# The Normalized Laplacians, Degree-Kirchhoff Index, and the Complexity of Möbius Graph of Linear Octagonal-Quadrilateral Networks

Authors: Jia-Bao Liu; Qian Zheng; Sakander Hayat
Journal: Journal of Mathematics
 (2021)

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## Abstract

The normalized Laplacian plays an indispensable role in exploring the structural properties of irregular graphs. LetLn8,4 represent a linear octagonal-quadrilateral network. Then, by identifying the opposite lateral edges of Ln8,4, we get the corresponding Möbius graph MQn8,4. In this paper, starting from the decomposition theorem of polynomials, we infer that the normalized Laplacian spectrum of MQn8,4 can be determined by the eigenvalues of two symmetric quasi-triangular matrices ℒA and ℒS of order 4n. Next, owing to the relationship between the two matrix roots and the coefficients mentioned above, we derive the explicit expressions of the degree-Kirchhoff indices and the complexity of MQn8,4.


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## Body

## 1. Introduction

It is well established that networks can be represented by graphs. The graphs we consider in this paper are simple, undirected, and connected. Let us first recall some definitions commonly used in graph theory. SupposeG represents a simple undirected graph with VG=n and EG=m. For more notations, the readers are referred to [1].Note thatDG=diagd1,d2,…,dn represents a degree matrix, where dp is the degree of vp. AG is the adjacency matrix of G. The Laplacian matrix of G is LG=DG−AG. The (p, q)-entry of the normalized Laplacian matrix is given by(1)ℒGpq=1,p=q,−1dpdq,p≠qandvp∽vq,0,otherwise.As a matter of fact, there are many parameters that can be used to describe the structure and properties of molecular graphs in graph networks. One of the parameters based on resistance distance is defined as the Wiener index [2, 3], which is(2)WG=∑i<jdij,where dij=dGvi,vj represents the length of the shortest path between two vertices vi and vj in G. The Wiener index is widely used in chemical and mathematical research. For details, see [4–7].The parameter of resistance distance was first proposed by Klein and Randic [8] in 1993. It means that if every edge of a graph G is regarded as a unit resistance, then the distance between any two vertices i and j in G is called resistance distance, which is denoted as rij. Similar to the Wiener index, we give the expression of the Kirchhoff index [9, 10] according to the resistance distance, namely,(3)KfG=∑i<jrij=n∑i=2n1μi.In 2007, Chen and Zhang [11] proposed that the eigenvalues and eigenvectors of normalized Laplacian spectrum can be used to describe the resistance distance, and the concept of Kirchhoff index is put forward. However, it is very difficult to calculate the degree-Kirchhoff index from the complexity division of graphs, so it is important to find the explicit expression of degree-Kirchhoff index. In recent years, many scholars have devoted themselves to the study of Kirchhoff index of various graphs. Huang et al. [12, 13] proved the Kirchhoff index of linear hexagonal chains and linear polyomino chains successively. Ma and Bian [14] determined the normalized Laplacians and degree-Kirchhoff index of cylinder phenylene chain. Liu et al. [15] described the normalized Laplacian and degree-Kirchhoff index of linear octagonal-quadrilateral networks. For more excellent results, refer to [16–21]. After learning the excellent works of scholars, in this paper, we use the correlation properties of Laplace matrix to calculate the degree-Kirchhoff index and the complexity of Möbius graph of linear octagonal-quadrilateral networks. The investigation of complex graph and network has gone through a spectacular development in the past decades. Especially in organic chemistry, more and more attention has been paid to the application of polyomino in polycyclic aromatic compounds. Many scholars are interested in the study of linear octagonal networks and related molecular graphs. We all know that linear octagonal network is an octagonal system without branch compression. It is constructed by regularly inserting some new points on the straight line of the linear polyomino network. The research on the structure and properties of this kind of natural graph network lays a solid foundation for the advancement of theoretical chemistry, as well as for the development of applied mathematics.LetLn8,4 be the linear octagonal-quadrilateral networks, and octagons and quadrilaterals are connected by a common edge, which are depicted in Figure 1. Then, the corresponding Möbius graph MQ38,4 of octagonal-quadrilateral networks is obtained by the reverse identification of the opposite edge by Ln8,4 (see Figure 2). Obviously, we can obtain that VMQn8,4=8n, EMQn8,4=10n.Figure 1
Linear octagonal-quadrilateral networks.Figure 2
GraphMQ38,4.The rest of the paper will be divided into the following sections. In Section2, we put forward some basic notations and related lemmas. In Section 3, we determine the normalized Laplacian spectrum of MQn8,4. In Section 4, we present Kemeny’s constant, the degree-Kirchhoff index, and the complexity of MQn8,4.


## 2. Preliminary

In this section, we introduce some common symbols and related calculation methods [1], which are applied to the rest of the article.The characteristic polynomial of matrixR of order n is defined as PRx=detxI−R. It is not difficult to find that π is an automorphism of G, and we can write the product of disjoint 1-cycles and transposition, namely,(4)π=1¯2¯,…,m¯1,1′2,2′,…,k,k′.Then, one hasVG=m+2k, and let v0=1¯,2¯,…,m¯,v1=1,2,…,k,v2=1′,2′,…,k′. Thus, the Laplacian matrix can be expressed in the form of block matrix, that is,(5)ℒG=ℒV0V0ℒV0V1ℒV0V2ℒV1V0ℒV1V1ℒV1V2ℒV2V0ℒV2V1ℒV2V2,where(6)ℒV0V1=ℒV0V2,ℒV1V2=ℒV2V1,ℒV1V1=ℒV2V2.Let(7)P=Im00012Ik12Ik012Ik−12Ik,and then(8)P′ℒGP=ℒAG00ℒSG.Note thatP′ is the transposition of P, where(9)ℒA=ℒV0V02ℒV0V12ℒV1V0ℒV1V1+ℒV1V2,ℒS=ℒV1V1−ℒV1V2.Lemma 1 (see [12]).
LetℒLnG,ℒAG,ℒSG be determined as above; then,(10)PℒLnG=PℒAGPℒSG.Lemma 2.
LetG be a graph with VG=n and EG=m, and 0=μ1<μ2≤⋯≤μnn≥2 are the eigenvalues of ℒG. Then, we can quickly confirm that the following formulas hold.(a)
(see [22]). Kemeny’s constant of G can be denoted as(11)KcG=∑i=2n1μi.(b)
(see [11]). The degree-Kirchhoff index of G is defined as(12)Kf∗G=2m∑k=2n1μk.(c)
(see [1]). The number of spanning trees of G can also be called the complexity of G. Then, the complexity of G is(13)∏i=1ndi∑k=2nλk=2mτG.


## 3. The Normalized Laplacian Spectrum ofMQn8,4

In this section, we focus on obtaining the normalized Laplacian spectrum ofMQn8,4 by Lemma 1.Given a square matrixT of order n. We will use Tp1,p2,…,pk to denote the submatrix obtained by deleting the p1th, p2th, …, pkth rows and corresponding columns of T. With a suitable labeling, the vertices of MQn8,4 are shown in Figure 2. Apparently, π=1,1′2,2′,…,4n,4n′ is an automorphism of MQn8,4. Then, v0=∅,v1=1,2,3,…,4nandv2=1′,2′,3′,…,4n′. Besides, we express ℒAMQn8,4 and ℒSMQn8,4 as ℒA and ℒS. Then, one can get(14)ℒA=ℒV1V1+ℒV1V2,ℒS=LV1V1−ℒV1V2.In view of equation (1), we have(15)ℒV1V1=1−16−161−12−121−16−161−13−131−16−161−12⋱−131−16−161−12−121−16−1614n×4n,ℒV1V2=−13−1300−1300⋱−1300−13−134n×4n.Hence,(16)ℒA=23−16−13−161−12−121−16−1623−13−1323−16−161−12⋱−1323−16−161−12−121−16−13−16234n×4n,ℒS=43−16−13−161−12−121−16−1643−13−1343−16−161−12⋱−1343−16−161−12−121−16−13−16434n×4n.Assume that0=η1<η2≤η3≤⋯≤η4n are the roots of PℒAx=0 and 0<φ1≤φ2≤φ3≤⋯≤φ4n are the roots of PℒSx=0, respectively. Then, according to Lemma 1, we can get that the spectrum of MQn is just η1,η2,…,η4n,φ1,φ2,…,φ4n, and it is easy to check that η1=0,ηp>0p=2,3,…,4n, and φq>0,q=1,2,…,4n.Next, we calculate some main results ofMQn related to the normalized Laplacian spectrum.


## 4. The Degree-Kirchhoff Index and the Complexity ofMQn8,4

In this section, we first introduce some theorems, which are obtained by describing the eigenvalues and eigenvectors of normalized Laplacian matrix. Then, we obtain Kemeny’s constant, the degree-Kirchhoff index, and the complexity ofMQn8,4 based on these theorems. where ϱn=41n14/2815+414n−15−414n/15+414n+15−414n+2.Theorem 1.
(17)∑p=24n1ηp=200n2−1160.Proof.
Let(18)PLSx=detxI−ℒA=x4n+a1x4n−1+⋯+a4n−1x+a4n=xx4n−1+a1x4n−2+⋯+a4n−2x+a4n−1,a4n−1≠0.
Then, we can exactly get thatη1,η2,…,η4n are the roots of the following equation:(19)x4n−1+a1x4n−2+⋯+a4n−2x+a4n−1=0.
Based on Vieta’s theorem ofPℒAx, it is easy to get(20)∑p=24n1ηp=−14n−2a4n−2−14n−1a4n−1.
Before calculating−14n−2a4n−2 and −14n−1a4n−1, we must determine pth order principal submatrices, Rp0,Rp1,Rp2, and Rp3, which consist of the first p rows and columns of the following matrices ℒA0,ℒA1,ℒA2, and ℒA3, respectively, p=1,2,…,4n. Let(21)ℒA0=23−16−161−12−121−16−1623−13−1323−16−161−12⋱−1323−16−161−12−121−16−16234n×4n,ℒA1=1−12−121−16−1623−13−1323−16−161−12−121−16⋱−161−12−121−16−1623−13−13234n×4n,ℒA2=1−16−1623−13−1323−16−161−12−121−16−1623−13⋱−121−16−1623−13−1323−16−1614n×4n,ℒA3=23−13−1323−16−161−12−121−16−1623−13−1323−16⋱−1623−13−1323−16−161−12−1214n×4n.
In this way, we can get four facts.Fact 1.
For1≤p≤4n,(22)rp0=p+1136p/4,ifp≡0mod4,13p+1136p−1/4,ifp≡1mod4,16p+1136p−2/4,ifp≡2mod4,112p+1136p−3/4,ifp≡3mod4.Fact 2.
For1≤p≤4n,(23)rp1=p+1136p/4,ifp≡0mod4,12p+1136p−1/4,ifp≡1mod4,14p+1136p−2/4,ifp≡2mod4,112p+1136p−3/4,ifp≡3mod4.Fact 3.
For1≤p≤4n,(24)rp2=12rp−20−19rp−31.Fact 4.
For1≤p≤4n,(25)rp3=23rp−10−19rp−21.Proof.
of Fact 1. Takerp0=detRp0,rp1=detRp1,rp2=detRp2, and rp3=detRp3. By a straightforward calculation, one can get the following values (see Table 1).
For4≤p≤4n−1, we can get expansion formula of det Rp0 with respect to its last row:(26)rp0=23rp−10−16rp−20,ifp≡0mod4,23rp−10−19rp−20,ifp≡1mod4,rp−10−16rp−20,ifp≡2mod4,rp−10−14rp−20,ifp≡3mod4.
For1≤p≤n−1, let ap=r4p0; for 0≤p≤n−1, let bp=r4p+10,cp=r4p+20,dp=r4p+30. Then, we can get a1=5/36,b0=2/3,c0=1/2,d0=5/36,b1=3/54,c1=7/216,d1=1/54, and for p≥2, we have(27)ap=23dp−1−16cp−1,bp=23ap−19dp−1,cp=bp−16ap,dp=cp−14bp.
Then, it is not difficult to obtain that(28)ap=18cp−24dp,bp=4cp−4dp,cp=118cp−1−11296cp−2,dp=118dp−1−11296dp−2.
According to the equation ofdp in (28), it is evident to see that x2−1/18x+1/1296=0, and its two roots are 1/36 and 1/36. Therefore, dp=xp+y1/36p is the general solution. Then, we can get(29)y=13,136x+y=154,x=13,y=13.
Thus, we can obtaindp=1/3p+11/36pp≥1. Similarly, we have cp=2p/3+1/21/36pp≥1;ap=4p+11/36pp≥1; and bp=2/32p+11/36pp≥1.
The result is obtained as desired.By similar consideration,Fact 2 is available. Then, based on the conclusion of Facts 1 and 2, we quickly get Facts 3 and 4.Now, we will further calculate−14n−1a4n−1 and −14n−2a4n−2 in equation (20). For the sake of discussion, it is assumed that r0=1.Table 1
Initial value.
rp0Valuerp0Valuerp0Valuerp0Valuer102/3r201/2r301/3r405/36r503/54r607/216r701/54r801/144Claim 1.
−14n−1a4n−1=40n21/36n.Proof.
of Claim 1. Since−14n−1a4n−1 is the total of all the principal minors of order 4n−1 of ℒA, we have(30)−14n−1a4n−1=∑p=14ndetℒAp=∑p=4,p≡0mod44ndetℒAp+∑p=1,p≡1mod44n−3detℒAp+∑p=2,p≡2mod44n−2detℒAp+∑p=3,p≡3mod44n−1detℒAp,where(31)∑p=4,p≡0mod44ndetℒAp=∑p=4,p≡0mod44nrp−10r4n−p0−19rp−21r4n−p−10,∑p=1,p≡1mod44n−3detℒAp=∑p=1,p≡1mod44n−3rp−10r4n−p1−19rp−21r4n−p−11,∑p=2,p≡2mod44n−2detℒAp=∑p=2,p≡2mod44n−2rp−10r4n−p2−19rp−21r4n−p−12,∑p=3,p≡3mod44n−1detℒAp=∑p=3,p≡3mod44n−1rp−10r4n−p3−19rp−21r4n−p−13.
ByFacts 1 and 2, we have(32)∑p=4,p≡0mod44nrp−10r4n−p0−19rp−21r4n−p−10=∑p=4,p≡0mod44np12136p−4/44n−p+11121364n−p/4−19p−114136p−4/41124n−p1364n−p−4/4=∑p=4,p≡0mod44n12n136n=12n2136n.
Similarly, byFacts 1–4, we can get(33)∑p=1,p≡1mod44n−3rp−10r4n−p1−19rp−21r4n−p−11=12n2136n,∑p=2,p≡2mod44n−2rp−10r4n−p2−19rp−21r4n−p−12=8n2136n,∑p=3,p≡3mod44n−1rp−10r4n−p3−19rp−21r4n−p−13=8n2136n.
Hence, according to the above results, we have(34)−14n−1a4n−1=∑p=14ndetℒAp=40n2136n.
The proof of Claim1 is completed.Claim 2.
−14n−2a4n−2=2/3200n4−11n21/36n.Proof.
of Claim 2. It is not hard to see that−14n−2a4n−2 is the total of those principal minors ℒA, which have (4n − 2) rows and columns. Thus, we have(35)−14n−2a4n−2=∑1≤i<j≤4ndetℒAp,q.
By equation (35), it can be seen that the change of i and j values will lead to different detℒAp,q results. Therefore, we will choose different p and q to list the following equations:(36)∑1≤p<q≤4ndetℒAp,q=∑p≡0mod44n−4∑q≡0mod44ndetℒAp,q+∑p≡0mod44n−4∑q≡1mod44n−3detℒAp,q+∑p≡0mod44n−4∑q≡2mod44n−2detℒAp,q+∑p≡0mod44n−4∑q≡3mod44n−1detℒAp,q+∑p≡1mod44n−3∑q≡0mod44ndetℒAp,q+∑p≡1mod44n−3∑q≡1mod44n−3detℒAp,q+∑p≡1mod44n−3∑q≡2mod44n−2detℒAp,q+∑p≡1mod44n−3∑q≡3mod44n−1detℒAp,q+∑p≡2mod44n−2∑q≡0mod44ndetℒAp,q+∑p≡2mod44n−2∑q≡1mod44n−3detℒAp,q+∑p≡2mod44n−2∑q≡2mod44n−2detℒAp,q+∑p≡2mod44n−2∑q≡3mod44n−1detℒAp,q+∑p≡3mod44n−1∑q≡0mod44ndetℒAp,q+∑p≡3mod44n−2∑q≡1mod44n−3detℒAp,q+∑p≡3mod44n−1∑q≡2mod44n−2detℒAp,q+∑p≡3mod44n−1∑q≡3mod44n−1detℒAp,q.
ByFacts 1–4, we can compute the following results.Case 1.
(37)∑p≡0mod44n−4∑q≡0mod44ndetℒAp,q=∑p≡0mod44n−4∑q≡0mod44nrp−10rq−p−10r4n−q0−19rp−21rq−p−10r4n−q−10=∑p≡0mod44n−4∑q≡0mod44n9q−p4n−q+p136n=12n4−n2136n.Case 2.
(38)∑p≡0mod44n−4∑q≡1mod44n−3detℒAp,q=∑p≡0mod44n−4∑q≡1mod44n−3rp−10rq−p−10r4n−q1−19rp−21rq−p−10r4n−q−11=∑p≡0mod44n−4∑q≡1mod44n−39q−p4n−q+p136n=328n4−12n3+n2+3n136n.Case 3.
(39)∑p≡0mod44n−4∑q≡2mod44n−2detℒAp,q=∑p≡0mod44n−4∑q≡2mod44n−2rp−10rq−p−10r4n−q2−19rp−21rq−p−10r4n−q−12=∑p≡0mod44n−4∑q≡2mod44n−26q−p4n−q+p136n=8n4+8n3+4n2+4n136n.Case 4.
(40)∑p≡0mod44n−4∑q≡3mod44n−1detℒAp,q=∑p≡0mod44n−4∑q≡3mod44n−1rp−10rq−p−10r4n−q3−19rp−21rq−p−10r4n−q−13=∑p≡0mod44n−4∑q≡3mod44n−16q−p4n−q+p136n=8n4−4n3+n2−5n136n.Case 5.
(41)∑p≡1mod44n−3∑q≡0mod44ndetℒAp,q=∑p≡1mod44n−3∑q≡0mod44nrp−10rq−p−11r4n−q0−19rp−21rq−p−11r4n−q−10=∑p≡1mod44n−3∑q≡0mod44n9q−p4n−q+p136n=328n4+12n3+n2−3n136n.Case 6.
(42)∑p≡1mod44n−3∑q≡1mod44n−3detℒAp,q=∑p≡1mod44n−3∑q≡1mod44n−3rp−10rq−p−11r4n−q1−19rp−21rq−p−11r4n−q−11=∑p≡1mod44n−3∑q≡1mod44n−39q−p4n−q+p136n=12n4−n2136n.Case 7.
(43)∑p≡1mod44n−3∑q≡2mod44n−2detℒAp,q=∑p≡1mod44n−3∑q≡2mod44n−2rp−10rq−p−11r4n−q2−19rp−21rq−p−11r4n−q−12=∑p≡1mod44n−3∑q≡2mod44n−29q−p4n−q+p136n=8n4+4n3+n2+5n136n.Case 8.
(44)∑p≡1mod44n−3∑q≡3mod44n−1detℒAp,q=∑p≡1mod44n−3∑q≡3mod44n−1rp−10rq−p−11r4n−q3−19rp−21rq−p−11r4n−q−13=∑p≡1mod44n−3∑q≡3mod44n−16q−p4n−q+p136n=8n4−8n3+4n2−4n136n.Case 9.
(45)∑p≡2mod44n−2∑q≡0mod44ndetℒAp,q=∑p≡2mod44n−2∑q≡0mod44nrp−10rq−p−12r4n−q0−19rp−21rq−p−12r4n−q−10=∑p≡2mod44n−2∑q≡0mod44n6q−p4n−q+p136n=8n4+8n3+4n2+4n136n.Case 10.
(46)∑p≡2mod44n−2∑q≡1mod44n−3detℒAp,q=∑p≡2mod44n−2∑q≡1mod44n−3rp−10rq−p−12r4n−q1−19rp−21rq−p−12r4n−q−11=∑p≡2mod44n−2∑q≡1mod44n−36q−p4n−q+p136n=8n4−4n3+n2−5n136n.Case 11.
(47)∑p≡2mod44n−2∑q≡2mod44n−2detℒAp,q=∑p≡2mod44n−2∑q≡2mod44n−2rp−10rq−p−12r4n−q2−19rp−21rq−p−12r4n−q−12=∑p≡2mod44n−2∑q≡2mod44n−24q−p4n−q+p136n=163n4−n2136n.Case 12.
(48)∑p≡2mod44n−2∑q≡3mod44n−1detℒAp,q=∑p≡2mod44n−2∑q≡3mod44n−1rp−10rq−p−12r4n−q3−19rp−21rq−p−12r4n−q−13=∑p≡2mod44n−2∑q≡3mod44n−14q−p4n−q+p136n=238n4+4n3+n2+5n136n.Case 13.
(49)∑p≡3mod44n−1∑q≡0mod44ndetℒAp,q=∑p≡3mod44n−1∑q≡0mod44nrp−10rq−p−13r4n−q0−19rp−21rq−p−13r4n−q−10=∑p≡3mod44n−1∑q≡0mod44n6q−p4n−q+p136n=8n4+4n3+n2+5n136n.Case 14.
(50)∑p≡3mod44n−1∑q≡1mod44n−3detℒAp,q=∑p≡3mod44n−1∑q≡1mod44n−3rp−10rq−p−13r4n−q1−19rp−21rq−p−13r4n−q−11=∑p≡3mod44n−1∑q≡1mod44n−36q−p4n−q+p136n=8n4−8n3+4n2−4n136n.Case 15.
(51)∑p≡3mod44n−1∑q≡2mod44n−2detℒAp,q=∑p≡3mod44n−1∑q≡2mod44n−2rp−10rq−p−13r4n−q2−19rp−21rq−p−13r4n−q−12=∑p≡3mod44n−1∑q≡2mod44n−24q−p4n−q+p136n=238n4−4n3+n2−5n136n.Case 16.
(52)∑p≡3mod44n−1∑q≡3mod44n−1detℒAp,q=∑p≡3mod44n−1∑q≡3mod44n−1rp−10rq−p−13r4n−q3−19rp−21rq−p−13r4n−q−13=∑p≡3mod44n−1∑q≡3mod44n−14q−p4n−q+p136n=163n4−n2136n.
Then, according to the value ofp, the above sixteen cases can be divided into the following four categories:(53)F0=∑p≡0mod44n−4∑q≡0mod44ndetℒAp,q+∑p≡0mod44n−4∑q≡1mod44n−3detℒAp,q+∑p≡0mod44n−4∑q≡2mod44n−2detℒAp,q+∑p≡0mod44n−4∑q≡3mod44n−1detℒAp,q=1280n4−28n3−11n2+7n136n,F1=∑p≡1mod44n−3∑q≡0mod44ndetℒAp,q+∑p≡1mod44n−3∑q≡1mod44n−3detℒAp,q+∑p≡1mod44n−3∑q≡2mod44n−2detℒAp,q+∑p≡1mod44n−3∑q≡3mod44n−1detℒAp,q=1280n4+28n3−11n2−7n136n.F2=∑p≡2mod44n−2∑q≡0mod44ndetℒAp,q+∑p≡2mod44n−2∑q≡1mod44n−3detℒAp,q+∑p≡2mod44n−2∑q≡2mod44n−2detℒAp,q+∑p≡2mod44n−2∑q≡3mod44n−1detℒAp,q=1380n4+20n3+n2+7n136n,F3=∑p≡3mod44n−1∑q≡0mod44ndetℒAp,q+∑p≡3mod44n−2∑q≡1mod44n−3detℒAp,q+∑p≡3mod44n−1∑q≡2mod44n−2detℒAp,q+∑p≡3mod44n−1∑q≡3mod44n−1detℒAp,q=1380n4−20n3+10n2−7n136n.
SubstitutingF0,F1,F2, and F3 into equation (35), one has(54)−14n−2a4n−2=F0+F1+F2+F3=23200n4−11n2136n.
This completes the proof.So, substituting the results of Claims1 and 2 into equation (20) yields(55)∑p=24n1ηp=−14n−2a4n−2−14n−1a4n−1=2/3200n4−11n21/36n40n21/36n=200n2−1160.Theorem 2.
(56)∑q=14n1φq=41n142815+414n−15−414n15+414n+15−414n+2.Proof.
Let(57)PLSx=detxI−ℒS=x4n+b1x4n−1+⋯+b4n−1x+b4n=xx4n−1+b1x4n−2+⋯+b4n−2x+b4n−1,b4n−1≠0.
Then, we can exactly get thatφ1,φ2,…,φ4n are the roots of the following equation:(58)x4n−1+b1x4n−2+⋯+b4n−2x+b4n−1=0.
Based on Vieta’s theorem ofPℒSx, one has(59)∑q=14n1φq=−14n−1b4n−1−14nb4n=−14n−1b4n−1detℒS.
Before calculating−14n−1b4n−1 and detℒS, we must determine ith order principal submatrices Sq0,Sq1,Sq2, and Sq3, which consist of the first q rows and columns of the matrices ℒS0,ℒS1,ℒS2, and ℒS3, respectively, q=1,2,…,4n. Let(60)ℒS0=43−16−161−12−121−16−1643−13−1343−16−161−12⋱−1343−16−161−12−121−16−16434n×4n,ℒS1=1−12−121−16−1643−13−1343−16−161−12−121−16⋱−161−12−121−16−1643−13−13434n×4n,ℒS2=1−16−1643−13−1343−16−161−12−121−16−1643−13⋱−121−16−1643−13−1343−16−1614n×4n,ℒS3=43−13−1343−16−161−12−121−16−1643−13−1343−16⋱−1643−13−1343−16−161−12−1214n×4n.
n this way, let us start with the following facts.Fact 5.
For1≤q≤4n,(61)sq0=48+91456512+149q/4+48−91456512−149q/4,ifq≡0mod4,23+3114168512+149q−1/4+23−3114168512−149q−1/4,ifq≡1mod4,712+5314336512+149q−2/4+712−5314336512−149q−2/4,ifq≡1mod4,512+2514224512+149q−3/4+512−2514224512−149q−3/4,ifq≡3mod4.Fact 6.
For1≤q≤4n,(62)sq1=12+111456512+149q/4+12−111456512−149q/4,ifq≡0mod4,12+1714112512+149q−1/4+12−1714112512−149q−1/4,ifq≡1mod4,38+2314224512+149q−2/4+38−2314224512−149q−2/4,ifq≡1mod4,512+2514224512+149q−3/4+512−2514224512−149q−3/4,ifq≡3mod4.Fact 7.
For1≤q≤4n,(63)sq2=76sq−20−19sq−31.Fact 8.
For1≤q≤4n,(64)sq3=43sq−10−19sq−21.Proof.
of Fact 5. Takesq0=detSq0,sq1=detSq1,sq2=detSq2, and sq3=detSq3. By direct calculation, it is not difficult to get the following values (see Table 2).
For4≤q≤4n, we have detSq0(65)sq0=43sq−10−16sq−20,ifq≡0mod4,43sq−10−19sq−20,ifq≡1mod4,sq−10−16sq−20,ifq≡2mod4,sq−10−14sq−20,ifq≡3mod4.
For1≤q≤n, let Aq=s4q; for 0≤q≤n−1, let Bq=s4q+1,Cq=s4q+2,Dq=s4q+3. Then, we may obtain that(66)Aq=43Dq−1−16Cq−1,Bq=43Aq−19Dq−1,Cq=Bq−16Aq,Dq=Cq−14Bq.
From the first three equations in (66), one can get Aq=12/13Cq+1/78Cq−1. Next, substituting Aq into the third equation, one has Bq=15/13Cq+1/468Cq−1. Then, substituting Bq into the fourth equation, we have Dq=37/52Cq−1/1872Cq−1. Finally, substituting Aq and dq into the first equation, one has cq−30cq−1+cq=2=0. Thus,(67)Cq=k1512+149q+k2512−149q.
In view ofC0=7/6,C1=211/216, we have(68)k1+k2=76,k1512+149+k2512−149=211216,k1=712+5314336,k2=712−5314336.
Thus, it is routine to deduce that(69)Aq=48+91456512+149q+48−91456512−149q,ifq≡0mod4,Bq=23+3114168512+149q+23−3114168512−149q,ifq≡1mod4,Cq=712+5314336512+149q+712−5314336512−149q,ifq≡1mod4,Dq=512+2514224512+149q+512−2514224512−149q,ifq≡3mod4.
In the same way, we can quickly prove the result ofFact 6.
Then, we expanddetsq2 and detsq3 according to the properties of determinant, and we can get Facts 7 and 8.
Now by exploiting the property of determinant, we can get(70)detℒS=43−160⋯0013−161−12⋯0000−121⋯000⋮⋮⋮⋱⋮⋮⋮000⋯1−120000⋯01−161300⋯0−16434n=43−160⋯000−161−12⋯0000−121⋯000⋮⋮⋮⋱⋮⋮⋮000⋯1−120000⋯01−161300⋯0−16434n+43−160⋯0013−161−12⋯0000−121⋯000⋮⋮⋮⋱⋮⋮⋮000⋯1−120000⋯0101300⋯0−1604n=s4n0−19s4n−21+2136n.
WithFacts 1 and 2, we can obtain one interesting claim.Table 2
Initial value.
sq0Valuesq0Valuesq0Valuesq0Values104/3s207/6s305/6s4033/36s5061/54s60211/216s7025/36s80989/1296Claim 3.
detℒS=5/12+14/9n+5/12−14/9n+21/36n.
Then, we are going to concentrate on calculating−14n−1b4n−1.Claim 4.
−14n−1b4n−1=41n14/2815+414n−15−414n/15+414n+15−414n+2.Proof.
Since−14n−1b4n−1 is the total of all the principal minors of order 4n−1 of ℒS, we have(71)−14n−1b4n−1=∑q=14ndetℒSq=∑q=4,q≡0mod44ndetℒSq+∑q=1,q≡1mod44n−3detℒSq+∑q=2,q≡2mod44n−2detℒSq+∑q=3,q≡3mod44n−1detℒSq,where(72)detℒSq=sq−10s4n−q0−19sq−21s4n−q−10,ifq≡0mod4,sq−10s4n−q1−19sq−21s4n−q−11,ifq≡1mod4,sq−10s4n−q2−19sq−21s4n−q−12,ifq≡2mod4,sq−10s4n−q3−19sq−21s4n−q−13,ifq≡3mod4.
Forq≡0mod4and4≤q≤4n−4, in view of (72) and Facts 5–8, one gets(73)detℒSq=sq−10s4n−q0−19sq−21s4n−q−10=512+2514224512+149j−4/4+512−2514224512−149j−4/4×48+251456512+1494n−j/4+48−251456512−1494n−j/4−1938+2314224512+149j−4/4+38−2314224512−149j−4/4×512+2514224512+1494n−j−4/4+512−2514224512−1494n−j−4/4=15n1456512+149n−512+149n.
Similarly, forq≡1mod4and1≤q≤4n−3, we have(74)∑q=1,q≡1mod44n−3detℒSq=sq−10s4n−q1−19sq−21s4n−q−11=15n1456512+149n−512+149n.
Forq≡2mod4and2≤q≤4n−2, we have(75)∑q=2,q≡2mod44n−2detℒSq=sq−10s4n−q2−19sq−21s4n−q−12=13n1428512+149n−512+149n.
Forq≡3mod4and3≤q≤4n−1, we have(76)∑q=3,q≡3mod44n−1detℒSq=sq−10s4n−q3−19sq−21s4n−q−13=13n1428512+149n−512+149n.
Thus, one has the following equation:(77)−14n−1b4n−1=41n1428512+149n−512+149n.
Therefore, substituting the results of Claims3 and 4 into (59) yields(78)∑q=14n1φq=41n142815+414n−15−414n15+414n+15−414n+2,as desired.Note thatEMQn8,4=10n. Taking the results of Theorems 1 and 2 to (a) and (b) of Lemma 2, we can immediately get the following two theorems.Theorem 3.
LetMQn8,4 be a Möbius graph with n octagons and n quadrilaterals. Then,(79)KcMQn8,4=∑p=24n1ηp+∑q=14n1φq=200n2−1160+41n142815+414n−15−414n15+414n+15−414n+2.Theorem 4.
LetMQn8,4 be a Möbius graph with n octagons and n quadrilaterals. Then,(80)Kf∗MQn8,4=20n∑p=24n1ηp+∑q=14n1φq=20n200n2−1160+41n142815+414n−15−414n15+414n+15−414n+2=200n3−11n3+20nϱn,Table3 shows the degree-Kirchhoff indices of Möbius graph of linear octagonal-quadrilateral networks.Table 3
The degree-Kirchhoff indices ofMQ18,4,MQ28,4,…,MQ168,4.
GKf∗GGKf∗GGKf∗GGKf∗GMQ18,4165.50MQ58,411054.43MQ98,457442.75MQ138,4164937.53MQ28,4963.33MQ68,418322.78MQ108,477587.71MQ148,4204359.11MQ38,42775.12MQ78,428210.28MQ118,4101951.83MQ158,4249599.85MQ48,46005.23MQ88,441116.93MQ128,4130935.10MQ168,4301059.74Finally, we will concentrate on calculating the complexity ofMQn8,4.Theorem 5.
LetMQn8,4 denote a Möbius graph of linear octagonal-quadrilateral networks of length n≥2. Then,(81)τMQn8,4=4n15+414n+15−14n+2.Proof.
By Claim1, one can get(82)∏p=24nηp=−14n−1a4n−1=40n2136n.
Similarly, according to Claim3, we have(83)∏q=14nφq=detℒS=512+149n+512−149n+2136n.
Note that∏i=18ndiMQn=24n34n and EMQn8,4=10n. By Lemma 2(c), one gets(84)τQn8,4=110n∏p=24nηp⋅∏q=14nφq=4n15+414n+15−14n+2.
This completes the proof.Thus, we can get the complexity ofMQn8,4, which is listed in Table 4.Table 4
The complexity ofQ1,Q2,…,Q10.
GτGGτGQ1128Q517379554400Q27200Q7607607778176Q3322944Q820809093939328Q412902464Q9701525710449792Q5483303040Q1023358178980900000



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