t+1}{2t+1-t}}\right)6^{2t+1-2t}\hfill \\ & =& {(-1)}^t\left(\begin{array}{c}t+1\\ t\end{array}\right)·{\displaystyle \frac{2t+1}{t+1}}·6^1={(-1)}^{t}6(t+1).\hfill \end{array}$$

Thus,

$$\begin{array}{ccc}\hfill K_n\left(⌊{\displaystyle \frac{n}{2}}⌋-1\right)& =& K_n\left(⌊{\displaystyle \frac{n}{2}}⌋\right)-{(-1)}^t\left(\begin{array}{c}2t+1-t\\ t\end{array}\right)\left({\displaystyle \frac{2t+1}{2t+1-t}}\right)6^{2t+1-2t}\hfill \\ & \stackrel{\mathrm{Equation}\left(14\right)}{=}& K_n-{(-1)}^{t}6(t+1),\hfill \end{array}$$

which ends the proof. □

3. Some Identities

In this section, we will provide some identities that involve the incomplete Edouard and incomplete Edouard–Lucas numbers, and their related sequences of numbers, such as balancing, Lucas-balancing, incomplete balancing, and incomplete Lucas-balancing numbers.

The next proposition is derived from Definition 1 and Equation (3).

Proposition 5.

Let n and k be positive integers with $0\le k<\lfloor {\displaystyle \frac{n-1}{2}}\rfloor .$ Then,

$$4E_n\left(k\right)=B_{n+1}\left(k\right)-B_n\left(k\right)-1,$$

(15)

In addition, for $k=\lfloor {\displaystyle \frac{n-1}{2}}\rfloor ,$ we have

$$4E_n\left(k\right)=\left\{\begin{array}{cc}{B}_{n+1}\left(k\right)-B_n\left(k\right)-1,& foroddn\\ B_{n+1}(k+1)-B_n\left(k\right)-1,& forevenn.\end{array}\right.$$

From Equations (9) and (15), we obtain the following proposition directly.

Proposition 6.

For $n\ge 3,$ the following identities hold:

$$4E_n\left(⌊{\displaystyle \frac{n-1}{2}}⌋-1\right)=\left\{\begin{array}{cc}4E_n\pm 3n\pm 1,& ifniseven,\\ 4E_n\pm 3(n+1)\pm 1,& ifnisodd.\end{array}\right.$$

The next result gives us the incomplete Edouard numbers in terms of balancing and Lucas-balancing numbers.

Proposition 7.

For incomplete Edouard numbers $E_n\left(k\right),$ the following identity holds:

$$4\sum _{k=0}^{⌊{\displaystyle \frac{n-1}{2}}⌋}{E}_n\left(k\right)=\left\{\begin{array}{cc}X,\hfill & \mathit{if}n\mathit{is}\mathit{even}\hfill \\ Y,\hfill & \mathit{if}n\mathit{is}\mathit{odd}\hfill \end{array}\right.$$

where $X=(3(n+1)C_{(n+1)}-B_{(n-1)})/12-(3nCn+7B_n)/16-(⌊{\displaystyle \frac{n-1}{2}}⌋+1)-6^n$ and $Y=(3(n+1)C_{(n+1)}-B_{(n-1)})/12-(3nCn+7B_n)/16-(⌊{\displaystyle \frac{n-1}{2}}⌋+1)$.

Proof. 

The proof follows by Equations (10) and (15). □

Recall Equation (11), namely,

$$B_{n+2}(k+1)=6B_{n+1}(k+1)-B_n\left(k\right),$$

for $0\le k<\lfloor {\displaystyle \frac{n-2}{2}}\rfloor .$ Then, it is possible to establish a recurrence relation to the incomplete Edouard numbers as follows.

Proposition 8.

Let n and k be positive integers with $0\le k<\lfloor {\displaystyle \frac{n-1}{2}}\rfloor .$ Then, the recurrence relation of the incomplete Edouard numbers $E_n\left(k\right)$ is given by

$$4E_n\left(k\right)=24E_{n-1}\left(k\right)-4E_{n-2}(k-1)+4,$$

Proof. 

For $0\le k\le \lfloor {\displaystyle \frac{n-2}{2}}\rfloor ,$ the proof follows by substituting Equation (11) into Equation (15). □

Now, from Definition 2 and Equation (4), we can find the relationship between the incomplete Lucas-balancing numbers and the incomplete Edouard–Lucas numbers.

Proposition 9.

Let n and k be positive integers with $0\le k\le \lfloor {\displaystyle \frac{n}{2}}\rfloor .$ Then,

$$\begin{array}{c}\hfill K_n\left(k\right)=2C_n\left(k\right)+1.\end{array}$$

Proposition 10.

Let n and k be positive integers with $1\le k<\lfloor {\displaystyle \frac{n}{2}}\rfloor .$ Then, the recurrence relation of the incomplete Edouard–Lucas numbers $K_n\left(k\right)$ is given by

$$\begin{array}{c}\hfill K_{n+2}(k+1)=6K_{n+1}(k+1)-K_n\left(k\right)+4,\end{array}$$

Proof. 

Since, for $1\le k<\lfloor {\displaystyle \frac{n}{2}}\rfloor ,$$6C_{n+1}(k+1)-C_n\left(k\right)=C_{n+2}(k+1)$ [12] (Proposition 3.10), holds by Proposition 9 we know that

$$\begin{array}{ccc}\hfill 6K_{n+1}(k+1)-K_n\left(k\right)& =& 6(2C_{n+1}(k+1)+1)-(2C_n\left(k\right)+1)\hfill \\ & =& 2(6C_{n+1}(k+1)-C_n\left(k\right))+5\hfill \\ & =& 2C_{n+2}(k+1)+5\hfill \\ & =& K_{n+2}(k+1)+4.\hfill \end{array}$$

□

Propositions 8 and 10 show us the recurrence relations of the incomplete Edouard and incomplete Edouard–Lucas numbers. Note that these sequences do not have the same recurrence relations as those which occur between the Edoaurd and Edouard–Lucas sequences of numbers.

The next result gives us the relation between the sum of incomplete Edouard–Lucas numbers and the balancing and Lucas Balancing numbers.

Proposition 11.

For incomplete Edouard–Lucas numbers,

$$\sum _{k=0}^{⌊{\displaystyle \frac{n-1}{2}}⌋}{K}_n\left(k\right)=\left\{\begin{array}{cc}2C_n+n(3B_n-2C_n)+⌊{\displaystyle \frac{n-1}{2}}⌋+1,\hfill & \mathit{if}n\mathit{is}\mathit{even},\hfill \\ C_n+n(3B_n-2C_n)+⌊{\displaystyle \frac{n-1}{2}}⌋+1,\hfill & \mathit{if}n\mathit{is}\mathit{odd}.\hfill \end{array}\right.$$

In addition, since $B_{n+1}\left(k\right)-B_{n-1}(k-1)=2C_n\left(k\right)$ for $0\le k\le ⌊{\displaystyle \frac{n}{2}}⌋,$ then using Proposition 3.9 in [12], we obtain the following result.

Proposition 12.

For incomplete Edouard numbers and incomplete Edouard–Lucas numbers, it holds that

$$4E_n\left(k\right)-K_n\left(k\right)=B_n\left(k\right)-B_{n-1}(k-1)-2,$$

for $0\le k\le \widehat{n},$ where $\widehat{n}=⌊{\displaystyle \frac{n}{2}}⌋$ and $B_n\left(k\right)$ is the n-th incomplete balancing number.

4. Generating Functions of Incomplete Edouard and Incomplete Edouard–Lucas Numbers

In this section, we provide generating functions for both incomplete Edouard and incomplete Edouard–Lucas numbers. First, we need to recall a result stated in [15], and then we will use it in the proof of the statement of each one of the generating functions. Consider the following result (see [15] (p. 592)):

Theorem 1

([15] (Lemma)). Let ${\left\{s_n\right\}}_{n\ge 0}$ be a complex sequence satisfying the following non-homogeneous recurrence relation:

$$s_n=a{s}_{n-1}+b{s}_{n-2}+r_n(n>1),$$

where a and b are complex numbers and ${\left\{r_n\right\}}_{n\ge 0}$ is a given complex sequence. Then, the generating function $U\left(t\right)$ of the sequence $\left\{s_n\right\}$ is

$$U\left(t\right)={\displaystyle \frac{G\left(t\right)+s_0-r_0+(s_1-s_{0}a-r_1)t}{1-at-b{t}^2}},$$

where $G\left(t\right)$ denotes the generating function of $\left\{r_n\right\}$.

This theorem can be obtained by using the standard method. By applying Theorem 1, we obtain the next generating functions of incomplete Edouard and incomplete Edouard–Lucas numbers, as follows.

Proposition 13.

The generating function for the incomplete Edouard sequence ${\left\{E_n\left(k\right)\right\}}_{n\ge 0}$ is

$$G{F}_E(k,t)={\displaystyle \frac{G\left(t\right)+E_0\left(k\right)-1+(E_1\left(k\right)-6E_0\left(k\right)-1)t}{1-6t+t^2}},$$

where $G\left(t\right)=1+t+t^2+\cdots$.

Proof. 

By Proposition 8 we know that

$$E_n\left(k\right)=6E_{n-1}\left(k\right)-E_{n-2}(k-1)+1,$$

where $a=6$, $b=-1$ and $\left\{r_n\right\}=\{1,1,\cdots \}$. So, the generating function of $\left\{r_n\right\}$ is $G\left(t\right)=1+t+t^2+\cdots$, and thus we obtain the result. □

Proposition 14.

The generating function for the incomplete Edouard–Lucas sequence ${\left\{K_n\left(k\right)\right\}}_{n\ge 0}$ is

$$G{F}_K(k,t)={\displaystyle \frac{G\left(t\right)+K_0\left(k\right)+4+(K_1\left(k\right)-6K_0\left(k\right)+4)t}{1-6t+t^2}},$$

where $G\left(t\right)=-4(1+t+t^2+\cdots )$.

Proof. 

By Proposition 10 we know that

$$K_n\left(k\right)=6K_{n-1}\left(k\right)-K_{n-2}(k-1)-4,$$

where $a=6$, $b=-1$ and $\left\{r_n\right\}=\{-4,-4,\cdots \}$. So, the generating function of $\left\{r_n\right\}$ is $G\left(t\right)=-4(1+t+t^2+\cdots )$, and thus we obtain the result. □

5. Conclusions

This study has introduced two new sequences of numbers: the incomplete Edouard and incomplete Edouard–Lucas numbers. We provided properties and recurrence relations for each sequence. Moreover, we established identities and relations between the incomplete Edouard and incomplete Edouard–Lucas numbers, and the balancing and Lucas-balancing numbers. Finally, we presented the generating functions of the incomplete Edouard and incomplete Edouard–Lucas numbers.

It seems to us that the results presented here are new and we believe that these sequences could be the subject of other future studies.