The Laplacians, Kirchhoff index and complexity of linear Möbius and cylinder
octagonal-quadrilateral networks

Jia-Bao Liu 1,∗, Lu-Lu Fang 1,∗, Qian Zheng 1, Xin-Bei Peng 1

1School of Mathematics and Physics, Anhui Jianzhu University, Hefei 230601,
P.R. China

††footnotetext: E-mail address<EMAIL_ADDRESS>fangluluajd<EMAIL_ADDRESS><EMAIL_ADDRESS>* Corresponding author.

Abstract. Spectrum graph theory not only facilitate comprehensively reflect
the topological structure and dynamic characteristics of networks, but also
offer significant and noteworthy applications in theoretical chemistry,
network science and other fields. Let $L_{n}^{8,4}$ represent a linear
octagonal-quadrilateral network, consisting of $n$ eight-member ring and $n$
four-member ring. The Möbius graph $Q_{n}(8,4)$ is constructed by reverse
identifying the opposite edges, whereas cylinder graph $Q^{\prime}_{n}(8,4)$
identifies the opposite edges by order. In this paper, the explicit formulas
of Kirchhoff indices and complexity of $Q_{n}(8,4)$ and $Q^{\prime}_{n}(8,4)$
are demonstrated by Laplacian characteristic polynomials according to
decomposition theorem and Vieta’s theorem. In surprise, the Kirchhoff index of
$Q_{n}(8,4)$($Q^{\prime}_{n}(8,4)$) is approximately one-third half of its
Wiener index as $n\to\infty$.
Keywords: Kirchhoff index; Complexity; Möbius graph; Cylinder graph.

##  1 Introduction

The graphs in this paper are simple, undirected and connected. Firstly recall
some definitions that most used in graph theory. Suppose $G$ as a simple
undirected graph with $\big{|}V_{G}\big{|}=n$ and $\big{|}E_{G}\big{|}=m$. The
adjacent matrix $A(G)=[a_{i,j}]_{n\times n}$ of $G$ is a symmetric matrix.
When the vertex $i$ is adjacent to $j$, we define $a_{i,j}=1$ and $a_{i,j}=0$
otherwise. For a vertex $i\in V_{G}$, $d_{i}$ is the number of edges arised
from $i$ and the diagonal matrix $D(G)=diag\\{d_{1},d_{2},\cdots,d_{n}\\}$ is
degree matrix. Subsequently, the Laplacian matrix is specified as
$L(G)=D(G)-A(G)$, which spectrum can be expressed by
$0=\mu_{1}<\mu_{2}\leq\cdots\leq\mu_{n}$. For more notations, one can be
referred to [1].

Alternatively, The $(m,n)$-entry of the Laplacian matrix can be noted by

$\displaystyle\big{(}L(G)\big{)}_{mn}=\begin{cases}d_{m},&m=n;\\\
-1,&m\neq{n}~{}and~{}v_{m}\backsim v_{n};\\\ 0,&otherwise.\end{cases}$ (1.1)

The traditional distance for every pairs of vertices $i$ and $j$ is
$d_{G}(v_{i},v_{j})=\\{d_{ij}|\ the\ shortset\ length\ of\ v_{i}\ \\\ and\
v_{j}\\}$. For many parameters used to describe the structure of a graph in
chemical and mathematical researches[2, 3, 4], the most commonly used one is
the distance-based parameter Wiener index[5, 6], which is known to

$\displaystyle W(G)=\sum_{i<j}d_{ij}.$

Suppose that each edge of a connected graph $G$ is regarded as a unit
resistor, and the resistance distance[7] between any two vertices $i$ and $j$
is recorded as $r_{ij}$. Similar to Wiener index, according to the resistance
distance, the expression of Kirchhoff index [8, 9] are given, namely

$\displaystyle Kf(G)=\sum_{i<j}r_{ij}.$

It’s well known that scholars in various fields have a strong interest in the
study of the Kirchhoff index, this has promoted researchers to try some
calculation methods to compute the Kirchhoff index of a given graph and get
its closed formula. As early as 1985, Y.L. Yang et al.[10] gave the
corresponding decomposition theorem for the corresponding matrix of linear
structure. Up to now, many scholars have used this method to describe the
Kirchhoff index of a series of linear hydrocarbons chain. For instances, in
2007, Yang et al.[11] calculated the Kirchhoff index of linear hexagonal chain
by using the decomposition of Laplacian matrix. In 2019, Geng et al.[12] used
this method to calculate the Möbius chain and cylinder chain of phenylenes
chain. Shi and Liu et al.[13] computed the Kirchhoff index of linear
octagonal-quadrilateral network in 2020. For more information, see [14, 15,
16, 17, 18, 19, 20]. After learning the excellent works of scholars, this
paper computes the Kirchhoff indices, Wiener indices and the complexity of
Möbius graph and its cylinder graph of linear octagonal-quadrilateral network.

Let $L_{n}^{8,4}$ be the linear octagonal-quadrilateral network as illustrated
in Figure 1, and octagons and quadrilaterals are connected by a common edge.
Then the corresponding Möbius graph $Q_{3}\big{(}8,4\big{)}$ of octagonal-
quadrilateral network is obtained by the reverse identification of the
opposite edge by $L_{n}^{8,4}$, and its cylinder graph
$Q^{\prime}_{3}\big{(}8,4\big{)}$ of octagonal-quadrilateral network is
obtained by identifying the opposite edge of $L_{n}^{8,4}$. An illustration of
this are given in Figure 2. Obviously, we can obtained that
$\big{|}V_{Q_{n}}(8,4)\big{|}=8n$, $\big{|}E_{Q_{n}}(8,4)\big{|}=10n$ and
$\big{|}V_{Q^{\prime}_{n}}(8,4)\big{|}=8n$,
$\big{|}E_{Q^{\prime}_{n}}(8,4)\big{|}=10n.$

Figure 1: Linear octagonal-quadrilateral networks.  Figure 2: A special class
of octagonal-quadrilateral networks for $n=3$.

The rest of this paper is organized as follows: In Section 2, we introduce
some basic notations and related lemmas. In Section 3, the Kirchhoff index,
the Wiener index and the complexity of $Q_{n}\big{(}8,4\big{)}$ and
$Q^{\prime}_{n}\big{(}8,4\big{)}$ are be computed. In Section 4, we
demonstrate the ratio of Wiener index and Kirchhoff index of
$Q_{n}\big{(}8,4\big{)}$ and $Q^{\prime}_{n}\big{(}8,4\big{)}$. Finally, we
provide insights into the subsequent extensions in Section 5.

##  2 Preliminary

We recommend the notion of symbols and related calculation methods. This in
turn would help to the proof of the rest of the article. The characteristic
polynomial of matrix $A$ of order $n$ is defined as
$P_{A}(x)=det\big{(}xI-A\big{)}$. We note that $\varpi$ is an automorphism of
$G$, we can write the product of disjoint 1-cycles and transposition, namely

$\mathscr{\varpi}=(\bar{1})(\bar{2})\cdots(\bar{m})(1,1^{\prime})(2,2^{\prime})\cdots(k,k^{\prime}).$

Then $\big{|}V(G)\big{|}=m+2k$, let
$v_{0}=\big{\\{}\bar{1},\bar{2},\cdots\bar{m}\big{\\}},~{}v_{1}=\big{\\{}1,2\cdots
k\big{\\}},~{}v_{2}=\big{\\{}1^{\prime},2^{\prime}\cdots k^{\prime}\big{\\}}.$
Thus the Laplacian matrix can be formed by block matrix, then

$L(G)=\left(\begin{array}[]{ccc}L_{V_{0}V_{0}}&L_{V_{0}V_{1}}&L_{V_{0}V_{2}}\\\
L_{V_{1}V_{0}}&L_{V_{1}V_{1}}&L_{V_{1}V_{2}}\\\
L_{V_{2}V_{0}}&L_{V_{2}V_{1}}&L_{V_{2}V_{2}}\\\ \end{array}\right),$ (2.2)

where

$L_{V_{0}V_{1}}=L_{V_{0}V_{2}},~{}L_{V_{1}V_{2}}=L_{V_{2}V_{1}},~{}and~{}L_{V_{1}V_{1}}=L_{V_{2}V_{2}}.$
(2.3)

Let

$P=\left(\begin{array}[]{ccc}I_{m}&0&0\\\
0&\frac{1}{\sqrt{2}}I_{k}&\frac{1}{\sqrt{2}}I_{k}\\\
0&\frac{1}{\sqrt{2}}I_{k}&-\frac{1}{\sqrt{2}}I_{k}\end{array}\right),$ (2.4)

where $I_{m}$ is the identity matrix of order $n$. Note that $P^{\prime}$ is
the transposition of $P$. Comprehensive consideration of Eqs.(2.2)-(2.4), thus
we employ

$P^{\prime}L(G)P=\left(\begin{array}[]{cc}L_{A}&0\\\
0&L_{S}\end{array}\right),$

where

$\displaystyle
L_{A}=\left(\begin{array}[]{cc}L_{V_{0}V_{0}}&\sqrt{2}L_{V_{0}V_{1}}\\\
\sqrt{2}L_{V_{1}V_{0}}&L_{V_{1}V_{1}}+L_{V_{1}V_{2}}\\\
\end{array}\right),~{}L_{S}=L_{V_{1}V_{1}}-L_{V_{1}V_{2}}.$

######  Lemma 2.1.

[10] $L_{A},~{}L_{S}$ thus obtained above can be used in $P_{L(L_{n})}$,
namely

$\displaystyle
P_{L(L_{n})}\big{(}x\big{)}=P_{L_{A}}\big{(}x\big{)}P_{L_{S}}\big{(}x\big{)}.$

######  Lemma 2.2.

[21] Gutman and mohar show a new calculation formula, namely

$\displaystyle Kf(G)=n\sum_{k=2}^{n}\frac{1}{\mu_{k}},$

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

######  Lemma 2.3.

[22] The number of spanning trees of $G$ can also be concluded to the
complexity, which is defined by

$\displaystyle\mathscr{L}(G)=\frac{1}{n}\prod_{i=1}^{n}\mu_{i}.$

######  Lemma 2.4.

[23] Assign $C_{n}$ for a cycle with $n$ vertices. The explicit formula of
Laplacian spectrum of $C_{n}$ can be constructed, which is expressed as

$\displaystyle S_{p}\big{(}L(C_{n})\big{)}=\Big{\\{}2-2\cos\frac{(2\pi
i)}{n}\Big{|}1\leq i\leq n\Big{\\}},~{}$

and the Kirchhoff index of $C_{n}$ is given by

$\displaystyle Kf\big{(}C_{n}\big{)}=\frac{n^{3}-n}{12}.$

##  3 Main results

In this section, we firstly focus on calculating the systematic formula for
the Kirchhoff index of $Q_{n}\big{(}8,4\big{)}$. By Lemma 2.1, we apply the
notion of the eigenvalues of Laplacian matrix to compute the
$Kf\big{(}Q_{n}(8,4)\big{)}$. Besides, we determine the complexity of
$Q_{n}\big{(}8,4\big{)}$, which is made up of the products of degree of all
vertices. The scenario is similar for $Q^{\prime}_{n}\big{(}8,4\big{)}$.

Given an $n\times n$ matrix $R$, and put deleting the
$i_{1}th,~{}i_{2}th,~{}\cdots,~{}i_{k}th$ rows and columns of $R$ are
expressed as $R\big{[}\\{i_{1},i_{2},\cdots,~{}i_{k}\\}\big{]}$. The vertices
of $Q_{n}\big{(}8,4\big{)}$ are tabulated in Figure 2. Evidently,
$\pi=(1,1^{\prime})(2,2^{\prime})\cdots(4n,(4n))^{\prime})$ is an automorphism
of $Q_{n}\big{(}8,4\big{)}$ and
$v_{0}=\emptyset,~{}v_{1}=\big{\\{}1,2,3,\cdots,4n\big{\\}},~{}v_{2}=\big{\\{}1^{\prime},2^{\prime},3^{\prime},\cdots,(4n)^{\prime}\big{\\}}.$
Maenwile, we assign $L_{A}\big{(}Q_{n}(8,4)\big{)}$ and
$L_{S}\big{(}Q_{n}(8,4)\big{)}$ as $L_{A}$ and $L_{S}$. Then one can get

$\displaystyle
L_{A}=L_{V_{1}V_{1}}+L_{V_{1}V_{2}},~{}L_{S}=L_{V_{1}V_{1}}-L_{V_{1}V_{2}}.$
(3.6)

By Equation (3.5), we have

$\displaystyle L_{V_{1}V_{1}}$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{ccccccccccc}3&-1&&&&&&&&&\\\
-1&2&-1&&&&&&&&\\\ &-1&2&-1&&&&&&&\\\ &&-1&3&-1&&&&&&\\\ &&&-1&3&-1&&&&&\\\
&&&&-1&2&-1&&&&\\\ &&&&&&\ddots&&&&\\\ &&&&&&-1&3&-1&&\\\ &&&&&&&-1&2&-1&\\\
&&&&&&&&-1&2&-1\\\ &&&&&&&&&-1&3\\\ \end{array}\right)_{(4n)\times(4n)},$
$\displaystyle L_{V_{1}V_{2}}$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{ccccccccccc}-1&&&&&&&&&&-1\\\
&0&&&&&&&&&\\\ &&0&&&&&&&&\\\ &&&-1&&&&&&&\\\ &&&&0&&&&&&\\\ &&&&&0&&&&&\\\
&&&&&&\ddots&&&&\\\ &&&&&&&-1&&&\\\ &&&&&&&&0&&\\\ &&&&&&&&&0&\\\
-1&&&&&&&&&&-1\\\ \end{array}\right)_{(4n)\times(4n)}.$

Hence,

$\displaystyle L_{A}$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{ccccccccccc}2&-1&&&&&&&&&-1\\\
-1&2&-1&&&&&&&&\\\ &-1&2&-1&&&&&&&\\\ &&-1&2&-1&&&&&&\\\ &&&-1&2&-1&&&&&\\\
&&&&-1&2&-1&&&&\\\ &&&&&&\ddots&&&&\\\ &&&&&&-1&2&-1&&\\\ &&&&&&&-1&2&-1&\\\
&&&&&&&&-1&2&-1\\\ -1&&&&&&&&&-1&2\\\ \end{array}\right)_{(4n)\times(4n)},$

and

$\displaystyle L_{S}$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{ccccccccccc}4&-1&&&&&&&&&1\\\
-1&2&-1&&&&&&&&\\\ &-1&2&-1&&&&&&&\\\ &&-1&4&-1&&&&&&\\\ &&&-1&4&-1&&&&&\\\
&&&&-1&2&-1&&&&\\\ &&&&&&\ddots&&&&\\\ &&&&&&-1&4&-1&&\\\ &&&&&&&-1&2&-1&\\\
&&&&&&&&-1&2&-1\\\ 1&&&&&&&&&-1&4\\\ \end{array}\right)_{(4n)\times(4n)}.$

Assuming that $0=\alpha_{1}<\alpha_{2}\leq\alpha_{3}\leq\cdots\leq\alpha_{4n}$
are the roots of $P_{L_{A}}\big{(}x\big{)}=0$, and
$0<\beta_{1}\leq\beta_{2}\leq\beta_{3}\leq\cdots\leq\beta_{4n}$ are the roots
of $P_{L_{S}}\big{(}x\big{)}=0$. By Lemma 2.2, we immediately have

$\displaystyle
Kf\big{(}Q_{n}(8,4)\big{)}=8n\Bigg{(}\sum_{i=2}^{4n}\frac{1}{\alpha_{i}}+\sum_{j=1}^{4n}\frac{1}{\beta_{j}}\Bigg{)}.$
(3.11)

Next, the prime aim of follows is to calculate
$\sum\limits_{i=2}^{4n}\frac{1}{\alpha_{i}}$ and
$\sum\limits_{j=1}^{4n}\frac{1}{\beta_{j}}$.

Combing with Lemma 2.1 and Lemma 2.4, one can get

$\displaystyle
Kf\big{(}Q_{n}(8,4)\big{)}=8n\Bigg{(}\frac{16n^{2}-1}{12}+\sum_{j=1}^{4n}\frac{1}{\beta_{j}}\Bigg{)}.$
(3.12)

######  Lemma 3.1.

Let $\beta_{j}$ be the eigenvalue of $L_{S}$. And assume that
$\mu=15+4\sqrt{14}$, $\nu=15-4\sqrt{14}$.

$\displaystyle\sum_{j=1}^{4n}\frac{1}{\beta_{j}}=\frac{(-1)^{4n-1}b_{4n-1}}{detL_{S}}.$
(3.13)

Proof.

$P_{L_{S}}(x)=det(xI-
L_{S})=x^{4n}+b_{1}x^{4n-1}+\cdots+b_{4n-1}x+b_{4n},~{}b_{4n}\neq 0.$

The key to get Eq.(3.8) is to determine $(-1)^{4n-1}b_{4n-1}$ and $detL_{S}$.
We demonstrate the $ith$ order principal submatrix
$M_{i}~{}and~{}M^{\prime}_{i}$ by from the first $i$ rows and columns of the
following matrices $L^{1}_{S}$ and $L^{2}_{S}$ respectively. For
$i=1,2,...,4n-1$, we have

$\displaystyle L_{S}^{1}$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{ccccccccccc}4&-1&&&&&&&&&\\\
-1&2&-1&&&&&&&&\\\ &-1&2&-1&&&&&&&\\\ &&-1&4&-1&&&&&&\\\ &&&-1&4&-1&&&&&\\\
&&&&-1&2&-1&&&&\\\ &&&&&&\ddots&&&&\\\ &&&&&&-1&4&-1&&\\\ &&&&&&&-1&2&-1&\\\
&&&&&&&&-1&2&-1\\\ &&&&&&&&&-1&4\\\ \end{array}\right)_{(4n)\times(4n)},$

and

$\displaystyle L_{S}^{2}$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{ccccccccccc}2&-1&&&&&&&&&\\\
-1&2&-1&&&&&&&&\\\ &-1&4&-1&&&&&&&\\\ &&-1&4&-1&&&&&&\\\ &&&-1&2&-1&&&&&\\\
&&&&-1&2&-1&&&&\\\ &&&&&&\ddots&&&&\\\ &&&&&&-1&2&-1&&\\\ &&&&&&&-1&2&-1&\\\
&&&&&&&&-1&4&-1\\\ &&&&&&&&&-1&4\\\ \end{array}\right)_{(4n)\times(4n)}.$

Applying the results above, we can get two Claims.

Claim 1. For $1\leq j\leq 4n$,

$\displaystyle
q_{j}=\begin{cases}\Big{(}\frac{1}{2}+\frac{9\sqrt{14}}{56}\Big{)}\cdot\mu^{\frac{j}{4}}+\Big{(}\frac{1}{2}-\frac{9\sqrt{14}}{56}\Big{)}\cdot\nu^{\frac{j}{4}},&if~{}j\equiv
0\ (mod\ 4);\\\
\Big{(}2+\frac{31\sqrt{14}}{56}\Big{)}\cdot\mu^{\frac{j-1}{4}}+\Big{(}2-\frac{31\sqrt{14}}{56}\Big{)}\cdot\nu^{\frac{j-1}{4}},&if~{}j\equiv
1\ (mod\ 4);\\\
\Big{(}\frac{7}{2}+\frac{53\sqrt{14}}{56}\Big{)}\cdot\mu^{\frac{j-2}{4}}+\Big{(}\frac{7}{2}-\frac{53\sqrt{14}}{56}\Big{)}\cdot\nu^{\frac{j-2}{4}},&if~{}j\equiv
2\ (mod\ 4);\\\
\Big{(}5+\frac{75\sqrt{14}}{56}\Big{)}\cdot\mu^{\frac{j-3}{4}}+\Big{(}5-\frac{75\sqrt{14}}{56}\Big{)}\cdot\nu^{\frac{j-3}{4}},&if~{}j\equiv
3\ (mod\ 4).\end{cases}$

Proof of Claim 1. Put $q_{j}:=detM_{j}$, with $q_{0}=1$, and
$q^{\prime}_{j}:=detM^{\prime}_{j}$, with $q_{0}=1$. Then by a direct
computing, we can get
$q_{1}=4,~{}q_{2}=7,~{}q_{3}=10,~{}q_{4}=33,~{}q_{5}=122,~{}q_{6}=211,~{}q_{7}=300,~{}q_{8}=989.$
For $4\leq j\leq 4n-1$, then

$\displaystyle q_{j}=\begin{cases}4q_{j-1}-q_{j-2},&if~{}j\equiv 0\ (mod\
4);\\\ 4q_{j-1}-q_{j-2},&if~{}j\equiv 1\ (mod\ 4);\\\
2q_{j-1}-q_{j-2},&if~{}j\equiv 2\ (mod\ 4);\\\ 2q_{j-1}-q_{j-2},&if~{}j\equiv
3\ (mod\ 4).\end{cases}$

For $1\leq j\leq n-1,$ let $a_{j}=m_{4j};~{}0\leq j\leq
n-1,~{}b_{j}=m_{4j+1},~{}c_{j}=m_{4j+2},~{}d_{j}=m_{4j+3}.$ Then we can get
$a_{1}=33,~{}b_{0}=4,~{}c_{0}=7,~{}d_{0}=10,~{}b_{1}=122,~{}c_{1}=211,~{}d_{1}=300,$
and for $j\geq 2$ , we have

$\displaystyle\begin{cases}a_{j}=4d_{j-1}-c_{j-1};\\\ b_{j}=4a_{j}-d_{j-1};\\\
c_{j}=2b_{j}-a_{j};\\\ d_{j}=2c_{j}-b_{j}.\end{cases}$ (3.16)

From the first three equations in (3.9), one can get
$a_{j}=\frac{2}{13}c_{j}+\frac{2}{13}c_{j-1}$. Next, substituting $a_{j}$ to
the third equation, one has $b_{j}=\frac{15}{26}c_{j}+\frac{1}{26}c_{j-1}$.
Then substituting $b_{j}$ to the fourth equation, we have
$d_{j}=\frac{37}{26}c_{j}-\frac{1}{26}c_{j-1}.$ Finally, Substituting $a_{j}$
and $d_{j}$ to the first equation, one has $c_{j}-30c_{j-1}+c_{j=2}=0.$ Thus

$\displaystyle c_{j}=k_{1}\cdot\mu^{j}+k_{2}\cdot\nu^{j}.$

According to $c_{0}=7,c_{1}=211,$ we have

$\displaystyle\begin{cases}k_{1}+k_{2}=7;\\\
k_{1}(15+4\sqrt{14})+k_{2}(15-4\sqrt{14})=211.\end{cases}$

and

$\displaystyle\begin{cases}k_{1}=(\frac{7}{2}+\frac{53\sqrt{14}}{56}\Big{)};\\\
k_{2}=(\frac{7}{2}-\frac{53\sqrt{14}}{56}\Big{)}.\end{cases}$

Thus

$\displaystyle\begin{cases}a_{j}=\Big{(}\frac{1}{2}+\frac{9\sqrt{14}}{56}\Big{)}\cdot\mu^{j}+\Big{(}\frac{1}{2}-\frac{9\sqrt{14}}{56}\Big{)}\cdot\nu^{j};\\\
b_{j}=\Big{(}2+\frac{31\sqrt{14}}{56}\Big{)}\cdot\mu^{j}+\Big{(}2-\frac{31\sqrt{14}}{56}\Big{)}\cdot\nu^{j};\\\
c_{j}=\Big{(}\frac{7}{2}+\frac{53\sqrt{14}}{56}\Big{)}\cdot\mu^{j}+\Big{(}\frac{7}{2}-\frac{53\sqrt{14}}{56}\Big{)}\cdot\nu^{j};\\\
d_{j}=\Big{(}5+\frac{75\sqrt{14}}{56}\Big{)}\cdot\mu^{j}+\Big{(}5-\frac{75\sqrt{14}}{56}\Big{)}\cdot\nu^{j},\end{cases}$

as desired.

Using a similar method of Claim 1, we can prove Claim 2.

Claim 2. For $1\leq j\leq 4n$,

$\displaystyle
q^{{}^{\prime}}_{j}=\begin{cases}\Big{(}\frac{1}{2}+\frac{11\sqrt{14}}{56}\Big{)}\cdot\mu^{\frac{j}{4}}+\Big{(}\frac{1}{2}-\frac{11\sqrt{14}}{56}\Big{)}\cdot\nu^{\frac{j}{4}},&if~{}j\equiv
0\ (mod\ 4);\\\
\Big{(}1+\frac{17\sqrt{14}}{56}\Big{)}\cdot\mu^{\frac{j-1}{4}}+\Big{(}1-\frac{17\sqrt{14}}{56}\Big{)}\cdot\nu^{\frac{j-1}{4}},&if~{}j\equiv
1\ (mod\ 4);\\\
\Big{(}\frac{3}{2}+\frac{23\sqrt{14}}{56}\Big{)}\cdot\mu^{\frac{j-2}{4}}+\Big{(}\frac{3}{2}-\frac{23\sqrt{14}}{56}\Big{)}\cdot\nu^{\frac{j-2}{4}},&if~{}j\equiv
2\ (mod\ 4);\\\
\Big{(}5+\frac{75\sqrt{14}}{56}\Big{)}\cdot\mu^{\frac{j-3}{4}}+\Big{(}5-\frac{75\sqrt{14}}{56}\Big{)}\cdot\nu^{\frac{j-3}{4}},&if~{}j\equiv
3\ (mod\ 4).\end{cases}$

By the related properties of determinant, it is evidently that

$\displaystyle\det L_{S}$ $\displaystyle=$
$\displaystyle\left|\begin{array}[]{ccccccc}4&-1&0&\cdots&0&0&1\\\
-1&2&-1&\cdots&0&0&0\\\ 0&-1&2&\cdots&0&0&0\\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\\ 0&0&0&\cdots&2&-1&0\\\
0&0&0&\cdots&0&2&-1\\\ 1&0&0&\cdots&0&-1&4\\\ \end{array}\right|_{4n}$
$\displaystyle=$
$\displaystyle\left|\begin{array}[]{ccccccc}4&-1&0&\cdots&0&0&0\\\
-1&2&-1&\cdots&0&0&0\\\ 0&-1&2&\cdots&0&0&0\\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\\ 0&0&0&\cdots&2&-1&0\\\
0&0&0&\cdots&0&2&-1\\\ 1&0&0&\cdots&0&-1&4\\\
\end{array}\right|_{4n}+\left|\begin{array}[]{ccccccc}4&-1&0&\cdots&0&0&1\\\
-1&2&-1&\cdots&0&0&0\\\ 0&-1&2&\cdots&0&0&0\\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\\ 0&0&0&\cdots&2&-1&0\\\
0&0&0&\cdots&0&2&0\\\ 1&0&0&\cdots&0&-1&0\\\ \end{array}\right|_{4n}$
$\displaystyle=$ $\displaystyle
q_{4n}+(-1)^{4n+1}\cdot(-1)^{4n-1}+(-1)^{4n+1}\cdot\Big{[}(-1)^{4n-1}+(-1)^{4n}q^{{}^{\prime}}_{4n-2}\Big{]}$
$\displaystyle=$ $\displaystyle q_{4n}-q^{{}^{\prime}}_{4n-2}+2.$

Together with Claims 1 and 2, we can get one interesting Claim (Claim 3).

Claim 3. $detL_{S}=\mu^{n}+\nu^{n}+2.$

So far, we have calculated the $detL_{S}$ in Eq.(3.8). Then the prime of the
rest is to calculate $(-1)^{4n-1}b_{4n-1}$.

Claim 4.
$(-1)^{4n-1}b_{4n-1}=\frac{9n\sqrt{14}\big{(}\mu^{n}-\nu^{n}\big{)}}{14}.$

Proof of Claim 4. We apply the notion that $(-1)^{4n-1}b_{4n-1}$ is the total
of all the principal minors of (4$n$-1)/order matix $L_{S}$, thus we obtain

$\displaystyle(-1)^{4n-1}b_{4n-1}$ $\displaystyle=$
$\displaystyle\sum_{j=1}^{4n}detL_{S}[i]$ $\displaystyle=$
$\displaystyle\sum_{j=4,j\equiv 0(mod\ 4)}^{4n}detL_{S}[j]+\sum_{j=1,j\equiv
1(mod\ 4)}^{4n-3}detL_{S}[j]$ $\displaystyle+\sum_{j=2,j\equiv 2(mod\
4)}^{4n-2}detL_{S}[j]+\sum_{j=3,j\equiv 3(mod\ 4)}^{4n-1}detL_{S}[j].$

Let $\left(\begin{array}[]{cc}O&P\\\ S&T\\\ \end{array}\right)$ be a block
matrix of $L_{S}[j]$ and N=$\left(\begin{array}[]{cc}0&-I_{j-1}\\\
I_{4n-j}&0\\\ \end{array}\right)$. It’s easy to get

$\displaystyle N^{\prime}L_{S}[j]N=\left(\begin{array}[]{cc}0&-I_{j-1}\\\
-I_{4n-j}&0\\\ \end{array}\right)^{\prime}\left(\begin{array}[]{cc}O&P\\\
S&T\\\ \end{array}\right)\left(\begin{array}[]{cc}0&-I_{j-1}\\\ -I_{4n-j}&0\\\
\end{array}\right)=\left(\begin{array}[]{cc}T&-S\\\ -P&O\\\
\end{array}\right).$ (3.27)

Combining with Eq.(3.10), based on the different value of $j$, we list the
following four cases.

Case 1. For $j\equiv 0(mod\ 4)$ and $4\leq j\leq 4n-4$.

$\displaystyle\ N^{\prime}L_{S}[j]N$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{ccccccc}4&-1&0&\cdots&0&0&0\\\
-1&2&-1&\cdots&0&0&0\\\ 0&-1&2&\cdots&0&0&0\\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\\ 0&0&0&\cdots&4&-1&0\\\
0&0&0&\cdots&-1&2&-1\\\ 0&0&0&\cdots&0&-1&2\\\
\end{array}\right)_{(4n-1)\times(4n-1)}=q_{4n-1},$

and

$\displaystyle\sum_{j=4,j\equiv 0(mod\
4)}^{4n}detL_{S}[j]=nq_{4n-1}=\frac{5n\sqrt{14}\big{(}\mu^{n}-\nu^{n}\big{)}}{56}.$

Case 2. For $j\equiv 1(mod\ 4)$ and $1\leq j\leq 4n-3$.

$\displaystyle\ N^{\prime}L_{S}[j]N$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{ccccccc}2&-1&0&\cdots&0&0&0\\\
-1&2&-1&\cdots&0&0&0\\\ 0&-1&4&\cdots&0&0&0\\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\\ 0&0&0&\cdots&2&-1&0\\\
0&0&0&\cdots&-1&2&-1\\\ 0&0&0&\cdots&0&-1&4\\\
\end{array}\right)_{(4n-1)\times(4n-1)}=q^{\prime}_{4n-1},$

and

$\displaystyle\sum_{j=1,j\equiv 1(mod\
4)}^{4n-3}detL_{S}[j]=nq^{\prime}_{4n-1}=\frac{5n\sqrt{14}\big{(}\mu^{n}-\nu^{n}\big{)}}{56}.$

Case 3. For $j\equiv 2(mod\ 4)$ and $2\leq j\leq 4n-2$.

$\displaystyle\ N^{\prime}L_{S}[j]N$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{ccccccc}2&-1&0&\cdots&0&0&0\\\
-1&4&-1&\cdots&0&0&0\\\ 0&-1&4&\cdots&0&0&0\\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\\ 0&0&0&\cdots&2&-1&0\\\
0&0&0&\cdots&-1&4&-1\\\ 0&0&0&\cdots&0&-1&4\\\
\end{array}\right)_{(4n-1)\times(4n-1)},$

and

$\displaystyle\sum_{j=2,j\equiv 2(mod\
4)}^{4n-2}detL_{S}[j]=n\Big{[}2\big{(}4q_{4n-3}-q^{\prime}_{4n-4}\big{)}-q_{4n-3}\Big{]}=\frac{13n\sqrt{14}\big{(}\mu^{n}-\nu^{n}\big{)}}{56}).$

Case 4. For $j\equiv 3(mod\ 4)$ and $3\leq j\leq 4n-1$.

$\displaystyle\ N^{\prime}L_{S}[j]N$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{ccccccc}4&-1&0&\cdots&0&0&0\\\
-1&4&-1&\cdots&0&0&0\\\ 0&-1&2&\cdots&0&0&0\\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\\ 0&0&0&\cdots&4&-1&0\\\
0&0&0&\cdots&-1&4&-1\\\ 0&0&0&\cdots&0&-1&2\\\
\end{array}\right)_{(4n-1)\times(4n-1)},$

and

$\displaystyle\sum_{j=3,j\equiv 3(mod\
4)}^{4n-1}detL_{S}[j]=n(4q_{4n-2}-q^{\prime}_{4n-3})=\frac{13n\sqrt{14}\big{(}\mu^{n}-\nu^{n}\big{)}}{56}.$

Thus, one has the following equation

$\displaystyle(-1)^{4n-1}b_{4n-1}=\frac{9n\sqrt{14}\big{(}\mu^{n}-\nu^{n}\big{)}}{14},$

as desired.

According to the results of facts 3 and 4, we immediately get

$\displaystyle\sum_{j=1}^{4n}\frac{1}{\beta_{j}}=\frac{9n\sqrt{14}}{14}\Bigg{(}\frac{(\mu^{n}-\nu^{n}\big{)}}{\mu^{n}+\nu^{n}\big{)}-2}\Bigg{)}.$
(3.32)

######  Theorem 3.2.

Suppose $Q_{n}\big{(}8,4\big{)}$ as a Möbius graph constructed by $n$
octagonals and $n$ quadrilaterals. Then

$\displaystyle
Kf\big{(}Q_{n}(8,4)\big{)}=\frac{32n^{3}-2n}{3}+\frac{36n^{2}\sqrt{14}}{7}\Bigg{(}\frac{\big{(}\mu^{n}-\nu^{n}\big{)}}{\big{(}\mu^{n}+\nu^{n}\big{)}+2}\Bigg{)}.$

Proof. Substituting Eqs.(3.7) and (3.11) into (3.6), the Kirchhoff index of
$Q_{n}(8,4)$ can be expressed

$\displaystyle Kf\big{(}Q_{n}(8,4)\big{)}$ $\displaystyle=$ $\displaystyle
8n\Bigg{(}\sum_{i=2}^{4n}\frac{1}{\alpha_{i}}+\sum_{j=1}^{4n}\frac{1}{\beta_{j}}\Bigg{)}$
$\displaystyle=$ $\displaystyle
8n\Bigg{(}\frac{16n^{2}-1}{12}+\frac{9n\sqrt{14}}{14}\cdot\frac{\big{(}\mu^{n}-\nu^{n}\big{)}}{\big{(}\mu^{n}+\nu\big{)}+2}\Bigg{)}$
$\displaystyle=$
$\displaystyle\frac{32n^{3}-2n}{3}+\frac{36n^{2}\sqrt{14}}{7}\Bigg{(}\frac{\big{(}\mu^{n}-\nu^{n}\big{)}}{\big{(}\mu^{n}+\nu\big{)}+2}\Bigg{)}.$

The result as desired.

Next, we aim to compute the Kirchhoff index for
$Q^{\prime}_{n}\big{(}8,4\big{)}$. By Figure 2, it is apparently to see
$\pi=(1,1^{\prime})(2,2^{\prime})\cdots(4n,(4n)^{\prime})$ is an automorphism
of $Q^{\prime}_{n}\big{(}8,4\big{)}$. In other words,
$v_{0}=\emptyset,~{}v_{1}=\big{\\{}1,2,3,\cdots,4n\big{\\}}~{}and~{}v_{2}=\big{\\{}1^{\prime},2^{\prime},3^{\prime},\cdots,(4n)^{\prime}\big{\\}}.$
In likewise, we express $L_{A}\big{(}Q^{\prime}_{n}(8,4)\big{)}$ and
$L_{S}\big{(}Q^{\prime}_{n}(8,4)\big{)}$ as $L^{\prime}_{A}$ and
$L^{\prime}_{S}$. Thus one can get $L^{\prime}_{A}=L_{A}$ and

$\displaystyle L^{\prime}_{S}$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{ccccccccccc}4&-1&&&&&&&&&-1\\\
-1&2&-1&&&&&&&&\\\ &-1&2&-1&&&&&&&\\\ &&-1&4&-1&&&&&&\\\ &&&-1&4&-1&&&&&\\\
&&&&-1&2&-1&&&&\\\ &&&&&&\ddots&&&&\\\ &&&&&&-1&4&-1&&\\\ &&&&&&&-1&2&-1&\\\
&&&&&&&&-1&2&-1\\\ -1&&&&&&&&&-1&4\\\ \end{array}\right)_{(4n)\times(4n)},$

######  Theorem 3.3.

Suppose $Q^{\prime}_{n}\big{(}8,4\big{)}$ is cylinder graph of the octagonal-
quadrilateral network, then

$\displaystyle
Kf\big{(}Q^{\prime}_{n}(8,4)\big{)}=\frac{32n^{3}-2n}{3}+\frac{36n^{2}\sqrt{14}}{7}\Bigg{(}\frac{\big{(}\mu^{n}-\nu^{n}\big{)}}{\big{(}\mu^{n}+\nu^{n}\big{)}-2}\Bigg{)}.$

Proof. Assuming that $0=\xi_{1}<\xi_{2}\leq\xi_{3}\leq\cdots\leq\xi_{4n}$ are
the roots of $P_{L_{A}}\big{(}x\big{)}=0$, and
$0<\lambda_{1}\leq\lambda_{2}\leq\lambda_{3}\leq\cdots\leq\lambda_{4n}$ are
the roots of $P_{L_{S}}\big{(}x\big{)}=0$. By Lemma 2.1, we have
$S_{p}\big{(}Q_{n}(8,4)\big{)}=\big{\\{}\xi_{1},\xi_{2},\cdots,\xi_{4n},\lambda_{1},\lambda_{2},\cdots,\lambda_{4n}\big{\\}}$.
Then we straightforwardly get that

$\displaystyle Kf\big{(}Q^{\prime}_{n}(8,4)\big{)}$ $\displaystyle=$
$\displaystyle
8n\Bigg{(}\sum_{i=2}^{4n}\frac{1}{\xi_{i}}+\sum_{j=1}^{4n}\frac{1}{\lambda_{j}}\Bigg{)}$
$\displaystyle=$ $\displaystyle
2Kf\big{(}C_{4n}\big{)}+8n\cdot\frac{(-1)^{4n-1}b^{\prime}_{4n-1}}{detL^{\prime}_{S}}.$

Using a method similar to the previous paper, it is can directly calculate
that $detL^{\prime}_{S}[j]=detL_{S}[j]$ and $b^{\prime}_{4n-1}=b_{4n-1}$. Note
that

$\displaystyle\det L^{\prime}_{S}$ $\displaystyle=$
$\displaystyle\left|\begin{array}[]{ccccccc}4&-1&0&\cdots&0&0&-1\\\
-1&2&-1&\cdots&0&0&0\\\ 0&-1&2&\cdots&0&0&0\\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\\ 0&0&0&\cdots&2&-1&0\\\
0&0&0&\cdots&0&2&-1\\\ -1&0&0&\cdots&0&-1&4\\\
\end{array}\right|_{4n}=q_{4n}-q^{\prime}_{4n-2}-2.$

Hence,

$\displaystyle Kf\big{(}Q^{\prime}_{n}(8,4)\big{)}$ $\displaystyle=$
$\displaystyle
8n\Bigg{(}\sum_{i=2}^{4n}\frac{1}{\xi_{i}}+\sum_{j=1}^{4n}\frac{1}{\lambda_{j}}\Bigg{)}$
$\displaystyle=$ $\displaystyle
2Kf\big{(}C_{4n}\big{)}+8n\frac{(-1)^{4n-1}b^{\prime}_{4n-1}}{detL^{\prime}_{S}}$
$\displaystyle=$
$\displaystyle\frac{32n^{3}-2n}{3}+\frac{36n^{2}\sqrt{14}}{7}\Bigg{(}\frac{\big{(}\mu^{n}-\nu^{n}\big{)}}{\big{(}\mu^{n}+\nu^{n}\big{)}-2}\Bigg{)}.$

This completes the proof.

The Kirchhoff indices of $Q_{n}\big{(}8,4\big{)}$ and
$Q^{\prime}_{n}\big{(}8,4\big{)}$ for $n=1,\cdot\cdot\cdot,15$ are tabulated
Table 1.

Table 1:  The Kirchhoff indices of $Q_{1}(8,4),Q_{2}(8,4),...,Q_{15}(8,4)$ and $Q^{\prime}_{1}(8,4),Q^{\prime}_{2}(8,4),...,Q^{\prime}_{15}(8,4).$ $G$ |  $Kf(G)$ |  $G$ |  $Kf(G)$
---|---|---|---
$Q_{1}(8,4)$ |  $28.00$ |  $Q^{\prime}_{1}(8,4)$ |  $30.57$
$Q_{2}(8,4)$ |  $160.8$ |  $Q^{\prime}_{2}(8,4)$ |  $161.14$
$Q_{3}(8,4)$ |  $459.17$ |  $Q^{\prime}_{3}(8,4)$ |  $459.20$
$Q_{4}(8,4)$ |  $987.88$ |  $Q^{\prime}_{4}(8,4)$ |  $987.89$
$Q_{5}(8,4)$ |  $1811.07$ |  $Q^{\prime}_{5}(8,4)$ |  $1811.07$
$Q_{6}(8,4)$ |  $2992.74$ |  $Q^{\prime}_{6}(8,4)$ |  $2992.74$
$Q_{7}(8,4)$ |  $4596.90$ |  $Q^{\prime}_{7}(8,4)$ |  $4596.90$
$Q_{8}(8,4)$ |  $6687.54$ |  $Q^{\prime}_{8}(8,4)$ |  $6687.54$
$Q_{9}(8,4)$ |  $9328.67$ |  $Q^{\prime}_{9}(8,4)$ |  $9328.67$
$Q_{10}(8,4)$ |  $12584.28$ |  $Q^{\prime}_{10}(8,4)$ |  $12584.28$
$Q_{11}(8,4)$ |  $16518.38$ |  $Q^{\prime}_{11}(8,4)$ |  $16518.38$
$Q_{12}$(8,4) |  $21194.96$ |  $Q^{\prime}_{12}(8,4)$ |  $21194.96$
$Q_{13}(8,4)$ |  $26678.03$ |  $Q^{\prime}_{13}(8,4)$ |  $26678.03$
$Q_{14}(8,4)$ |  $33031.59$ |  $Q^{\prime}_{14}(8,4)$ |  $33031.59$
$Q_{15}(8,4)$ |  $40319.63$ |  $Q^{\prime}_{15}(8,4)$ |  $40319.63$

In the sequel, the prime aim is to concentrate on calculate the complexity of
$Q_{n}\big{(}8,4\big{)}$ and $Q^{\prime}_{n}\big{(}8,4\big{)}$.

######  Theorem 3.4.

Let $Q_{n}\big{(}8,4\big{)}$ and $Q^{\prime}_{n}\big{(}8,4\big{)}$ be Möbius
graph and cylinder graph of the linear octagonal-quadrilateral networks,
respectively. Then

$\displaystyle\mathscr{L}\big{(}Q_{n}(8,4)\big{)}=2n\big{(}\mu^{n}+\nu^{n}+2\big{)},$

and

$\displaystyle\mathscr{L}\big{(}Q^{\prime}_{n}(8,4)\big{)}=2n\big{(}\mu^{n}+\nu^{n}-2\big{)}.$

Proof. With initial conditions as
$L_{A}=L^{\prime}_{A}=L(C_{4n});~{}\tau(C_{4n})=4n.$ Based on Lemma 2.3 and
Lemma 2.4, one has the following equation.

$\displaystyle\prod_{i=2}^{4n}\alpha_{i}=\prod_{i=2}^{4n}\xi_{i}=4n\tau(C_{4n})=16n^{2}.$

Note that

$\displaystyle\prod_{j=1}^{4n}\beta_{j}=detL_{S}=\big{(}\mu^{n}+\nu^{n}+2\big{)},$
$\displaystyle\prod_{j=1}^{4n}\lambda_{j}=detL^{\prime}_{S}=\big{(}\mu^{n}+\nu^{n}-2\big{)}.$

Hence,

$\displaystyle\mathscr{L}\big{(}Q_{n}(8,4)\big{)}=\frac{1}{8n}\prod_{i=2}^{4n}\alpha_{i}\cdot\prod_{j=1}^{4n}\beta_{j}=2n\big{(}\mu^{n}+\nu^{n}+2\big{)},$

and

$\displaystyle\mathscr{L}\big{(}Q^{\prime}_{n}(8,4)\big{)}=\frac{1}{8n}\prod_{i=2}^{4n}\xi_{i}\cdot\prod_{j=1}^{4n}\lambda_{j}=2n\big{(}\mu^{n}+\nu^{n}-2\big{)}.$

The complexity of $Q_{n}(8,4)$ and $Q^{\prime}_{n}(8,4)$ have been computed,
which are tabulated in Table 2.

Table 2: The complexity of $Q_{1},Q_{2},...,Q_{8}$ and $Q^{\prime}_{1},Q^{\prime}_{2},...,Q^{\prime}_{8}$. $G$ |  $\mathscr{L}(G)$ |  $G$ |  $\mathscr{L}(G)$
---|---|---|---
$Q_{1}$ | $64$ | $Q^{\prime}_{1}$ | $56$
$Q_{2}$ | $3600$ | $Q^{\prime}_{2}$ | $3584$
$Q_{3}$ | $161472$ | $Q^{\prime}_{3}$ | $161448$
$Q_{4}$ | $6451232$ | $Q^{\prime}_{4}$ | $6451200$
$Q_{5}$ | $241651520$ | $Q^{\prime}_{5}$ | $241651480$
$Q_{6}$ | $8689777200$ | $Q^{\prime}_{6}$ | $8689777152$
$Q_{7}$ | $303803889088$ | $Q^{\prime}_{7}$ | $303803889032$
$Q_{8}$ | $10404546969664$ | $Q^{\prime}_{8}$ | $10404546969600$

##  4 Relation between Kirchhoff index and Wiener index of $Q_{n}(8,4)$ and
$Q^{\prime}_{n}(8,4)$

First part of this section is a certification of the Wiener index of
$Q_{n}\big{(}8,4\big{)}$ and $Q^{\prime}_{n}\big{(}8,4\big{)}$, and the rest
is Wiener index and Kirchhoff index for comparison.

######  Theorem 4.1.

For the network $Q_{n}\big{(}8,4\big{)}$,

$\displaystyle W\big{(}Q_{n}(8,4)\big{)}=32n^{3}+16n^{2}+4n.$

Proof. For the Möbius graph $Q_{n}\big{(}8,4\big{)}$, two types of vertices
can be considered $(n\geq 2)$.

(a) vertex with degree two,

(b) vertex with degree three.

Hence, for type a, we obtain

$\displaystyle\omega_{1}(k)=4n\Bigg{[}\Big{(}\sum_{k=1}^{2n}k+\sum_{k=1}^{2n}k+\sum_{k=2}^{2n}k+\sum_{k=3}^{2n}k\Big{)}+3+4\Bigg{]},$
(4.35)

and for type b, we have

$\displaystyle\omega_{2}(k)=4n\Bigg{(}\sum_{k=1}^{2n}k+\sum_{k=1}^{2n}k+\sum_{k=1}^{2n}k+\sum_{k=2}^{2n}k\Bigg{)}.$
(4.36)

Summing the Eqs.(4.12), (4.13) together, we get the Eq.(4.14).

$\begin{split}W\big{(}Q_{n}(8,4)\big{)}&=\frac{\omega_{1}(k)+\omega_{2}(k)}{2}\\\
&=\frac{4n\Bigg{(}\sum\limits_{k=1}^{2n}k+\sum\limits_{k=1}^{2n}k+\sum\limits_{k=1}^{2n}k+\sum\limits_{k=2}^{2n}k+\sum\limits_{k=1}^{2n}k+\sum\limits_{k=1}^{2n}k+\sum\limits_{k=2}^{2n}k+\sum\limits_{k=3}^{2n}k+7\Bigg{)}}{2}\\\
&=32n^{3}+16n^{2}+4n.\end{split}$ (4.37)

######  Theorem 4.2.

Turning to the cylinder graph $Q^{\prime}_{n}(8,4)$, then

$\displaystyle W\big{(}Q^{\prime}_{n}(8,4)\big{)}=32n^{3}+16n^{2}-8n.$

Proof. According to the proof of $W\big{(}Q_{n}(8,4)\big{)}$, we have the
Eqs.(4.15), (4.16) by similar approach.

$\displaystyle\omega_{1}(k)=4n\Big{[}\Big{(}\sum_{k=1}^{2n-1}k+\sum_{k=1}^{2n}k+\sum_{k=2}^{2n}k+\sum_{k=3}^{2n+1}k\Big{)}+3+4\Big{]}.$
(4.38)

$\displaystyle\omega_{2}(k)=4n\Bigg{[}1+\Big{(}\sum_{k=1}^{2n-1}k+\sum_{k=1}^{2n}k+\sum_{k=2}^{2n}k+\sum_{k=2}^{2n+1}k\Big{)}\Bigg{]}.$
(4.39)

Combining the Eqs.(4.15), (4.16) together, we derive the Eq.(4.17).

$\begin{split}W\big{(}Q^{\prime}_{n}(8,4)\big{)}&=\frac{\omega_{1}(k)+\omega_{2}(k)}{2}\\\
&=\frac{4n\Bigg{(}\sum\limits_{k=1}^{2n-1}k+\sum\limits_{k=1}^{2n}k+\sum\limits_{k=2}^{2n}k+\sum\limits_{k=2}^{2n+1}k+\sum\limits_{k=1}^{2n-1}k+\sum\limits_{k=1}^{2n}k+\sum\limits_{k=2}^{2n}k+\sum\limits_{k=3}^{2n+1}k+8\Bigg{)}}{2}\\\
&=32n^{3}+16n^{2}-8n.\end{split}$ (4.40)

Whatever $n$ is even or odd, we deduced the Wiener indices of the Möbius and
cylinder graph of the linear octagonal-quadrilateral networks must be
tantamount.

Corollary 4.3. For different values of $n$, we illustrate the ratios of the
Wiener index to the Kirchhoff index for both $Q_{n}(8,4)$ and
$Q^{\prime}_{n}(8,4)$ are gradually closing to $3$.

$\lim_{n\to\infty}\frac{W\big{(}Q_{n}(8,4)\big{)}}{Kf\big{(}Q_{n}(8,4)\big{)}}=3,$

and

$\lim_{n\to\infty}\frac{W\big{(}Q^{\prime}_{n}(8,4)\big{)}}{Kf\big{(}Q^{\prime}_{n}(8,4)\big{)}}=3.$

##  5 Conclusion

A central result of this paper is a systematic expression for the Kirchhoff
indices and complexity of the linear Möbius and cylinder octagonal-
quadrilateral networks in terms of the reciprocal of the eigenvalues of the
Laplacian matrix. By comparatively analyzing the correlation ability of
topological indices with Wiener indices and Kirchhoff indices, we have
discovered the underlying relationship between them that the ratio value
consistently being three. Finally, we expect the results developed here could
become useful part of the research in other polygon chains and their variants.

## Funding

This work was supported in part by Anhui Provincial Natural Science Foundation
under Grant 2008085J01, and by Natural Science Fund of Education Department of
Anhui Province under Grant KJ2020A0478 and National Natural Science Foundation
of China Grant 11601006, 12001008.

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