Topological indices of linear crossed phenylenes with respect to their Laplacian and normalized Laplacian spectrum

Zhi-Yu Shi; Jia-Bao Liu; Zhi-Yu Shi; Jia-Bao Liu

doi:10.3934/math.2024262

AIMS Mathematics

2024, Volume 9, Issue 3: 5431-5450. doi: 10.3934/math.2024262

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Research article

Topological indices of linear crossed phenylenes with respect to their Laplacian and normalized Laplacian spectrum

Zhi-Yu Shi ^{1
,
,},
Jia-Bao Liu ²

1.
Engineering Department, Anhui Sanlian University, Hefei 230601, China
2.
School of Mathematics and Physics, Anhui Jianzhu University, Hefei 230601, China

Received: 23 November 2023 Revised: 31 December 2023 Accepted: 10 January 2024 Published: 29 January 2024
MSC : 05C50, 05C90

As a powerful tool for describing and studying the properties of networks, the graph spectrum analyses and calculations have attracted substantial attention from the scientific community. Let $C_{n}$ represent linear crossed phenylenes. Based on the Laplacian (normalized Laplacian, resp.) polynomial of $C_{n}$ , we first investigated the Laplacian (normalized Laplacian, resp) spectrum of $C_{n}$ in this paper. Furthermore, the Kirchhoff index, multiplicative degree-Kirchhoff, index and complexity of $C_{n}$ were obtained through the relationship between the roots and the coefficients of the characteristic polynomials. Finally, it was found that the Kirchhoff index and multiplicative degree-Kirchhoff index of $C_{n}$ were approximately one quarter of their Wiener index and Gutman index, respectively.

Keywords:

Citation: Zhi-Yu Shi, Jia-Bao Liu. Topological indices of linear crossed phenylenes with respect to their Laplacian and normalized Laplacian spectrum[J]. AIMS Mathematics, 2024, 9(3): 5431-5450. doi: 10.3934/math.2024262

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Abstract

1. Introduction

In recent years, researchers have been interested in the study of complex networks ^[1,2,3,4]. Three common characteristics of complex networks are: small-world, scale-free, and fractal. Yang and Huang et al. ^[5,6] have determined the Kirchhoff index and multiplicative degree-Kirchhoff index of hexagonal chains, and they obtained that the Kirchhoff index and multiplicative degree-Kirchhoff index of hexagonal chains are approximately half of their Wiener index and Gutman index, respectively. In particular, Peng et al. ^[7] studied the Kirchhoff index and complexity for linear phenylenes, and determined that the Kirchhoff index of linear phenylenes is approximately half of its Wiener index. In addition, Z. Zhu and J.-B. Liu ^[8] obtained the multiplicative degree-Kirchhoff index and complexity of generalized phenylenes. In 2018, Pan and Li ^[9] determined the Kirchhoff index, multiplicative degree-Kirchhoff index, and complexity of linear crossed hexagonal networks, and obtained that the Kirchhoff index and multiplicative degree-Kirchhoff index of linear crossed hexagonal chains are approximately one quarter of their Wiener index and Gutman index, respectively. For other networks, see ^{[10,11,12,13,14]}.

Motivated by these, we investigate the Laplacian and normalized Laplacian spectra of linear crossed phenylenes. We also obtain that the Kirchhoff index and multiplicative degree-Kirchhoff index of linear crossed phenylenes are approximately one quarter of their Wiener index and Gutman index, respectively.

In this paper, we suppose $G = (E_{G}, V_{G})$ is a graph with edge set $E_{G} = \{e_1, e_2, \cdots, e_m\}$ and vertex set $V_{G} = \{v_1, v_2, \cdots, v_n\}$ . For more notations, one can be referred to ^[15].

Let $D(G) = diag\{d_{1}, d_{2}, \cdots, d_{n}\}$ represent a degree matrix, and $A(G)$ be the adjacency matrix, where $d_{i}$ is the degree of $v_{i}$ . Therefore, we can calculate the Laplacian matrix and normalized Laplacian matrix, which are defined as $L(G) = D(G)-A(G)$ and $\mathcal{L}(G) = D(G)^{-\frac{1}{2}}LD(G)^{-\frac{1}{2}}$ , respectively. The Laplacian matrix is

$\begin{eqnarray} (L(G))_{ij} = \begin{cases} d_{i}, & i = j;\\ -1, & i\neq{j}, \; v_{i} \; and \; v_{j} \; are \; adjacent;\\ 0, & otherwise. \end{cases} \end{eqnarray}$

(1.1)

The normalized Laplacian matrix is

$\begin{eqnarray} (\mathcal{L}(G))_{ij} = \begin{cases} 1, & i = j, \; d_{i}\neq 0;\\ -\frac{1}{\sqrt{d_{i}d_{j}}}, & i\neq{j}, \; v_{i} \; and \; v_{j} \; are \; adjacent; \\ 0, & otherwise. \end{cases} \end{eqnarray}$

(1.2)

The distance between vertices $v_i$ and $v_j$ , denoted by $d_{ij}$ , is defined as the length of the shortest path between vertices $v_i$ and $v_j$ . The Wiener index ^[16,17] is defined as

$\begin{eqnarray} W(G) = \sum\limits_{i < j}d_{ij}. \end{eqnarray}$

(1.3)

In 1994, the Gutman index ^[18] is defined as

$\begin{eqnarray} Gut(G) = \sum\limits_{i < j}d_{i}d_{j}d_{ij}. \end{eqnarray}$

(1.4)

Klein and Randić ^[19] were the first to put forward the concept of resistance distance, and the resistance distance between vertices $v_i$ and $v_j$ is denoted by $r_{ij}$ . Klein et al. ^[20,21] introduced the Kirchhoff index as $Kf(G) = \sum_{i < j}r_{ij}$ . In 2007, Chen et al. ^[22] proposed the multiplicative degree-Kirchhoff index as $Kf^{*}(G) = \sum_{i < j}d_{i}d_{j}r_{ij}$ . Gutman and Mohar ^[23] introduced the Kirchhoff index as

$\begin{eqnarray} Kf(G) = n\sum\limits_{k = 2}^{n}\frac{1}{\mu_{k}}, \end{eqnarray}$

(1.5)

where $0 = \mu_{1} < \mu_{2}\leq\cdots\leq\mu_{n}(n\geq2)$ are the eigenvalues of $L(G)$ .

According to the normalized Laplacian, Chen et al. ^[22] proposed the multiplicative degree-Kirchhoff index as

$\begin{eqnarray} Kf^{*}(G) = 2m\sum\limits_{k = 2}^{n}\frac{1}{\lambda_{k}}, \end{eqnarray}$

(1.6)

where $\lambda_{1}\leq\lambda_{2}\leq\cdots\leq\lambda{n}$ are the normalized Laplacian eigenvalues of $\mathcal{L}(G)$ .

The number of spanning trees of $G$ can also be called the complexity of $G$ ^[15], denoted by $\tau(G)$ .

In Section 2, we mainly introduce some notations and theorems. Next, applying the relationship between the roots and coefficients of $C_n$ , the Laplacian spectrum of $C_{n}$ is determined in Section 3. In Section 4, we obtain the normalized Laplacian spectrum of $C_{n}$ in the same way as in Section 3. The conclusion is summarized in Section 5.

2. Preliminary

First, we state some notations and theorems, which will be used later.

Given an $n\times n$ matrix $M$ , the submatrix of $M$ is represented by $M[i_{1}, \cdots, i_{k}]$ , where $M[i_{1}, \cdots, i_{k}]$ is formed by removing the $i_{1}$ -th, $\cdots$ , $i_{k}$ -th rows and columns of $M$ . Let $P_{M}(x) = det(xI-M)$ represent the characteristic polynomial of $M$ .

Label linear crossed phenylenes as shown in . Evidently, $|V(C_{n})| = 6n+2, |E(C_{n})| = 14n+1$ and $\pi = (1, 1')(2, 2')\cdots\big(3n+1, (3n+1)'\big)$ is an automorphism of $C_{n}$ . Set $V_{1} = \{1, 2, \cdots, 3n+1\}, V_{2} = \{1', 2', \cdots, (3n+1)'\}$ .

Figure 1. Linear crossed phenylenes

$C_n$ .

DownLoad: Full-Size Img PowerPoint

Thus, $L(C_n)$ and $\mathcal{L}(C_n)$ can be expressed by

$\begin{equation*} L(C_{n}) = \left( \begin{array}{cc} L_{V_{1}V_{1}}& L_{V_{1}V_{2}}\\ L_{V_{2}V_{1}}& L_{V_{2}V_{2}}\\ \end{array} \right),\; \mathcal{L}(C_{n}) = \left( \begin{array}{cc} \mathcal{L}_{V_{1}V_{1}}& \mathcal{L}_{V_{1}V_{2}}\\ \mathcal{L}_{V_{2}V_{1}}& \mathcal{L}_{V_{2}V_{2}}\\ \end{array} \right), \end{equation*}$

where

$\begin{equation*} L_{V_{1}V_{1}} = L_{V_{2}V_{2}},\; L_{V_{1}V_{2}} = L_{V_{2}V_{1}},\; \mathcal{L}_{V_{1}V_{1}} = \mathcal{L}_{V_{2}V_{2}},\; \mathcal{L}_{V_{1}V_{2}} = \mathcal{L}_{V_{2}V_{1}}. \end{equation*}$

Let

$\begin{equation*} T = \left( \begin{array}{cc} \frac{1}{\sqrt{2}}I_{3n+1}& \frac{1}{\sqrt{2}}I_{3n+1}\\ \frac{1}{\sqrt{2}}I_{3n+1}& -\frac{1}{\sqrt{2}}I_{3n+1} \end{array} \right), \end{equation*}$

then

$\begin{equation*} TL(C_{n})T' = \left( \begin{array}{cc} L_{A}& 0\\ 0& L_{S} \end{array} \right),\; T\mathcal{L}(C_{n})T' = \left( \begin{array}{cc} \mathcal{L}_{A}& 0\\ 0& \mathcal{L}_{S} \end{array} \right), \end{equation*}$

where $T'$ is the transpose of the matrix $T$ and

$\begin{eqnarray*} L_{A} = L_{V_{1}V_{1}}+L_{V_{1}V_{2}}, L_{S} = L_{V_{1}V_{1}}-L_{V_{1}V_{2}}, \mathcal{L}_{A} = \mathcal{L}_{V_{1}V_{1}}+\mathcal{L}_{V_{1}V_{2}}, \mathcal{L}_{S} = \mathcal{L}_{V_{1}V_{1}}-\mathcal{L}_{V_{1}V_{2}}. \end{eqnarray*}$

Theorem 2.1. ^[24] If $L_{A}, L_{S}, \mathcal{L}_{A}, \mathcal{L}_{S}$ are defined as above, the following formula can be obtained:

$\begin{eqnarray*} P_{L(C_{n})}(x) = P_{L_{A}}(x)P_{L_{S}}(x), P_{\mathcal{L}(C_{n})}(x) = P_{\mathcal{L}_{A}}(x)P_{\mathcal{L}_{S}}(x). \end{eqnarray*}$

Theorem 2.2. ^[15] If $G$ is a graph with $|V_G| = n$ and $|E_G| = m$ , then

$\begin{eqnarray*} 2m\tau(G) = \prod\limits_{i = 1}^{n}d_i\cdot\prod\limits_{i = 2}^{n}\lambda_i, \end{eqnarray*}$

where $\tau(G)$ is the complexity of $G$ , and $\lambda_{i}$ is the normalized Laplacian eigenvalue of $\mathcal{L}(G)$ .

3. The Kirchhoff index and Wiener index of $C_{n}$ in terms of the Laplacian spectrum

In this section, we mainly calculate the Kirchhoff index and Wiener index of $C_{n}$ .

According to (1.1), we can get $L_{V_1 V_1}$ and $L_{V_1 V_2}$ :

$\begin{eqnarray*} L_{V_1 V_1}& = & \left( \begin{array}{ccccccccccc} 3 & -1 & & & & & & & & &\\ -1 & 4 & -1 & & & & & & & &\\ & -1 & 5 & -1 & & & & & & &\\ & & -1 & 5 & -1 & & & & & &\\ & & & -1 & 4 & -1 & & & & &\\ & & & & -1 & 5 & -1 & & & &\\ & & & & & & \ddots & & & &\\ & & & & & & -1 & 5 & -1 & &\\ & & & & & & & -1 & 4 & -1 &\\ & & & & & & & & -1 & 5 & -1\\ & & & & & & & & & -1 & 3\\ \end{array} \right)_{(3n+1)\times(3n+1)}, \end{eqnarray*}$

$\begin{eqnarray*} L_{V_1 V_2}& = & \left( \begin{array}{ccccccccccc} -1 & -1 & & & & & & & & &\\ -1 & 0 & -1 & & & & & & & &\\ & -1 & -1 & -1 & & & & & & &\\ & & -1 & -1 & -1 & & & & & &\\ & & & -1 & 0 & -1 & & & & &\\ & & & & -1 & -1 & -1 & & & &\\ & & & & & & \ddots & & & &\\ & & & & & & -1 & -1 & -1 & &\\ & & & & & & & -1 & 0 & -1 &\\ & & & & & & & & -1 & -1 & -1\\ & & & & & & & & & -1 & -1\\ \end{array} \right)_{(3n+1)\times(3n+1)}. \end{eqnarray*}$

Based on Theorem 2.1, the Laplacian spectrum consists of the eigenvalues of $L_{A}$ , and $L_{S}$ of $C_{n}$ can be obtained.

$\begin{eqnarray*} L_A& = & \left( \begin{array}{ccccccccccc} 2 & -2 & & & & & & & & &\\ -2 & 4 & -2 & & & & & & & &\\ & -2 & 4 & -2 & & & & & & &\\ & & -2 & 4 & -2 & & & & & &\\ & & & -2 & 4 & -2 & & & & &\\ & & & & -2 & 4 & -2 & & & &\\ & & & & & & \ddots & & & &\\ & & & & & & -2 & 4 & -2 & &\\ & & & & & & & -2 & 4 & -2 &\\ & & & & & & & & -2 & 4 & -2\\ & & & & & & & & & -2 & 2\\ \end{array} \right)_{(3n+1)\times(3n+1)}, \end{eqnarray*}$

$\begin{eqnarray*} L_S = diag(4,4,6,6,4,6,\cdots,6,4,6,4)_{(3n+1)}. \end{eqnarray*}$

Assume that $0 = \alpha_{1} < \alpha_{2}\leq\alpha_{3}\leq\cdots\leq\alpha_{3n+1}$ are the roots of $P_{L_{A}}(x)$ , and $0 < \beta_{1}\leq\beta_{2}\leq\beta_{3}\leq\cdots\leq\beta_{3n+1}$ are the roots of $P_{L_{S}}(x)$ . By (1.5), we have

$\begin{eqnarray} Kf(C_{n}) = (6n+2)\Big(\sum\limits_{i = 2}^{3n+1}\frac{1}{\alpha_{i}}+\sum\limits_{i = 1}^{3n+1}\frac{1}{\beta_{i}}\Big). \end{eqnarray}$

(3.1)

It is obvious from the matrix $L_S$ that the following expression can be obtained:

$\begin{eqnarray} \sum\limits_{i = 1}^{3n+1}\frac{1}{\beta_{i}} = \frac{1}{4}\times(n+2)+\frac{1}{6}\times(2n-1) = \frac{7n+4}{12}. \end{eqnarray}$

(3.2)

Thus, we need to calculate the first sum in (3.1).

Let

$\begin{eqnarray*} P_{L_{A}}(x) = det(xI-L_{A}) = x^{3n+1}+a_{1}x^{3n}+\cdots+a_{3n}x. \end{eqnarray*}$

Based on the Vieta $^{'}$ s Theorem of $P_{L_{A}}(x)$ , we can get

$\begin{eqnarray} \sum\limits_{i = 2}^{3n+1}\frac{1}{\alpha_{i}} = \frac{(-1)^{3n-1}a_{3n-1}}{(-1)^{3n}a_{3n}}. \end{eqnarray}$

(3.3)

Obviously, we obtain that $(-1)^{3n}a_{3n}$ is the sum of all the principal minors of order 3n of $L_{A}$ and $(-1)^{3n-1}a_{3n-1}$ is the sum of all the principal minors of order $3n-1$ of $L_{A}$ . So, let $F_{k}$ be the $k$ -th order principal submatrix, which consists of the first $k$ columns and $k$ rows of $L_{A}$ , and $f_{k} = det(F_{k})$ , $k = 1, 2, \cdots, 3n$ . Thus, we can get $f_{1} = 2, f_{2} = 4$ , and for $1\leq{i}\leq3n$ ,

$\begin{eqnarray*} f_{i} = 4f_{i-1}-4f_{i-2}. \end{eqnarray*}$

The solution of the previous recurrence relation is

$\begin{eqnarray*} f_{i} = 2^{i}. \end{eqnarray*}$

Now, let $H_{k}$ be the $k$ -th order principal submatrix, which consists of the last $k$ columns and $k$ rows of $L_{A}$ , and $h_{k} = det(H_{k})$ , $k = 1, 2, \cdots, 3n$ . Based on the symmetry matrix $L_A$ , one gets $h_k = f_k$ , and let $f_0 = 1$ .

Fact 1. $(-1)^{3n}a_{3n} = (3n+1)2^{3n}.$

Proof. Since $(-1)^{3n}a_{3n}$ is the sum of all the principal minors of order 3n of $L_{A}$ , we have

$\begin{eqnarray*} (-1)^{3n}a_{3n}& = &2f_{3n}+\sum\limits_{i = 2}^{3n}f_{i-1}f_{3n+1-i}\\ & = &\sum\limits_{i = 1}^{3n+1}f_{i-1}f_{3n+1-i}\\ & = &(3n+1)2^{3n}. \end{eqnarray*}$

This completes the proof. ■

Fact 2. $(-1)^{3n-1}a_{3n-1} = n(3n+1)(3n+2)2^{3n-2}$

Proof. Since $(-1)^{3n-1}a_{3n-1}$ is the sum of all the principal minors of order $3n-1$ of $L_{A}$ , we obtain

$\begin{eqnarray*} (-1)^{3n-1}a_{3n-1} = \sum\limits_{1\leq{i} < j\leq3n+1}N_{ij}f_{i-1}f_{3n+1-j}, \end{eqnarray*}$

where

$\begin{eqnarray*} N_{ij}& = & \left| \begin{array}{ccccccc} 4 & -2 & & & & &\\ -2 & 4 & -2 & & & &\\ & -2 & 4 & -2 & & &\\ & & & \ddots & & &\\ & & & -2 & 4 & -2 &\\ & & & & -2 & 4 & -2\\ & & & & & -2 & 4 \\ \end{array} \right|_{(j-i-1)\times(j-i-1)}\\ & = &(j-i)2^{j-i-1}. \end{eqnarray*}$

Therefore, we can have

$\begin{eqnarray*} (-1)^{3n-1}a_{3n-1}& = &\sum\limits_{1\leq{i} < j\leq3n+1}(j-i)2^{j-i-1}f_{i-1}f_{3n+1-j}\\ & = &\sum\limits_{1\leq{i} < j\leq3n+1}(j-i)2^{3n-1}\\ & = &n(3n+1)(3n+2)2^{3n-2}. \end{eqnarray*}$

The result is as desired. ■

Together with (3.3) and Facts 1 - 2, we obtain the following lemma.

Lemma 3.1. If $0 = \alpha_{1} < \alpha_{2}\leq\alpha_{3}\leq\cdots\leq\alpha_{3n+1}$ are the eigenvalues of $L_A$ , one gets

$\begin{eqnarray*} \sum\limits_{i = 2}^{3n+1}\frac{1}{\alpha_{i}} = \frac{n(3n+2)}{4}. \end{eqnarray*}$

Applying (3.1) - (3.2) and Lemma 3.1, we obtain the following theorem.

Theorem 3.2. For linear crossed phenylenes $C_n$ , we have

$\begin{eqnarray*} Kf(C_{n}) = \frac{27n^{3}+48n^{2}+25n+4}{6}. \end{eqnarray*}$

The Kirchhoff indices of $C_{n}$ are shown in , where $1\leq{n}\leq15$ .

Table 1. Kirchhoff indices from

$C_{1}$ to

$C_{15}$ .

$G$	$Kf(G)$	$G$	$Kf(G)$	$G$	$Kf(G)$	$G$	$Kf(G)$	$G$	$Kf(G)$
$C_{1}$	$17.33$	$C_{4}$	$433.33$	$C_{7}$	$1965.33$	$C_{10}$	$5342.33$	$C_{13}$	$11293.33$
$C_{2}$	$77.00$	$C_{5}$	$784.00$	$C_{8}$	$2850.00$	$C_{11}$	$7004.00$	$C_{14}$	$13975.00$
$C_{3}$	$206.67$	$C_{6}$	$1285.67$	$C_{9}$	$3966.67$	$C_{12}$	$8978.67$	$C_{15}$	$17050.67$

| Show Table

DownLoad: CSV

Theorem 3.3. Assume that $C_n$ are the linear crossed phenylenes, then

$\begin{eqnarray*} \lim\limits_{n\rightarrow \infty}\frac{Kf(C_{n})}{W(C_{n})} = \frac{1}{4}. \end{eqnarray*}$

Proof. By first classifying and discussing the following cases of vertices, the Wiener index of $C_n$ is obtained.

● Vertex $3j-1(j = 1, 2, \cdots, n)$ of $C_n$ :

$\begin{eqnarray*} w_{1}(i) = 2+2\Big(\sum\limits_{k = 1}^{i-1}k+\sum\limits_{k = 1}^{3n+1-i}k\Big),i = 3j-1. \end{eqnarray*}$

● Vertex $3j(j = 1, 2, \cdots, n)$ of $C_n$ :

$\begin{eqnarray*} w_{2}(i) = 1+2\Big(\sum\limits_{k = 1}^{i-1}k+\sum\limits_{k = 1}^{3n+1-i}k\Big),i = 3j. \end{eqnarray*}$

● Vertex $3j+1(j = 1, 2, \cdots, n-1)$ of $C_n$ :

$\begin{eqnarray*} w_{3}(i) = 1+2\Big(\sum\limits_{k = 1}^{i-1}k+\sum\limits_{k = 1}^{3n+1-i}k\Big),i = 3j+1. \end{eqnarray*}$

● Vertex $1$ of $C_n$ :

$\begin{eqnarray*} w_{4}(i) = 1+2\Big(\sum\limits_{k = 1}^{3n}k\Big). \end{eqnarray*}$

In view of (1.3), we have

$\begin{eqnarray*} W(C_{n})& = &\frac{2\sum\nolimits_{i = 3j-1}w_{1}(i)+2\sum\nolimits_{i = 3j}w_{2}(i)+2\sum\nolimits_{i = 3j+1}w_{3}(i)+4w_{4}(i)}{2}\\ & = &\frac{2\sum\nolimits_{j = 1}^{n}[2+2(\sum\nolimits_{k = 1}^{3j-2}k+\sum\nolimits_{k = 1}^{3n-3j+2}k)]+2\sum\nolimits_{j = 1}^{n}[1+2(\sum\nolimits_{k = 1}^{3j-1}k+\sum\nolimits_{k = 1}^{3n-3j+1}k)]}{2}\\ &&+\frac{2\sum\nolimits_{j = 1}^{n-1}[1+2(\sum\nolimits_{k = 1}^{3j}k+\sum\nolimits_{k = 1}^{3n-3j}k)]+4(1+2\sum\nolimits_{k = 1}^{3n}k)}{2}\\ & = &18n^{3}+18n^{2}+8n+1. \end{eqnarray*}$

Combining with $Kf(C_n)$ and $W(C_n)$ , we have

$\begin{eqnarray*} \lim\limits_{n\rightarrow \infty}\frac{Kf(C_{n})}{W(C_{n})} = \frac{1}{4}, \end{eqnarray*}$

as desired. ■

4. The multiplicative degree-Kirchhoff index, Gutman index, and complexity of $C_{n}$ in terms of the normalized Laplacian spectrum

In this section, we mainly calculate the multiplicative degree-Kirchhoff index, Gutman index, and complexity of $C_{n}$ .

According to (1.2), we can get $\mathcal{L}_{V_1 V_1}$ and $\mathcal{L}_{V_1 V_2}$ :

$\begin{eqnarray*} \mathcal{L}_{V_1 V_1}& = & \left( \begin{array}{cccccccccc} 1 & \frac{-1}{\sqrt{12}} & & & & & & & &\\ \frac{-1}{\sqrt{12}} & 1 & \frac{-1}{\sqrt{20}} & & & & & & &\\ & \frac{-1}{\sqrt{20}} & 1 & \frac{-1}{\sqrt{25}} & & & & & &\\ & & \frac{-1}{\sqrt{25}} & 1 & \frac{-1}{\sqrt{20}} & & & & &\\ & & & \frac{-1}{\sqrt{20}} & 1 & \frac{-1}{\sqrt{20}} & & & &\\ & & & & & \ddots & & & &\\ & & & & & \frac{-1}{\sqrt{25}} & 1 & \frac{-1}{\sqrt{20}} & &\\ & & & & & & \frac{-1}{\sqrt{20}} & 1 & \frac{-1}{\sqrt{20}} &\\ & & & & & & & \frac{-1}{\sqrt{20}} & 1 & \frac{-1}{\sqrt{15}}\\ & & & & & & & & \frac{-1}{\sqrt{15}} & 1\\ \end{array} \right)_{(3n+1)\times(3n+1)}, \end{eqnarray*}$

$\begin{eqnarray*} \mathcal{L}_{V_1 V_2}& = & \left( \begin{array}{cccccccccc} \frac{-1}{3} & \frac{-1}{\sqrt{12}} & & & & & & & &\\ \frac{-1}{\sqrt{12}} & 0 & \frac{-1}{\sqrt{20}} & & & & & & &\\ & \frac{-1}{\sqrt{20}} & \frac{-1}{5} & \frac{-1}{\sqrt{25}} & & & & & &\\ & & \frac{-1}{\sqrt{25}} & \frac{-1}{5} & \frac{-1}{\sqrt{20}} & & & & &\\ & & & \frac{-1}{\sqrt{20}} & 0 & \frac{-1}{\sqrt{20}} & & & &\\ & & & & & \ddots & & & &\\ & & & & & \frac{-1}{\sqrt{25}} & \frac{-1}{5} & \frac{-1}{\sqrt{20}} & &\\ & & & & & & \frac{-1}{\sqrt{20}} & 0 & \frac{-1}{\sqrt{20}} &\\ & & & & & & & \frac{-1}{\sqrt{20}} & \frac{-1}{5} & \frac{-1}{\sqrt{15}}\\ & & & & & & & & \frac{-1}{\sqrt{15}} & \frac{-1}{3}\\ \end{array} \right)_{(3n+1)\times(3n+1)}. \end{eqnarray*}$

Based on Theorem 2.1, the normalized Laplacian spectrum consists of the eigenvalues of $\mathcal{L}_{A}$ , and $\mathcal{L}_{S}$ of $C_{n}$ can be obtained.

$\begin{eqnarray*} \mathcal{L}_A& = & \left( \begin{array}{ccccccccccc} \frac{2}{3} & \frac{-1}{\sqrt{3}} & & & & & & & & &\\ \frac{-1}{\sqrt{3}} & 1 & \frac{-1}{\sqrt{5}} & & & & & & & &\\ & \frac{-1}{\sqrt{5}} & \frac{4}{5} & \frac{-2}{\sqrt{25}} & & & & & & &\\ & & \frac{-2}{\sqrt{25}} & \frac{4}{5} & \frac{-1}{\sqrt{5}} & & & & & &\\ & & & \frac{-1}{\sqrt{5}} & 1 & \frac{-1}{\sqrt{5}} & & & & &\\ & & & & \frac{-1}{\sqrt{5}} & \frac{4}{5} & \frac{-2}{\sqrt{25}} & & & &\\ & & & & & & \ddots & & & &\\ & & & & & & \frac{-2}{\sqrt{25}} & \frac{4}{5} & \frac{-1}{\sqrt{5}} & &\\ & & & & & & & \frac{-1}{\sqrt{5}} & 1 & \frac{-1}{\sqrt{5}} &\\ & & & & & & & & \frac{-1}{\sqrt{5}} & \frac{4}{5} & \frac{-2}{\sqrt{15}}\\ & & & & & & & & & \frac{-2}{\sqrt{15}} & \frac{2}{3}\\ \end{array} \right)_{(3n+1)\times(3n+1)}, \end{eqnarray*}$

$\begin{eqnarray*} \mathcal{L}_S = diag\Big(\frac{4}{3},1,\frac{6}{5},\frac{6}{5},1,\frac{6}{5},\cdots,\frac{6}{5},1,\frac{6}{5},\frac{4}{3}\Big)_{(3n+1)}. \end{eqnarray*}$

Assume that $0 = \gamma_{1} < \gamma_{2}\leq\gamma_{3}\leq\cdots\leq\gamma_{3n+1}$ are the roots of $P_{\mathcal{L}_{A}}(x)$ , and $0 < \delta_{1}\leq\delta_{2}\leq\delta_{3}\leq\cdots\leq\delta_{3n+1}$ are the roots of $P_{\mathcal{L}_{S}}(x)$ . By (1.6), we have

$\begin{eqnarray} Kf^{*}(C_{n}) = (28n+2)\Big(\sum\limits_{i = 2}^{3n+1}\frac{1}{\gamma_{i}}+\sum\limits_{i = 1}^{3n+1}\frac{1}{\delta_{i}}\Big). \end{eqnarray}$

(4.1)

It is obvious from the matrix $\mathcal{L}_S$ that the following expression can be obtained:

$\begin{eqnarray} \sum\limits_{i = 1}^{3n+1}\frac{1}{\delta_{i}} = 1\times{n}+\frac{5}{6}\times(2n-1)+\frac{3}{4}\times2 = \frac{8n+2}{3}. \end{eqnarray}$

(4.2)

Therefore, we need to calculate the first sum in (4.1).

Let

$\begin{eqnarray*} P_{\mathcal{L}_{A}}(x) = det(xI-\mathcal{L}_{A}) = x^{3n+1}+b_{1}x^{3n}+\cdots+b_{3n}x. \end{eqnarray*}$

Based on the Vieta $^{'}$ s Theorem of $P_{\mathcal{L}_{A}}(x)$ , we can get

$\begin{eqnarray} \sum\limits_{i = 2}^{3n+1}\frac{1}{\gamma_{i}} = \frac{(-1)^{3n-1}b_{3n-1}}{(-1)^{3n}b_{3n}}. \end{eqnarray}$

(4.3)

Obviously, we obtain that $(-1)^{3n}b_{3n}$ is the sum of all the principal minors of order $3n$ of $\mathcal{L}_{A}$ and $(-1)^{3n-1}b_{3n-1}$ is the sum of the principal minors of order $3n-1$ of $\mathcal{L}_{A}$ . So, let $T_{k}$ be the $k$ -th order principal submatrix, which consists of the first $k$ columns and $k$ rows of $\mathcal{L}_{A}$ , and $t_{k} = det(T_{k})$ , $k = 1, 2, \cdots, 3n$ . Thus, we can get $t_{1} = \frac{2}{3}, t_{2} = \frac{1}{3}, t_{3} = \frac{2}{15}$ , and for $1\leq{i}\leq{n-1}$ ,

$\begin{eqnarray*} \begin{cases} t_{3i+1} = \frac{4}{5}t_{3i}-\frac{4}{25}t_{3i-1};\\ t_{3i+2} = t_{3i+1}-\frac{1}{5}t_{3i};\\ t_{3i+3} = \frac{4}{5}t_{3i+2}-\frac{1}{5}t_{3i+1}. \end{cases} \end{eqnarray*}$

The solution of the previous recurrence relation is

$\begin{eqnarray*} \begin{cases} t_{3i-2} = \frac{25}{3}\cdot(\frac{2}{25})^{i};\\ t_{3i-1} = \frac{25}{6}\cdot(\frac{2}{25})^{i};\\ t_{3i} = \frac{5}{3}\cdot(\frac{2}{25})^{i}; \end{cases} \end{eqnarray*}$

where $1\leq{i}\leq{n}$ .

Now, let $S_{k}$ be the $k$ -th order principal submatrix, which consists of the last $k$ columns and $k$ rows of $\mathcal{L}_{A}$ , and $s_{k} = det(S_{k})$ , $k = 1, 2, \cdots, 3n$ . Thus, we can get $s_{1} = \frac{2}{3}, s_{2} = \frac{4}{15}, s_{3} = \frac{2}{15}$ , and for $1\leq{i}\leq{n-1}$ ,

$\begin{eqnarray*} \begin{cases} s_{3i+1} = \frac{4}{5}s_{3i}-\frac{1}{5}s_{3i-1};\\ s_{3i+2} = \frac{4}{5}s_{3i+1}-\frac{4}{25}s_{3i};\\ s_{3i+3} = s_{3i+2}-\frac{1}{5}s_{3i+1}. \end{cases} \end{eqnarray*}$

The solution of the previous recurrence relation is

$\begin{eqnarray*} \begin{cases} s_{3i-2} = \frac{25}{3}\cdot(\frac{2}{25})^{i};\\ s_{3i-1} = \frac{10}{3}\cdot(\frac{2}{25})^{i};\\ s_{3i} = \frac{5}{3}\cdot(\frac{2}{25})^{i}; \end{cases} \end{eqnarray*}$

where $1\leq{i}\leq{n}$ .

Without loss of generality, let $t_0 = 1$ and $s_0 = 1$ .

Fact 1. $(-1)^{3n}b_{3n} = \frac{5}{9}(14n+1)(\frac{2}{25})^{n}.$

Proof. By a similar discussion as in Section 3, we obtain that $(-1)^{3n}b_{3n}$ is the sum of all the principal minors of order $3n$ of $\mathcal{L}_{A}$ ,

$\begin{eqnarray*} (-1)^{3n}b_{3n}& = &\sum\limits_{j = 1}^{3n+1}t_{j-1}s_{3n+1-j}\\ & = &t_{3n}+s_{3n}+\sum\limits_{j = 2}^{3n}t_{j-1}s_{3n+1-j}\\ & = &2\cdot\frac{5}{3}\cdot\Big(\frac{2}{25}\Big)^{n}+\sum\limits_{i = 1}^{n}\frac{25}{3}\cdot\Big(\frac{2}{25}\Big)^{i}\cdot\frac{10}{3}\cdot\Big(\frac{2}{25}\Big)^{n-i+1} +\sum\limits_{i = 1}^{n}\frac{25}{6}\cdot\Big(\frac{2}{25}\Big)^{i}\cdot\frac{25}{3}\cdot\Big(\frac{2}{25}\Big)^{n-i+1}\\ &&+\sum\limits_{i = 1}^{n-1}\frac{5}{3}\cdot\Big(\frac{2}{25}\Big)^{i}\cdot\frac{5}{3}\cdot\Big(\frac{2}{25}\Big)^{n-i}\\ & = &\frac{5}{9}(14n+1)\Big(\frac{2}{25}\Big)^{n}. \end{eqnarray*}$

The result is as desired. ■

Fact 2. $(-1)^{3n-1}b_{3n-1} = \frac{98n^{3}+21n^{2}+9n}{45}(\frac{2}{25})^{n-1}.$

Proof. Since $(-1)^{3n-1}b_{3n-1}$ is the sum of all the principal minors of order $3n-1$ of $\mathcal{L}_{A}$ , one has

$\begin{eqnarray} (-1)^{3n-1}b_{3n-1} = \sum\limits_{1\leq{i} < j\leq3n+1}det\mathcal{L}_{A}[i,j]t_{i-1}s_{3n+1-j}. \end{eqnarray}$

(4.4)

In view of (4.4), all possible cases are listed.

Case 1. $i = 3p, j = 3q, 1\leq{p} < q\leq{n},$

$\begin{eqnarray*} det\mathcal{L}_{A}[3p,3q]& = & \left| \begin{array}{cccccccc} \frac{4}{5} & \frac{-1}{\sqrt{5}} & & & & & &\\ \frac{-1}{\sqrt{5}} & 1 & \frac{-1}{\sqrt{5}} & & & & &\\ & \frac{-1}{\sqrt{5}} & \frac{4}{5} & \frac{-2}{\sqrt{25}} & & & &\\ & & & \ddots & & & &\\ & & & \frac{-1}{\sqrt{5}} & 1 & \frac{-1}{\sqrt{5}} & &\\ & & & & \frac{-1}{\sqrt{5}} & \frac{4}{5} & \frac{-2}{\sqrt{25}} &\\ & & & & & \frac{-2}{\sqrt{25}} & \frac{4}{5} & \frac{-1}{\sqrt{5}}\\ & & & & & & \frac{-1}{\sqrt{5}} & 1\\ \end{array} \right|_{(3q-3p-1)\times (3q-3p-1)}\\ & = &\frac{3}{5}(q-p)\Big(\frac{2}{25}\Big)^{q-p-1}. \end{eqnarray*}$

Case 2. $i = 3p, j = 3q+1, 1\leq{p}\leq{q}\leq{n},$

$\begin{eqnarray*} det\mathcal{L}_{A}[3p,3q+1]& = & \left| \begin{array}{cccccccc} \frac{4}{5} & \frac{-1}{\sqrt{5}} & & & & & &\\ \frac{-1}{\sqrt{5}} & 1 & \frac{-1}{\sqrt{5}} & & & & &\\ & \frac{-1}{\sqrt{5}} & \frac{4}{5} & \frac{-2}{\sqrt{25}} & & & &\\ & & & \ddots & & & &\\ & & & \frac{-1}{\sqrt{5}} & \frac{4}{5} & \frac{-2}{\sqrt{25}} & &\\ & & & & \frac{-2}{\sqrt{25}} & \frac{4}{5} & \frac{-1}{\sqrt{5}} &\\ & & & & & \frac{-1}{\sqrt{5}} & 1 & \frac{-1}{\sqrt{5}}\\ & & & & & & \frac{-1}{\sqrt{5}} & \frac{4}{5}\\ \end{array} \right|_{(3q-3p)\times (3q-3p)}\\ & = &\frac{2}{25}[3(q-p)+1]\Big(\frac{2}{25}\Big)^{q-p-1}. \end{eqnarray*}$

Case 3. $i = 3p, j = 3q+2, 1\leq{p}\leq{q}\leq{n-1},$

$\begin{eqnarray*} det\mathcal{L}_{A}[3p,3q+2]& = & \left| \begin{array}{cccccccc} \frac{4}{5} & \frac{-1}{\sqrt{5}} & & & & & &\\ \frac{-1}{\sqrt{5}} & 1 & \frac{-1}{\sqrt{5}} & & & & &\\ & \frac{-1}{\sqrt{5}} & \frac{4}{5} & \frac{-2}{\sqrt{25}} & & & &\\ & & & \ddots & & & &\\ & & & \frac{-2}{\sqrt{25}} & \frac{4}{5} & \frac{-1}{\sqrt{5}} & &\\ & & & & \frac{-1}{\sqrt{5}} & 1 & \frac{-1}{\sqrt{5}} &\\ & & & & & \frac{-1}{\sqrt{5}} & \frac{4}{5} & \frac{-2}{\sqrt{25}}\\ & & & & & & \frac{-2}{\sqrt{25}} & \frac{4}{5}\\ \end{array} \right|_{(3q-3p+1)\times (3q-3p+1)}\\ & = &\frac{2}{5}[3(q-p)+2]\Big(\frac{2}{25}\Big)^{q-p}. \end{eqnarray*}$

Case 4. $i = 3p+1, j = 3q, 0\leq{p} < q\leq{n},$

$\begin{eqnarray*} det\mathcal{L}_{A}[3p+1,3q]& = & \left| \begin{array}{cccccccc} 1 & \frac{-1}{\sqrt{5}} & & & & & &\\ \frac{-1}{\sqrt{5}} & \frac{4}{5} & \frac{-2}{\sqrt{25}} & & & & &\\ & \frac{-2}{\sqrt{25}} & \frac{4}{5} & \frac{-1}{\sqrt{5}} & & & &\\ & & & \ddots & & & &\\ & & & \frac{-1}{\sqrt{5}} & 1 & \frac{-1}{\sqrt{5}} & &\\ & & & & \frac{-1}{\sqrt{5}} & \frac{4}{5} & \frac{-2}{\sqrt{25}} &\\ & & & & & \frac{-2}{\sqrt{25}} & \frac{4}{5} & \frac{-1}{\sqrt{5}}\\ & & & & & & \frac{-1}{\sqrt{5}} & 1\\ \end{array} \right|_{(3q-3p-2)\times (3q-3p-2)}\\ & = &\frac{1}{2}[3(q-p)-1]\Big(\frac{2}{25}\Big)^{q-p-1}. \end{eqnarray*}$

Case 5. $i = 3p+1, j = 3q+1, 0\leq{p} < q\leq{n},$

$\begin{eqnarray*} det\mathcal{L}_{A}[3p+1,3q+1]& = & \left| \begin{array}{cccccccc} 1 & \frac{-1}{\sqrt{5}} & & & & & &\\ \frac{-1}{\sqrt{5}} & \frac{4}{5} & \frac{-2}{\sqrt{25}} & & & & &\\ & \frac{-2}{\sqrt{25}} & \frac{4}{5} & \frac{-1}{\sqrt{5}} & & & &\\ & & & \ddots & & & &\\ & & & \frac{-1}{\sqrt{5}} & \frac{4}{5} & \frac{-2}{\sqrt{25}} & &\\ & & & & \frac{-2}{\sqrt{25}} & \frac{4}{5} & \frac{-1}{\sqrt{5}} &\\ & & & & & \frac{-1}{\sqrt{5}} & 1 & \frac{-1}{\sqrt{5}}\\ & & & & & & \frac{-1}{\sqrt{5}} & \frac{4}{5}\\ \end{array} \right|_{(3q-3p-1)\times (3q-3p-1)}\\ & = &\frac{3}{5}(q-p)\Big(\frac{2}{25}\Big)^{q-p-1}. \end{eqnarray*}$

Case 6. $i = 3p+1, j = 3q+2, 0\leq{p}\leq{q}\leq{n-1},$

$\begin{eqnarray*} det\mathcal{L}_{A}[3p+1,3q+2]& = & \left| \begin{array}{cccccccc} 1 & \frac{-1}{\sqrt{5}} & & & & & &\\ \frac{-1}{\sqrt{5}} & \frac{4}{5} & \frac{-2}{\sqrt{25}} & & & & &\\ & \frac{-2}{\sqrt{25}} & \frac{4}{5} & \frac{-1}{\sqrt{5}} & & & &\\ & & & \ddots & & & &\\ & & & \frac{-2}{\sqrt{25}} & \frac{4}{5} & \frac{-1}{\sqrt{5}} & &\\ & & & & \frac{-1}{\sqrt{5}} & 1 & \frac{-1}{\sqrt{5}} &\\ & & & & & \frac{-1}{\sqrt{5}} & \frac{4}{5} & \frac{-2}{\sqrt{25}}\\ & & & & & & \frac{-2}{\sqrt{25}} & \frac{4}{5}\\ \end{array} \right|_{(3q-3p)\times (3q-3p)}\\ & = &[3(q-p)+1]\Big(\frac{2}{25}\Big)^{q-p}. \end{eqnarray*}$

Case 7. $i = 3p+2, j = 3q, 0\leq{p} < q\leq{n},$

$\begin{eqnarray*} det\mathcal{L}_{A}[3p+2,3q]& = & \left| \begin{array}{cccccccc} \frac{4}{5} & \frac{-2}{\sqrt{25}} & & & & & &\\ \frac{-2}{\sqrt{25}} & \frac{4}{5} & \frac{-1}{\sqrt{5}} & & & & &\\ & \frac{-1}{\sqrt{5}} & 1 & \frac{-1}{\sqrt{5}} & & & &\\ & & & \ddots & & & &\\ & & & \frac{-1}{\sqrt{5}} & 1 & \frac{-1}{\sqrt{5}} & &\\ & & & & \frac{-1}{\sqrt{5}} & \frac{4}{5} & \frac{-2}{\sqrt{25}} &\\ & & & & & \frac{-2}{\sqrt{25}} & \frac{4}{5} & \frac{-1}{\sqrt{5}}\\ & & & & & & \frac{-1}{\sqrt{5}} & 1\\ \end{array} \right|_{(3q-3p-3)\times (3q-3p-3)}\\ & = &[3(q-p)-2]\Big(\frac{2}{25}\Big)^{q-p-1}. \end{eqnarray*}$

Case 8. $i = 3p+2, j = 3q+1, 0\leq{p} < q\leq{n},$

$\begin{eqnarray*} det\mathcal{L}_{A}[3p+2,3q+1]& = & \left| \begin{array}{cccccccc} \frac{4}{5} & \frac{-2}{\sqrt{25}} & & & & & &\\ \frac{-2}{\sqrt{25}} & \frac{4}{5} & \frac{-1}{\sqrt{5}} & & & & &\\ & \frac{-1}{\sqrt{5}} & 1 & \frac{-1}{\sqrt{5}} & & & &\\ & & & \ddots & & & &\\ & & & \frac{-1}{\sqrt{5}} & \frac{4}{5} & \frac{-2}{\sqrt{25}} & &\\ & & & & \frac{-2}{\sqrt{25}} & \frac{4}{5} & \frac{-1}{\sqrt{5}} &\\ & & & & & \frac{-1}{\sqrt{5}} & 1 & \frac{-1}{\sqrt{5}}\\ & & & & & & \frac{-1}{\sqrt{5}} & \frac{4}{5}\\ \end{array} \right|_{(3q-3p-2)\times (3q-3p-2)}\\ & = &\frac{2}{5}[3(q-p)-1]\Big(\frac{2}{25}\Big)^{q-p-1}. \end{eqnarray*}$

Case 9. $i = 3p+2, j = 3q+2, 0\leq{p} < q\leq{n-1},$

$\begin{eqnarray*} det\mathcal{L}_{A}[3p+2,3q+2]& = & \left| \begin{array}{cccccccc} \frac{4}{5} & \frac{-2}{\sqrt{25}} & & & & & &\\ \frac{-2}{\sqrt{25}} & \frac{4}{5} & \frac{-1}{\sqrt{5}} & & & & &\\ & \frac{-1}{\sqrt{5}} & 1 & \frac{-1}{\sqrt{5}} & & & &\\ & & & \ddots & & & &\\ & & & \frac{-2}{\sqrt{25}} & \frac{4}{5} & \frac{-1}{\sqrt{5}} & &\\ & & & & \frac{-1}{\sqrt{5}} & 1 & \frac{-1}{\sqrt{5}} &\\ & & & & & \frac{-1}{\sqrt{5}} & \frac{4}{5} & \frac{-2}{\sqrt{25}}\\ & & & & & & \frac{-2}{\sqrt{25}} & \frac{4}{5}\\ \end{array} \right|_{(3q-3p-1)\times (3q-3p-1)}\\ & = &\frac{12}{25}(q-p)\Big(\frac{2}{25}\Big)^{q-p-1}. \end{eqnarray*}$

Therefore, we can get

$\begin{eqnarray*} (-1)^{3n-1}b_{3n-1}& = &\sum\limits_{1\leq{i} < j\leq3n+1}det\mathcal{L}_{A}[i,j]t_{i-1}s_{3n+1-j}\\ & = &X_{1}+X_{2}+X_{3}, \end{eqnarray*}$

where

$\begin{eqnarray*} X_{1}& = &\sum\limits_{1\leq{p} < q\leq{n}}det\mathcal{L}_{A}[3p,3q]t_{3p-1}s_{3n-3q+1}+\sum\limits_{1\leq{p}\leq{q}\leq{n}}det\mathcal{L}_{A}[3p,3q+1]t_{3p-1}s_{3n-3q}\\ &&+\sum\limits_{1\leq{p}\leq{q}\leq{n-1}}det\mathcal{L}_{A}[3p,3q+2]t_{3p-1}s_{3n-3q-1}\\ & = &\frac{14n^{3}-5n^{2}-3n}{18}\Big(\frac{2}{25}\Big)^{n-1}, \end{eqnarray*}$

$\begin{eqnarray*} X_{2}& = &\sum\limits_{0\leq{p} < q\leq{n}}det\mathcal{L}_{A}[3p+1,3q]t_{3p}s_{3n-3q+1}+\sum\limits_{0\leq{p} < q\leq{n}}det\mathcal{L}_{A}[3p+1,3q+1]t_{3p}s_{3n-3q}\\ &&+\sum\limits_{0\leq{p}\leq{q}\leq{n-1}}det\mathcal{L}_{A}[3p+1,3q+2]t_{3p}s_{3n-3q-1}\\ & = &\frac{70n^{3}+31n^{2}+37n}{90}\Big(\frac{2}{25}\Big)^{n-1}, \end{eqnarray*}$

$\begin{eqnarray*} X_{3}& = &\sum\limits_{0\leq{p} < q\leq{n}}det\mathcal{L}_{A}[3p+2,3q]t_{3p+1}s_{3n-3q+1}+\sum\limits_{0\leq{p} < q\leq{n}}det\mathcal{L}_{A}[3p+2,3q+1]t_{3p+1}s_{3n-3q}\\ &&+\sum\limits_{0\leq{p} < q\leq{n-1}}det\mathcal{L}_{A}[3p+2,3q+2]t_{3p+1}s_{3n-3q-1}\\ & = &\frac{2}{45}(14n^{3}+9n^{2}-n)\Big(\frac{2}{25}\Big)^{n-1}. \end{eqnarray*}$

Thus, we can obtain

$\begin{eqnarray*} (-1)^{3n-1}b_{3n-1} = X_{1}+X_{2}+X_{3} = \frac{98n^{3}+21n^{2}+9n}{45}\Big(\frac{2}{25}\Big)^{n-1}, \end{eqnarray*}$

which is the desired result. ■

Together with (4.3) and Facts 1 - 2, one can get the following lemma.

Lemma 4.1. Assume that $0 = \gamma_{1} < \gamma_{2}\leq\gamma_{3}\leq\cdots\leq\gamma_{3n+1}$ are the eigenvalues of $\mathcal{L}_A$ , then one gets

$\begin{eqnarray*} \sum\limits_{i = 2}^{3n+1}\frac{1}{\gamma_{i}} = \frac{98n^{3}+21n^{2}+9n}{28n+2}. \end{eqnarray*}$

According to (4.1) - (4.2) and Lemma 4.1, we obtain the following theorem.

Theorem 4.2. For linear crossed phenylenes $C_n$ , we have

$\begin{eqnarray*} Kf^{*}(C_{n}) = \frac{294n^{3}+287n^{2}+99n+4}{3}. \end{eqnarray*}$

The multiplicative degree-Kirchhoff indices of $C_{n}$ are shown in , where $1\leq{n}\leq15$ .

Theorem 4.3. Assume that $C_n$ are the linear crossed phenylenes, then

$\begin{eqnarray*} \lim\limits_{n\rightarrow \infty}\frac{Kf^{*}(C_{n})}{Gut(C_{n})} = \frac{1}{4}. \end{eqnarray*}$

Proof. By first classifying and discussing the following cases of vertices, the Gutman index of $C_n$ is obtained.

Table 2. Multiplicative degree-Kirchhoff indices from

$C_{1}$ to

$C_{15}$ .

$G$	$Kf^{*}(G)$	$G$	$Kf^{*}(G)$	$G$	$Kf^{*}(G)$	$G$	$Kf^{*}(G)$	$G$	$Kf^{*}(G)$
$C_{1}$	$208.00$	$C_{4}$	$7688.00$	$C_{7}$	$37806.00$	$C_{10}$	$106438.00$	$C_{13}$	$229460.00$
$C_{2}$	$1166.00$	$C_{5}$	$14428.00$	$C_{8}$	$55620.00$	$C_{11}$	$140618.00$	$C_{14}$	$285298.00$
$C_{3}$	$3463.33$	$C_{6}$	$24271.33$	$C_{9}$	$78301.33$	$C_{12}$	$181429.33$	$C_{15}$	$349531.33$

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● Vertex $3i-1(i = 1, 2, \cdots, n)$ of $C_n$ :

$\begin{eqnarray*} g_{3i-1}& = &2\sum\limits_{i = 1}^{n}\big[4\times4\times2+2\times3\times4\times(3i-2)+2\times3\times4\times(3n-3i+2)\\ &&+2\sum\limits_{k = 1}^{i-1}4\times4\times3\times(i-k)+2\sum\limits_{k = i+1}^{n}4\times4\times3\times(k-i)\\ &&+2\sum\limits_{k = 2}^{i}4\times5\times(3i-3k+1)+2\sum\limits_{k = i+1}^{n}4\times5\times(3k-3i-1)\\ &&+2\sum\limits_{k = 1}^{i-1}4\times5\times(3i-k-1)+2\sum\limits_{k = i}^{n}4\times5\times(3k-3i+1)\big]\\ & = &8(28n^{3}+3n^{2}+5n). \end{eqnarray*}$

● Vertex $3i(i = 1, 2, \cdots, n)$ of $C_n$ :

$\begin{eqnarray*} g_{3i}& = &2\sum\limits_{i = 1}^{n}\big[5\times5\times1+2\times3\times5\times(3i-1)+2\times3\times5\times(3n-3i+1)\\ &&+2\sum\limits_{k = 1}^{i}4\times5\times(3i-3k+1)+2\sum\limits_{k = i+1}^{n}4\times5\times(3k-3i-1)\\ &&+2\sum\limits_{k = 2}^{i}5\times5\times(3i-3k+2)+2\sum\limits_{k = i+1}^{n}5\times5\times(3k-3i-2)\\ &&+2\sum\limits_{k = 1}^{i-1}5\times5\times3\times(i-k)+2\sum\limits_{k = i+1}^{n}5\times5\times3\times(k-i)\big]\\ & = &10(28n^{3}+3n^{2}). \end{eqnarray*}$

● Vertex $3i-2(i = 2, 3, \cdots, n)$ of $C_n$ :

$\begin{eqnarray*} g_{3i-2}& = &2\sum\limits_{i = 2}^{n}\big[5\times5\times1+2\times3\times5\times(3i-3)+2\times3\times5\times(3n-3i+3)\\ &&+2\sum\limits_{k = 1}^{i-1}4\times5\times(3i-3k-1)+2\sum\limits_{k = i}^{n}4\times5\times(3k-3i+1)\\ &&+2\sum\limits_{k = 2}^{i-1}5\times5\times3\times(i-k)+2\sum\limits_{k = i+1}^{n}5\times5\times3\times(k-i)\\ &&+2\sum\limits_{k = 1}^{i-1}5\times5\times(3i-3k-2)+2\sum\limits_{k = i}^{n}5\times5\times(3k-3i+2)\big]\\ & = &10(28n^{3}-39n^{2}+16n-5). \end{eqnarray*}$

● Corner vertex of $C_n$ :

$\begin{eqnarray*} g_o& = &2\big[3\times3\times1+2\times3\times3\times3n+2(\sum\limits_{i = 1}^{n}3\times4(3i-2)\\ &&+\sum\limits_{i = 1}^{n}3\times5\times(3i-1)+\sum\limits_{i = 2}^{n}3\times5\times(3i-3))\big]\\ &&+2\big[3\times3\times1+2\times3\times3\times3n+2(\sum\limits_{i = 1}^{n}3\times4\times(3n-3i+2)\\ &&+\sum\limits_{i = 1}^{n}3\times5\times(3n-3i+1)+\sum\limits_{i = 2}^{n}3\times5\times(3n-3i+3))\big]\\ & = &504n^{2}+36n+36. \end{eqnarray*}$

Applying (1.4), we obtain

$\begin{eqnarray*} Gut(C_{n}) = \frac{g_{o}+g_{3i-1}+g_{3i-2}+g_{3i}}{2} = 392n^{3}+84n^{2}+118n-7. \end{eqnarray*}$

Combining with $Kf^{*}(C_n)$ and $Gut(C_n)$ , one has

$\begin{eqnarray*} \lim\limits_{n\rightarrow \infty}\frac{Kf^{*}(C_{n})}{Gut(C_{n})} = \frac{1}{4}. \end{eqnarray*}$

This completes the proof. ■

In the following, we can calculate the complexity of $C_{n}$ .

Theorem 4.4. For linear crossed phenylenes $C_{n}$ , we have

$\begin{eqnarray*} \tau(C_{n}) = 2^{7n+2}\cdot3^{2n-1}. \end{eqnarray*}$

Proof. Based on Theorem 2.2, we can get $\prod_{i = 1}^{6n+2}d_{i}\prod_{i = 2}^{3n+1}\gamma_{i}\prod_{i = 1}^{3n+1}\delta_{i} = 2(14n+1)\tau(C_{n}),$

where

$\begin{eqnarray*} \prod\limits_{i = 1}^{6n+2}d_{i} = 3^{4}\cdot4^{2n}\cdot5^{4n-2}, \end{eqnarray*}$

$\begin{eqnarray*} \prod\limits_{i = 2}^{3n+1}\gamma_{i} = (-1)^{3n}b_{3n} = \frac{5}{9}\cdot(14n+1)\cdot\Big(\frac{2}{25}\Big)^{n}, \end{eqnarray*}$

$\begin{eqnarray*} \prod\limits_{i = 1}^{3n+1}\delta_{i} = \Big(\frac{4}{3}\Big)^{2}\cdot\Big(\frac{6}{5}\Big)^{2n-1}. \end{eqnarray*}$

Hence,

$\begin{eqnarray*} \tau(C_{n}) = 2^{7n+2}\cdot3^{2n-1}. \end{eqnarray*}$

The result is as desired. ■

The complexity of $C_{n}$ is shown in , where $1\leq{n}\leq12$ .

Table 3. The complexity from

$C_{1}$ to

$C_{12}$ .

$G$	$\tau(G)$	$G$	$\tau(G)$
$C_{1}$	$1536$	$C_{7}$	$3590096234354105647104$
$C_{2}$	$1769472$	$C_{8}$	$4135790861975929705463808$
$C_{3}$	$2038431744$	$C_{9}$	$4764431072996271020694306816$
$C_{4}$	$2348273369088$	$C_{10}$	$5488624596091704215839841452032$
$C_{5}$	$2705210921189376$	$C_{11}$	$6322895534697643256647497352740864$
$C_{6}$	$3116402981210161152$	$C_{12}$	$7283975655971685031657916950357475328$

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5. Conclusions

Based on the Laplacian (normalized Laplacian, resp) polynomial of $C_{n}$ , we determined the Kirchhoff index, multiplicative degree-Kirchhoff index, and complexity of linear crossed phenylenes through the decomposition theorem and Vieta $^{'}$ s Theorem. In addition, we found that the Kirchhoff index and multiplicative degree-Kirchhoff index of linear crossed phenylenes were approximately one quarter of their Wiener index and Gutman index, respectively, which further enriched the results of the Kirchhoff index, multiplicative degree-Kirchhoff index, and complexity for the linear crossed chains.

Use of AI tools declaration

The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

Acknowledgments

This work was supported in part by Anhui Provincial Natural Science Foundation Project under Grant KJ2021A1175, Funding project for cultivating top-notch talents in universities under Grant gxgnfx2022096, Anhui Provincial Education Reform Research Project under Grant 2022jyxm481, Anhui Provincial Natural Science Foundation Project under Grant 2023AH051697.

Conflict of interest

All authors declare no conflicts of interest in this paper.

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Topological indices of linear crossed phenylenes with respect to their Laplacian and normalized Laplacian spectrum

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3. The Kirchhoff index and Wiener index of $C_{n}$ in terms of the Laplacian spectrum

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Topological indices of linear crossed phenylenes with respect to their Laplacian and normalized Laplacian spectrum

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