On the reflexive edge strength of the circulant graphs

Mohamed Basher; Mohamed Basher

doi:10.3934/math.2021543

AIMS Mathematics

2021, Volume 6, Issue 9: 9342-9365. doi: 10.3934/math.2021543

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Research article

On the reflexive edge strength of the circulant graphs

Mohamed Basher ^,

Department of Mathematics and Computer Science, Faculty of Science, Suez University, Suez, Egypt

Received: 04 March 2021 Accepted: 11 June 2021 Published: 23 June 2021
MSC : 05C78, 05C12, 05C90

A labeling of a graph is an assignment that carries some sets of graph elements into numbers (usually the non negative integers). The total $k$ -labeling is an assignment $f_{e}$ from the edge set to the set $\{1, 2, ..., k_{e} \}$ and assignment $f_{v}$ from the vertex set to the set $\{0, 2, 4, ..., 2k_{v} \}$ , where $k = max\{k_{e}, 2k_{v} \}$ . An edge irregular reflexive $k$ -labeling of the graph $G$ is the total $k$ -labeling, if distinct edges have distinct weights, where the edge weight is defined as the sum of label of that edge and the labels of the end vertices. The minimum $k$ for which the graph $G$ has an edge irregular reflexive $k$ -labeling is called the reflexive edge strength of the graph $G$ , denoted by $res(G)$ . In this paper we study the edge reflexive irregular $k$ -labeling for some cases of circulant graphs and determine the exact value of the reflexive edge strength for several classes of circulant graphs.

Keywords:

Citation: Mohamed Basher. On the reflexive edge strength of the circulant graphs[J]. AIMS Mathematics, 2021, 6(9): 9342-9365. doi: 10.3934/math.2021543

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Abstract

1. Introduction

Let $G$ be a connected, simple, and undirected graph with a vertex set $V(G)$ and edge set $E(G)$ . In graph theory, graph labeling is a topic that is interesting to study. Graph labelings were introduced for first time by Sedláček in 1963 ^[32], the various methods for labeling of the edges of graph have been studied by stewart in ^[33], and then formulated by Kotig and Rosa in ^[27]. In 1988, Chartrand et al. ^[20] defined irregular labeling for a graph $G$ . Informally, the irregular labeling defined as follows. Let $G(V, E)$ be a graph, the mapping $f:E\rightarrow \{1, 2, 3, ..., k\}$ is called irregular $k$ -labeling of $G$ , if every different vertices $u$ and $v$ have distinct weights, that is:

$\begin{align*} \sum\limits_{\substack{x\in V}}f(ux)\neq \sum\limits_{\substack{y\in V}}f(vy). \end{align*}$

The irregularity strength of $G$ , denoted by $s(G)$ , is known as the minimum positive natural number $k$ which $G$ has an irregular $k$ -labeling. Therefore, there are many researchers who give certain results of the irregular labeling (see ^{[1,7,9,18,21,22,26,27,32,33,36]}). The natural extension of irregularity strength of a graph $G$ , Baća, Jendrol, Miller, and Ryan ^[11] introduced the other types of irregular labeling based on the total labeling, namely the vertex irregular total k-labeling and edge irregular total k-labeling. Some results on the vertex irregular total k-labeling and edge irregular total k-labeling can be found in ^{[2,3,4,5,8,10,12,13,16,23,28,30]}. In ^[11] Baća et al. defined the concept of edge irregular total $k$ -labeling as a labeling of vertices and edges of $G$ , $f:V\bigcup E\rightarrow \{1, 2, ..., k\}$ such that for any different edges $xy$ , $x^{'}y^{'}$ have distinct weights i.e. $wt(xy)\neq wt(x^{'}y^{'})$ where $wt(xy) = f(x)+f(xy)+f(y)$ . The minimum $k$ for which $G$ has an edge irregular total $k$ -labeling is namely the total edge irregularity strength of $G$ , denoted by $tes(G)$ . Furthermore, Ryan et al. ^[31] generalized the concept of edge irregular total $k$ -labeling of a graph to be an edge irregular reflexive $k$ -labeling.

For a graph $G$ the total $k$ -labeling defined the map $f_{e}:E(G)\rightarrow \{1, 2, 3, ..., k_{e}\}$ and $f_{v}:V(G)\rightarrow \{0, 2, 4, ..., 2k_{v}\}$ , where $k = max\{k_{e}, 2k_{v}\}$ . The total $k$ -labeling is called an edge irregular reflexive $k$ -labeling, if for every two distinct edges $xy$ and $x^{'}y^{'}$ of $G$ , $wt(xy)\neq wt(x^{'}y^{'})$ , where $wt(xy) = f_{v}(x)+f_{e}(xy)+f_{v}(y)$ . The minimum value of $k$ for which such labeling exists is known as the reflexive edge strength of graph $G$ and is denoted by $res(G)$ . Over the last years, $res(G)$ has been investigated for different family of graphs (see ^[6,14,35,38]). The notion of reflexive edge strength was defined in ^[35]. The following lemma estimate the lower bound of the reflexive edge strength of graph $G$ .

Lemma 1.1. (^[35]) For every graph $G$ ,

$\begin{align*} res(G)\geq\left\{ \begin{array}{l} \big\lceil\frac{|E(G)|}{3}\big\rceil , {{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |E(G)|\not\equiv 2,3 \; \mathit{(mod \;6),}}}\\ \big\lceil\frac{|E(G)|}{3}\big\rceil +1, {{\ \ \ \ \ \ \ \ \ \ |E(G)|\equiv 2,3 \;\mathit{(mod\; 6).}}} \end{array} \right. \end{align*}$

Also, Baća et al. ^[14] suggested the following conjecture:

Conjecture 1.1. (^[14]) Let $G$ be a simple graph with maximum degree $\triangle = \triangle(G)$ . Then:

$\begin{equation*} res(G) = max\{\big\lfloor\frac{\triangle+2}{2}\big\rfloor,\big\lceil\frac{|E(G)|}{3}\big\rceil+r\} \end{equation*}$

where $r = 1$ for $|E(G)|\equiv2, 3$ (mod 6), and zero otherwise.

2. Circulant graphs

The circulant graphs are another important family of graphs that can be used to construct local area networks see ^[17]. Suppose $1\leq s_{1} \leq s_{2} \leq...\leq s_{m}\leq \frac{n}{2}$ , where $n$ and $s_{j}$ , $j = 1, 2, 3, ..., m$ are positive numbers. The circulant graph $C_{n}(s_{1}, s_{2}, ..., s_{m})$ is the graph on vertex set $V = \{x_{0}, x_{1}, ..., x_{n-1}\}$ and edge set $E = \{x_{i}x_{i+s_{j}}: i = 0, 1, 2, ..., n-1 { \ and \ } j = 1, 2, ..., m\}$ where $i+s_{j}$ is taken modulo $n$ see ^[15]. It is clear to see that $C_{n}(s_{1}, s_{2}, ..., s_{m})$ is a $r$ -regular, where $r = 2m-1$ if $s_{m} = \frac{n}{2}$ with $n(2m-1)/2$ edges on other hand $r = 2m$ with $mn$ edges if $s_{m} < \frac{n}{2}$ . The circulant graphs have a large number of applications and uses in telecommunication network VLSI design, parallel and distributed computing. The properties of circulant graphs; such as bipartitness, planarity, Hamiltonicity and colourability have been studied in ^{[19,24,25,29,34,37]}. In this paper we investigate the existence of the reflexive edge irregularity strength for some classes of circulant graphs.

3. Main results

In this section, we discuss the edge irregular reflexive k-labeling for some classes of circulant graphs.

Theorem 1. Let $C_{n}(1, 2)$ , be a circulant graph on $n\geq4$ vertices. Then

$\begin{align*} res(C_{n}(1,2)) = \left\{ \begin{array}{l} 4, {{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ n = 4 ,}}\\ 5,{{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ n = 5,6 ,}}\\ \lceil\frac{2n}{3}\rceil , {{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ n\equiv 0,2 \;\mathit{(mod \;3)} \;and \; n\geq8 ,}}\\ \lceil\frac{2n}{3}\rceil +1, {{\ \ \ \ \ \ \ \ \ \ \ n\equiv 1 \;\mathit{(mod\; 3)}\; and \; n\geq7 }}. \end{array} \right. \end{align*}$

Proof. Let $C_{n}(1, 2)$ be a circulant graph with vertex set $V(C_{n}(1, 2)) = \{x_{i}:1\leq i\leq n\}$ and edge set $E(C_{n}(1, 2)) = \{x_{i}x_{i+j}:1\leq i\leq n, 1\leq j\leq 2 \}$ , where indices are taken modulo $n$ . First note that $C_{n}(1, 2)$ isomorphic to the wheel $W_{3}$ and so from ^[22] Ryan proved that $res(W_{3}) = 4$ , so $res(C_{n}(1, 2)) = 4$ . By the same arguments that used in ^[22], we can find that $res(C_{5}(1, 2))\geq5$ and $res(C_{6}(1, 2))\geq5$ . The corresponding labelings for $C_{n}(1, 2)$ , $n = 4, 5, 6, 7, 8, 9$ , are shown in Figure 1.

Figure 1. The reflexive edge irregular

$k$ -labeling of

$C_{n}(1, 2), n = 4, 5, 6, 7, 8, 9$ .

DownLoad: Full-Size Img PowerPoint

For $n\geq10$ and based on Lemma 1.1, we show the following lower bound for the circulant $C_{n}(1, 2)$ : $res(C_{n}(1, 2))\geq k = \lceil\frac{2n}{3}\rceil$ if $n\equiv 0, 2$ (mod 3) and $res(C_{n}(1, 2))\geq k = \lceil\frac{2n}{3}\rceil+1$ if $n\equiv 1$ (mod 3). Then, to convince our proof we define a total labeling of $C_{n}(1, 2)$ such as:

$\begin{align*} f(x_{i}) = \left\{ \begin{array}{l} 0, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1\leq i\leq \lfloor\frac{n}{3}\rfloor , }\\ 2\lfloor\frac{k}{4}\rfloor,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{3}\rfloor+1\leq i\leq \lfloor\frac{n}{2}\rfloor , }\\ 2\lfloor\frac{k}{2}\rfloor,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor+1\leq i\leq\lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor , }\\ 2\lfloor\frac{k}{6}\rfloor, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor+1\leq i\leq n }. \end{array} \right. \end{align*}$

$\begin{align*} f(x_{i}x_{i+1}) = \left\{ \begin{array}{l} i, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1\leq i\leq \lfloor\frac{n}{3}\rfloor-1 , }\\ 2n-2\lfloor\frac{n}{2}\rfloor-\\-2\lfloor\frac{k}{4}\rfloor-2,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ i = \lfloor\frac{n}{3}\rfloor , }\\ 2n-2\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor-\\-4\lfloor\frac{k}{4}\rfloor+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{3}\rfloor+1\leq i\leq \lfloor\frac{n}{2}\rfloor-1 ,}\\ 2n-2\lfloor\frac{n}{3}\rfloor-\\-2(\lfloor\frac{k}{2}\rfloor+\lfloor\frac{k}{4}\rfloor)+1,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ i = \lfloor\frac{n}{2}\rfloor ,}\\ 2n-\lfloor\frac{n}{2}\rfloor-2\lfloor\frac{n}{3}\rfloor-\\-4\lfloor\frac{k}{2}\rfloor+3+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor+1\leq i\leq \lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor-1 ,}\\ 2n-2\lfloor\frac{n}{3}\rfloor-\\-2(\lfloor\frac{k}{2}\rfloor+\lfloor\frac{k}{6}\rfloor)-2,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ i = \lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor ,}\\ \lfloor\frac{n}{3}\rfloor-\lfloor\frac{n}{2}\rfloor-\\-4\lfloor\frac{k}{6}\rfloor+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor+1\leq i\leq n-1 ,}\\ 2\lfloor\frac{n}{3}\rfloor-2\lfloor\frac{k}{6}\rfloor-2,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ i = n . } \end{array} \right. \end{align*}$

$\begin{align*} f(x_{i}x_{i+2}) = \left\{ \begin{array}{l} \lfloor\frac{n}{3}\rfloor-1+i, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1\leq i\leq \lfloor\frac{n}{3}\rfloor-2 , }\\ 2n-2\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor-\\-2\lfloor\frac{k}{4}\rfloor+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{3}\rfloor-1\leq i\leq \lfloor\frac{n}{3}\rfloor , }\\ 2n-\lfloor\frac{n}{2}\rfloor-2\lfloor\frac{n}{3}\rfloor-\\-4\lfloor\frac{k}{4}\rfloor-1+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{3}\rfloor+1\leq i\leq \lfloor\frac{n}{2}\rfloor-2 } \;\text{and} \\ {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor\geq2 ,}\\ 2n-2\lfloor\frac{n}{3}\rfloor-\lfloor\frac{n}{2}\rfloor-\\-2(\lfloor\frac{k}{2}\rfloor+\lfloor\frac{k}{4}\rfloor)+3+i,{\ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor-1\leq i\leq \lfloor\frac{n}{2}\rfloor ,}\\ 2n-\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor-\\-4\lfloor\frac{k}{2}\rfloor+2+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor+1\leq i\leq \lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor-2 ,}\\ 2n-3\lfloor\frac{n}{3}\rfloor-\lfloor\frac{n}{2}\rfloor-\\-2(\lfloor\frac{k}{2}\rfloor+\lfloor\frac{k}{6}\rfloor)+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor-1\leq i\leq \lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor ,}\\ i-4\lfloor\frac{k}{6}\rfloor,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor+1\leq i\leq n-2 }\; \text{and}\\ {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor\geq3 , }\;\text{where} \; n \text{ is even}, \\ 2\lfloor\frac{n}{3}\rfloor-2\lfloor\frac{n}{2}\rfloor-\\-2\lfloor\frac{k}{6}\rfloor+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ n-1\leq i\leq n . } \end{array} \right. \end{align*}$

Evidently, the vertices of $C_{n}(1, 2)$ labeled with even numbers. Now we will compute the weights of edges under the labeling $f$ :

The edge set of $C_{n}(1, 2)$ can be split into the eight mutually disjoint subsets, $A_{\ell}, 1\leq \ell \leq 8$ as follows: \\For $1\leq j\leq2$ .

● $A_{1} = \{x_{i}x_{i+j}:1\leq i \leq \lfloor\frac{n}{3}\rfloor-j \}$ the set of all edges which has end vertices labeled with 0.

● $A_{2} = \{x_{i}x_{i+j}:\lfloor\frac{n}{3}\rfloor-j+1\leq i \leq \lfloor\frac{n}{3}\rfloor \}$ the set of all edges which has end vertices labeled with 0 and $2\lfloor\frac{k}{4}\rfloor$ .

● $A_{3} = \left\{ \begin{array}{ll} \{x_{\lfloor\frac{n}{3}\rfloor+1}x_{\lfloor\frac{n}{3}\rfloor+2} \}, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{if}\; \lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor = 2. } & \\\\ \{x_{i}x_{i+j}:\lfloor\frac{n}{3}\rfloor+1\leq i \leq \lfloor\frac{n}{2}\rfloor-j, \}, {\ \ \ \ \ \ \ \ \text{if}\; \lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor\geq3 } & \end{array} \right.$

the set of all edges which has end vertices labeled with $2\lfloor\frac{k}{4}\rfloor$ .

● $A_{4} = \{x_{i}x_{i+j}:\lfloor\frac{n}{2}\rfloor+1-j\leq i\leq \lfloor\frac{n}{2}\rfloor \}$ the set of all edges which has end vertices labeled with $2\lfloor\frac{k}{4}\rfloor$ and $2\lfloor\frac{k}{2}\rfloor$ .

● $A_{5} = \{x_{i}x_{i+j}:\lfloor\frac{n}{2}\rfloor+1\leq i\leq \lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor-j \}$ the set of all edges which has end vertices labeled with $2\lfloor\frac{k}{2}\rfloor$ .

● $A_{6} = \{x_{i}x_{i+j}:\lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor-j+1\leq i\leq \lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor \}$ the set of all edges which has end vertices labeled with $2\lfloor\frac{k}{2}\rfloor$ and $2\lfloor\frac{k}{6}\rfloor$ .

● $A_{7} = \left\{ \begin{array}{ll} \{x_{n-1}x_{n} \}, {\ \ \ \ \ \ \ \ \ \ \ \text{if}\; n \; \text{is even and} \lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor = 2. } & \\\\ \{x_{i}x_{i+j}:\lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor+1\leq i\leq n-j \}, {\ \ \ \ \text{otherwise}.} & \end{array} \right.$

the set of all edges which has end vertices labeled with $2\lfloor\frac{k}{6}\rfloor$ .

● $A_{8} = \{x_{i}x_{i+j}: n-j+1\leq i\leq n \}$ the set of all edges which has end vertices labeled with 0 and $2\lfloor\frac{k}{6}\rfloor$ .

Hence, the edge weights under the labeling $f$ are the following:

1. The weights of the edges of the set $A_{1}$ , admit the successive integers from the set $A_{1} = \{1, 2, ..., 2\lfloor\frac{n}{3}\rfloor-3 \}$ .

2. The weights of the edges of the set $A_{2}$ , admit the successive integers from the set $A_{2} = \{2n-2\lfloor\frac{n}{2}\rfloor-2, 2n-2\lfloor\frac{n}{2}\rfloor-1, 2n-2\lfloor\frac{n}{2}\rfloor \}$

3. The weight of the edge of the set $A_{3}$ , admit the integer $\{2n-2\lfloor\frac{n}{2}\rfloor+1, ..., 2n-2\lfloor\frac{n}{3}\rfloor-3 \}$ ,

4. The weights of the edges of the set $A_{4}$ , admit the successive integers from the set $A_{4} = \{2n-2\lfloor\frac{n}{3}\rfloor+1, 2n-2\lfloor\frac{n}{3}\rfloor+2, 2n-2\lfloor\frac{n}{3}\rfloor+3 \}$

5. The weights of the edges of the set $A_{5}$ , admit the successive integers from the set $A_{5} = \{2n-2\lfloor\frac{n}{3}\rfloor+4, ..., 2n \}$

6. The weights of the edges of the set $A_{6}$ , admit the successive integers from the set $A_{6} = \{2n-2\lfloor\frac{n}{3}\rfloor-2, 2n-2\lfloor\frac{n}{3}\rfloor-1, 2n-2\lfloor\frac{n}{3}\rfloor\}$

7. The weights of the edges of the set $A_{7}$ , admit the successive integers from the set $A_{7} = \{2\lfloor\frac{n}{3}\rfloor+1, ..., 2n-2\lfloor\frac{n}{2}\rfloor-3 \}$

8. The weights of the edges of the set $A_{8}$ , admit the successive integers from the set $A_{8} = \{2\lfloor\frac{n}{3}\rfloor-2, 2\lfloor\frac{n}{3}\rfloor-1, 2\lfloor\frac{n}{3}\rfloor \}$

This mean that for $n\geq10$ , the edge weights are from the set $W = \{1, 2, 3, ..., 2n \}$ . Thus, we can easily see that all integers in $W$ are distinct.

Theorem 2. Let $C_{n}(1, 2, 3)$ , be a circulant graph on $n\geq6$ vertices. Then

$\begin{align*} res(C_{n}(1,2,3)) = \left\{ \begin{array}{l} n , {{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ n \; is\; even,}}\\ n+1, {{\ \ \ \ \ \ \ \ \ \ \ n \; is \;odd}}. \end{array} \right. \end{align*}$

Proof. According to the fact that the circulant $C_{n}(1, 2, 3)$ has $3n$ edges if $n\geq7$ , then using the Lemma 1.1, we obtain the lower bound for the circulant $C_{n}(1, 2, 3)$ as following: $res(C_{n}(1, 2, 3))\geq n$ for $n$ is even and $res(C_{n}(1, 2, 3))\geq n+1$ for $n$ is odd. To prove the equality, it suffices to prove the existence of an optimal total labeling of $C_{n}(1, 2, 3)$ such as:

The corresponding labelings for $C_{n}(1, 2, 3)$ , $n = 6, 7, 8, 9, 10$ , are shown in . Suppose $k = n$ for $n$ is even and $k = n+1$ for $n$ is odd. Hence, for $n\geq11$ we have the following labelings:

$\begin{align*} f(x_{i}) = \left\{ \begin{array}{l} 0, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1\leq i\leq \lfloor\frac{n}{3}\rfloor , }\\ 2\lfloor\frac{k}{4}\rfloor+2,{\ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{3}\rfloor+1\leq i\leq \lfloor\frac{n}{2}\rfloor , }\\ k ,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor+1\leq i\leq\lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor , }\\ 2\lfloor\frac{k}{8}\rfloor, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor+1\leq i\leq n }. \end{array} \right. \end{align*}$

Figure 2. The reflexive edge irregular

$k$ -labeling of

$C_{n}(1, 2, 3), n = 6, 7, 8, 9, 10$ .

DownLoad: Full-Size Img PowerPoint

Case 1. If $\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor = 2$ .

Firstly, for $n$ is even, it is only realizable when $n = 12$ , then the Figure 3, shows labelings of vertices and edges along with their weights.

Figure 3. A reflexive irregular 12-labeling of

$C_{12}(1, 2, 3)$ and its edge weights.

DownLoad: Full-Size Img PowerPoint

Secondly, for $n$ is odd.

$\begin{align*} f(x_{i}x_{i+1}) = \left\{ \begin{array}{l} i, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1\leq i\leq \lfloor\frac{n}{3}\rfloor-1 , }\\ 3\lfloor\frac{n}{3}\rfloor-2\lfloor\frac{k}{4}\rfloor+2,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ i = \lfloor\frac{n}{3}\rfloor , }\\ 3\lfloor\frac{n}{3}\rfloor-4\lfloor\frac{k}{4}\rfloor+6,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ i = \lfloor\frac{n}{3}\rfloor+1 ,}\\ 3\lfloor\frac{n}{3}\rfloor-k-2\lfloor\frac{k}{4}\rfloor+15,{\ \ \ \ \ \ \ \ \ \ \ \ \ i = \lfloor\frac{n}{3}\rfloor+2 ,}\\ 2\lfloor\frac{n}{3}\rfloor-2k+19+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{3}\rfloor+3\leq i\leq 2\lfloor\frac{n}{3}\rfloor+1 ,}\\ 3\lfloor\frac{n}{3}\rfloor-k-2\lfloor\frac{k}{8}\rfloor+11,{\ \ \ \ \ \ \ \ \ \ \ \ \ i = 2\lfloor\frac{n}{3}\rfloor+2 ,}\\ \lfloor\frac{n}{3}\rfloor-4\lfloor\frac{k}{8}\rfloor-2+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2\lfloor\frac{n}{3}\rfloor+3\leq i\leq 2\lfloor\frac{n}{3}\rfloor+4 ,}\\ 3\lfloor\frac{n}{3}\rfloor-2\lfloor\frac{k}{8}\rfloor-5,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ i = n . } \end{array} \right. \end{align*}$

$\begin{align*} f(x_{i}x_{i+2}) = \left\{ \begin{array}{l} \lfloor\frac{n}{3}\rfloor-1+i, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1\leq i\leq \lfloor\frac{n}{3}\rfloor-2 , }\\ 2\lfloor\frac{n}{3}\rfloor-2\lfloor\frac{k}{4}\rfloor+4+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{3}\rfloor-1\leq i\leq \lfloor\frac{n}{3}\rfloor , }\\ 2\lfloor\frac{n}{3}\rfloor-k-2\lfloor\frac{k}{4}\rfloor+\\+15+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{3}\rfloor+1\leq i\leq \lfloor\frac{n}{3}\rfloor+2 ,}\\ 3\lfloor\frac{n}{3}\rfloor-2k+18+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{3}\rfloor+3\leq i\leq 2\lfloor\frac{n}{3}\rfloor ,}\\ \lfloor\frac{n}{3}\rfloor-k-2\lfloor\frac{k}{8}\rfloor+\\+11+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2\lfloor\frac{n}{3}\rfloor+1\leq i\leq2\lfloor\frac{n}{3}\rfloor+2 ,}\\ 3\lfloor\frac{n}{3}\rfloor-4\lfloor\frac{k}{8}\rfloor+3,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ i = 2\lfloor\frac{n}{3}\rfloor+3 ,}\\ 3\lfloor\frac{n}{3}\rfloor-2\lfloor\frac{k}{8}\rfloor-10+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2\lfloor\frac{n}{3}\rfloor+4\leq i\leq n . } \end{array} \right. \end{align*}$

$\begin{align*} f(x_{i}x_{i+3}) = \left\{ \begin{array}{l} 2\lfloor\frac{n}{3}\rfloor-3+i, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1\leq i\leq \lfloor\frac{n}{3}\rfloor-3 \;\text{and}\; \lfloor\frac{n}{3}\rfloor\geq4 , }\\ 2\lfloor\frac{n}{3}\rfloor-2\lfloor\frac{k}{4}\rfloor+7+i,{\ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{3}\rfloor-2\leq i\leq \lfloor\frac{n}{3}\rfloor-1 , }\\ 3\lfloor\frac{n}{3}\rfloor-k+9,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ i = \lfloor\frac{n}{3}\rfloor , }\\ 2\lfloor\frac{n}{3}\rfloor-k-2\lfloor\frac{k}{4}\rfloor+\\+17+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{3}\rfloor+1\leq i\leq \lfloor\frac{n}{3}\rfloor+2 ,}\\ 4\lfloor\frac{n}{3}\rfloor-2k+16+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{3}\rfloor+3\leq i\leq 2\lfloor\frac{n}{3}\rfloor-1 ,}\\ \lfloor\frac{n}{3}\rfloor-k-2\lfloor\frac{k}{8}\rfloor+\\+14+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2\lfloor\frac{n}{3}\rfloor\leq i\leq2\lfloor\frac{n}{3}\rfloor+2 ,}\\ \lfloor\frac{n}{3}\rfloor-2\lfloor\frac{k}{8}\rfloor-5+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ 2\lfloor\frac{n}{3}\rfloor+3\leq i\leq n ,}\\ \end{array} \right. \end{align*}$

Case 2. If $\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor\geq3$

For $1\leq i\leq \lfloor\frac{n}{2}\rfloor$

$\begin{align*} f(x_{i}x_{i+1}) = \left\{ \begin{array}{l} i, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1\leq i\leq \lfloor\frac{n}{3}\rfloor-1 , }\\ 3n-3\lfloor\frac{n}{2}\rfloor-\\-2\lfloor\frac{k}{4}\rfloor-7,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ i = \lfloor\frac{n}{3}\rfloor , }\\ 3n-3\lfloor\frac{n}{2}\rfloor-\\-\lfloor\frac{n}{3}\rfloor-4\lfloor\frac{k}{4}\rfloor-4+i,{\ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{3}\rfloor+1\leq i\leq\lfloor\frac{n}{2}\rfloor-1 ,}\\ 3n-3\lfloor\frac{n}{3}\rfloor-\\k-2\lfloor\frac{k}{4}\rfloor-1,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ i = \lfloor\frac{n}{2}\rfloor .} \end{array} \right. \end{align*}$

For $\lfloor\frac{n}{2}\rfloor+1\leq i\leq n$

$\begin{align*} f(x_{i}x_{i+1}) = \left\{ \begin{array}{l} 3n-3\lfloor\frac{n}{3}\rfloor-\\-\lfloor\frac{n}{2}\rfloor-2k+6+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor+1\leq i\leq \lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor-1 ,}\\ 3n-3\lfloor\frac{n}{3}\rfloor-\\-k-2\lfloor\frac{k}{8}\rfloor-5,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ i = \lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor ,}\\ 2\lfloor\frac{n}{3}\rfloor-\lfloor\frac{n}{2}\rfloor-\\-4\lfloor\frac{k}{8}\rfloor+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor+1\leq i\leq 2\lfloor\frac{n}{2}\rfloor ,}\\ 3\lfloor\frac{n}{3}\rfloor-2\lfloor\frac{k}{8}\rfloor-5,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ i = n . } \end{array} \right. \end{align*}$

$\begin{align*} f(x_{i}x_{i+2}) = \left\{ \begin{array}{l} \lfloor\frac{n}{3}\rfloor-1+i, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1\leq i\leq \lfloor\frac{n}{3}\rfloor-2 , }\\ 3n-3\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor-\\-2\lfloor\frac{k}{4}\rfloor-5+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{3}\rfloor-1\leq i\leq\lfloor\frac{n}{3}\rfloor , }\\ 3n-2\lfloor\frac{n}{2}\rfloor-2\lfloor\frac{n}{3}\rfloor-\\-4\lfloor\frac{k}{4}\rfloor-5+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{3}\rfloor+1\leq i\leq\lfloor\frac{n}{2}\rfloor-2 ,}\\ 3n-3\lfloor\frac{n}{3}\rfloor-\lfloor\frac{n}{2}\rfloor-\\-k-2\lfloor\frac{k}{4}\rfloor+1+i,{\ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor-1\leq i\leq\lfloor\frac{n}{2}\rfloor ,}\\ 3n-2\lfloor\frac{n}{3}\rfloor-\lfloor\frac{n}{2}\rfloor-\\-2k+5+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor+1\leq i\leq \lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor-2 ,}\\ 3n-4\lfloor\frac{n}{3}\rfloor-\lfloor\frac{n}{2}\rfloor-\\-k-2\lfloor\frac{k}{8}\rfloor-3+i,{\ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor-1\leq i\leq\lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor ,}\\ \lfloor\frac{n}{3}\rfloor-4\lfloor\frac{k}{8}\rfloor+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor+1\leq i\leq 2\lfloor\frac{n}{2}\rfloor-1 ,}\\ 3\lfloor\frac{n}{3}\rfloor-2\lfloor\frac{n}{2}\rfloor-\\-2\lfloor\frac{k}{8}\rfloor-4+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2\lfloor\frac{n}{2}\rfloor\leq i\leq n . } \end{array} \right. \end{align*}$

$\begin{align*} f(x_{i}x_{i+3}) = \left\{ \begin{array}{l} 2\lfloor\frac{n}{3}\rfloor-3+i, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1\leq i\leq \lfloor\frac{n}{3}\rfloor-3 , }\\ 3n-3\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor-\\-2\lfloor\frac{k}{4}\rfloor-2+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{3}\rfloor-2\leq i\leq\lfloor\frac{n}{3}\rfloor , }\\ 3n-\lfloor\frac{n}{2}\rfloor-3\lfloor\frac{n}{3}\rfloor-\\-4\lfloor\frac{k}{4}\rfloor-7+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{3}\rfloor+1\leq i\leq\lfloor\frac{n}{2}\rfloor-3 }\;\text{and} \\ {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor\geq4,\\ 3n-3\lfloor\frac{n}{3}\rfloor-\lfloor\frac{n}{2}\rfloor-\\-k-2\lfloor\frac{k}{4}\rfloor+4+i,{\ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor-2\leq i\leq\lfloor\frac{n}{2}\rfloor ,}\\ 3n-\lfloor\frac{n}{3}\rfloor-\lfloor\frac{n}{2}\rfloor-\\-2k+3+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor+1\leq i\leq \lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor-3 ,}\\ 3n-4\lfloor\frac{n}{3}\rfloor-\lfloor\frac{n}{2}\rfloor-\\-k-2\lfloor\frac{k}{8}\rfloor+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor-2\leq i\leq\lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor ,}\\ 2\lfloor\frac{n}{2}\rfloor-2\lfloor\frac{n}{3}\rfloor-\\-4\lfloor\frac{k}{8}\rfloor-1+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor+1\leq i\leq 2\lfloor\frac{n}{2}\rfloor-2 ,}\\ n+3\lfloor\frac{n}{3}\rfloor-4\lfloor\frac{n}{2}\rfloor-\\-2\lfloor\frac{k}{8}\rfloor-2+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2\lfloor\frac{n}{2}\rfloor-1\leq i\leq n . } \end{array} \right. \end{align*}$

Obviously $f$ is $k$ -labeling and the vertices are labeled with even numbers. For the edge weights of $x_{i}x_{i+j}, 1\leq i\leq n, 1\leq j\leq 3$ in $C_{n}(1, 2, 3)$ under the labeling $f$ we have:

In Case 1, the edge set of $C_{n}(1, 2, 3)$ can be split into the nine mutually disjoint subsets, $A_{\ell}, 1\leq \ell \leq 9$ as follows:

1. $A_{1} = \{x_{i}x_{i+j}:1\leq i \leq \lfloor\frac{n}{3}\rfloor-j, 1\leq j \leq3 \}$ the set of all edges which has end vertices labeled with 0.

2. $A_{2} = \{x_{i}x_{i+j}:\lfloor\frac{n}{3}\rfloor-j+1\leq i\leq \lfloor\frac{n}{3}\rfloor, 1\leq j \leq2 \}$ the set of all edges which has end vertices labeled with 0 and $2\lfloor\frac{k}{4}\rfloor+2$ .

3. $A_{3} = \{x_{\lfloor\frac{n}{3}\rfloor+1}x_{\lfloor\frac{n}{3}\rfloor+2} \}$ the set of only one edge which has end vertices labeled with $2\lfloor\frac{k}{4}\rfloor+2$ ,

4. $A_{4} = \{x_{i}x_{i+j}:\lfloor\frac{n}{3}\rfloor+3-j\leq i\leq \lfloor\frac{n}{3}\rfloor+2, 1\leq j \leq2 \}$ the set of all edges which has end vertices labeled with $2\lfloor\frac{k}{4}\rfloor+2$ and $2\lfloor\frac{k}{2}\rfloor$ .

5. $A_{5} = \{x_{i}x_{i+j}:\lfloor\frac{n}{3}\rfloor+3\leq i\leq 2\lfloor\frac{n}{3}\rfloor, 1\leq j \leq2 \}$ the set of all edges which has end vertices labeled with $2\lfloor\frac{k}{2}\rfloor$ .

6. $A_{6} = \{x_{i}x_{i+j}:2\lfloor\frac{n}{3}\rfloor+3-j\leq i\leq 2\lfloor\frac{n}{3}\rfloor+2, 1\leq j \leq 3 \}$ the set of all edges which has end vertices labeled with $2\lfloor\frac{k}{2}\rfloor$ and $2\lfloor\frac{k}{8}\rfloor$ .

7. $A_{7} = \{x_{i}x_{i+j}:2\lfloor\frac{n}{3}\rfloor+3\leq i\leq 2\lfloor\frac{n}{3}\rfloor+5-j, 1\leq j \leq2 \}$ the set of all edges which has end vertices labeled with $2\lfloor\frac{k}{8}\rfloor$ .

8. $A_{8} = \{x_{i}x_{i+j}:2\lfloor\frac{n}{3}\rfloor+6-j\leq i\leq 2\lfloor\frac{n}{3}\rfloor+5, 1\leq j \leq 3 \}$ the set of all edges which has end vertices labeled with 0 and $2\lfloor\frac{k}{8}\rfloor$ .

9. $A_{9} = \{x_{\lfloor\frac{n}{3}\rfloor}x_{\lfloor\frac{n}{3}\rfloor+3} \}$ the set of only edge which has end vertices labeled with 0 and $2\lfloor\frac{k}{2}\rfloor$ .

Thus, we obtain the edge weights as follows:

1. The weights of the edges of the set $A_{1}$ , admit the successive integers from the set $A_{1} = \{1, 2, ..., 3\lfloor\frac{n}{3}\rfloor-6 \}$ .

2. The weights of the edges of the set $A_{2}$ , admit the successive integers from the set $A_{2} = \{3\lfloor\frac{n}{3}\rfloor+4, ..., 3\lfloor\frac{n}{3}\rfloor+8 \}$

3. The weight of the edge of the set $A_{3}$ , admits the integer $\{3\lfloor\frac{n}{3}\rfloor+10 \}$ ,

4. The weights of the edges of the set $A_{4}$ , admit the successive integers from the set $A_{4} = \{3\lfloor\frac{n}{3}\rfloor+17, ..., 2\lfloor\frac{n}{3}\rfloor+21 \}$

5. The weights of the edges of the set $A_{5}$ , admit the successive integers from the set $A_{5} = \{3\lfloor\frac{n}{3}\rfloor+22, ..., 6\lfloor\frac{n}{3}\rfloor+15 \}$

6. The weights of the edges of the set $A_{6}$ , admit the successive integers from the set $A_{6} = \{3\lfloor\frac{n}{3}\rfloor+11, ..., 3\lfloor\frac{n}{3}\rfloor+16 \}$

7. The weights of the edges of the set $A_{7}$ , admit the successive three integers from the set $A_{7} = \{3\lfloor\frac{n}{3}\rfloor+1, 3\lfloor\frac{n}{3}\rfloor+2, 2\lfloor\frac{n}{3}\rfloor+3 \}$

8. The weights of the edges of the set $A_{8}$ , admit the successive integers from the set $A_{8} = \{3\lfloor\frac{n}{3}\rfloor-5, ..., 3\lfloor\frac{n}{3}\rfloor \}$

9. The weight of the edge of the set $A_{9}$ , admit the integer $\{ 3\lfloor\frac{n}{3}\rfloor+9 \}$ .

In Case 2, the set of edges can be divided into eight mutually disjoint subsets as in the proof of Theorem 1.1. Therefore, the edge weights under the labeling $f$ are the following:

1. The weights of the edges of the set $A_{1}$ , admit the successive integers from the set $A_{1} = \{1, 2, ..., 3\lfloor\frac{n}{3}\rfloor-6 \}$ .

2. The weights of the edges of the set $A_{2}$ , admit the successive integers from the set $A_{2} = \{3n-3\lfloor\frac{n}{2}\rfloor-5, ..., 3n-3\lfloor\frac{n}{2}\rfloor \}$

3. The weights of the edges of the set $A_{3}$ , admit the successive integers from the set $\{3n-3\lfloor\frac{n}{2}\rfloor+1, ..., 3n-3\lfloor\frac{n}{3}\rfloor-6 \}$ ,

4. The weights of the edges of the set $A_{4}$ , admit the successive integers from the set $A_{4} = \{3n-3\lfloor\frac{n}{3}\rfloor+1, ..., 3n-3\lfloor\frac{n}{3}\rfloor+6 \}$

5. The weights of the edges of the set $A_{5}$ , admit the successive integers from the set $A_{5} = \{3n-3\lfloor\frac{n}{3}\rfloor+7, ..., 3n \}$

6. The weights of the edges of the set $A_{6}$ , admit the successive integers from the set $A_{6} = \{3n-3\lfloor\frac{n}{3}\rfloor-5, ..., 3n-3\lfloor\frac{n}{3}\rfloor \}$

7. The weights of the edges of the set $A_{7}$ , admit the successive integers from the set $A_{7} = \{3\lfloor\frac{n}{3}\rfloor+1, ..., 3n-3\lfloor\frac{n}{2}\rfloor-6 \}$

8. The weights of the edges of the set $A_{8}$ , admit the successive integers from the set $A_{8} = \{3\lfloor\frac{n}{3}\rfloor-5, ..., 3\lfloor\frac{n}{3}\rfloor \}$

Moreover, for $n$ is odd and $\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor = 2$ the edge weights are from the set $W = \{1, 2, 3, ..., 6\lfloor\frac{n}{3}\rfloor+15\}$ and for $\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor\geq3$ , the edge weights are from the set $W = \{1, 2, 3, ..., 3n\}$ . It is not difficult to see that all numbers in set $W$ are different.

Theorem 3. Let $C_{n}(1, 2, 3, 4, 5)$ , be a circulant graph on $n\geq15$ vertices. Then

$\begin{align*} res(C_{n}(1,2,3,4,5)) = \left\{ \begin{array}{l} \lceil\frac{5n}{3}\rceil, {{\ \ \ \ \ \ \ \ \ \ \ n\equiv 0,1,2,5 \; \mathit{(mod \;6),}}}\\ \lceil\frac{5n}{3}\rceil+1, {{\ \ \ \ \ \ n\equiv 3,4 \;\mathit{ (mod \;6).}}} \end{array} \right. \end{align*}$

Proof. The corresponding labelings for $C_{n}(1, 2, 3, 4, 5)$ , $n = 10, 11, 12, 13, 14$ , are shown in . From Lemma 1.1, we get the following lower bound for the circulant $C_{n}(1, 2, 3, 4, 5)$ : $res(C_{n}(1, 2, 3, 4, 5))\geq k = \lceil\frac{5n}{3}\rceil$ if $n\equiv 0, 1, 2, 5$ (mod 6) and $res(C_{n}(1, 2, 3, 4, 5))\geq k = \lceil\frac{5n}{3}\rceil+1$ if $n\equiv 3, 4$ (mod 6). Now, we will prove that:

$\begin{align*} res(C_{n}(1,2,3,4,5))\leq\left\{ \begin{array}{l} \lceil\frac{5n}{3}\rceil, { \ \ \ \ \ \ \ \ \ \ \ n\equiv 0,1,2,5 \; \text{(mod 6)}, }\\ \lceil\frac{5n}{3}\rceil+1, { \ \ \ \ \ \ n\equiv 3,4 \; \text{(mod 6)} }. \end{array} \right. \end{align*}$

Figure 4. The reflexive edge irregular

$k$ -labeling of

$C_{n}(1, 2, 3, 4, 5), n = 10, 11, 12, 13, 14$ .

DownLoad: Full-Size Img PowerPoint

For this we construct the $f$ -labeling on $C_{n}(1, 2, 3, 4, 5)$ as follows:

For $n\geq15$ we have the following labelings:

$\begin{align*} f(x_{i}) = \left\{ \begin{array}{l} 0, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1\leq i\leq \lfloor\frac{n}{3}\rfloor , }\\ 2\lfloor\frac{k}{4}\rfloor+4,{\ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{3}\rfloor+1\leq i\leq \lfloor\frac{n}{2}\rfloor , }\\ 2\lfloor\frac{k}{2}\rfloor ,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor+1\leq i\leq\lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor , }\\ 2\lfloor\frac{k}{8}\rfloor, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor+1\leq i\leq n }. \end{array} \right. \end{align*}$

Moreover, for the labeling of the edges $x_{i}x_{i+1}, 1\leq i\leq n$ we distinguish two cases.

Case 1. If $\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor\leq4$ .

Firstly, for $n$ is odd.

$\begin{align*} f(x_{i}x_{i+1}) = \left\{ \begin{array}{l} i, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1\leq i\leq \lfloor\frac{n}{3}\rfloor-1 , }\\ 5\lfloor\frac{n}{2}\rfloor-2\lfloor\frac{k}{4}\rfloor-13,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ i = \lfloor\frac{n}{3}\rfloor , }\\ 4\lfloor\frac{n}{3}\rfloor+\frac{\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor}{2}(\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor+1)-\\-4\lfloor\frac{k}{4}\rfloor+7+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{3}\rfloor+1\leq i\leq\lfloor\frac{n}{2}\rfloor-1 ,}\\ 5\lfloor\frac{n}{3}\rfloor+\frac{\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor}{2}(\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor+9)-\\-2(\lfloor\frac{k}{2}\rfloor+\lfloor\frac{k}{4}\rfloor)+17,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ i = \lfloor\frac{n}{2}\rfloor .} \end{array} \right. \end{align*}$

For $\lfloor\frac{n}{2}\rfloor+1\leq i\leq n.$

$\begin{align*} f(x_{i}x_{i+1}) = \left\{ \begin{array}{l} 9\lfloor\frac{n}{2}\rfloor-5\lfloor\frac{n}{3}\rfloor-4\lfloor\frac{k}{2}\rfloor+20+i,{\ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor+1\leq i\leq \lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor-1 ,}\\ 5\lfloor\frac{n}{3}\rfloor+(\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor)^{2}-\\-2(\lfloor\frac{k}{2}\rfloor+\lfloor\frac{k}{8}\rfloor)+16,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ i = \lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor ,}\\ 3\lfloor\frac{n}{3}\rfloor-\frac{\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor}{2}(\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor-7)-\\-4\lfloor\frac{k}{8}\rfloor-10+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor+1\leq i\leq 2\lfloor\frac{n}{2}\rfloor ,}\\ 5\lfloor\frac{n}{3}\rfloor-2\lfloor\frac{k}{8}\rfloor-14,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ i = n . } \end{array} \right. \end{align*}$

$\begin{align*} f(x_{i}x_{i+2}) = \left\{ \begin{array}{l} \lfloor\frac{n}{3}\rfloor-1+i, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1\leq i\leq \lfloor\frac{n}{3}\rfloor-2 , }\\ 5\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor-2\lfloor\frac{k}{4}\rfloor-11+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{3}\rfloor-1\leq i\leq\lfloor\frac{n}{3}\rfloor . } \end{array} \right. \end{align*}$

For $\lfloor\frac{n}{3}\rfloor+1\leq i\leq\lfloor\frac{n}{2}\rfloor$ .

If $\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor = 2$ .

$\begin{align*} f(x_{i}x_{i+2}) = 4\lfloor\frac{n}{3}\rfloor-2(\lfloor\frac{k}{2}\rfloor+\lfloor\frac{k}{4}\rfloor)+28+i. \\ \end{align*}$

If $\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor\geq3$ .

$\begin{align*} f(x_{i}x_{i+2}) = \left\{ \begin{array}{l} 4\lfloor\frac{n}{3}\rfloor+\frac{\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor}{2}(\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor+3)-\\-4\lfloor\frac{k}{4}\rfloor+6+i, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{3}\rfloor+1\leq i\leq \lfloor\frac{n}{2}\rfloor-2 , }\\ 4\lfloor\frac{n}{3}\rfloor+\frac{\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor}{2}(\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor+7)-\\-2(\lfloor\frac{k}{4}\rfloor+\lfloor\frac{k}{2}\rfloor)+19+i, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor-1\leq i\leq \lfloor\frac{n}{2}\rfloor . } \end{array} \right. \end{align*}$

For $\lfloor\frac{n}{2}\rfloor+1\leq i\leq n.$

$\begin{align*} f(x_{i}x_{i+2}) = \left\{ \begin{array}{l} 9\lfloor\frac{n}{2}\rfloor-4\lfloor\frac{n}{3}\rfloor-4\lfloor\frac{k}{2}\rfloor+19+i, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor+1\leq i\leq\lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor-2 , }\\ 4\lfloor\frac{n}{3}\rfloor-\lfloor\frac{n}{2}\rfloor+(\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor)^{2}-\\-2(\lfloor\frac{k}{2}\rfloor+\lfloor\frac{k}{8}\rfloor)+18+i, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor-1\leq i\leq\lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor , }\\ 3\lfloor\frac{n}{3}\rfloor+\frac{\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor}{2}(9-\lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor)-\\-4\lfloor\frac{k}{8}\rfloor-10+i, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor+1\leq i\leq2\lfloor\frac{n}{2}\rfloor-1 , }\\ 5\lfloor\frac{n}{3}\rfloor-2\lfloor\frac{n}{2}\rfloor-2\lfloor\frac{k}{8}\rfloor-13+i, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2\lfloor\frac{n}{2}\rfloor\leq i\leq n . } \end{array} \right. \end{align*}$

$\begin{align*} f(x_{i}x_{i+3}) = 2\lfloor\frac{n}{3}\rfloor-3+i, {\ \ \ \ \ \ \ \ \ \ \ \ \ 1\leq i\leq \lfloor\frac{n}{3}\rfloor-3 . } \end{align*}$

For $\lfloor\frac{n}{3}\rfloor-2\leq i\leq\lfloor\frac{n}{3}\rfloor$ .

If $\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor = 2$ .

$\begin{align*} f(x_{i}x_{i+3}) = \left\{ \begin{array}{l} 4\lfloor\frac{n}{3}\rfloor-2\lfloor\frac{k}{4}\rfloor+2+i, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{3}\rfloor-2\leq i\leq \lfloor\frac{n}{3}\rfloor-1 , }\\ 5\lfloor\frac{n}{3}\rfloor-2\lfloor\frac{k}{2}\rfloor+10, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ i = \lfloor\frac{n}{3}\rfloor . } \end{array} \right. \end{align*}$

If $\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor\geq3$ .

$\begin{align*} f(x_{i}x_{i+3}) = 5\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor-2\lfloor\frac{k}{4}\rfloor-8+i. \end{align*}$

For $\lfloor\frac{n}{3}\rfloor+1\leq i\leq\lfloor\frac{n}{2}\rfloor$ .

If $\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor = 2$ .

$\begin{align*} f(x_{i}x_{i+3}) = 4\lfloor\frac{n}{3}\rfloor-2(\lfloor\frac{k}{2}\rfloor+\lfloor\frac{k}{4}\rfloor)+30+i. \end{align*}$

If $\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor = 3$

$\begin{align*} f(x_{i}x_{i+3}) = 4\lfloor\frac{n}{3}\rfloor-2(\lfloor\frac{k}{2}\rfloor+\lfloor\frac{k}{4}\rfloor)+37+i. \end{align*}$

If $\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor = 4$

$\begin{align*} f(x_{i}x_{i+3}) = \left\{ \begin{array}{l} 4\lfloor\frac{n}{3}\rfloor-4\lfloor\frac{k}{4}\rfloor+22+i, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{3}\rfloor+1\leq i\leq \lfloor\frac{n}{2}\rfloor-3 , }\\ 4\lfloor\frac{n}{3}\rfloor-2(\lfloor\frac{k}{2}\rfloor+\lfloor\frac{k}{4}\rfloor)+44+i, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor-2\leq i\leq \lfloor\frac{n}{2}\rfloor . } \end{array} \right. \end{align*}$

For $\lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor+1\leq i\leq2\lfloor\frac{n}{2}\rfloor+1$ .

If $\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor = 2$

$\begin{align*} f(x_{i}x_{i+3}) = 3\lfloor\frac{n}{3}\rfloor-2\lfloor\frac{k}{8}\rfloor-14+i. \end{align*}$

If $\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor\geq3$ .

$\begin{align*} f(x_{i}x_{i+3}) = \left\{ \begin{array}{l} 3\lfloor\frac{n}{3}\rfloor+\frac{(\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor)}{2}(11-\lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor)-\\-4\lfloor\frac{k}{8}\rfloor-11+i, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor+1\leq i\leq 2\lfloor\frac{n}{2}\rfloor-2 , }\\ 5\lfloor\frac{n}{3}\rfloor-2\lfloor\frac{n}{2}\rfloor-2\lfloor\frac{k}{8}\rfloor-10+i, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2\lfloor\frac{n}{2}\rfloor-1\leq i\leq n . } \end{array} \right. \end{align*}$

$\begin{align*} f(x_{i}x_{i+4}) = 3\lfloor\frac{n}{3}\rfloor-6+i, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1\leq i\leq \lfloor\frac{n}{3}\rfloor-4 . } \end{align*}$

For $\lfloor\frac{n}{3}\rfloor-3\leq i\leq\lfloor\frac{n}{3}\rfloor.$

If $\lfloor\frac{n}{3}\rfloor-\lfloor\frac{n}{2}\rfloor = 2.$

$\begin{align*} f(x_{i}x_{i+4}) = \left\{ \begin{array}{l} 4\lfloor\frac{n}{3}\rfloor-2\lfloor\frac{k}{4}\rfloor+5+i, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{3}\rfloor-3\leq i\leq\lfloor\frac{n}{3}\rfloor-2 , }\\ 4\lfloor\frac{n}{3}\rfloor-2\lfloor\frac{k}{2}\rfloor+12+i, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{3}\rfloor-1\leq i\leq\lfloor\frac{n}{3}\rfloor . } \end{array} \right. \end{align*}$

If $\lfloor\frac{n}{3}\rfloor-\lfloor\frac{n}{2}\rfloor = 3.$

$\begin{align*} f(x_{i}x_{i+4}) = \left\{ \begin{array}{l} 4\lfloor\frac{n}{2}\rfloor-2\lfloor\frac{k}{4}\rfloor-1+i, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{3}\rfloor-3\leq i\leq\lfloor\frac{n}{3}\rfloor-1 , }\\ 5\lfloor\frac{n}{3}\rfloor-2\lfloor\frac{k}{2}\rfloor+18, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ i = \lfloor\frac{n}{3}\rfloor . } \end{array} \right. \end{align*}$

If $\lfloor\frac{n}{3}\rfloor-\lfloor\frac{n}{2}\rfloor = 4.$

$\begin{align*} f(x_{i}x_{i+4}) = 4\lfloor\frac{n}{2}\rfloor-2\lfloor\frac{k}{4}\rfloor+i. \end{align*}$

For $\lfloor\frac{n}{3}\rfloor+1\leq i\leq\lfloor\frac{n}{2}\rfloor.$

$\begin{align*} f(x_{i}x_{i+4}) = \left\{ \begin{array}{l} 4\lfloor\frac{n}{3}\rfloor-2(\lfloor\frac{k}{2}\rfloor+\lfloor\frac{k}{4}\rfloor)+32+i, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{if}\; \lfloor\frac{n}{3}\rfloor-\lfloor\frac{n}{2}\rfloor = 2 , }\\ 4\lfloor\frac{n}{3}\rfloor-2(\lfloor\frac{k}{2}\rfloor+\lfloor\frac{k}{4}\rfloor)+40+i, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{if}\; \lfloor\frac{n}{3}\rfloor-\lfloor\frac{n}{2}\rfloor = 3 , }\\ 4\lfloor\frac{n}{3}\rfloor-2(\lfloor\frac{k}{2}\rfloor+\lfloor\frac{k}{4}\rfloor)+48+i, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{if}\; \lfloor\frac{n}{3}\rfloor-\lfloor\frac{n}{2}\rfloor = 4 . } \end{array} \right. \end{align*}$

For $\lfloor\frac{n}{2}\rfloor+1\leq i\leq\lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor$ .

$\begin{align*} f(x_{i}x_{i+4}) = 9\lfloor\frac{n}{2}\rfloor-2\lfloor\frac{n}{3}\rfloor-4\lfloor\frac{k}{2}\rfloor+14+i,{\ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor+1\leq i\leq\lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor-4 . } \end{align*}$

If $\lfloor\frac{n}{3}\rfloor-\lfloor\frac{n}{2}\rfloor = 2.$

$\begin{align*} f(x_{i}x_{i+4}) = \left\{ \begin{array}{l} 3\lfloor\frac{n}{3}\rfloor-2(\lfloor\frac{k}{2}\rfloor+\lfloor\frac{k}{8}\rfloor)+27+i, {\ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor-3\leq i\leq\lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor , }\\ 5\lfloor\frac{n}{3}\rfloor-2\lfloor\frac{k}{2}\rfloor+13, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ i = \lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor . } \end{array} \right. \end{align*}$

If $\lfloor\frac{n}{3}\rfloor-\lfloor\frac{n}{2}\rfloor\geq3.$

$\begin{align*} &f(x_{i}x_{i+4}) = 4\lfloor\frac{n}{3}\rfloor-\lfloor\frac{n}{2}\rfloor+(\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor)^{2}-\\& \ \ \ \ \ \ \ \ \ \ \ \ \ \ -2(\lfloor\frac{k}{2}\rfloor+\lfloor\frac{k}{8}\rfloor)+25+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor-3\leq i\leq\lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor . } \end{align*}$

For $\lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor+1\leq i\leq n$ .

If $\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor = 2$ .

$\begin{align*} f(x_{i}x_{i+4}) = 3\lfloor\frac{n}{3}\rfloor-2\lfloor\frac{k}{8}\rfloor-11+i. \end{align*}$

If $\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor = 3$ .

$\begin{align*} f(x_{i}x_{i+4}) = 3\lfloor\frac{n}{3}\rfloor-2\lfloor\frac{k}{8}\rfloor-12+i. \end{align*}$

If $\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor = 4$ .

$\begin{align*} f(x_{i}x_{i+4}) = \left\{ \begin{array}{l} 3\lfloor\frac{n}{3}\rfloor-4\lfloor\frac{k}{8}\rfloor+5+i, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor+1\leq i\leq2\lfloor\frac{n}{2}\rfloor-3 , }\\ 3\lfloor\frac{n}{3}\rfloor-2\lfloor\frac{k}{8}\rfloor-14+i, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2\lfloor\frac{n}{2}\rfloor-2\leq i\leq n . } \end{array} \right. \end{align*}$

For $1\leq i\leq\lfloor\frac{n}{3}\rfloor$ .

If $\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor = 2$ .

$\begin{align*} f(x_{i}x_{i+5}) = \left\{ \begin{array}{l} 5\lfloor\frac{n}{3}\rfloor-2\lfloor\frac{k}{4}\rfloor+3+i, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1\leq i\leq2 , }\\ 5\lfloor\frac{n}{3}\rfloor-2\lfloor\frac{k}{2}\rfloor+11+i, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3\leq i\leq\lfloor\frac{n}{3}\rfloor . } \end{array} \right. \end{align*}$

If $\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor = 3$ .

$\begin{align*} f(x_{i}x_{i+5}) = \left\{ \begin{array}{l} 4\lfloor\frac{n}{3}\rfloor-10+i, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1\leq i\leq\lfloor\frac{n}{3}\rfloor-5 , \;\text{and}\; \lfloor\frac{n}{3}\rfloor\geq6 , }\\ 4\lfloor\frac{n}{2}\rfloor-2\lfloor\frac{k}{4}\rfloor+3+i, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{3}\rfloor-4\leq i\leq\lfloor\frac{n}{3}\rfloor-2 , }\\ 4\lfloor\frac{n}{3}\rfloor-2\lfloor\frac{k}{2}\rfloor+20+i, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{3}\rfloor-1\leq i\leq\lfloor\frac{n}{3}\rfloor . } \end{array} \right. \end{align*}$

If $\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor = 4$ .

$\begin{align*} f(x_{i}x_{i+5}) = \left\{ \begin{array}{l} 4\lfloor\frac{n}{3}\rfloor-10+i, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1\leq i\leq\lfloor\frac{n}{3}\rfloor-5 , }\\ 4\lfloor\frac{n}{2}\rfloor-2\lfloor\frac{k}{4}\rfloor+5+i, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{3}\rfloor-4\leq i\leq\lfloor\frac{n}{3}\rfloor-1 , }\\ 5\lfloor\frac{n}{3}\rfloor-2\lfloor\frac{k}{2}\rfloor+25, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ i = \lfloor\frac{n}{3}\rfloor . } \end{array} \right. \end{align*}$

For $\lfloor\frac{n}{3}\rfloor+1\leq i\leq\lfloor\frac{n}{2}\rfloor$ .

$\begin{align*} f(x_{i}x_{i+5}) = \left\{ \begin{array}{l} 4\lfloor\frac{n}{3}\rfloor-2(\lfloor\frac{k}{2}\rfloor+\lfloor\frac{k}{4}\rfloor)+34+i, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{if}\; \lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor = 2 , }\\ 4\lfloor\frac{n}{3}\rfloor-2(\lfloor\frac{k}{2}\rfloor+\lfloor\frac{k}{4}\rfloor)+43+i, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{if}\; \lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor = 3 , }\\ 4\lfloor\frac{n}{3}\rfloor-2(\lfloor\frac{k}{2}\rfloor+\lfloor\frac{k}{4}\rfloor)+52+i, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{if}\; \lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor = 4 . }\\ \end{array} \right. \end{align*}$

For $\lfloor\frac{n}{2}\rfloor+1\leq i\leq\lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor$ .

If $\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor = 2$

$\begin{align*} f(x_{i}x_{i+5}) = \left\{ \begin{array}{l} 4\lfloor\frac{n}{3}\rfloor-2(\lfloor\frac{k}{2}\rfloor+\lfloor\frac{k}{8}\rfloor)+26+i, {\ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor+1\leq i\leq\lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor-2 , }\\ 4\lfloor\frac{n}{3}\rfloor-2\lfloor\frac{k}{2}\rfloor+11+i, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor-1\leq i\leq\lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor . } \end{array} \right. \end{align*}$

If $\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor = 3$

$\begin{align*} f(x_{i}x_{i+5}) = \left\{ \begin{array}{l} 8\lfloor\frac{n}{3}\rfloor-4\lfloor\frac{k}{2}\rfloor+\\+37+i, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor+1\leq i\leq\lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor-5 \;\text{and}\; \lfloor\frac{n}{3}\rfloor\geq6 , }\\ 3\lfloor\frac{n}{3}\rfloor-2\lfloor\frac{k}{2}\rfloor-\\-2\lfloor\frac{k}{8}\rfloor+36+i, {\ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor-4\leq i\leq\lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor-1 , }\\ 5\lfloor\frac{n}{3}\rfloor-2\lfloor\frac{k}{2}\rfloor+21, {\ \ \ \ \ \ \ \ \ \ i = \lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor . } \end{array} \right. \end{align*}$

If $\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor = 4$

$\begin{align*} f(x_{i}x_{i+5}) = \left\{ \begin{array}{l} 8\lfloor\frac{n}{3}\rfloor-4\lfloor\frac{k}{2}\rfloor+46+i, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor+1\leq i\leq\lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor-5 , }\\ 3\lfloor\frac{n}{3}\rfloor-2(\lfloor\frac{k}{2}\rfloor+\lfloor\frac{k}{8}\rfloor)+42+i, {\ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor-4\leq i\leq\lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor . } \end{array} \right. \end{align*}$

For $\lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor+1\leq i\leq n$ .

$\begin{align*} f(x_{i}x_{i+5}) = \left\{ \begin{array}{l} 3\lfloor\frac{n}{3}\rfloor-2\lfloor\frac{k}{8}\rfloor-8+i, {\ \ \ \ \ \ \ \ \ \ \ \ \ \text{if}\; \lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor = 2,3 , }\\ 3\lfloor\frac{n}{3}\rfloor-2\lfloor\frac{k}{8}\rfloor-9+i, {\ \ \ \ \ \ \ \ \ \ \ \ \text{if}\; \lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor = 4 . } \end{array} \right. \end{align*}$

Secondly, for $n$ is even.

$\begin{align*} f(x_{i}x_{i+1}) = \left\{ \begin{array}{l} i, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1\leq i\leq \lfloor\frac{n}{3}\rfloor-1 , }\\ \frac{5n}{2}-2\lfloor\frac{k}{4}\rfloor-18,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ i = \lfloor\frac{n}{3}\rfloor , }\\ 4\lfloor\frac{n}{3}\rfloor-\frac{\frac{n}{2}-\lfloor\frac{n}{3}\rfloor}{2}(1-\frac{n}{2}+\lfloor\frac{n}{3}\rfloor)-\\-4\lfloor\frac{k}{4}\rfloor+7+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{3}\rfloor+1\leq i\leq \frac{n}{2}-1 ,}\\ 5\lfloor\frac{n}{3}\rfloor+\frac{\frac{n}{2}-\lfloor\frac{n}{3}\rfloor}{2}(\frac{n}{2}-\lfloor\frac{n}{3}\rfloor+9)-\\-2(\lfloor\frac{k}{2}\rfloor+\lfloor\frac{k}{4}\rfloor)+12,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ i = \frac{n}{2} ,}\\ \frac{9n}{2}-5\lfloor\frac{n}{3}\rfloor-4\lfloor\frac{k}{2}\rfloor+15+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{n}{2}+1\leq i\leq \frac{n}{2}+\lfloor\frac{n}{3}\rfloor-1 ,}\\ 6\lfloor\frac{n}{3}\rfloor-\frac{n}{2}+(\frac{n}{2}-\lfloor\frac{n}{3}\rfloor)^{2}-\\-2(\lfloor\frac{k}{2}\rfloor+\lfloor\frac{k}{8}\rfloor)+16,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ i = \frac{n}{2}+\lfloor\frac{n}{3}\rfloor ,}\\ 3\lfloor\frac{n}{3}\rfloor+\frac{\frac{n}{2}-\lfloor\frac{n}{3}\rfloor}{2}(9-\frac{n}{2}+\lfloor\frac{n}{3}\rfloor)-\\-4\lfloor\frac{k}{8}\rfloor-15+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{n}{2}+\lfloor\frac{n}{3}\rfloor+1\leq i\leq n-1 ,}\\ 5\lfloor\frac{n}{3}\rfloor-2\lfloor\frac{k}{8}\rfloor-14,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ i = n . } \end{array} \right. \end{align*}$

$\begin{align*} f(x_{i}x_{i+2}) = \left\{ \begin{array}{l} \lfloor\frac{n}{3}\rfloor-1+i, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1\leq i\leq\lfloor\frac{n}{3}\rfloor-2 , }\\ \frac{5n}{2}-\lfloor\frac{n}{3}\rfloor-2\lfloor\frac{k}{4}\rfloor-16+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{3}\rfloor-1\leq i\leq\lfloor\frac{n}{3}\rfloor , }\\ 4\lfloor\frac{n}{3}\rfloor+\frac{\frac{n}{2}-\lfloor\frac{n}{3}\rfloor}{2}(\frac{n}{2}-\lfloor\frac{n}{3}\rfloor +1)-\\-4\lfloor\frac{k}{4}\rfloor+6+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{3}\rfloor+1\leq i\leq\frac{n}{2}-2 ,}\\ 4\lfloor\frac{n}{3}\rfloor+\frac{\frac{n}{2}-\lfloor\frac{n}{3}\rfloor}{2}(\frac{n}{2}-\lfloor\frac{n}{3}\rfloor+7)-\\-2(\lfloor\frac{k}{2}\rfloor+\lfloor\frac{k}{4}\rfloor)+14+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{n}{2}-1\leq i\leq\frac{n}{2} ,}\\ \frac{9n}{2}-4\lfloor\frac{n}{3}\rfloor-4\lfloor\frac{k}{2}\rfloor+14+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{n}{2}+1\leq i\leq \frac{n}{2}+\lfloor\frac{n}{3}\rfloor-2 ,}\\ 5\lfloor\frac{n}{3}\rfloor-n+(\frac{n}{2}-\lfloor\frac{n}{3}\rfloor)^{2}-\\-2(\lfloor\frac{k}{2}\rfloor+\lfloor\frac{k}{8}\rfloor)+18+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{n}{2}+\lfloor\frac{n}{3}\rfloor-1\leq i\leq\frac{n}{2}+\lfloor\frac{n}{3}\rfloor ,}\\ 3\lfloor\frac{n}{3}\rfloor+\frac{\frac{n}{2}-\lfloor\frac{n}{3}\rfloor}{2}(11-\frac{n}{2}+\lfloor\frac{n}{3}\rfloor)-\\-4\lfloor\frac{k}{8}\rfloor-16+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{n}{2}+\lfloor\frac{n}{3}\rfloor+1\leq i\leq n-2 ,}\\ 5\lfloor\frac{n}{3}\rfloor-n-2\lfloor\frac{k}{8}\rfloor-11+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ n-2\leq i\leq n . } \end{array} \right. \end{align*}$

For $1\leq i\leq\lfloor\frac{n}{3}\rfloor$ .

$\begin{align*} f(x_{i}x_{i+3}) = \left\{ \begin{array}{l} 2\lfloor\frac{n}{3}\rfloor-3+i, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1\leq i\leq\lfloor\frac{n}{3}\rfloor-3 , }\\ \frac{5n}{2}-\lfloor\frac{n}{3}\rfloor-2\lfloor\frac{k}{4}\rfloor-13+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{3}\rfloor-2\leq i\leq\lfloor\frac{n}{3}\rfloor . } \end{array} \right. \end{align*}$

For $\lfloor\frac{n}{3}\rfloor+1\leq i\leq\frac{n}{2}$ .

If $\frac{n}{2}-\lfloor\frac{n}{3}\rfloor = 3$ .

$\begin{align*} f(x_{i}x_{i+3}) = 4\lfloor\frac{n}{3}\rfloor-2(\lfloor\frac{k}{4}\rfloor+\lfloor\frac{k}{2}\rfloor)+32+i. \end{align*}$

If $\frac{n}{2}-\lfloor\frac{n}{3}\rfloor = 4$ .

$\begin{align*} f(x_{i}x_{i+3}) = \left\{ \begin{array}{l} 4\lfloor\frac{n}{3}\rfloor-4\lfloor\frac{k}{4}\rfloor+18+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{3}\rfloor+1\leq i\leq\frac{n}{2}-3 }\\ 4\lfloor\frac{n}{3}\rfloor-2(\lfloor\frac{k}{4}\rfloor+\lfloor\frac{k}{2}\rfloor)+39+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor-2\leq i\leq\frac{n}{2} }. \end{array} \right. \end{align*}$

For $\frac{n}{2}+1\leq i\leq\frac{n}{2}+\lfloor\frac{n}{3}\rfloor$ .

$\begin{align*} f(x_{i}x_{i+3}) = \left\{ \begin{array}{l} \frac{9n}{2}-3\lfloor\frac{n}{3}\rfloor-4\lfloor\frac{k}{2}\rfloor+12+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{n}{2}+1\leq i\leq\frac{n}{2}+\lfloor\frac{n}{3}\rfloor-3 }\\ 5\lfloor\frac{n}{3}\rfloor-n+(\frac{n}{2}-\lfloor\frac{n}{3}\rfloor)^{2}-\\-2(\lfloor\frac{k}{2}\rfloor+\lfloor\frac{k}{8}\rfloor)+21+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{n}{2}+\lfloor\frac{n}{3}\rfloor-2\leq i\leq\frac{n}{2}+\lfloor\frac{n}{3}\rfloor }. \end{array} \right. \end{align*}$

For $\frac{n}{2}+\lfloor\frac{n}{3}\rfloor+1\leq i\leq n$ .

If $\frac{n}{2}-\lfloor\frac{n}{3}\rfloor = 3$ .

$\begin{align*} f(x_{i}x_{i+3}) = 3\lfloor\frac{n}{3}\rfloor-2\lfloor\frac{k}{8}\rfloor-15+i. \end{align*}$

If $\frac{n}{2}-\lfloor\frac{n}{3}\rfloor = 4$ .

$\begin{align*} f(x_{i}x_{i+3}) = \left\{ \begin{array}{l} 3\lfloor\frac{n}{3}\rfloor-4\lfloor\frac{k}{8}\rfloor+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor+1\leq i\leq n-3 }\\ 3\lfloor\frac{n}{3}\rfloor-2\lfloor\frac{k}{8}\rfloor-17+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ n-2\leq i\leq n }. \end{array} \right. \end{align*}$

For $1\leq i\leq\lfloor\frac{n}{3}\rfloor$ .

$\begin{align*} f(x_{i}x_{i+4}) = 3\lfloor\frac{n}{3}\rfloor-6+i,{\ \ \ \ \ \ \ \ \ \ \ 1\leq i\leq\lfloor\frac{n}{3}\rfloor-4 . } \end{align*}$

If $\frac{n}{2}-\lfloor\frac{n}{3}\rfloor = 3$ .

$\begin{align*} f(x_{i}x_{i+4}) = \left\{ \begin{array}{l} 4\lfloor\frac{n}{3}\rfloor-2\lfloor\frac{k}{4}\rfloor+6+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{3}\rfloor-3\leq i\leq\lfloor\frac{n}{3}\rfloor-1 }\\ 5\lfloor\frac{n}{3}\rfloor-2\lfloor\frac{k}{2}\rfloor+13,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ i = \lfloor\frac{n}{3}\rfloor . } \end{array} \right. \end{align*}$

If $\frac{n}{2}-\lfloor\frac{n}{3}\rfloor = 4$ .

$\begin{align*} f(x_{i}x_{i+4}) = 4\lfloor\frac{n}{3}\rfloor-2\lfloor\frac{k}{4}\rfloor+11+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{3}\rfloor-3\leq i\leq\lfloor\frac{n}{3}\rfloor . } \end{align*}$

For $\lfloor\frac{n}{3}\rfloor+1\leq i\leq\frac{n}{2}+\lfloor\frac{n}{3}\rfloor$ .

$\begin{align*} f(x_{i}x_{i+4}) = \left\{ \begin{array}{l} 4\lfloor\frac{n}{3}\rfloor-2(\lfloor\frac{k}{2}\rfloor+\lfloor\frac{k}{4}\rfloor)+35+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{if}\; \frac{n}{2}-\lfloor\frac{n}{3}\rfloor = 3 }\\ 4\lfloor\frac{n}{3}\rfloor-2(\lfloor\frac{k}{2}\rfloor+\lfloor\frac{k}{4}\rfloor)+43+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{if}\; \frac{n}{2}-\lfloor\frac{n}{3}\rfloor = 4 . } \end{array} \right. \end{align*}$

For $\frac{n}{2}+1\leq i\leq\frac{n}{2}+\lfloor\frac{n}{3}\rfloor$ .

If $\frac{n}{2}-\lfloor\frac{n}{3}\rfloor = 3$ .

$\begin{align*} f(x_{i}x_{i+4}) = \left\{ \begin{array}{l} 7\lfloor\frac{n}{3}\rfloor-4\lfloor\frac{k}{2}\rfloor+36+i,{\ \ \ \ \ \ \ \ \ \ \ \frac{n}{2}+1\leq i\leq\frac{n}{2}+\lfloor\frac{n}{3}\rfloor-4 }\\ 3\lfloor\frac{n}{3}\rfloor-2(\lfloor\frac{k}{2}\rfloor+\lfloor\frac{k}{8}\rfloor)+\\+28+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{n}{2}+\lfloor\frac{n}{3}\rfloor-3\leq i\leq\frac{n}{2}+\lfloor\frac{n}{3}\rfloor-1 }\\ 5\lfloor\frac{n}{3}\rfloor-2\lfloor\frac{k}{2}\rfloor+14,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ i = \frac{n}{2}+\lfloor\frac{n}{3}\rfloor . } \end{array} \right. \end{align*}$

If $\frac{n}{2}-\lfloor\frac{n}{3}\rfloor = 4$ .

$\begin{align*} f(x_{i}x_{i+4}) = \left\{ \begin{array}{l} 7\lfloor\frac{n}{3}\rfloor-4\lfloor\frac{k}{2}\rfloor+45+i,{\ \ \ \ \ \ \ \ \ \ \frac{n}{2}+1\leq i\leq\frac{n}{2}+\lfloor\frac{n}{3}\rfloor-4 }\\ 3\lfloor\frac{n}{3}\rfloor-2(\lfloor\frac{k}{2}\rfloor+\lfloor\frac{k}{8}\rfloor)+\\+33+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{n}{2}+\lfloor\frac{n}{3}\rfloor-3\leq i\leq\frac{n}{2}+\lfloor\frac{n}{3}\rfloor . } \end{array} \right. \end{align*}$

For $\frac{n}{2}+\lfloor\frac{n}{3}\rfloor+1\leq i\leq n$ .

$\begin{align*} f(x_{i}x_{i+4}) = \left\{ \begin{array}{l} 3\lfloor\frac{n}{3}\rfloor-2\lfloor\frac{k}{8}\rfloor-12+i,{\ \ \ \ \ \ \ \ \ \ \ \ \text{if}\; \frac{n}{2}-\lfloor\frac{n}{3}\rfloor = 3 }\\ 3\lfloor\frac{n}{3}\rfloor-2\lfloor\frac{k}{8}\rfloor-13+i,{\ \ \ \ \ \ \ \ \ \ \ \ \text{if}\; \frac{n}{2}-\lfloor\frac{n}{3}\rfloor = 4 . } \end{array} \right. \end{align*}$

For $1\leq i\leq\lfloor\frac{n}{3}\rfloor$ .

If $\frac{n}{2}-\lfloor\frac{n}{3}\rfloor = 3$ .

$\begin{align*} f(x_{i}x_{i+5}) = \left\{ \begin{array}{l} 4\lfloor\frac{n}{3}\rfloor-10+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1\leq i\leq\lfloor\frac{n}{3}\rfloor-5 \;\text{and}\; \lfloor\frac{n}{3}\rfloor\geq6 , }\\ 4\lfloor\frac{n}{3}\rfloor-2\lfloor\frac{k}{4}\rfloor+10+i,{\ \ \ \ \ \ \ \ \ \lfloor\frac{n}{3}\rfloor-4\leq i\leq\lfloor\frac{n}{3}\rfloor-2 , }\\ 4\lfloor\frac{n}{3}\rfloor-2\lfloor\frac{k}{2}\rfloor+16+i,{\ \ \ \ \ \ \ \ \ \lfloor\frac{n}{3}\rfloor-1\leq i\leq\lfloor\frac{n}{3}\rfloor . } \end{array} \right. \end{align*}$

If $\frac{n}{2}-\lfloor\frac{n}{3}\rfloor = 4$ .

$\begin{align*} f(x_{i}x_{i+5}) = \left\{ \begin{array}{l} 4\lfloor\frac{n}{3}\rfloor-10+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1\leq i\leq\lfloor\frac{n}{3}\rfloor-5 , }\\ 4\lfloor\frac{n}{3}\rfloor-2\lfloor\frac{k}{4}\rfloor+16+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{3}\rfloor-4\leq i\leq\lfloor\frac{n}{3}\rfloor-1 , }\\ 5\lfloor\frac{n}{3}\rfloor-2\lfloor\frac{k}{2}\rfloor+20,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ i = \lfloor\frac{n}{3}\rfloor . } \end{array} \right. \end{align*}$

For $\lfloor\frac{n}{3}\rfloor+1\leq i\leq \frac{n}{2}$ .

$\begin{align*} f(x_{i}x_{i+5}) = \left\{ \begin{array}{l} 4\lfloor\frac{n}{3}\rfloor-2(\lfloor\frac{k}{2}\rfloor+\lfloor\frac{k}{4}\rfloor)+38+i,{\ \ \ \ \ \ \ \ \ \ \ \ \text{if}\; \frac{n}{2}-\lfloor\frac{n}{3}\rfloor = 3 }\\ 4\lfloor\frac{n}{3}\rfloor-2(\lfloor\frac{k}{2}\rfloor+\lfloor\frac{k}{4}\rfloor)+47+i,{\ \ \ \ \ \ \ \ \ \ \ \ \text{if}\; \frac{n}{2}-\lfloor\frac{n}{3}\rfloor = 4 . } \end{array} \right. \end{align*}$

For $\frac{n}{2}+1\leq i\leq \frac{n}{2}+\lfloor\frac{n}{3}\rfloor$ .

If $\frac{n}{2}-\lfloor\frac{n}{3}\rfloor = 3$ .

$\begin{align*} f(x_{i}x_{i+5}) = \left\{ \begin{array}{l} 8\lfloor\frac{n}{3}\rfloor-4\lfloor\frac{k}{2}\rfloor+32+i,{\ \ \ \ \ \ \ \ \ \frac{n}{2}+1\leq i\leq\frac{n}{2}+\lfloor\frac{n}{3}\rfloor-5 \;\text{and}\; \lfloor\frac{n}{3}\rfloor\geq6 , }\\ 3\lfloor\frac{n}{3}\rfloor-2(\lfloor\frac{k}{2}\rfloor+\lfloor\frac{k}{8}\rfloor)+\\+32+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{n}{2}+\lfloor\frac{n}{3}\rfloor-4\leq i\leq\frac{n}{2}+\lfloor\frac{n}{3}\rfloor-2 , }\\ 3\lfloor\frac{n}{3}\rfloor-2\lfloor\frac{k}{2}\rfloor+15+i,{\ \ \ \ \ \ \ \ \ \ \ \frac{n}{2}+\lfloor\frac{n}{3}\rfloor-1\leq i\leq\frac{n}{2}+\lfloor\frac{n}{3}\rfloor . } \end{array} \right. \end{align*}$

If $\frac{n}{2}-\lfloor\frac{n}{3}\rfloor = 4$ .

$\begin{align*} f(x_{i}x_{i+5}) = \left\{ \begin{array}{l} 8\lfloor\frac{n}{3}\rfloor-4\lfloor\frac{k}{2}\rfloor+41+i,{\ \ \ \ \ \ \ \ \ \ \ \frac{n}{2}+1\leq i\leq\frac{n}{2}+\lfloor\frac{n}{3}\rfloor-5 , }\\ 3\lfloor\frac{n}{3}\rfloor-2(\lfloor\frac{k}{2}\rfloor+\lfloor\frac{k}{8}\rfloor)+\\+38+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{n}{2}+\lfloor\frac{n}{3}\rfloor-4\leq i\leq\frac{n}{2}+\lfloor\frac{n}{3}\rfloor-1 , }\\ 5\lfloor\frac{n}{3}\rfloor-2\lfloor\frac{k}{2}\rfloor+21,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ i = \frac{n}{2}+\lfloor\frac{n}{3}\rfloor . } \end{array} \right. \end{align*}$

For $\frac{n}{2}+\lfloor\frac{n}{3}\rfloor+1\leq i\leq n$ .

$\begin{align*} f(x_{i}x_{i+5}) = 3\lfloor\frac{n}{3}\rfloor-2\lfloor\frac{k}{8}\rfloor-9+i. \end{align*}$

Case 2. If $\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor\geq5$ .

$\begin{align*} f(x_{i}x_{i+1}) = \left\{ \begin{array}{l} i, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1\leq i\leq \lfloor\frac{n}{3}\rfloor-1 , }\\ 5n-5\lfloor\frac{n}{2}\rfloor-2\lfloor\frac{k}{4}\rfloor-18,{\ \ \ \ \ \ \ \ \ \ i = \lfloor\frac{n}{3}\rfloor , }\\ 5n-5\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor-\\-4\lfloor\frac{k}{4}\rfloor-8+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{3}\rfloor+1\leq i\leq\lfloor\frac{n}{2}\rfloor-1 ,}\\ 5n-5\lfloor\frac{n}{3}\rfloor-\\-2(\lfloor\frac{k}{2}\rfloor+\lfloor\frac{k}{4}\rfloor)-3,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ i = \lfloor\frac{n}{2}\rfloor ,}\\ 5n-5\lfloor\frac{n}{3}\rfloor-\lfloor\frac{n}{2}\rfloor-\\-4\lfloor\frac{k}{2}\rfloor+15+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor+1\leq i\leq \lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor-1 ,}\\ 5n-5\lfloor\frac{n}{3}\rfloor-\\-2(\lfloor\frac{k}{2}\rfloor+\lfloor\frac{k}{8}\rfloor)-14,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ i = \lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor ,}\\ 4\lfloor\frac{n}{3}\rfloor-\lfloor\frac{n}{2}\rfloor-\\-4\lfloor\frac{k}{8}\rfloor+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor+1\leq i\leq n-1 ,}\\ 5\lfloor\frac{n}{3}\rfloor-2\lfloor\frac{k}{8}\rfloor-14,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ i = n . } \end{array} \right. \end{align*}$

$\begin{align*} f(x_{i}x_{i+2}) = \left\{ \begin{array}{l} \lfloor\frac{n}{3}\rfloor-1+i, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1\leq i\leq \lfloor\frac{n}{3}\rfloor-2 , }\\ 5n-5\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor-\\-2\lfloor\frac{k}{4}\rfloor-16+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{3}\rfloor-1\leq i\leq\lfloor\frac{n}{3}\rfloor , }\\ 5n-4\lfloor\frac{n}{2}\rfloor-2\lfloor\frac{n}{3}\rfloor-\\-4\lfloor\frac{k}{4}\rfloor-9+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{3}\rfloor+1\leq i\leq\lfloor\frac{n}{2}\rfloor-2 ,}\\ 5n-\lfloor\frac{n}{2}\rfloor-5\lfloor\frac{n}{3}\rfloor-\\-2(\lfloor\frac{k}{2}\rfloor+\lfloor\frac{k}{4}\rfloor)-1+i,{\ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor-1\leq i\leq\lfloor\frac{n}{2}\rfloor ,}\\ 5n-\lfloor\frac{n}{2}\rfloor-4\lfloor\frac{n}{3}\rfloor-\\-4\lfloor\frac{k}{2}\rfloor+14+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor+1\leq i\leq \lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor-2 ,}\\ 5n-\lfloor\frac{n}{2}\rfloor-6\lfloor\frac{n}{3}\rfloor-\\-2(\lfloor\frac{k}{2}\rfloor+\lfloor\frac{k}{8}\rfloor)-12+i,{\ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor-1\leq i\leq\lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor ,}\\ n-2\lfloor\frac{n}{3}\rfloor+3\lfloor\frac{n}{3}\rfloor-\\-4\lfloor\frac{k}{8}\rfloor-1+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor+1\leq i\leq n-2 ,}\\ -n+5\lfloor\frac{n}{3}\rfloor-2\lfloor\frac{k}{8}\rfloor-\\-12+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ n-1\leq i\leq n . } \end{array} \right. \end{align*}$

For $1\leq i\leq \lfloor\frac{n}{2}\rfloor$

$\begin{align*} f(x_{i}x_{i+3}) = \left\{ \begin{array}{l} 2\lfloor\frac{n}{3}\rfloor-3+i, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1\leq i\leq \lfloor\frac{n}{3}\rfloor-3 , }\\ 5n-5\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor-\\-2\lfloor\frac{k}{4}\rfloor-13+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{3}\rfloor-2\leq i\leq\lfloor\frac{n}{3}\rfloor , }\\ 5n-3\lfloor\frac{n}{2}\rfloor-3\lfloor\frac{n}{3}\rfloor-\\-4\lfloor\frac{k}{4}\rfloor-11+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{3}\rfloor+1\leq i\leq\lfloor\frac{n}{2}\rfloor-3 ,}\\ 5n-\lfloor\frac{n}{2}\rfloor-5\lfloor\frac{n}{3}\rfloor-\\-2(\lfloor\frac{k}{2}\rfloor+\lfloor\frac{k}{4}\rfloor)+2+i,{\ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor-2\leq i\leq\lfloor\frac{n}{2}\rfloor .} \end{array} \right. \end{align*}$

For $\lfloor\frac{n}{2}\rfloor+1\leq i\leq n$

$\begin{align*} f(x_{i}x_{i+3}) = \left\{ \begin{array}{l} 5n-\lfloor\frac{n}{2}\rfloor-3\lfloor\frac{n}{3}\rfloor-\\-4\lfloor\frac{k}{2}\rfloor+15+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor+1\leq i\leq \lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor-3 ,}\\ 5n-\lfloor\frac{n}{2}\rfloor-6\lfloor\frac{n}{3}\rfloor-\\-2(\lfloor\frac{k}{2}\rfloor+\lfloor\frac{k}{8}\rfloor)-9+i,{\ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor-2\leq i\leq\lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor ,}\\ 2n-3\lfloor\frac{n}{2}\rfloor+2\lfloor\frac{n}{3}\rfloor-\\-4\lfloor\frac{k}{8}\rfloor-3+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor+1\leq i\leq n-3 ,}\\ -n+5\lfloor\frac{n}{3}\rfloor-\\-2\lfloor\frac{k}{8}\rfloor-9+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ n-2\leq i\leq n . } \end{array} \right. \end{align*}$

$\begin{align*} f(x_{i}x_{i+4}) = \left\{ \begin{array}{l} 3\lfloor\frac{n}{3}\rfloor-6+i, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1\leq i\leq \lfloor\frac{n}{3}\rfloor-4 , }\\ 5n-5\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor-\\-2\lfloor\frac{k}{4}\rfloor-9+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{3}\rfloor-3\leq i\leq\lfloor\frac{n}{3}\rfloor , }\\ 5n-2\lfloor\frac{n}{2}\rfloor-4\lfloor\frac{n}{3}\rfloor-\\-4\lfloor\frac{k}{4}\rfloor-14+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{3}\rfloor+1\leq i\leq\lfloor\frac{n}{2}\rfloor-4 ,}\\ 5n-\lfloor\frac{n}{2}\rfloor-5\lfloor\frac{n}{3}\rfloor-\\-2(\lfloor\frac{k}{2}\rfloor+\lfloor\frac{k}{4}\rfloor)+6+i,{\ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor-3\leq i\leq\lfloor\frac{n}{2}\rfloor ,}\\ 5n-\lfloor\frac{n}{2}\rfloor-2\lfloor\frac{n}{3}\rfloor-\\-4\lfloor\frac{k}{2}\rfloor+9+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor+1\leq i\leq \lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor-4 ,}\\ 5n-\lfloor\frac{n}{2}\rfloor-6\lfloor\frac{n}{3}\rfloor-\\-2(\lfloor\frac{k}{2}\rfloor+\lfloor\frac{k}{8}\rfloor)-5+i,{\ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor-3\leq i\leq\lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor ,}\\ 3n-4\lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor-\\-4\lfloor\frac{k}{8}\rfloor-6+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor+1\leq i\leq n-4 ,}\\ -n+5\lfloor\frac{n}{3}\rfloor-\\-2\lfloor\frac{k}{8}\rfloor-5+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ n-3\leq i\leq n . } \end{array} \right. \end{align*}$

$\begin{align*} f(x_{i}x_{i+5}) = \left\{ \begin{array}{l} 4\lfloor\frac{n}{3}\rfloor-10+i, {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1\leq i\leq \lfloor\frac{n}{3}\rfloor-5 , }\\ 5n-5\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor-\\-2\lfloor\frac{k}{4}\rfloor-4+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{3}\rfloor-4\leq i\leq\lfloor\frac{n}{3}\rfloor , }\\ 5n-\lfloor\frac{n}{2}\rfloor-5\lfloor\frac{n}{3}\rfloor-\\-4\lfloor\frac{k}{4}\rfloor-18+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{3}\rfloor+1\leq i\leq\lfloor\frac{n}{2}\rfloor-5 \;\text{and}\; \lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor\geq6 ,}\\ 5n-6\lfloor\frac{n}{3}\rfloor-\\-2(\lfloor\frac{k}{2}\rfloor+\lfloor\frac{k}{4}\rfloor)+6+i,{\ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor-4\leq i\leq\lfloor\frac{n}{2}\rfloor ,}\\ 5n-\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor-\\-4\lfloor\frac{k}{2}\rfloor+5+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor+1\leq i\leq \lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor-5 ,}\\ 5n-\lfloor\frac{n}{2}\rfloor-6\lfloor\frac{n}{3}\rfloor-\\-2(\lfloor\frac{k}{2}\rfloor+\lfloor\frac{k}{8}\rfloor)+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor-4\leq i\leq\lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor ,}\\ 4n-5\lfloor\frac{n}{2}\rfloor-\\-4\lfloor\frac{k}{8}\rfloor-10+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor+1\leq i\leq n-5 \;\text{and}\;}\\ {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor\geq6 , \;\text{where}\; n \;\text{is even}\;,}\\ -n+5\lfloor\frac{n}{3}\rfloor-\\-2\lfloor\frac{k}{8}\rfloor+i,{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ n-4\leq i\leq n . } \end{array} \right. \end{align*}$

Hence, we can see that $f$ is $k$ -labeling and all vertices are labeled with even numbers.

Furthermore, we will estimate the weights of edges under the labeling $f$ as follows:

In Case 1, the edge set of $C_{n}(1, 2, 3, 4, 5), n\geq15$ can be split into nine mutually disjoint subsets, $A_{\ell}, 1\leq \ell \leq 9$ as follows:

● $A_{1} = \left\{ \begin{array}{ll} \{x_{i}x_{i+j}:1\leq i \leq 5-j, 1\leq j\leq4 \}, {\ \ \ \ \ \ \ \ \ \ \ \text{if}\; \lfloor\frac{n}{3}\rfloor = 5. } & \\\\ \{x_{i}x_{i+j}:1\leq i \leq \lfloor\frac{n}{3}\rfloor-j, 1\leq j\leq5 \}, {\ \ \ \ \ \ \ \ \text{if}\; \lfloor\frac{n}{3}\rfloor\geq6 } & \end{array} \right.$

the set of all edges which has end vertices labeled with 0.

● $A_{2} = \{x_{i}x_{i+j}:\lfloor\frac{n}{3}\rfloor-j+1\leq i\leq \lfloor\frac{n}{3}\rfloor, 1\leq j\leq\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor \}$ the set of all edges which has end vertices labeled with 0 and $2\lfloor\frac{k}{4}\rfloor+2$ ,

● $A_{3} = \{x_{i}x_{i+j}:\lfloor\frac{n}{3}\rfloor+1\leq i \leq \lfloor\frac{n}{2}\rfloor-j, 1\leq j\leq \lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor-1 \}$ the set of all edges which has end vertices labeled with $2\lfloor\frac{k}{4}\rfloor+2$ ,

● $A_{4} = \{x_{i}x_{i+j}:\lfloor\frac{n}{2}\rfloor+1-j\leq i\leq \lfloor\frac{n}{2}\rfloor, 1\leq j\leq \lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor \}$ the set of all edges which has end vertices labeled with $2\lfloor\frac{k}{4}\rfloor+2$ and $2\lfloor\frac{k}{2}\rfloor$ .

● $A_{5} = \left\{ \begin{array}{ll} \{x_{i}x_{i+j}:\lfloor\frac{n}{2}\rfloor+1\leq i \leq \lfloor\frac{n}{2}\rfloor+5-j, 1\leq j\leq4 \}, {\ \ \ \ \ \ \ \ \text{if}\; \lfloor\frac{n}{3}\rfloor = 5. } & \\\\ \{x_{i}x_{i+j}:\lfloor\frac{n}{2}\rfloor+1\leq i \leq \lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor-j, 1\leq j\leq5 \}, {\ \ \ \ \ \text{if}\; \lfloor\frac{n}{3}\rfloor\geq6 } & \end{array} \right.$

the set of all edges which has end vertices labeled with $2\lfloor\frac{k}{2}\rfloor$ .

● $A_{6} = \{x_{i}x_{i+j}:n-j\leq i\leq \lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor, 1\leq j\leq n-\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3} \}$ the set of all edges which has end vertices labeled with $2\lfloor\frac{k}{2}\rfloor$ and $2\lfloor\frac{k}{8}\rfloor$ .

● $A_{7} = \{x_{i}x_{i+j}:\lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor+1\leq i\leq n-j, 1\leq j\leq n-\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor-1 \}$ the set of all edges which has end vertices labeled with $2\lfloor\frac{k}{8}\rfloor$ .

● $A_{8} = \{x_{i}x_{i+j}:n-j+2\leq i\leq n, 1\leq j\leq n-\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor \}$ the set of all edges which has end vertices labeled with 0 and $2\lfloor\frac{k}{8}\rfloor$ .

● $A_{9} = \{x_{i}x_{i+j}:\lfloor\frac{n}{2}\rfloor-j+1\leq i\leq \lfloor\frac{n}{3}\rfloor, \lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor+1\leq j\leq 5\}\cup\{x_{i}x_{i+j}:\lfloor\frac{n}{2}\rfloor-j+2\leq i\leq n, n-\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor+1\rfloor\leq j\leq 5\}$ the set of all edges which has end vertices labeled with 0 and $2\lfloor\frac{k}{2}\rfloor$ .

Thus, for $n$ is odd the edge weights under the labeling $f$ are the following:

1. The weights of the edges of the set $A_{1}$ , receive the consecutive integers from the set $A_{1} = \{1, 2, ..., 5\lfloor\frac{n}{3}\rfloor-15 \}$ .

2. The weights of the edges of the set $A_{2}$ , receive the consecutive integers from the set $A_{2} = \{5\lfloor\frac{n}{2}\rfloor-9, ..., 5\lfloor\frac{n}{2}\rfloor+\frac{\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor}{2}(11-\lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor)-10 \}$ .

3. The weights of the edges of the set $A_{3}$ , receive the consecutive integers from the set $\{5\lfloor\frac{n}{3}\rfloor+\frac{\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor}{2}(\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor+1)+16, ..., 5\lfloor\frac{n}{3}\rfloor+(\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor)^{2}+15\}$ .

4. The weights of the edges of the set $A_{4}$ , receive the consecutive integers from the set $A_{4} = \{5\lfloor\frac{n}{2}\rfloor+\frac{\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor}{2}(\lfloor\frac{n}{2}-\lfloor\frac{n}{3}\rfloor-1)+21, ..., 10\lfloor\frac{n}{2}\rfloor-5\lfloor\frac{n}{3}\rfloor+20 \}$ .

5. The weights of the edges of the set $A_{5}$ , receive the consecutive integers from the set $A_{5} = \{10\lfloor\frac{n}{2}\rfloor-5\lfloor\frac{n}{3}\rfloor+21, ..., 10\lfloor\frac{n}{2}\rfloor+5\}$ .

6. The weights of the edges of the set $A_{6}$ , receive the consecutive integers from the set $A_{6} = \{5\lfloor\frac{n}{3}\rfloor+(\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor)^{2}+16, ..., 5\lfloor\frac{n}{2}\rfloor+\frac{\lfloor\frac{n}{2}-\lfloor\frac{n}{3}\rfloor}{2}(\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor-1)+20 \}$ .

7. The weights of the edges of the set $A_{7}$ , receive the consecutive integers from the set $A_{7} = \{5\lfloor\frac{n}{2}\rfloor+\frac{\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor}{2}(\lfloor\frac{n}{3}\rfloor-\lfloor\frac{n}{2}\rfloor-1)-9, ..., 5\lfloor\frac{n}{2}\rfloor-10 \}$ .

8. The weights of the edges of the set $A_{8}$ , receive the consecutive integers from the set $A_{8} = \{5\lfloor\frac{n}{3}\rfloor-14, ..., 5\lfloor\frac{n}{2}\rfloor+\frac{\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor}{2}(\lfloor\frac{n}{3}\rfloor-\lfloor\frac{n}{2}\rfloor-1)-10 \}$ .

9. The weights of the edges of the set $A_{9}$ , receive the consecutive integers from the set $A_{9} = \{ 5\lfloor\frac{n}{3}\rfloor+\frac{\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor}{2}(11-\lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{3}\rfloor)-9, ..., 5\lfloor\frac{n}{3}\rfloor+\frac{\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor}{2}(\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor+1)+15 \}$ .

Also, for $n$ is even the edge weights under the labeling $f$ are the following:

1. The weights of the edges of the set $A_{1}$ , receive the consecutive integers from the set $A_{1} = \{1, 2, ..., 5\lfloor\frac{n}{3}\rfloor-15 \}$ .

2. The weights of the edges of the set $A_{2}$ , receive the consecutive integers from the set $A_{2} = \{\frac{5n}{2}-14, ..., 5\lfloor\frac{n}{3}\rfloor+\frac{\frac{n}{2}-\lfloor\frac{n}{3}\rfloor}{2}(\lfloor\frac{n}{3}\rfloor -\frac{n}{2}+21)-15 \}$ .

3. The weights of the edges of the set $A_{3}$ , receive the consecutive integers from the set $\{5\lfloor\frac{n}{3}\rfloor+\frac{\frac{n}{2}-\lfloor\frac{n}{3}\rfloor}{2}(\frac{n}{2}-\lfloor\frac{n}{3}\rfloor-1)+16, ..., 6\lfloor\frac{n}{3}\rfloor-\frac{n}{2}+(\frac{n}{2}-\lfloor\frac{n}{3}\rfloor)^{2}+15\}$ .

4. The weights of the edges of the set $A_{4}$ , receive the consecutive integers from the set $A_{4} = \{5\lfloor\frac{n}{3}\rfloor+\frac{\frac{n}{2}-\lfloor\frac{n}{3}\rfloor}{2}(\frac{n}{2}-\lfloor\frac{n}{3}\rfloor+9)+16, ..., 5n-5\lfloor\frac{n}{3}\rfloor+15 \}$ .

5. The weights of the edges of the set $A_{5}$ , receive the consecutive integers from the set $A_{5} = \{5n-5\lfloor\frac{n}{3}\rfloor+16, ..., 5n \}$ .

6. The weights of the edges of the set $A_{6}$ , receive the consecutive integers from the set $A_{6} = \{6\lfloor\frac{n}{3}\rfloor-\frac{n}{2}+(\frac{n}{2}-\lfloor\frac{n}{3}\rfloor)^{2}+16, ..., 5\lfloor\frac{n}{3}\rfloor+\frac{\frac{n}{2}-\lfloor\frac{n}{3}\rfloor}{2}(\frac{n}{2}-\lfloor\frac{n}{3}\rfloor+9)+15 \}$ .

7. The weights of the edges of the set $A_{7}$ , receive the consecutive integers from the set $A_{7} = \{\frac{n}{2}+4\lfloor\frac{n}{3}\rfloor+\frac{\frac{n}{2}-\lfloor\frac{n}{3}\rfloor}{2}(\lfloor\frac{n}{3}\rfloor-\frac{n}{2}+9)-14, ..., \frac{5n}{2}-15 \}$ .

8. The weights of the edges of the set $A_{8}$ , receive the consecutive integers from the set $A_{8} = \{5\lfloor\frac{n}{3}\rfloor-14, ..., \frac{n}{2}+4\lfloor\frac{n}{3}\rfloor+\frac{\frac{n}{2}-\lfloor\frac{n}{3}\rfloor}{2}(\lfloor\frac{n}{3}\rfloor-\frac{n}{2}+9)-15 \}$ .

9. The weights of the edges of the set $A_{9}$ , receive the consecutive integers from the set $A_{9} = \{5\lfloor\frac{n}{3}\rfloor+\frac{\frac{n}{2}-\lfloor\frac{n}{3}\rfloor}{2}(\lfloor\frac{n}{3}\rfloor -\frac{n}{2}+21)-14, ..., 5\lfloor\frac{n}{3}\rfloor+\frac{\frac{n}{2}-\lfloor\frac{n}{3}\rfloor}{2}(\frac{n}{2}-\lfloor\frac{n}{3}\rfloor-1)+15\}$ .

In Case 2, using the similar arguments as in the proof of Theorem 1.1, the edge set of $C_{n}(1, 2, 3, 4, 5)$ can be divided into eight mutually disjoint subsets, so we can find the edge weights as follows:

1. The weights of the edges of the set $A_{1}$ , receive the consecutive integers from the set $A_{1} = \{1, 2, ..., 5\lfloor\frac{n}{3}\rfloor-15 \}$ .

2. The weights of the edges of the set $A_{2}$ , receive the consecutive integers from the set $A_{2} = \{5n-5\lfloor\frac{n}{2}\rfloor-14, ..., 5n-\lfloor\frac{n}{2}\rfloor \}$ .

3. The weights of the edges of the set $A_{3}$ , receive the consecutive integers from the set $\{5n-5\lfloor\frac{n}{2}\rfloor+1, ..., 5n-5\lfloor\frac{n}{3}\rfloor-15\}$ .

4. The weights of the edges of the set $A_{4}$ , receive the consecutive integers from the set $A_{4} = \{5n-5\lfloor\frac{n}{3}\rfloor+1, ..., 5n-5\lfloor\frac{n}{3}\rfloor+15 \}$ .

5. The weights of the edges of the set $A_{5}$ , receive the consecutive integers from the set $A_{5} = \{5n-5\lfloor\frac{n}{3}\rfloor-5\lfloor\frac{n}{3}\rfloor+16, ..., 5n \}$ .

6. The weights of the edges of the set $A_{6}$ , receive the consecutive integers from the set $A_{6} = \{5n-5\lfloor\frac{n}{3}\rfloor-14, ..., 5n-5\lfloor\frac{n}{3}\rfloor \}$ .

7. The weights of the edges of the set $A_{7}$ , receive the consecutive integers from the set $A_{7} = \{5\lfloor\frac{n}{3}\rfloor+1, ..., 5n-5\lfloor\frac{n}{2}\rfloor-15 \}$ .

8. The weights of the edges of the set $A_{8}$ , receive the consecutive integers from the set $A_{8} = \{5\lfloor\frac{n}{3}\rfloor-14, ..., 5\lfloor\frac{n}{3}\rfloor \}$ .

We can see that all vertex labels are even numbers. Furthermore, for $\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor\leq4$ the edge weights are from the set $W = \{1, 2, ..., 10\lfloor\frac{n}{2}\rfloor+5\}$ , when $n$ is odd and from the set $W = \{1, 2, ..., 5n\}$ , when $n$ is even. Also for $\lfloor\frac{n}{2}\rfloor-\lfloor\frac{n}{3}\rfloor\geq5$ , the edge weights are from the set $W = \{1, 2, ..., 5n\}$ . Finally we can easily check that all numbers in the set $W$ are distinct.

4. Conclusions

In this paper we discuss an edge irregular reflexive k-labeling for three families of the circulant graphs $C_{n}(1, 2), C_{n}(1, 2, 3)$ and $C_{n}(1, 2, 3, 4, 5)$ including their exact values of the reflexive edge strength. For $n\geq8$ we proved that $res(C_{n}(1, 2)) = \lceil\frac{2n}{3}\rceil$ for $n\equiv0, 2$ (mod 3) and $res(C_{n}(1, 2)) = \lceil\frac{2n}{3}\rceil$ +1 for $n\equiv1$ (mod 3). Also we proved that $res(C_{n}(1, 2, 3)) = n$ for $n$ is even and $res(C_{n}(1, 2, 3)) = n+1$ for $n$ is odd, where $n\geq6$ . Moreover we showed that $res(C_{n}(1, 2, 3, 4, 5)) = \lceil\frac{5n}{3}\rceil+1$ for $n\equiv3, 4$ (mod 6), $n\geq15$ and $res(C_{n}(1, 2, 3, 4, 5)) = \lceil\frac{5n}{3}\rceil$ , otherwise.

Acknowledgments

The author would like to thank the editors and anonymous reviewers for their careful reading, recommendations, and comments, which helped to improve the paper presentation.

Conflict of interest

The authors declare that they have no competing interest.

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