# Statistical Analyses of a Class of Random Pentagonal Chain Networks with respect to Several Topological Properties

Authors: Jia-Bao Liu; Qing Xie; Jiao-Jiao Gu
Journal: Journal of Function Spaces
 (2023)

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## Abstract

There has been an upsurge of research on complex networks in recent years. The purpose of this paper is to study the mathematical properties of the random pentagonal chain networksPECn with the help of graph theory. Based on the networks PECn, we first obtain the expected value expressions of the Gutman index, Schultz index, multiplicative degree-Kirchhoff index, and additive degree-Kirchhoff index, and then, we get the explicit expression formulas of their variances. Finally, we find that their limiting distributions all have the probabilistic and statistical significance of normal distribution.


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## Body

## 1. Introduction

Complex networks have become increasingly important in scientific research, especially in statistical physics and information science [1]. In the research of complex networks, the analysis and synthesis issues of complex dynamic networks have received extensive attention in recent years [2]. Therefore, researchers began to try to use this new theoretical tool to study some complex systems in real life, among which the structure of complex systems and the relationship between system structure and function are the focuses of researches [3, 4]. The application of graph theory, as the name implies, is to solve the corresponding problems in real life by combining the knowledge of graph theory, widely used in engineering design. Graph theory is a basic but very effective tool to study complex networks. There are a lot of complex systems in nature which can be described by network graphs. It is composed of many nodes and the edges between the nodes, where the nodes represent different individuals and the edges represent the relationships between individuals. Topological index is one of the tools in graph theory, which can reflect the related properties and connected information of some kinds of topological structures. It has been extensively researched in graph theory. Therefore, this paper combines the three concepts to make a quantitative analysis of a kind of random pentagonal chain networks.Suppose thatG represents a simple undirected graph with edge set EG and vertex set VG. In G, the number of edges associated with a vertex i is called the degree of it, and we express it by dgi. The distance dsi,j between any two vertices i and j in the graph represents the length of the shortest path between them. To learn more about the relevant definitions, one can be referred to [5]. The Wiener index WG is the sum of the distances between all pairs of vertices in a graph G, denoted by
(1)WG=∑i,j∈VGdsi,j.This graph invariant related to the distance of a graph was introduced into chemistry in 1947 [6] and mathematics 30 years later [7]. It is a very important parameter in certain applications of chemistry and network theory, demonstrating the connectivity of the structure. Binu et al. [8] discussed the Wiener index of a fuzzy graph and application to illegal immigration networks. Asir and Rabikka [9] presented a method for calculating the Wiener index of zero-divisor graph of any positive integer n. Wei and Shiu [10] obtained the expected values and the average values of the Wiener index in random polygonal chains. Chen et al. [11, 12] studied and compared the Wiener index in the catacondensed hexagonal system and the tree-like polyphenyl system. At present, the Wiener index is still broadly used in the fields of physics, molecular biology, and mathematics. For more applications, please refer to [13–18].Based on the concept of distance and electrical network theory, the resistance distancerdi,j is derived, which was put forward by Klein and Randić [19]. When each edge in the graph is equivalent to a unit resistance, it can be considered as the effective resistance between two vertices. We can learn more details through [20]. The Kirchhoff index KfG, also known as full effective resistance, or effective graph resistance, is defined as the sum of the resistance distances of all vertex pairs in the graph [21–23], that is,
(2)KfG=∑i,j∈VGrdi,j.Through the efforts of the researchers, it has been founded that the Kirchhoff index has played an important role in chemistry. For linear hexagonal chains, some special molecular graphs, such as the bipartite unicyclic graphs, bicyclic graphs, and Cayley graphs, we can calculate the Kirchhoff indices of these polycyclic structures to evaluate their cyclicity [24–28]. At the same time, the Kirchhoff index is also of great significance in mathematics. For the details of the mathematical properties, please refer to [29–31].As for the Wiener index, researchers improved it by weighting it. That is, a graphG=VG,EG related to the weight function ωi∈VG, expressed by G,ω. Suppose ⊕ represents a calculation symbol in +, −, ×, and ÷, then the transformed Wiener index WG,ω can be expressed as
(3)WG,ω=12∑i∈VG∑j∈VGωi⊕ωjdsi,j.When⊕ is the arithmetic operation × and set ω·≡dg·, the Gutman index is obtained. Then, Equation (3) is equivalent to
(4)GutG=12∑i∈VG∑j∈VGdgidgjdsi,j=∑i,j∈VGdgidgjdsi,j.When⊕ is the arithmetic operation + and set ω·≡dg·, the Schultz index is obtained. Then, Equation (3) is equivalent to
(5)SG=12∑i∈VG∑j∈VGdgi+dgjdsi,j=∑i,j∈VGdgi+dgjdsi,j.In the field of chemistry, the Gutman index and Schultz index can accurately reflect the similar molecular structure characteristics as the Wiener index. For more details, properties, and applications of the Gutman index and Schultz index, if you are interested, you can refer to the paper [32–37].Similarly, researchers also improved and optimized the Kirchhoff index. Considering the degree of vertices in a graph, the multiplicative degree-Kirchhoff index was put forward in 2007 by Chen and Zhang [38], which was expressed as
(6)Kf∗G=∑i,j∈VGdgidgjrdi,j.In 2012, the additive degree-Kirchhoff index was proposed by Gutman et al. [39], denoted by
(7)Kf+G=∑i,j∈VGdgi+dgjrdi,j.The Kirchhoff index and degree-Kirchhoff index are used as graph parameters based on resistance distance in physics and chemistry. And network science also has a wide range of applications. For an electrical network, the smaller the Kirchhoff index is, the less power the network consumes per unit time. Chemically, Kirchhoff index can be used to describe the structural characteristics of molecules and define the topological radius of polymers. In network science, the smaller the Kirchhoff index is, the stronger the robustness of the network is. These fields have attracted the attention of more and more researchers. Readers can refer to references [40–42] for more information.The random chain networksPECn studied in this paper are finite 2-connected network graphs, which can be seen as being formed by sequentially connecting n pentagons. That is, PECn is obtained by PECn−1 and then randomly connected a pentagon at the end, as shown in Figure 1. Whenn=1,2,3, as shown in Figure 2. For n≥3, there are two ways to add pentagons at the end, and the results can be expressed as PECn1 and PECn2, as shown in Figure 3. For such a random chain networks, any step for δ=3,4,⋯,n is stochastic, and their probabilities are p and p~, respectively, that is, from PECn−1 to PECn1 with probability p and to PECn2 with probability p~=1−p, in which the probabilities p and p~ are constants and independent to the step δ at the same time.Figure 1
The generation process of the random pentagonal chain networksPECn.Figure 2
Random pentagonal chain networks whenn=1,2,3.Figure 3
Two types of connections for the random pentagonal chain networks.Motivated by [43], we use two random variables Xn1 and Xn2 to represent our choices. If our choice is PECni, we put Xni=1; otherwise, it is equal to 0 (i=1,2). One holds that
(8)PXn1=1=p,PXn2=1=p~=1−p,and Xn1+Xn2=1.Yang and Zhang [44] and Ma et al. [45] determined explicit formulas of WG and EWG for a class of random chain networks composed of hexagons, respectively. Similarly, Huang et al. [46] obtained EKfG for a class of spiro random chain networks composed of hexagons. Wei and Shiu [10] put forward the expression of EWG in random polygonal chains and proved its asymptotic property. At the same way, Zhang et al. [47] obtained the simple formulas of EGutG, ESG, EKf∗G, and EKf+G for a class of random polyphenylene chain networks.The research motivation of this paper is to investigate the structural properties of random pentagonal chain networks and compare the similarities and differences between their corresponding different topological indices. Motivated by [43, 45–48], for this paper, in Section 2, we determine the explicit formulas of EGutPECn, ESPECn, EKf∗PECn, and EKf+PECn for the random pentagonal chain networks and show their differences visually through the function images. In Section 3, we continue to calculate VarGutPECn, VarSPECn, VarKf∗PECn, and VarKf+PECn and to prove that the four indices of PECn asymptotically obey normal distributions. In order to accomplish the purpose of this paper, we must put forward the following hypotheses.Hypothesis 1.
It randomly and independently chooses a way attaching the new terminal pentagonOn+1toPECn,n=2,3,⋯. To be more precise, the sequences of random variables Xn1,Xn2n=2∞ are independent and must satisfy Equation (8).Hypothesis 2.
We put0<p<1.
Under the condition that both Hypotheses1 and 2 are true, we can get the following:
(1)
The analytical expressions of the variances ofGutPECn, SPECn, Kf∗PECn, and Kf+PECn are obtained(2)
Whenn⟶∞, we verify that the random variables GutPECn, SPECn, Kf∗PECn, and Kf+PECn asymptotically obey normal distributions. Replace them with Xn. It is evident to see that(9)limn⟶∞supε∈ℝPXn−EXnVarXn≤ε−∫−∞εe−x2/22πdx=0,
whereEXn and VarXn represent the expectation and variance of this random variable Xn, respectively
In this paper, assume thatfx and hx are two functions of x. We put fx=Ohx if limx⟶∞fx/hx=0 and put fx≍hx if limx⟶∞fx/hx=1.


## 2. The Expected Values ofGutPECn, SPECn, Kf∗PECn, and Kf+PECn

For the random chain networksPECn, we find that GutPECn, SPECn, Kf∗PECn, and Kf+PECn are random variables. Then, we determine the analytic expressions of EGutPECn, ESPGn, EKf∗PECn, and EKf+PECn in this section.In fact,PECn is organized by adding a new terminal pentagon On to PECn−1 by an edge, where the vertices of On are labelled as v1, v2, v3, v4, and v5 in clockwise direction. For all i∈VPECn−1, one has
(10)dsv1,i=dswn−1,i+1,dsv2,i=dswn−1,i+2,dsv3,i=dswn−1,i+3,dsv4,i=dswn−1,i+3,dsv5,i=dswn−1,i+2,∑i∈VPECn−1dgPECni=12n−13,∑i∈VPECndgPECn+1i=12n−1.Meanwhile,(11)∑k=15dgvkdsv1,vk=22,∑k=15dgvkdsv2,vk=13,∑k=15dgvkdsv3,vk=14,∑k=15dgvkdsv4,vk=15,∑k=15dgvkdsv5,vk=13.Theorem 3.
Forn≥1, the analytic expression of EGutPECn is
(12)EGutPECn=72−24pn3+72p−12n2+1−48pn−1.Proof of Theorem 3.
By Equation (4), we use VPECn\wn to denote the set of all other vertices in PECn after the vertex wn is removed. Then, one can be convinced that
(13)GutPECn+1=∑i,j∈VPECndgidgjdsi,j+∑i∈VPECn∑vk∈VOn+1dgidgvkdsi,vk+∑vk,vl∈VOn+1dgvkdgvldsvk,vl.Note that(14)∑i,j∈VPECndgidgjdsi,j=∑i,j∈VPECnwndgidgjdsi,j+∑i∈VPECnwndgPECn+1wndgidswn,i=∑i,j∈VPECnwndgidgjdsi,j+∑i∈VPECnwndgPECnwn+1dgidswn,i=GutPECn+∑i∈VPECndgidswn,i.Recall thatdgv1=3 and dgvk=2 for k∈2,3,4,5. From Equation (10), we have
(15)∑i∈VPECn∑vk∈VOn+1dgidgvkdsi,vk=∑i∈VPECndgi3dswn,i+1+2dswn,i+2+2dswn,i+3+2dswn,i+3+2dswn,i+2=∑i∈VPECndgi11dswn,i+23=11∑i∈VPECndgidswn,i+276n−23.From Equation (11), one follows that
(16)∑vk,vl∈VOn+1dgvkdgvldsvk,vl=12∑k=15dgvk∑l=15dgvldsvk,vl=123×12+2×2×13+2×2×14=72.Then,(17)GutPECn+1=GutPECn+12∑i∈VPECndgidswn,i+276n+49.For the random chain networksPECn, we obtain that ∑i∈VPECndgidswn,i is a stochastic variate. Let
(18)An=E∑i∈VPECndgidswn,i.By using the above formula and Equation (17), we can get the following relation for EGutPECn+1. One sees
(19)EGutPECn+1=EGutPECn+12An+276n+49.Then, we go on to consider the following two possible cases.Case 1.
PECn⟶PECn+11.
In this case,wn (of PECn) overlaps with v2 or v5 (of On). Therefore, ∑i∈VPECndgidswn,i is rewritten as ∑i∈VPECndgidsv2,i or ∑i∈VPECndgidsv5,i with probability p.Case 2.
PECn⟶PECn+12.
In this case,wn (of PECn) overlaps with v3 or v4 (of On). Therefore, ∑i∈VPECndgidswn,i is rewritten as ∑i∈VPECndgidsv3,i or ∑i∈VPECndgidsv4,i with probability p~=1−p.Together with the above cases, by applying the expectation operator and Equation (10), one follows that
(20)An=p∑i∈VPECndgidsv2,i+1−p∑i∈VPECndgidiv3,i,where
(21)∑i∈VPECndgidsv2,i=∑i∈VPECn−1dgidswn−1,i+2+∑i∈VOndgidsv2,i=∑i∈VPECn−1dgidswn−1,i+2∑i∈VPECn−1dgi+13,∑i∈VPECndgidsv3,i=∑i∈VPECn−1dgidswn−1,i+3+∑i∈VOndgidsv3,i=∑i∈VPECn−1dgidswn−1,i+3∑i∈VPECn−1dgi+14.Then, we obtain(22)An=p∑i∈VPECn−1dgidswn−1,i+2∑i∈VPECn−1dgi+13+1−p∑i∈VPECn−1dgidswn−1,i+3∑i∈VPECn−1dgi+14=pAn−1+212n−13+13+1−pAn−1+312n−13+14=An−1+36−12pn+12p−25.Meanwhile, forn=1, the boundary condition is
(23)A1=∑i∈VPEC1dgidsw1,i=12.Using above condition and the recurrence relation with respect toAn, it is no hard to obtain
(24)An=18−6pn2+6p−7n+1.From Equation (19), it holds that
(25)EGutPECn+1=EGutPECn+12An+276n+49=EGutPECn+1218−6pn2+6p−7n+1+276n+49.Forn=1, we obtain EGutPEC1=60. Similarly, according to the recurrence relation related to EGutPECn, we have
(26)EGutPECn=72−24pn3+72p−12n2+1−48pn−1.Theorem 4.
Forn≥1, the analytic expression of ESPECn is
(27)ESPECn=60−20pn3+60p+7n2−40p+7n.Proof of Theorem 4.
Notice that the random chain networksPECn+1 are organized by adding a new terminal pentagon On+1 to PECn by an edge. By Equation (5), one has
(28)SPECn+1=Δ1+Δ2+Δ3,where
(29)Δ1=∑i,j∈VPECndgi+dgjdsi,j,Δ2=∑i∈VPECn∑vk∈VOn+1dgi+dgvkdsi,vk,Δ3=∑vk,vl∈VOn+1dgvk+dgvkdsvk,vl.
LetdswnPECn=∑i∈VPECndswn,i. It is routine to check that
(30)Δ1=∑i,j∈VPECnwndgi+dgjdsi,j+∑i∈VPECnwndgPECn+1wn+dgidswn,i=∑i,j∈VPECnwndgi+dgjdsi,j+∑i∈VPECnwndgPECnwn+1+dgidswn,i=SPECn+dswnPECn.BecausePECn has 5n vertices and dgv1=3, dgvk=2 for k∈2,3,4,5. By Equation (10), we can know
(31)Δ2=∑i∈VPECn∑vk∈VOn+1dgidsi,vk+∑i∈VPECn∑vk∈VOn+1dgvkdsvk,i=∑i∈VPECndgi∑vk∈VOn+1dsi,vk+∑i∈VPECn∑vk∈VOn+1dgvkdsi,vk=∑i∈VPECndgi5dswn,i+11+∑i∈VPECn11dgwn+23=5∑i∈VPECndgidswn,i+1112n−1+11∑i∈VPECndswn,i+23×5n=5∑i∈VPECndgidswn,i+11dswnPECn+247n−11.Since that,∑k=15dsvk,vl=6 for l=1,2,⋯,5. From Equation (11), one sees that
(32)Δ3=∑vk,vl∈VOn+1dgvk+dgvldsvk,vl=12∑k=15∑l=15dgvk+dgvldsvk,vl=∑k=15∑l=15dgvkdsvk,vl=6×3+2×4=66.Then, Equation (28) can be rewritten as
(33)SPECn+1=SPECn+5∑i∈VPECndgidswn,i+12dswnPECn+247n+55.We know thatdswnPECn is a random variable. Let
(34)Bn=EdswnPECn.According to the above formula and Equation (33), we have the following relation for ESPECn+1. It holds that
(35)ESPECn+1=ESPECn+5An+12Bn+247n+55.We proceed by taking into account the following two cases.Case 1.
PECn⟶PECn+11.
In this case,wn (of PECn) coincides with v2 or v5 (of On). Then, dswnPECn is given by dsv2PECn or dsv5PECn with probability p.Case 2.
PECn⟶PECn+12.
In this case,wn (of PECn) coincides with v3 or v4 (of On). Then, dswnPECn is given by dsv3PECn or dsv4PECn with probability p~=1−p.Together with Case1 and Case 2, we obtain
(36)Bn=p·dsv2PECn+1−p·dsv3PECn.where
(37)dsv2PECn=dswn−1PECn−1+2×5n−1+6=dswn−1PECn−1+10n−4,dsv3PECn=dswn−1PECn−1+3×5n−1+6=dswn−1PECn−1+15n−9.Then, as an immediate consequence, we have(38)Bn=pBn−1+10n−4+1−pBn−1+15n−9=Bn−1+15−5pn+5p−9.Whenn=1, we can easily find that
(39)B1=Edsw1PEC1=6.Then, using above formula and the recurrence relation, it is routine to check that(40)Bn=152−52pn2+52p−32n.By using Equation (24) and Equation (35), we can get
(41)ESPECn+1=ESPECn+5An+12Bn+247n+55=ESPECn+518−6pn2+6p−7n+1+127.5−2.5pn2+2.5p−1.5n+247n+55=ESPECn+180−60pn2+60p+194n+60.According to the initial conditions of the difference equation, we findESPEC1=60. Then, we arrive at
(42)ESPECn=60−20pn3+60p+7n2−40p+7n.Next to start, we determine the analytic expressions ofEKf∗PECn and EKf+PECn. We continue to follow the original conditions, PECn+1 is organized by adding a new terminal pentagon On+1 to PECn by an edge, where the vertices of On+1 are labelled as v1, v2, v3, v4, and v5 in clockwise direction. For all i∈VPECn, one has
(43)rdv1,i=rdwn,i+1,rdv2,i=rdwn,i+1+45,rdv3,i=rdwn,i+1+65,rdv4,i=rdwn,i+1+65,rdv5,i=rdwn,i+1+45,∑i∈VPECndgPECn+1i=12n−1.Therefore,(44)∑k=15dgvkrdv1,vk=8,∑k=15dgvkrdv2,vk=445,∑k=15dgvkrdv3,vk=465,∑k=15dgvkrdv4,vk=465,∑k=15dgvkrdv5,vk=445.Theorem 5.
Forn≥1, the analytic expression of EKf∗PECn is
(45)EKf∗PECn=2645−485pn3+1445p−125n2+1935−965pn−49.Proof of Theorem 5.
By Equation (6), one can be convinced that
(46)Kf∗PECn+1=∑i,j∈VPECndgidgjrdi,j+∑i∈VPECn∑vk∈VOn+1dgidgvkrdi,vk+∑vk,vl∈VOn+1dgvkdgvlrdvk,vl.
Note that(47)∑i,j∈VPECndgidgjrdi,j=Kf∗PECn+∑i∈VPECndgirdwn,i.
Because ofdgv1=3 and dgvk=2 for k∈2,3,4,5, from Equation (43), we have
(48)∑i∈VPECn∑vk∈VOn+1dgidgvkrdi,vk=∑i∈VPECndgi3rdwn,i+1+4rdwn,i+1+45+4rdwn,i+1+65=∑i∈VPECndgi11rdwn,i+19=11∑i∈VPECndgirdwn,i+1912n−1.
From Equation (44), we follow that
(49)∑vk,vl∈VOn+1dgvkdgvlrdvk,vl=12∑k=15dgvk∑l=15dgvlrdvk,vl=123×8+2×2×445+2×2×465=96.
Then,(50)Kf∗PECn+1=Kf∗PECn+12∑i∈VPECndgirdwn,i+228n+77.
For thePECn, we know that ∑i∈VPECndgirdwn,i is a stochastic variate. Let
(51)Cn=E∑i∈VPECndgirdwn,i.Then, let us continue to consider the two probable cases shown below.Case 1.
PECn⟶PECn+11.
In this case,wn (of PECn) overlaps with v2 or v5 (of On). Therefore, ∑i∈VPECndgirdwn,i is rewritten as ∑i∈VPECndgvrdv2,i or ∑i∈VPECndgirdv5,i with probability p.Case 2.
PECn⟶PECn+12.
In this case,wn (of PECn) overlaps with v3 or v4 (of On). Therefore, ∑i∈VPECndgirdwn,i is rewritten as ∑i∈VPECndgirdv3,i or ∑i∈VPECndgirdv4,i with probability p~=1−p.Together with the above cases, by applying the expectation operator and Equation (43) and Equation (44), we can get that
(52)Cn=p∑i∈VPECndgirdv2,i+1−p∑i∈VPECndgirdv3,i=p∑i∈VPECn−1dgirdwn−1,i+1+45×12n−13+445+1−p∑i∈VPECn−1dgirdwn−1,i+1+65×12n−13+465.By taking the expectation of both sides of the above equation, and noting thatECn=Cn, we obtain
(53)Cn=Cn−1+p95×12n−13+445+1−p115×12n−13+465=Cn−1+1325−245pn+245p−975.Whenn=1, the boundary condition is
(54)C1=E∑i∈VPEC1dgirdw1,i=8.By using the above results to solve the difference equation, we can obtain(55)Cn=665−125pn2+125p−315n+1.So combining formulas (50) and (55), we can know
(56)EKf∗PECn+1=EKf∗PECn+12Cn+228n+77=EKf∗PECn+12665−125pn2+125p−315n+1+228n+77=EKf∗PECn+7925−1445pn2+1445p+7685n+89.We get thatEKf∗PEC1=40 for n=1. Similarly, EKf∗PECn can be obtained by solving the equation
(57)EKf∗PECn=2645−485pn3+1445p−125n2+1935−965pn−49.Theorem 6.
Forn≥1, the analytic expression of EKf+PECn is
(58)EKf+PECn=44−8pn3+48p+11n2−88p+15n+48p.Proof of Theorem 6.
According to the known conditions and the composition rules of the networksPECn+1, by Equation (7), one has
(59)Kf+PECn+1=Θ1+Θ2+Θ3,where
(60)Θ1=∑i,j∈VPECndgi+dgjrdi,j,Θ2=∑i∈VPECn∑vk∈VOn+1dgi+dgvkrdi,vk,Θ3=∑vk,vl∈VOn+1dgvk+dgvlrdvk,vl.
LetrdwnPECn=∑i∈VPECnrdwn,i. It is routine to check that
(61)Θ1=∑i,j∈VPECndgi+dgjrdi,j=Kf+PECn+rdwnPECn.
Note thatPECn has 5n vertices and dgv1=3 and dgvk=2 for k∈2,3,4,5. By Equation (43), we can get
(62)Θ2=∑i∈VPECn∑vk∈VOn+1dgi+dgvkrdi,vk=∑i∈VPECn∑vk∈VOn+1dgirdi,vk+∑i∈VPECn∑vk∈VOn+1dgvkrdi,vk=∑i∈VPECndgirdwn,i+1+rdwn,i+1+45+rdwn,i+1+65+rdwn,i+1+65+rdwn,i+1+45+∑i∈VPECn3rdwn,i+1+2rdwn,i+1+45+2rdwn,i+1+65+2rdwn,i+1+65+2rdwn,i+1+45=5∑i∈VPECndgirdwn,i+912n−1+11rdwnPECn+19×5n.
And due to∑k=15rdvk,xl=4 for l=1,2,⋯,5, from Equation (44), one sees that
(63)Θ3=∑vk,vl∈VOn+1dgvk+dgvlrdvk,vl=12∑k=15∑l=15dgvk+dgvlrdvk,vl=∑k=15∑l=15dgvkrdvk,vl=4×3+2×4=44.
Then, Equation (59) can be rewritten as
(64)Kf+PECn+1=Kf+PECn+5∑i∈VPECndgirdwn,i+12rdwnPECn+203n+35.For that reason,rdwnPECnis a random variable. Let
(65)Dn=ErdwnPECn.And we proceed by taking into account the following two cases.Case 1.
PECn⟶PECn+11.
In this case,wn (of PECn) coincides with v2 or v5 (of On). Then, rdwnPECn is given by rdv2PECn or rdv5PECn with probability p.Case 2.
PECn⟶PECn+12.
In this case,wn (of PECn) coincides with v3 or v4 (of On). Then, rdwnPECn is given by rdv3PECn or rdv4PECn with probability p~=1−p.Together with Case1 and Case 2, we have
(66)Dn=p·rdv2PECn+1−p·rdv3PECn=p·rdwn−1PECn−1+5n−11+45+4+1−p1·rdwn−1PECn−1+5n−11+65+4.By taking the expectation of both sides of the above equation, and noting thatEDn=Dn, we obtain
(67)Dn=pDn−1+9n−1+4+1−pDn−1+11n−1+4=Dn−1+11−2pn+6p−7.Whenn is equal to 1, the initial condition is
(68)D1=Erdw1PEC1=4.Combining the above conditions, through the calculation, we can get(69)Dn=112−pn2+5p−32n−4p.From Equations (64) and (69), we can get
(70)EKf+PECn=EKf+PECn+12Dn+5Cn+203n+35=EKf+PECn+12112−pn2+5p−32n−4p+5665−125pn2+125p−315n+1+203n+35=EKf+PECn+132−24pn2+72p+154n+40−48p.Then according toEKf+PEC1=40 and the recurrence relation related to EKf+PECn, we have
(71)EKf+PECn=44−8pn3+48p+11n2−88p+15n+48p.Next, we use the computer to generate the images of expectation functions of the above four topological indices about the networksPECn, as shown in Figure 4.Remark 7.
Through the figure, we can see that the expected values of the four indices are related to two variablesp and n, and all of them show a positive correlation. Among them, the growth rate of EGutPECn is the fastest and EKf∗PECn grows the slowest, while the growth rates of ESPECn and EKf+PECn are between them. This further illustrates the relationships between these four topological indices and vertex distance and degree.Figure 4
Images of the expectation function for the four topological indices.


## 3. The Variances and Limiting Behaviours forGutPECn, SPECn, Kf∗PECn, and Kf+PECn

In this section, we will calculate the explicit analytical expressions forVarGutPECn, VarSPECn, VarKf∗PECn, and VarKf+PECn. And we will prove that the Gutman, Schultz, multiplicative degree-Kirchhoff, and additive degree-Kirchhoff indices asymptotically obey normal distributions. We use the same notation as those used at Section 2.Theorem 8.
Suppose Hypotheses1 and 2 are correct, then we can get the following results:
(i)
The variance of the Gutman index is denoted by(72)VarGutPECn=130α12n5−5γ1n4+10β12n3−30α12+45β12−65γ1n2+59α12+65β12−120γ1n−30α12+30β12−60γ1,where
(73)α12=2882p+43221−p−288p+4321−p2,β12=1562p+30021−p−156p+3001−p2,γ1=288·156·p+432·300·1−p−288p+4321−p·156p+3001−p.(ii)
Forn⟶∞, GutPECn asymptotically obeys normal distributions. One has(74)limn⟶∞supε∈ℝPGutPECn−EGutPECnVarGutPECn≤ε−∫−∞εe−x2/22πdx=0.Proof of Theorem 8.
LetJn=12∑i∈VPECndgidswn,i. Then by Equation (17), we have
(75)GutPECn+1=GutPECn+Jn+276n+49.Recalling thatXn1 and Xn2 are random variables which stand for our choice to construct PECn+1 by PECn, we have the next two facts.Fact 9.
JnXn1=Jn−1+288n−156Xn1.Proof.
IfXn1=0, the result is obvious. Then, we only take into account Xn1=1, which implies PECn⟶PECn+11. In this case, wn (of PECn) overlaps with v2 or v5 (of On) (see Figure 3). In this situation, by using Equation (10) and Equation (11),
(76)Jn=12∑i∈VPECndgidsv2,i=12∑i∈VPECn−1dgidsv2,i+12∑i∈VOndgidsv2,i=12∑i∈VPECn−1dgidsi,wn−1+dsv2,wn−1+12×13=12∑i∈VPECn−1dgidsi,wn−1+2+12×13=Jn−1+24×12n−13+156=Jn−1+288n−156.Thus, we conclude the desired fact.Fact 10.
JnXn2=Jn−1+432n−300Xn2.Similar to the proof of Fact9, we only consider the fact Xn2=1, that is, PECn⟶PECn+12. In the same way, we omit the details.Noting thatXn1+Xn2=1, by the above discussions, it holds that
(77)Jn=JnXn1+Xn2=Jn−1+288n−156Xn1+Jn−1+432n−300Xn2=Jn−1+288Xn1+432Xn2n−156Xn1+300Xn2=Jn−1+nλn−μn,where for each n,
(78)λn=288Xn1+432Xn2,μn=156Xn1+300Xn2.Therefore, by Equation (75), it follows that
(79)GutPECn=GutPEC1+∑r=1n−1Jr+∑r=1n−1276r+49=GutPEC1+∑r=1n−1∑s=1r−1Js+1−Js+J1+∑r=1n−1276r+49=GutPEC1+∑r=1n−1∑s=1r−1Js+1−Js+n−1J1+∑r=1n−1276r+49=GutPEC1+∑r=1n−1∑s=1r−1s+1λs+1−μs+1+On2.By direct calculation, we have(80)Varλs=α12,Varμs=β12,Covλs,μs=γ1,where for any two stochastic variates X and Y, CovX,Y=EXY−EXEY (refer to ref. [48]); using the properties of variance, Equation (79), and exchanging the order of r and s, we can directly find out that
(81)VarGutPECn=Var∑r=1n−1∑s=1r−1s+1λs+1−μs+1=Var∑s=1n−2∑r=s+1n−1s+1λs+1−μs+1=Var∑s=1n−2s+1λs+1−μs+1n−s−1=∑s=1n−2n−s−12Vars+1λs+1−μs+1=∑s=1n−2n−s−12Covs+1λs+1−μs+1,s+1λs+1−μs+1=∑s=1n−2n−s−12s+12Covλs+1,λs+1−2s+1Covλs+1,μs+1+Covμs+1,μs+1=∑s=1n−2n−s−12s+12α12−2s+1γ1+β12.By using a computer, the above expression indicates the result in Theorem8 (i).Now, we will go on to the proof of Theorem8 (ii). Firstly, for any n∈ℕ, let
(82)Rn=∑r=1n−1∑s=1r−1s+1λs+1,Tn=∑r=1n−1∑s=1r−1μs+1,ξ=Eλs,φt=Eetλs−ξ.By these notations, obviously, we have(83)etRn−ERn=et∑r=1n−1∑s=1r−1s+1λs+1−ξ=et∑s=1n−2∑r=s+1n−1s+1λs+1−ξ=et∑s=1n−2n−s−1s+1λs+1−ξ.Then,
(84)EetRn−ERn=Eet∑s=1n−2n−s−1s+1λs+1−ξ=∏s=1n−2Eetn−s−1s+1λs+1−ξ=∏s=1n−2φtn−s−1s+1,and for some ρ>0,
(85)Tn=∑r=1n−1∑s=1r−1μs+1≤ρn2.Note that(86)VarGutPECn≍130α12n5,φt=1+α122t2+Ot2,∑s=1n−2s+12n−s−12≍n530.By Taylor’s formula and Equations (79)–(85), one holds that
(87)limn⟶∞EexptGutPECn−EGutPECnVarGutPECn=limn⟶∞EexptGutPEC1+Rn−Tn+On2−EGutPEC1+Rn−Tn+On2α1n5/2/30=limn⟶∞Eexpt30Rn−ERnα1n5/2=limn⟶∞∏s=1n−2φ30tn−s−1s+1α1n5/2=limn⟶∞exp∑s=1n−2lnφ30tn−s−1s+1α1n5/2=limn⟶∞exp∑s=1n−2ln1+α122·30t2n−s−12s+12α12n5+O1n=limn⟶∞exp∑m=1n−2α122·30t2n−s−12s+12α12n5+O1n=et2/2.Assume thati is a complex number with i2=−1. We use it instead of t, and one has
(88)limn⟶∞EexpitGutPECn−EGutPECnVarGutPECn=e−t2/2.According to the above formula ([49], Chapter 1) and the theory of continuity of probability characteristic functions ([50], Chapter 15), we complete the proof of Theorem 8 (ii).Theorem 11.
Suppose Hypotheses1 and 2 are true, then the next results hold.
(i)
The variance of the Schultz index is denoted by(89)VarSPECn=130α22n5−5γ2n4+10β22n3−30α22+45β22−65γ2n2+59α22+65β22−120γ2n−30α22+30β22−60γ2,where
(90)α22=2402p+36021−p−240p+3601−p2,β22=1132p+23321−p−113p+2331−p2,γ2=240·113p+360·2331−p−240p+3601−p113p+2331−p.(ii)
Forn⟶∞, SPECn asymptotically obeys normal distributions. One has(91)limn⟶∞supε∈ℝPSPECn−ESPECnVarSPECn≤ε−∫−∞εe−x2/22πdx=0.Proof of Theorem 11.
LetFn=∑i∈VPECn5dgi+12dswn,i. Then by Equation (33), we get
(92)SPECn+1=SPECn+Fn+247n+55.According to the previous Proof of Theorem8, the two facts are obtained.Fact 12.
FnXn1=Fn−1+240n−113Xn1.Proof.
IfXn1=0, the above result is distinct. So we take into account Xn1=1, which indicates PECn⟶PECn+11. In this case, wn (of PECn) coincides with v2 or v5 (of On) (see Figure 3). In this situation, by using Equation (10) and Equation (11), Fn becomes
(93)∑i∈VPECn5dgi+12dsv2,i=∑i∈VPECn−15dgi+12dsv2,i+∑i∈VOn5dgi+12dsv2,i=∑i∈VPECn−15dgi+12dswn−1,i+2+5×13+12×6=∑i∈VPECn−15dgi+12dswn−1,i+2×5×12n−13+12×5n−1+137=Fn−1+240n−113.Thus, we conclude the desired Fact.Fact 13.
FnXn2=Fn−1+360n−233Xn2.As that in the proof of Fact12, we only consider the fact Xn2=1, that is, PECn⟶PECn+12. The proof is similar and details are omitted.Noting thatXn1+Xn2=1, by the above discussions, it holds that
(94)Fn=FnXn1+Xn2=Fn−1+240n−113Xn1+Fn−1+360n−233Xn2=Fn−1+240Xn1+360Xn2n−113Xn1+233Xn2=Fn−1+nλn′−μn′,where for each n,
(95)λn′=240Xn1+360Xn2,μn′=113Xn1+233Xn2.Therefore, by Equation (92),
(96)SPECn=SPEC1+∑r=1n−1Fr+∑r=1n−1247r+55=SPEC1+∑r=1n−1∑s=1r−1Fs+1−Fs+F1+∑r=1n−1247r+55=SPEC1+∑r=1n−1∑s=1r−1Fs+1−Fs+n−1F1+∑r=1n−1247r+55=SPEC1+∑r=1n−1∑s=1r−1s+1λs+1′−μs+1′+On2.Suppose that(97)Varλs′=α22,Varμs′=β22,Covλs′,μs′=γ2.If we substituteGutPECn by SPECn in the Proof of Theorem 8, the rest of the proof is identical to the Proof of Theorem 8 and the details are omitted here.Theorem 14.
Suppose Hypotheses1 and 2 are true, there are the next main results.
(i)
The variance ofKf∗PECn is denoted by(98)VarKf∗PECn=130α32n5−5γ3n4+10β32n3−30α32+45β32−65γ3n2+59α32+65β32−120γ3n−30α32+30β32−60γ3,where
(99)α32=129652p+1584521−p−12965p+158451−p2,β32=87652p+1164521−p−8765p+116451−p2,γ3=12965·8765p+15845·116451−p−12965p+158451−p8765p+116451−p.(ii)
Forn⟶∞, Kf∗PECn asymptotically obeys normal distributions. One has(100)limn⟶∞supε∈ℝPKf∗PECn−EKf∗PECnVarKf∗PECn≤ε−∫−∞εe−x2/22πdx=0.Proof of Theorem 14.
Obtained by formula (50), we see
(101)Kf∗PECn+1=Kf∗PECn+12∑i∈VPECndgirdwn,i+288n+77.
LetHn=12∑i∈VPECndgirdwn,i; we have
(102)Kf∗PECn+1=Kf∗PECn+Hn+288n+77.Recalling thatXn1 and Xn2 are random variables which show the way to construct PECn+1 from PECn, we get the following two facts.Fact 15.
HnXn1=Hn−1+1296/5n−876/5Xn1.Proof.
IfXn1=0, the result is obvious. So we just take into account Xn1=1, which implies PECn⟶PECn+11. In this case, wn (of PECn) overlaps with v2 or v5 (of On) (see Figure 3). In this situation, by using Equation (43) and Equation (44), Hn becomes
(103)12∑i∈VPECndgirdv2,i=12∑i∈VPECn−1dgirdv2,i+12∑i∈VOndgirdv2,i=12∑i∈VPECn−1dgirdwn−1,i+1+45+12×445=Hn−1+12×95×12n−13+12×445=Hn−1+12965n−8765.Thus, we conclude the desired fact.Fact 16.
HnXn2=Hn−1+1584/5n−1164/5Xn2.As that in the proof of Fact15, we only consider the fact Xn2=1, that is, PECn⟶PECn+12. The proof is similar and details are omitted.Noting thatXn1+Xn2=1, by the above discussions, it holds that
(104)Hn=HnXn1+Xn2=Hn−1+12965n−8765Xn1+Pn−1+15845n−11645Xn2=Hn−1+12965Xn1+15845Xn2n−8765Xn1+11645Xn2=Hn−1+nλ^n−μ^n,where for each n,
(105)λ^n=12965Xn1+15845Xn2,μ^n=8765Xn1+11645Xn2.Therefore, by Equation (102),
(106)Kf∗PECn=Kf∗PEC1+∑r=1n−1Hr+∑r=1n−1228r+77=Kf∗PEC1+∑r=1n−1∑s=1r−1Hs+1−Hs+H1+∑r=1n−1228r+77=Kf∗PEC1+∑r=1n−1∑s=1r−1s+1λ^s+1−μ^s+1+n−1p+∑r=1n−1228r+77=Kf∗PEC1+∑r=1n−1∑s=1r−1s+1λ^s+1−μ^s+1+On2.Through direct calculation, we have(107)Varλ^s=α32,Varμ^s=β32,Covλ^s,μ^s=γ3.If we substituteGutPECn by SPECn in the Proof of Theorem 8, the rest of the proof is identical to the Proof of Theorem 8 and the details are omitted here.We proceed by showing the following result about the expatiatory formula of the variance ofKf+PECn.Theorem 17.
Suppose Hypotheses1 and 2 are true, we can get the following two results:
(i)
The variance ofKf+PECn is denoted by(108)VarKf+PECn=130α42n5−5γ4n4+10β42n3−30α42+45β42−65γ4n2+59α42+65β42−120γ4n−30α42+30β42−60γ4,where
(109)α42=2162p+26421−p−216p+2641−p2,β42=1332p1+18121−p−133p+1811−p2,γ4=216·133p+264·1811−P−216p+2641−p133p+1811−p.(ii)
Forn⟶∞, Kf+PECn asymptotically obeys normal distributions. One has(110)limn⟶∞supε∈ℝPKf+PECn−EKf+PECnVarKf+PECn≤ε−∫−∞εe−x2/22πdx=0.Proof of Theorem 17.
Obtained by formula (64) yields, one sees that
(111)Kf+PECn+1=Kf+PECn+∑i∈VPECn12+5dgirdwn,i+203n+35.
LetIn=∑i∈VPECn12+5dgirdwn,i. Then, we obtain
(112)Kf+PECn+1=Kf+PECn+In+203n+35.Recalling thatXn1 and Xn2 are random variables, we obtain the following facts.Fact 18.
InXn1=In−1+216n−133Xn1.Proof.
IfXn1=0, the result is distinct. Then, we just take into account Xn1=1, which indicates PECn⟶PECn+11. In this fact, wn (of PECn) coincides with v2 or v5 (of On) (see Figure 3). In this situation, In becomes
(113)∑i∈VPECn12+5dgirdv2,i=∑i∈VPECn−112+5dgirdv2,i+∑i∈VOn12+5dgirdv2,i=∑i∈VPECn−112+5dgirdwn−1,i+1+45+12∑i∈VOnrdv2,i+8∑i∈VOndgirdv2,i=∑i∈VPECn−112+5dgirdwn−1,i+95∑i∈VPECn−112+5dgi+12×4+5×445=In−1+9512×5n−1+5×12n−13+48+44=In−1+216n−133.Thus, we obtain the desired fact.Fact 19.
InXn2=In−1+264n−181Xn2.Similar to the proof of Fact18, we only consider the fact Xn2=1, that is, PECn⟶PECn+12. In the same way, we omit the details.Noting thatXn1+Xn2=1, by the above discussions, it holds that
(114)In=In−1+216Xn1+264Xn2n−133Xn1+181Xn2=In−1+nλ~n−μ~n,where for each n,
(115)λ~n=216Xn1+264Xn2,μ~n=133Xn1+181Xn2.Therefore, by Equation (112),
(116)Kf+PECn=Kf+PEC1+∑r=1n−1Ir+∑r=1n−1203r+35=Kf+PEC1+∑r=1n−1∑s=1r−1Is+1−Is+I1+∑r=1n−1203r+35=Kf+PEC1+∑r=1n−1∑s=1r−1s+1λ~s+1−μ~s+1+n−1I1+∑r=1n−1203l+35=Kf+PEC1+∑r=1n−1∑s=1l−1s+1λ~s+1−μ~s+1+On2.Similarly, by calculation, we can obtain the following:(117)Varα~s=σ42,Varμ~m=β42,Covλ~s,μ~s=γ4.The rest of the proof is similar to the above theorems, and the details are omitted here.Remark 20.
Through the above calculation, it can be found that in the random pentagonal chain networks, whenn⟶∞, the four topological indices all obey the normal distribution; that is, combined with the expected values obtained in the second part, we can obtain their probability density functions, which have great influence in many aspects of statistics and are also very important for the application of complex networks in the fields of mathematics, physics, and engineering.


## 4. Conclusion

In this paper, we obtained the expected values and variances of the Gutman index, Schultz index, multiplicative degree-Kirchhoff index, and additive degree-Kirchhoff index about a class of the random pentagonal chain networks. It was calculated and observed that under the same conditions, the expected value of the Gutman index was the largest and that of multiplicative degree-Kirchhoff index was the lowest. Meanwhile, we found that for the networksPECn, these indices all approximately obeyed the normal distribution.The conclusion of this paper expands the research field of graph parameters related to vertex distance and degree and connects graph theory, topological indices, and statistical properties, which can more intuitively and accurately study the relevant characteristics of complex networks in various fields. In the field of chemistry, these results can be used as the basis for the study of the properties, functions, and structures of chemical molecules. In transportation, they can be used as the theoretical basis for route selection and positioning. In the field of computer applications, it can be assisted in solving problems such as network transportation and algorithm design. In a word, the research of this paper can bring some convenience to our life and study.In the future, we will not only focus on random chain networks but also combine the complex networks in some areas of real life, take the results of this paper as the basis, and use more diverse indicators and algorithms to study and solve some practical problems.



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