# Computing the Normalized Laplacian Spectrum and Spanning Tree of the Strong Prism of Octagonal Network

Authors: Yasir Ahamad; Umar Ali; Imran Siddique; Aiyared Iampan; Walaa A. Afifi; Hamiden Abd-El-Wahed Khalifa
Journal: Journal of Mathematics
 (2022)

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## Abstract

Spectrum analysis and computing have expanded in popularity in recent years as a critical tool for studying and describing the structural properties of molecular graphs. LetOn2 be the strong prism of an octagonal network On. In this study, using the normalized Laplacian decomposition theorem, we determine the normalized Laplacian spectrum of On2 which consists of the eigenvalues of matrices ℒA and ℒS of order 3n+1. As applications of the obtained results, the explicit formulae of the degree-Kirchhoff index and the number of spanning trees for On2 are on the basis of the relationship between the roots and coefficients.


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## Body

## 1. Introduction

Graphs are a convenient way to depict chemical structures, where atoms are associated with vertices, while chemical bonds are associated with edges. This manifestation carries a wealth of knowledge about the molecule’s chemical characteristics. In quantitative structure-activity/property relationship (QSAR/QSPR) studies, one may see that many chemical and physical properties of molecules are closely correlated with graph-theoretical parameters known as topological indices. One such graph-theoretical parameter is the multiplicative degree-Kirchhoff index (see [1]). In statistical physics (see [2]), the enumeration of spanning trees in a graph is a crucial problem. It is interesting to note that the multiplicative degree-Kirchhoff index is closely related to the number of spanning trees in a graph. The normalized Laplacian acts as a link between them.LetG be an n-vertex simple, undirected, and connected graph with the vertex set of VG and an edge set of EG. For standard notation and terminology, one may refer to the recent papers (see [3, 4]). The (combinatorial) Laplacian matrix of graph G is specified as LG=DG−AG, where DG is the vertex degree diagonal matrix of order n and AG is an adjacency matrix of order n.The normalized Laplacian is defined by(1)ℒGij=1,ifi=j,−1dvidvj,ifi≠j,vi∼vj,0,otherwise.Evidently,LG=DG−AG and ℒG=DG−1/2LGDG−1/2. As we all know, the normalized Laplacian technique is useful for analyzing the structural features of nonregular graphs. In reality, the interaction between a graph’s structural features and its eigenvalues is the focus of spectral graph theory. For more information, see recent articles [5–8] or the book [9].Many parameters were used to characterize and describe the structural features of graphs in chemical graph theory. The Wiener index [10, 11] was a well-known distance-based index, as it is known as WG=∑i<jdij. Eventually, Gutman [12] defined the Gutman index as follows:(2)GutG=∑i<jdidjdij.In accordance with electrical network theory, Klein and Randić [13] presented a new distance function called resistance distance that is denoted as rij. The resistance distance in electrical networks is between two arbitrary vertices i and j when every edge is replaced by a unit resistor. Klein and Ivanciuc [14] called it the Kirchhoff index, the total sum of resistance distances between each pair of vertices of G, which is KfG=∑i<jrij. Later, the degree-Kirchhoff index was established by Chen and Zhang [1] and denoted by Kf∗G=∑i<jdidjrij.Because of their practical uses in physics, chemistry, and other sciences, the Kirchhoff index and the degree-Kirchhoff index have gained a lot of attention. Klein and Lovász [15, 16] separately established that(3)KfG=n∑k=2n1νk,where 0=ν1<ν2≤⋯≤νn are the eigenvalues of LG. According to Chen [17], the degree-Kirchhoff index is,(4)Kf∗G=2m∑k=1n1νk,where ν1≤ν2≤⋯≤νn are the eigenvalues of ℒG.Since the Kirchhoff index and multiplicative degree-Kirchhoff index have been widely used in the domains of physics, chemistry, and network science. During the previous few decades, many scientists have been working on explicit formulae for the Kirchhoff and degree-Kirchhoff indices of graphs with particular structures, such as cycles [18], complete multipartite graphs [19], generalized phenylene [20], crossed octagonal [21], hexagonal chains [22], pentagonal-quadrilateral network [23], and so on. Other research on the Kirchhoff index and the multiplicative degree-Kirchhoff index of a graph has been published (see [24–31]). In organic chemistry, polyomino systems have received a lot of attention, especially in polycyclic aromatic compounds. Tree-like octagonal networks are condensed into octagonal networks that belong to the polycyclic conjugated hydrocarbons’ family. The octagonal system without any branches is known as a linear octagonal network [32]. As shown in Figure 1, a linear octagonal network could also be created from a linear polyomino network by adding additional points to the line according to specified rules.Figure 1
GraphOn with labeled vertices.Thestrong product between the graphs G and H is denoted by G⊠H, where the vertex set VG⊠H is VG×VH and a,xb,y is an edge of G⊠H if a=b and x is adjacent to y in H or x=y and a is adjacent to b in G or xy∈EH and ab∈EG. In particular, the strong product of K2 and G is known as the strong prism of G. Recently, Li [33] and Ali [34] calculated the resistance distance-based parameters of the strong prism of unique graphs, such as strong prism of Sn and Ln⊠K2, respectively. Let On2 be the strong prism of K2 and On, denoted by On2=K2⊠On, as shown in Figure 2. Obviously, EOn2=34n+6 and VOn2=12n+4.Figure 2
GraphOn2 with labeled vertices.In this paper, motivated by [34–36], we derive an explicit analytical expression for the multiplicative degree-Kirchhoff index and also spanning trees of On2.


## 2. Preliminaries

In this section, we start by going over some basic notation and then introduce a suitable technique. Given the square matrixR having order n, we refer to Ri1,i2,…,ik as the submatrix of R that results from deleting the i1th, i2th, …, ikth columns and rows. Let ΦR=detxIn−R be the characteristic polynomial of the square matrix R. The labeled vertices of On2 are as depicted in Figure 2 and V1=u1,…,u3n+1,v1,…,v3n+1 and V2=u1′,…,u3n+1′,v1′,…,v3n+1′. The normalized Laplacian matrix ℒOn2 could be represented as a block matrix below:(5)ℒOn2=ℒV11On2ℒV12On2ℒV21On2ℒV22On2.It is simple to verify thatℒV12On2=ℒV21On2 and ℒV11On2=ℒV22On2.Let(6)T=12I6n+212I6n+212I6n+2−12I6n+2.Then,(7)TℒOn2T′=ℒAOn200ℒSOn2,where(8)ℒAOn2=ℒV11+ℒV12,ℒSOn2=ℒV11−ℒV12.Huang et al. obtained the following lemma.Lemma 1 (see [8]).
LetG be a graph and let ℒAOn2 and ℒSOn2 be as described above. Then, we have ΦℒOn2=ΦℒA·ΦℒS.Lemma 2 (see [1]).
Letρ1≤ρ2≤⋯≤ρn be the eigenvalues of ℒG; then, the degree-Kirchhoff index can also be written as Kf∗G=2m∑i=2n1/ρi.Lemma 3 (see [17]).
LetG be n-vertex connected graph of size m; then, the spanning trees is τG=1/2m∏i=1ndi∏k=2nρk.


## 3. Main Results

In this section, we are committed to the explicit analytical solution for the multiplicative degree-Kirchhoff index, as well as the spanning tree ofOn2. In terms of the role of normalized Laplacian ℒ, the following block matrices of ℒV11On2 and ℒV12On2 are obtained according to equation (8).(9)LV11On2=1−1500⋯00−15000⋯00−151−150⋯000000⋯000−151−135⋯000000⋯0000−1351⋯00000−17⋯00⋮⋮⋮⋮⋱⋮⋮⋮⋮⋮⋮⋱⋮⋮0000⋯1−150000⋯000000⋯−1510000⋯0−15−15000⋯001−1500⋯000000⋯00−151−150⋯000000⋯000−151−135⋯00000−17⋯0000−1351⋯00⋮⋮⋮⋮⋱⋮⋮⋮⋮⋮⋮⋱⋮⋮0000⋯000000⋯1−150000⋯0−150000⋯−1516n+2×6n+2,.LV12On2=−15−1500⋯00−15000⋯00−15−15−150⋯000000⋯000−15−15−135⋯000000⋯0000−135−17⋯00000−17⋯00⋮⋮⋮⋮⋱⋮⋮⋮⋮⋮⋮⋱⋮⋮0000⋯−15−150000⋯000000⋯−15−150000⋯0−15−15000⋯00−15−1500⋯000000⋯00−15−15−150⋯000000⋯000−15−15−135⋯00000−17⋯0000−135−17⋯00⋮⋮⋮⋮⋱⋮⋮⋮⋮⋮⋮⋱⋮⋮0000⋯000000⋯−17−150000⋯0−150000⋯−15−156n+2×6n+2.By equation (8), we have a matrix of order 6n+2:(10)LAOn2=225−1500⋯00−15000⋯00−1525−150⋯000000⋯000−1525−135⋯000000⋯0000−13537⋯00000−17⋯00⋮⋮⋮⋱⋮⋮⋮⋮⋮⋮⋱⋮⋮⋮0000⋯25−150000⋯000000⋯−15250000⋯0−15−15000⋯0025−1500⋯000000⋯00−1525−150⋯000000⋯000−1525−135⋯00000−17⋯0000−13537⋯00⋮⋮⋮⋱⋮⋮⋮⋮⋮⋮⋮⋱⋮⋮0000⋯000000⋯25−150000⋯0−150000⋯−1525,and ℒSOn2=diag6/5,6/5,6/5,8/7,…,6/5,6/5,8/7,6/5,6/5,8/7,…,8/7,6/5,6/5,6/5Q, a diagonal matrix with order 6n+2.The normalized Laplacian spectrum ofOn2 is constructed by the eigenvalues of ℒAOn2 and ℒSOn2, according to Lemma 1. Given the fact that ℒSOn2 is just a diagonal matrix of order 6n+2, it is obvious that 6/5 with multiplicity 4n+4 and 8/7 with multiplicity 2n−2 are the eigenvalues of ℒSOn2.Let(11)A=25−1500⋯00−1525−150⋯000−1525−135⋯0000−13537⋯00⋮⋮⋮⋮⋱⋮⋮0000⋯15−150000⋯−15253n+1×3n+1,C=−15000⋯000000⋯000000⋯00000−17⋯00⋮⋮⋮⋱⋮⋮⋮0000⋯000000⋯0−153n+1×3n+1.Thus,1/2ℒA could be represented by the block matrix below:(12)12ℒA=ACCA.Let(13)T=12I3n+112I3n+112I3n+1−12I3n+1.Then,(14)T12ℒAT′=A+C00A−C,where T′ indicates the transposition of T. Let P=A+C and Q=A−C. Then,(15)P=15−1500⋯000−1525−150⋯0000−1525−135⋯00000−13527⋯000⋮⋮⋮⋮⋱⋮⋮⋮0000⋯25−1500000⋯−1525−150000⋯0−15153n+1×3n+1,Q=35−1500⋯000−1525−150⋯0000−1525−135⋯00000−13547⋯000⋮⋮⋮⋮⋱⋮⋮⋮0000⋯25−1500000⋯−1525−150000⋯0−15353n+1×3n+1.By Lemma1, it is simple to verify that the eigenvalues of 1/2ℒA consist of those of P and Q. Suppose that the eigenvalues of P and Q are denoted by γi and ξji,j=1,2,…,3n+1 with γ1≤γ2≤⋯≤γ3n+1 and ξ1≤ξ2≤⋯≤ξ3n+1, respectively. Then, the eigenvalues of ℒA are 2γ1,2γ2,…,2γ3n+1 and 2ξ1,2ξ2,…,2ξ3n+1.where 0=γ1<γ2≤⋯≤γ3n+1 and 0<ξ1≤ξ2≤⋯≤ξ3n+1 are eigenvalues of P and Q, respectively.Lemma 4.
Suppose thatOn2 is the strong product of octagonal network. Then,(16)Kf∗On2=234n+64n+456+2n−278+12∑i=23n+11γi+12∑j=13n+11ξj,On the basis of the relation between the coefficients and roots ofΦP (resp. ΦQ), the formulae of ∑i=23n+11/γi (resp. ∑j=13n+11/ξj) are obtained in the next lemmas.Lemma 5.
Suppose that0=γ1<γ2≤⋯≤γ3n+1 are described as above. Then,(17)∑i=23n+11γi=1359n3+1115n2+434n1417n+3.Suppose thatΦP=x3n+1+a1x3n+⋯+a3n−1x2+a3nx=xx3n+a1x3n−1+⋯+a3n−1x+a3n. Then, γ2,γ3,…,γ3n+1 satisfy the equation below:(18)x3n+a1x3n−1+⋯+a3n−1x+a3n=0,so 1/γ2,1/γ3,…,1/γ3n+1 satisfy the equation below:(19)a3nx3n+a3n−1x3n−1+⋯+a1x+1=0.Hence, by Vieta’s theorem, we obtain(20)∑i=23n+11γi=−13n−1a3n−1−13na3n.For the sake of convenience, considerWi of P, which is the ith order principal submatrix generated by the first i columns and rows, i=1,2,…,3n. Let wi=detWi. Then,(21)w1=15,w2=125,w3=1125,w4=1875,w5=14375,w6=121875,w3i=25w3i−1−125w3i−2,for1≤i≤n,w3i+1=27w3i−135w3i−1,for1≤i≤n−1,w3i+2=25w3i+1−135w3i,for1≤i≤n−1.By explicit calculation, these general formulae can be obtained as follows:(22)w3i=751175i,for1≤i≤n,w3i+1=151175i,for0≤i≤n−1,w3i+2=1251175i,for0≤i≤n−1.The structure and determinant of matrixℒA are preserved by a permutation similarity transformation of a square matrix, and one gets detU3n+1−i=detW3n+1−i. We have(23)−13na3n=∑i=13n+1detPi=∑i=23ndetPi+2w3n=∑k=1ndetP3k+∑k=1n−1detP3k+1+∑k=0n−1detP3k+2+2w3n=∑k=1nw3k−1+2·w3n−k+1+∑k=1n−1w3k·w3n−k+∑k=0n−1w3k+1·w3n−k−1+2+2w3n=17n+36251175n−1,as desired.Claim 1.
−13n−1a3n−1=1359n3+1115n2+434n/87501/175n−1.Proof .
of Claim 1. Noticing that−13n−1a3n−1 is equal to the sum of all principal minors of P with 3n−1 columns and rows, we have(24)−13n−1a3n−1=∑1≤i<j3n+1Wi−1000Z000U,1≤i<j≤3n+1,where(25)Z=ki+1,i+1−135⋯0−135ki+2,i+2⋯0⋮⋮⋱⋮00⋯kj−1,j−1,U=kj+1,j+1⋯00⋮⋱⋮⋮0⋯k3n,3n−150⋯−15k3n+1,3n+1.
Note that(26)−13n−1a3n−1=∑1≤i<j3n+1detPi,j=∑1≤i<j3n+1wi−1·w3n+1−j·detZ.Remark 1.
If1≤i<i+1=j≤3n+1, then Z is an empty matrix and let detZ=1. By equation (26), there are different possibilities which can be selected for i and j. Therefore, all these cases are classified as follows.Case 1.
Leti=3p and j=3q, for 1≤i<j≤3n+1. So, 1≤p<q≤n:(27)detZ=27−13500⋯00−13525−150⋯000−1525−135⋯0000−13527⋯00⋮⋮⋮⋮⋱⋮⋮0000⋯27−1350000⋯−135253q−3p−1=3q−3p+11175q−p.Case 2.
Leti=3p and j=3q+1, for 1≤i<i+1<j≤3n+1. So, 1≤p≤q≤n−1:(28)detZ=27−13500⋯00−13525−150⋯000−1525−135⋯0000−13527⋯00⋮⋮⋮⋮⋱⋮⋮0000⋯25−150000⋯−15253q−3p=3q−3p+271175q−p.Case 3.
Leti=3p and j=3q+2, for 1≤i<j≤3n+1. So, 1≤p≤q≤n−1:(29)detZ=27−13500⋯00−13525−150⋯000−1525−135⋯0000−13527⋯00⋮⋮⋮⋮⋮⋮⋮0000⋯25−1350000⋯−135273q−3p+1=3q−3p+3351175q−p.Case 4.
Leti=3p+1 and j=3q, for 1≤i<j≤3n+1. So, 0≤p<q≤n:(30)detZ=25−1500⋯00−1525−1350⋯000−13527−135⋯0000−13525⋯00⋮⋮⋮⋮⋱⋮⋮0000⋯27−1350000⋯−135253q−3p−2=353q−3p−11175q−p.Case 5.
Leti=3p+1 and j=3q+1, for 1≤i<j≤3n+1. So, 0≤p<q≤n−1:(31)detZ=25−1500⋯00−1525−1350⋯000−13527−135⋯0000−13525⋯00⋮⋮⋮⋮⋱⋮⋮0000⋯25−150000⋯−15253q−3p−1=73q−3p1175q−p.Case 6.
Leti=3p+1 and j=3q+2, for 1≤i<i+1<j≤3n+1. So, 0≤p≤q≤n−1:(32)detZ=25−1500⋯00−1525−1350⋯000−13527−135⋯0000−13525⋯00⋮⋮⋮⋮⋱⋮⋮0000⋯25−1350000⋯−135273q−3p=3q−3p+11175q−p.Case 7.
Leti=3p+2 and j=3q, for 1≤i<i+1<j≤3n+1. So, 0≤p<p+1<q≤n:(33)detZ=25−13500⋯00−13527−1350⋯000−13525−15⋯0000−1525⋯00⋮⋮⋮⋮⋱⋮⋮0000⋯27−1350000⋯−135253q−3p−3=3q−3p+11175q−p.Case 8.
Leti=3p+2 and j=3q+1, for 1≤i<j≤3n+1. So, 0≤p<q≤n−1:(34)detZ=25−13500⋯00−13527−1350⋯000−13525−15⋯0000−1525⋯00⋮⋮⋮⋮⋱⋮⋮0000⋯25−150000⋯−15253q−3p−2=353q−3p−11175q−p.Case 9.
Leti=3p+2 and j=3q+2, for 1≤i<j≤3n+1. So, 0≤p<q≤n−1:(35)detZ=25−13500⋯00−13527−1350⋯000−13525−15⋯0000−1525⋯00⋮⋮⋮⋮⋱⋮⋮0000⋯25−1350000⋯−135273q−3p−1=53q−3p1175q−p.
Combining these results with equation (26) and Cases 1–9 yields(36)−13n−1a3n−1=∑1≤i<j≤3n+1wi−1·w3n+1−j·detZj−1−i=E1+E2+E3,where(37)E1=∑1≤p<q≤ndetP3p,3q+∑1≤p≤q≤n−1detP3p,3q+1+∑1≤p≤q≤n−1detP3p,3q+2+∑1≤p≤ndetP3p,3n+1=nn−1n+22501175n−1+n2n−12501175n−1+nn2−12501175n−1+3n2+n3501175n−1=21n3+15n2−16n17501175n−1,E2=∑1≤p<q≤ndetP3p+1,3q+∑1≤p<q≤n−1detP3p+1,3q+1+∑1≤p≤q≤n−1detP3p+1,3q+2+∑1≤p≤n−1detP3p+1,3n+1+∑1≤q≤ndetP1,3q+∑1≤q≤n−1detP1,3q+1+∑0≤q≤n−1detP1,3q+2+Z3n−1=7n2n−12501175n−1+49nn−1n−212501175n−1+7nn−122501175n−1+21nn−12501175n−1+n3n+1501175n−1+21nn−12501175n−1+n3n−1501175n−1+3n251175n−1=119n3+108n2+73n12501245n−1,E3=∑0≤p<q≤ndetP3p+2,3q+∑0≤p<q≤n−1detP3p+2,3q+1+∑0≤p<q≤n−1detP3p+2,3q+2+∑0≤p≤n−1detP3p+2,3n+1=nn+1n+387501175n−1+7n2n−12501175n−1+nn2−1501175n−1+n3n+1501175n−1=421n3+284n2+3n87501175n−1.
SubstitutingE1, E2, and E3 in Equation (36), we get Claim 2.
Also, we can get Lemma5 by combining Claims 1 and 2.Lemma 6.
Let0<ξ1<ξ2≤⋯≤ξ3n+1 be the eigenvalues of Q as above. Then,(38)∑j=13n+11ξj=5η1+η245+13154+15n−1+45−13154−15n−1,where η1=1500+40115+n1605+397154+15n−1 and η2=1500−40115+n1605−397154−15n−1.Proof .
Suppose thatΦQ=x3n+1+b1x3n+⋯+b3nx+b3n+1.
So,1/ξ1,1/ξ2,…,1/ξ3n+1 satisfy the equation below:(39)b3n+1x3n+1+b3nx3n+⋯+b1x+1=0.
By Vieta’s theorem, we obtain(40)∑j=13n+11ξj=∑j=13n+1ξ1⋯ξj−1ξj+1⋯ξ3n+1∏j=13n+1ξj=−13nb3n−13n+1b3n+1=−13nb3ndetQ.
In order to find−13nb3n and detQ in (40), consider Ri of Q, which is the ith order principal submatrix generated by the first i columns and rows, 1≤i≤3n. Let ri=detRi. Then, r1=3/5, r2=1/5, r3=7/125, r4=23/875, r5=39/4375, r6=11/4375, and(41)r3i=25r3i−1−125r3i−2,for1≤i≤n,r3i+1=47r3i−135r3i−1,for1≤i≤n−1,r3i+2=25r3i+1−135r3i,for1≤i≤n−1.
Similar to the method used as described above, we have(42)r3i=35+715504+15175i+35−715504−15175i,for1≤i≤n,r3i+1=75+19157504+15175i+75−19157504−15175i,for0≤i≤n−1,r3i+2=45+11151504+15175i+45−11151504−15175i,for0≤i≤n−1.Fact 1.
detQ=45+1315/93754+15/175n−1+45−1315/93754−15/175n−1.Proof .
Fact 1. ExpandingdetQ along the last row, we have(43)detQ=35detr3n−125detr3n−1=3535+715504+15175n+35−715504−15175n−12545+11151504+15175n−1+45−11151504−15175n−1=45+131593754+15175n−1+45−131593754−15175n−1.Fact 2.
(44)−13nb3n=1500+40115+n1605+3971518754+15175n−1+1500−40115+n1605−3971518754−15175n−1.Proof .
of Fact 2. Noting that−13nb3n is the summation of all principal minors of Q with 3n columns and rows, we have(45)−13nb3n=∑i=13n+1detQi=∑i=13n+1detRi−100S3n+1−i=∑i=13n+1detri−1·dets3n+1−i,where(46)S3n+1−i=li+1,i+1⋯00⋮⋱⋮⋮0⋯l3n,3n−1350⋯−135l3n+1,3n+1.
The structure and determinant of matrixℒA are preserved by a permutation similarity transformation of a square matrix, and one gets detS3n+1−i=detR3n+1−i. In line with Equation (45), we have(47)−13nb3n=∑i=13n+1detQi=∑p=1ndetQ3p+∑p=0n−1detQ3p+1+∑p=0n−1detQ3p+2+r3n=∑p=1nr3p−1+2·r3n−p+1+∑p=1n−1r3p·r3n−p+∑p=0n−1r3p+1·r3n−p−1+2+2r3n.
The following forms can also be generated by using the above equations:(48)∑p=1nr3p−1+2·r3n−p+1=n150+37153754+15175n−1+150−37153754−15175n−1+141534+15175n−141534−15175n,(49)∑p=1n−1r3p·r3n−p=n−135+715254+15175n+35−715254−15175n+1518754+15175n−1−1518754−15175n−1,(50)∑p=0n−1r3p+1·r3n−p−1+2=n150+37153754+15175n−1+150−37153754−15175n−1+141534+15175n−141534−15175n,(51)2r3n=35+715254+15175n+35−715254−15175n.
We can obtain the desired result ofFact 2 by substituting equations (48)–(51) into (47).
In view of (40), Facts 1 and 2 and Lemma 6 hold immediately.The following theorem is derived from Lemmas4–6.Theorem 1.
LetOn2=K2⊠On. Then,(52)Kf∗On2=4077n3+10604n2+4844n+39921+34n+6−13nb3ndetQ,where(53)−13nb3n=1500+40115+n1605+3971518754+15175n−1+1500−40115+n1605−3971518754−15175n−1detQ=45+131593754+15175n−1+45−131593754−15175n−1.The explicit formulae of the spanning trees ofOn2 are given below.Theorem 2.
LetOn2=K2⊠On. Then,(54)τOn2=216n−3·34n+33545+13154+15n−1+45−13154−15n−1.Proof .
By Lemma2, we have 6/54n+4·8/72n−2∏i=23n+12γi∏i=13n+12ξi∏v∈VOn2dOn2=2EOn2τOn2. Note that(55)∏v∈VOn2dOn2=58n+8·74n−4,EOn2=34n+6,∏i=23n+1γi=17n+36251175n−1,∏i=13n+1ξi=detQ=45+131593754+15175n−1+45−131593754−15175n−1.
Hence, Theorem2 immediately follows, along with Lemma 2.


## 4. Conclusion

In this study, we considerOn2, which is the strong prism of the octagonal network. Using the normalized Laplacian theorems, we have determined the multiplicative degree-Kirchhoff index and the spanning tree of On2. New discoveries, developments, and advancements in research are still required. In the near future, we will be exploring a more complex chemistry network.



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