On the Number of Spanning Trees in Augmented Triangular Prism Graphs

Abstract

In computer science and graph theory, prism and antiprism graphs are crucial for network modeling, optimization, and network connectivity comprehension. Applications such as social network analysis, fault-tolerant circuit design, and parallel and distributed computing all make use of them. Their structured nature makes them important, since it offers a framework for researching intricate characteristics, including resilient design, communication patterns, and network efficiency. This work uses the electrically equivalent transformations technique to compute the explicit formulas for the number of spanning trees of three novel families of graphs that have been produced using triangular prisms with their distinctive iteration feature. Additionally, the relationship between these graphs’ average degree and entropy is examined and contrasted with the entropy of additional graphs that share the same average degree as these previously studied graphs.

1. Introduction

A subgraph of a connected, undirected graph $G$ with $n$ vertices that contains all the vertices and $n-1$ edges without creating any cycles is called a spanning tree. The number of spanning trees in graph $G$, denoted by $\tau \left(G\right)$, is also referred to as the complexity of $G$ [1]. This is an important and well-studied quantity with a wide range of applications. Listing specific chemical isomers [2], counting the number of Eulerian circuits in a graph [3], solving computationally difficult problems like the wandering salesman and Steiner tree problems [4], and deriving formulas for different graph types can be helpful in figuring out which graphs have the most spanning trees and are the most notable application domains.

Spanning trees are used in network reliability to evaluate a network’s performance, improve its fault resistance, and understand its structure. They form the foundation for protocols like the Spanning Tree Protocol (STP), which provide redundancy, prevent network loops, and determine how to add new links for optimal reliability [5].

A traditional Kirchhoff matrix from 1847 can be used to determine the number of spanning trees for a connected graph $G=(V,E)$ with $n$ vertices [6]. The Kirchhoff matrix $L$ is an $n\times n$ characteristic matrix $L=D-A$, where $A$ is the adjacency matrix of $G$ and $D$ is the diagonal matrix of the degrees of $G$, such that $L=\left[a_{ij}\right]$, defined as follows:

$$L=[a_{ij}]=\left\{\begin{array}{ll}\mathit{deg}(v_i)& \mathrm{if}i=j\\ -1& \mathrm{if}i\ne j\mathrm{and}(v_i,v_j)\in E\left(G\right)\\ 0& \mathrm{if}i\ne j\mathrm{and}(v_i,v_j)\notin E\left(G\right)\end{array}\right.$$

Each co-factor of $L$ corresponds to the number of spanning trees in a graph $G$.

Another way to determine the number of spanning trees is to assume that the eigenvalues of the Kirchhoff matrix $L$ of a graph $G$ with $n$ vertices are ${\mu }_1\ge {\mu }_2\ge \dots \ge {\mu }_n=0$.

In 1974, Kelmans and Chelnokov [7] demonstrated the following:

$$\tau (G)={\displaystyle \frac{1}{n}}{\prod }_{i=1}^{n-1}{\mu }_i.$$

(1)

This method has been used to calculate the number of spanning trees of cartesian and composition products of two graphs [8].

The deletion–contraction strategy is a popular technique for determining the number of spanning trees $\tau \left(G\right)$. The number of spanning trees in a multigraph can be reliably determined using this method. This method uses the following formula:

$$\tau \left(G\right)=\tau \left(G-e\right)+\tau \left(G/e\right),$$

(2)

where $G-e$ is the graph that results from removing an arbitrary edge $e$ and $G/e$ is the graph that results from contracting an arbitrary edge [9].

1.1. Electrically Equivalent Transformations

The technique of reducing a complicated electrical circuit to a simpler, equivalent one without altering its overall electrical behavior is known as electrical equivalent transformation. Kirchhoff was inspired by the study of graph theory, which concerns the possibility of representing an electrical network as a graph with edge weights corresponding to connectivity.

The ratio of weighted spanning trees to weighted thickets, where a thicket is a particular kind of spanning forest, can be used to determine the effective conductance between two vertices $u$ and $v$ [10]. That is the effect conductance between two vertices $u$ and $v$, which can be expressed as the quotient of the (weighted) number of spanning trees and the (weighted) number of thickets.

We list the impact of a few basic modifications on the quantity of spanning trees below.

Let $G$ represent the weighted number of spanning trees and let $H$ be the associated electrically equivalent graph. Let $\tau \left(G\right)$ be the weighted number of spanning trees of $G$ and $\tau \left(H\right)$ be the weighted number of spanning trees of $H$.

• Parallel edges: When two parallel edges in $G$ with conductances $u$ and $v$ are joined to form a single edge in $H$ with a conductance of $u+v$, the number of spanning trees in $H$, $\tau \left(H\right)$, remains equal to $\tau \left(G\right)$.
• Serial edges: If two serial edges in $G$ with conductances $u$ and $v$ are linked to generate a single edge in $H$ with a conductance of $uv/(u+v)$, the number of spanning trees in $H$, $\tau \left(H\right)$, can be calculated as $(1/(u+v\left)\right)$ multiplied by $\tau \left(G\right)$.
• Δ-Y Transformation: When a triangle in $G$ with conductances $u,v$, and $w$ is transformed into an electrically equivalent star graph in $H$ with conductances $x=(uv+vw+wu)/u,y=(uv+vw+wu)/v$, and $z=(uv+vw+wu)/w$, the number of spanning trees in $H$, $\tau \left(H\right)$, can be computed as $1/(u+v+w)$ multiplied by $\tau \left(G\right)$.
• Y-Δ Transformation: One may obtain the number of spanning trees in $H$, $\tau \left(H\right)$, by multiplying $\tau \left(G\right)$ by $1/(u+v+w)$, when an electrically equivalent triangle in $H$ with conductances $x=vw/(u+v+w),y=uw/(u+v+w)$ and $z=uv/(u+v+w)$ is created from a star graph in $G$ with conductances $u,v$, and $w$.

This method has been used to calculate the number of spanning trees in specific graph sequences generated by a Johnson skeleton graph [11]. Johnson graph 63, also known as the six-tetrahedral graph, is a highly symmetric graph with unique properties that make it a subject of interest in graph theory and related fields.

The “63 Johnson skeleton graph” is a particular example of a graph from the larger family of Johnson graphs, which are defined by subsets and intersections rather than by the structure of a polyhedron.

1.2. Some Types of Triangular Prism Graphs

The constant goal in mathematics is to create a new structure from an old one. This is also true for graphs, where a given set of graphs can be used to construct a huge number of new graphs.

We will begin by introducing three new categories of triangular prism graphs. The number of trees that extend into each of the three augmented families that these three graphs establish will then be determined, along with their entropy, in the next section.

The triangular prism graph’s use as a model for technical and physical systems, including those in network science, engineering, and optics, makes it significant. It illustrates the dispersion of light principles in optics. In engineering, it depicts the stable triangular construction utilized in building frames and bridges. The triangular prism graph is used in graph theory and network research to examine the intricacy of structures and network characteristics like connection and signal transmission.

In this work, we use the electrically equivalent transformations to determine the number of spanning trees for three new graph families based on a triangular prism graph, or $TP$, as illustrated below in Figure 1:

(1)

The triangular prism graph, or $TPS$, is a graph that is made by substituting a star $\left(\mathrm{3,3}\right)$-gon graph for the middle triangle of a triangular prism, $TP$. See Figure 1b.

(2)

The triangular prism graph, or $TPP$, is a graph that is made by substituting another triangular prism for the middle triangle of a triangular prism, $TP$. See Figure 1c.

(3)

The triangular prism graph, or $TPA$, is a graph that is made by substituting a triangular antiprism graph for the middle triangle of a triangular prism, $TP$. See Figure 1d.

A triangular prism graph is a graph that depicts the vertices and edges of a three-dimensional triangle prism. On the other hand, a Johnson graph is a family of graphs made from set systems in which the edges join subsets of a given size and the vertices are subsets of a particular size. Their origins are crucially different: a Johnson graph is defined algebraically from combinations of sets, while a triangular prism graph is directly based on geometric geometry.

2. Main Results

2.1. Complexity of the Graph Family Generated by Triangular Prism Graph, $TPS$

The graph family generated by the triangular prism graph $TPS$ and denoted by $F_{TPS}^{\left(n\right)}$ is a recursive definition using the graphs $F_{TPS}^{\left(2\right)}$ and $F_{TPS}^{\left(1\right)}$ (triangle or $K_3$): a replica of $F_{TPS}^{\left(2\right)}$ is used in place of the middle triangle of $F_{TPS}^{\left(2\right)}$ to create the graph $F_{TPS}^{\left(3\right)}$. The middle triangle in the graph $F_{TPS}^{(n-1)}$ is typically swapped out for $F_{TPS}^{\left(2\right)}$ to make $F_{TPS}^{\left(n\right)}$, as shown in Figure 2. $\left|V\left(F_{TPS}^{\left(n\right)}\right)\right|=27n-24$ and $\left|E\left(F_{TPS}^{\left(n\right)}\right)\right|=54n-51,n=1,2,\dots$ are the total vertices and edges of $F_{TPS}^{\left(n\right)}$, respectively. According to this architecture in the large $n$ limit, the average degree of $F_{TPS}^{\left(n\right)}$ is $4$.

Theorem 1.

For $n\ge 1$, the number of spanning trees in the graph family, $F_{TPS}^{\left(n\right)}$, generated by the triangular prism graph, $TPS$, is given by the following:

$$3\times {\left(8208\right)}^{n-1}\times z_1^2\times {\left(\lambda {{\rho }_1}^{n-2}+\mu {{\rho }_2}^{n-2}\right)}^2$$

where $z_1={\displaystyle \frac{{\left(\frac{233762561+15293\sqrt{233649273}}{113288}\right)}^{n-1}\left(\frac{863673+59\sqrt{233649273}}{408072}\right)-\frac{2217-\sqrt{233649273}}{23772}}{{\left(\frac{233762561+15293\sqrt{233649273}}{113288}\right)}^{n-1}\left(\frac{116377883+7185\sqrt{233649273}}{38494792}\right)-1}}$,

${\rho }_1={\displaystyle \frac{15293+\sqrt{233649273}}{2}}$, ${\rho }_2={\displaystyle \frac{15293-\sqrt{233649273}}{2}}$,

$\lambda ={\displaystyle \frac{717459034292+46940824\sqrt{233649273}}{77883091}}$, and $\mu ={\displaystyle \frac{717459034292-46940824\sqrt{233649273}}{77883091}}$.

Proof. 

The electrically equivalent transformation is used to convert $F_{TPS}^{\left(i\right)}$ to $F_{TPS}^{(i-1)}$. The process of transforming $F_{TPS}^{\left(2\right)}$ into $F_{TPS}^{\left(1\right)}$ is shown in Figure 3.

The result of combining these fourteen transformations is the following:

$$\begin{array}{l}\tau (F_{TPS}^{(2)})={\displaystyle \frac{1}{9^4{z}_2}}\times 5^9\times {(3z_2+2)}^3\times {({\displaystyle \frac{18}{5}})}^3\times {\displaystyle \frac{18z_2}{(3z_2+2)}}\times ({\displaystyle \frac{5^3}{3^{11}}})\times {\displaystyle \frac{3z_2+2}{(2z_2+1)}}\times {({\displaystyle \frac{14}{5}})}^9\times {({\displaystyle \frac{9z_2+5}{3z_2+2}})}^3\times {({\displaystyle \frac{27}{14}})}^3\times {\displaystyle \frac{9(2z_2+1)}{9z_2+5}}\times \\ {\displaystyle \frac{126z_2+70}{9(181z_2+99)}}\times {({\displaystyle \frac{849z_2+467}{14(9z_2+5)}})}^3\times {\displaystyle \frac{153(181z_2+99)}{7(849z_2+467)}}\tau (F_{TPS}^{(1)}).\end{array}$$

Thus, we have

$$\tau \left(F_{TPS}^{\left(2\right)}\right)=7616(11886z_2+6538{)}^2\tau (F_{TPS}^{\left(1\right)}).$$

(3)

Moreover,

$$\begin{array}{ll}\tau \left(F_{TPS}^{\left(n\right)}\right)& ={\displaystyle \prod _{i=2}^n}7616(11886z_i+6538{)}^2\tau (F_{TPS}^{\left(1\right)})\\ & =3\times 7616^{n-1}{z}_1^2{\left[{\prod }_{i=2}^n\left(11886z_i+6538\right)\right]}^2.\end{array}$$

(4)

where $z_{i-1}={\displaystyle \frac{8755z_i+4811}{11886z_i+6538}},i=2,3,\dots ,n.$

Its characteristic equation is $11886$ ${\omega }^2-2217\omega -4811=0,$ with the following roots:

$${\omega }_1={\displaystyle \frac{2217-\sqrt{233649273}}{23772}}\mathrm{and}{\omega }_2={\displaystyle \frac{2217+\sqrt{233649273}}{23772}}.$$

These two roots can be subtracted from both sides of $z_{i-1}={\displaystyle \frac{8755z_i+4811}{11886z_i+6538}}$ to obtain the following:

$$\begin{array}{ll}{z}_{i-1}-{\omega }_1& =z_{i-1}-{\displaystyle \frac{2217-\sqrt{233649273}}{23772}}={\displaystyle \frac{8755z_i+4811}{11886z_i+6538}}-{\displaystyle \frac{2217-\sqrt{233649273}}{23772}}\\ & =\left(15293+\sqrt{233649273}\right)\times {\displaystyle \frac{z_i-\frac{2217-\sqrt{233649273}}{23772}}{2\left(11886z_i+6538\right)}},\end{array}$$

(5)

$$\begin{array}{ll}{z}_{i-1}-{\omega }_2& =z_{i-1}-{\displaystyle \frac{2217+\sqrt{233649273}}{23772}}={\displaystyle \frac{8755z_i+4811}{11886z_i+6538}}-{\displaystyle \frac{2217+\sqrt{233649273}}{23772}}\\ & =(15293-\sqrt{233649273})\times {\displaystyle \frac{z_i-\frac{2217+\sqrt{233649273}}{23772}}{2(11886z_i+6538)}}.\end{array}$$

(6)

Let $t_i={\displaystyle \frac{z_i-\frac{2217-\sqrt{233649273}}{23772}}{z_i-\frac{2217+\sqrt{233649273}}{23772}}}$, then $t_{i-1}={\displaystyle \frac{z_{i-1}-\frac{2217-\sqrt{233649273}}{23772}}{z_{i-1}-\frac{2217+\sqrt{233649273}}{23772}}}{t}_i$

Then, by Equations (5) and (6), we obtain the following:

$$t_{i-1}=\left({\displaystyle \frac{15293+\sqrt{233649273}}{15293-\sqrt{233649273}}}\right)t_i\mathrm{or}{t}_{i-1}=({\displaystyle \frac{233672561+15293\sqrt{233649273}}{113288}})t_i.$$

Therefore,

$$t_i=({\displaystyle \frac{233672561+15293\sqrt{233649273}}{113288}}{)}^{n-i}{t}_n$$

From the expression $t_i={\displaystyle \frac{z_i-\frac{2217-\sqrt{233649273}}{23772}}{z_i-\frac{2217+\sqrt{233649273}}{23772}}}$, we have the following:

$$t_i\left(z_i-{\displaystyle \frac{2217+\sqrt{233649273}}{23772}}\right)=z_i-{\displaystyle \frac{2217-\sqrt{233649273}}{23772}},$$

Then,

$$z_i={\displaystyle \frac{(\frac{233672561+15293\sqrt{233649273}}{113288}{)}^{n-i}\times \left(\frac{2217+\sqrt{233649273}}{23772}\right)t_n-(\frac{2217-\sqrt{233649273}}{23772})}{(\frac{233672561+15293\sqrt{233649273}}{113288}{)}^{n-i}{t}_n-1}}.$$

Therefore,

$$z_1={\displaystyle \frac{(\frac{233672561+15293\sqrt{233649273}}{113288}{)}^{n-1}\times \left(\frac{2217+\sqrt{233649273}}{23772}\right)t_n-(\frac{2217-\sqrt{233649273}}{23772})}{(\frac{233672561+15293\sqrt{233649273}}{113288}{)}^{n-1}{t}_n-1}}.$$

If $z_n=1$, we obtain the following:

$$\begin{array}{ll}{t}_n& ={\displaystyle \frac{1-\frac{2217-\sqrt{233649273}}{23772}}{1-\frac{2217+\sqrt{233649273}}{23772}}}={\displaystyle \frac{21555+\sqrt{233649273}}{21555-\sqrt{233649273}}}\\ & ={\displaystyle \frac{116377883+7185\sqrt{233649273}}{38494792}}.\end{array}$$

Thus, we obtain the following:

$$z_1={\displaystyle \frac{{\left(\frac{233762561+15293\sqrt{233649273}}{113288}\right)}^{n-1}\left(\frac{863673+59\sqrt{233649273}}{408072}\right)-\frac{2217-\sqrt{233649273}}{23772}}{{\left(\frac{233762561+15293\sqrt{233649273}}{113288}\right)}^{n-1}\left(\frac{116377883+7185\sqrt{233649273}}{38494792}\right)-1}},n\ge 1$$

(7)

With the expression $z_{n-1}={\displaystyle \frac{8755z_n+4811}{11886z_n+6538}}$ and the coefficients of $8755z_n+4811$ and $11886z_n+6538$, represented as ${\alpha }_n$ and ${\beta }_n$, respectively, we obtain the following:

$$\begin{gathered} 11886z_n+6538={\alpha }_0(8755z_n+4811)+{\beta }_0(11886z_n+6538), \\ 11886z_{n-1}+6538={\displaystyle \frac{{\alpha }_1(8755z_n+4811)+{\beta }_1(11886z_n+6538)}{{\alpha }_0(8755z_n+4811)+{\beta }_0(11886z_n+6538)}}, \\ 11886z_{n-2}+6538={\displaystyle \frac{{\alpha }_2(8755z_n+4811)+{\beta }_2(11886z_n+6538)}{{\alpha }_1(8755z_n+4811)+{\beta }_1(11886z_n+6538)}}, \\ ⋮ \\ 11886z_{n-i}+6538={\displaystyle \frac{{\alpha }_i(8755z_n+4811)+{\beta }_i(11886z_n+6538)}{{\alpha }_{i-1}(8755z_n+4811)+{\beta }_{i-1}(11886z_n+6538)}}, \end{gathered}$$

(8)

$$\begin{gathered} 11886z_{n-(i+1)}+6538={\displaystyle \frac{{\alpha }_{i+1}(8755z_n+4811)+{\beta }_{i+1}(11886z_n+6538)}{{\alpha }_i(8755z_n+4811)+{\beta }_i(11886z_n+6538)}} \\ ⋮ \\ 11886z_2+6538={\displaystyle \frac{{\alpha }_{n-2}(8755z_n+4811)+{\beta }_{n-2}(11886z_n+6538)}{{\alpha }_{n-3}(8755z_n+4811)+{\beta }_{n-3}(11886z_n+6538)}}, \end{gathered}$$

(9)

So, we obtain the following:

$$\tau \left(F_{TPS}^{\left(n\right)}\right)=3\times 7616^{n-1}{z}_1^2\left[{\alpha }_{n-2}\right(8755z_n+4811)+{\beta }_{n-2}(11886z_n+6538){]}^2$$

(10)

Using the expression $z_{n-1}={\displaystyle \frac{8755z_n+4811}{11886z_n+6538}}$ with Equations (8) and (9), we obtain the following:

$${\alpha }_{i+1}=15293{\alpha }_i-56644{\alpha }_{i-1};{\beta }_{i+1}=15293{\beta }_i-56644{\beta }_{i-1}$$

(11)

Equation (11) has the characteristic equation ${\rho }^2-15293\rho +56644=0$, with the roots ${\rho }_1={\displaystyle \frac{15293+\sqrt{233649273}}{2}}$ and ${\rho }_2={\displaystyle \frac{15293-\sqrt{233649273}}{2}}$.

The general solutions of Equation (11) are ${\alpha }_i=a_1{\rho }_1^i+a_2{\rho }_2^i;{\beta }_i=b_1{\rho }_1^i+b_2{\rho }_2^i$.

Using the initial conditions ${\alpha }_0=0,{\beta }_0=1$ and ${\alpha }_1=11886,{\beta }_1=6538$ yields the following:

$$\begin{array}{l}{\alpha }_i={\displaystyle \frac{11886\sqrt{233649273}}{233649273}}({\displaystyle \frac{15293+\sqrt{233649273}}{2}}{)}^i-{\displaystyle \frac{11886\sqrt{233649273}}{233649273}}({\displaystyle \frac{15293-\sqrt{233649273}}{2}}{)}^i;\\ {\beta }_i=({\displaystyle \frac{233649273-2217\sqrt{233649273}}{467298546}})({\displaystyle \frac{15293+\sqrt{233649273}}{2}}{)}^i+\\ ({\displaystyle \frac{233649273+2217\sqrt{233649273}}{467298546}})({\displaystyle \frac{15293-\sqrt{233649273}}{2}}{)}^i\end{array}$$

(12)

If $z_n=1$, $F_{TPS}^{\left(n\right)}$ is devoid of any electrically equivalent transformation. By entering Equation (12) into Equation (10), we obtain the following:

$$\begin{array}{ll}\tau \left(F_{TPS}^{\left(n\right)}\right)& =3\times {\left(7616\right)}^{n-1}{z}_1^2\times \\ & {\left(\begin{array}{c}\left({\displaystyle \frac{7717459034292+\mathrm{46,940,824}\sqrt{233649273}}{77883091}}\right){\left({\displaystyle \frac{15293+\sqrt{233649273}}{2}}\right)}^{n-2}+\\ \left({\displaystyle \frac{7717459034292-\mathrm{46,940,824}\sqrt{233649273}}{77883091}}\right){\left({\displaystyle \frac{15293-\sqrt{233649273}}{2}}\right)}^{n-2}\end{array}\right)}^2,n\ge 2.\end{array}$$

(13)

Equation (13) is satisfied for $n=1$, and $\tau \left(F_{TPS}^{\left(1\right)}\right)=3$. Thus, the number of spanning trees in the graph family $F_{TPS}^{\left(n\right)}$ generated by triangular prism graph $TPS$ is determined by the following:

$$\begin{array}{ll}\tau \left(F_{TPS}^{\left(n\right)}\right)& =3\times {\left(7616\right)}^{n-1}{z}_1^2\times \\ & {\left(\begin{array}{c}\left({\displaystyle \frac{7717459034292+\mathrm{46,940,824}\sqrt{233649273}}{77883091}}\right){\left({\displaystyle \frac{15293+\sqrt{233649273}}{2}}\right)}^{n-2}+\\ \left({\displaystyle \frac{7717459034292-\mathrm{46,940,824}\sqrt{233649273}}{77883091}}\right){\left({\displaystyle \frac{15293-\sqrt{233649273}}{2}}\right)}^{n-2}\end{array}\right)}^2,n\ge 1.\end{array}$$

(14)

By substituting Equation (7) into Equation (14), we achieve the intended outcome. □

2.2. Complexity of the Graph Family Generated by Triangular Prism Graph, $TPA$

The graph family generated by triangular prism graph $TPA$ and denoted by $F_{TPA}^{\left(n\right)}$ is a recursive definition using the graphs $F_{TPA}^{\left(2\right)}$ and $F_{TPA}^{\left(1\right)}$ (triangle or $K_3$): a replica of $F_{TPA}^{\left(2\right)}$ is used in place of the middle triangle of $F_{TPA}^{\left(2\right)}$ to create the graph $F_{TPA}^{\left(3\right)}$. The middle triangle in the graph $F_{TPA}^{(n-1)}$ is typically swapped out for $F_{TPA}^{\left(2\right)}$ to make $F_{TPA}^{\left(n\right)}$, as shown in Figure 4. $\left|V\left(F_{TPA}^{\left(n\right)}\right)\right|=27n-24$ and $\left|E\left(F_{TPA}^{\left(n\right)}\right)\right|=66n-63,n=1,2,\dots$ are the total vertices and edges of $F_{TPA}^{\left(n\right)}$, respectively. According to this architecture in the large $n$ limit, the average degree of $F_{TPA}^{\left(n\right)}$ is $4.89$.

Theorem 2.

For $n\ge 1$, the number of spanning trees in the graph family, $F_{TPA}^{\left(n\right)},$ generated by the triangular prism graph $TPA$ is given by the following:

$$3\times {\left(34336\right)}^{n-1}\times z_1^2\times {\left(\mathit{\lambda }{{\rho }_1}^{n-2}+\mu {{\rho }_2}^{n-2}\right)}^2$$

where

$$\begin{array}{l}{z}_1={\displaystyle \frac{{\left(\frac{11435617471+\mathrm{453,720}\sqrt{635248119}}{1151329}\right)}^{n-1}\left(\frac{131015223+5360\sqrt{635248119}}{47917657}\right)-\frac{3012-\sqrt{635248119}}{36395}}{{\left(\frac{11435617471+\mathrm{453,720}\sqrt{635248119}}{1151329}\right)}^{n-1}(\frac{3485404+133\sqrt{635248119}}{954535})-1}},\\ {\rho }_1=75620+3\sqrt{635248119},{\rho }_2=75620-3\sqrt{635248119},\\ \lambda ={\displaystyle \frac{889696626522+17650067\sqrt{635248119}}{10123476}}\mathrm{and}\mu ={\displaystyle \frac{889696626522-17650067\sqrt{635248119}}{10123476}}.\end{array}$$

Proof. 

The electrically equivalent transformation is used to convert the graph $F_{TPA}^{\left(i\right)}$ to the graph $F_{TPA}^{(i-1)}$. The process of transforming the graph $F_{TPA}^{\left(2\right)}$ into the graph $F_{TPA}^{\left(1\right)}$ is shown in Figure 5.

These fourteen transformations are combined to produce the following:

$$\begin{array}{l}\tau (F_{TPA}^{\left(2\right)})={\displaystyle \frac{1}{9^4{z}_2}}\times 5^9\times {\left(3z_2+2\right)}^3\times {\left({\displaystyle \frac{18}{5}}\right)}^3\times ({\displaystyle \frac{18z_2}{3z_2+2}})\times {\left({\displaystyle \frac{5}{8}}\right)}^3\times {\displaystyle \frac{1}{9^4}}\times {\displaystyle \frac{3z_2+2}{5z_2+3}}\times {\left({\displaystyle \frac{29}{5}}\right)}^9\times {\left({\displaystyle \frac{18z_2+11}{3z_2+2}}\right)}^3\times {\left({\displaystyle \frac{72}{29}}\right)}^3\times \\ ({\displaystyle \frac{9\left(5z_2+3\right)}{\left(18z_2+11\right)}})\times ({\displaystyle \frac{\left(522z_2+319\right)}{9\left(811z_2+494\right)}})\times {\left({\displaystyle \frac{3765z_2+2296}{522z_2+319}}\right)}^3\times {\displaystyle \frac{666\left(811z_2+494\right)}{29\left(3765z_2+2296\right)}}\tau \left(F_{TPA}^{\left(1\right)}\right).\end{array}$$

Thus, we have the following:

$$\tau \left(F_{TPA}^{\left(2\right)}\right)=34336(109185z_2+66584{)}^2\tau \left(F_{TPA}^{\left(1\right)}\right).$$

(15)

Furthermore,

$$\begin{array}{ll}\tau \left(F_{TPA}^{\left(n\right)}\right)& ={\displaystyle \prod _{i=2}^n}34336(109185z_i+66584{)}^2\tau (F_{TPA}^{\left(1\right)})\\ & =3\times 34336^{n-1}{z}_1^2{\left[{\prod }_{i=2}^n\right(109185z_i+66584\left)\right]}^2.\end{array}$$

(16)

where $z_{i-1}={\displaystyle \frac{84656z_i+51615}{109185z_i+66584}},i=2,3,\dots ,n.$

Its characteristic equation is $\mathrm{109,185}$ ${\omega }^2-18072\omega -51615=0$, having the following roots:

$${\omega }_1={\displaystyle \frac{3012-\sqrt{635248119}}{36395}}\mathrm{and}{\omega }_2={\displaystyle \frac{3012+\sqrt{635248119}}{36395}}.$$

When we subtract these two roots from each side of $z_{i-1}={\displaystyle \frac{84656z_i+51615}{109185z_i+66584}}$, we obtain the following:

$$\begin{array}{ll}{z}_{i-1}-{\omega }_1& =z_{i-1}-{\displaystyle \frac{3012-\sqrt{635248119}}{36395}}={\displaystyle \frac{84656z_i+51615}{109185z_i+66584}}-{\displaystyle \frac{3012-\sqrt{635248119}}{36395}}\\ & =(75620+3\sqrt{635248119})\times {\displaystyle \frac{z_i-\frac{3012-\sqrt{635248119}}{36395}}{109185z_i+66584}},\end{array}$$

(17)

$$\begin{array}{ll}{z}_{i-1}-{\omega }_2& {=z}_{i-1}-{\displaystyle \frac{3012+\sqrt{635248119}}{36395}}={\displaystyle \frac{84656z_i+51615}{109185z_i+66584}}-{\displaystyle \frac{3012+\sqrt{635248119}}{36395}}\\ & =(75620-3\sqrt{635248119})\times {\displaystyle \frac{z_i-\frac{3012+\sqrt{635248119}}{36395}}{109185z_i+66584}}.\end{array}$$

(18)

Let $t_i={\displaystyle \frac{z_i-\frac{3012-\sqrt{635248119}}{36395}}{z_i-\frac{3012+\sqrt{635248119}}{36395}}}$; then, $t_{i-1}={\displaystyle \frac{z_{i-1}-\frac{3012-\sqrt{635248119}}{36395}}{z_{i-1}-\frac{3012+\sqrt{635248119}}{36395}}}{t}_i$.

Then, using Equations (17) and (18), we obtain the following:

$$t_{i-1}=\left({\displaystyle \frac{75620+\sqrt{635248119}}{75620-\sqrt{635248119}}}\right)t_i\mathrm{or}{t}_{i-1}=({\displaystyle \frac{11435617471+453720\sqrt{635248119}}{1151329}})t_i.$$

Therefore,

$$t_i=({\displaystyle \frac{11435617471+453720\sqrt{635248119}}{1151329}}{)}^{n-i}{t}_n$$

From the expression $t_i={\displaystyle \frac{z_i-\frac{3012-\sqrt{635248119}}{36395}}{z_i-\frac{3012+\sqrt{635248119}}{36395}}}$, we have the following:

$$t_i\left(z_i-{\displaystyle \frac{3012+\sqrt{635248119}}{36395}}\right)=z_i-{\displaystyle \frac{3012-\sqrt{635248119}}{36395}},$$

Then,

$$z_i={\displaystyle \frac{(\frac{11435617471+\mathrm{453,720}\sqrt{635248119}}{1151329}{)}^{n-i}\times \left(\frac{3012+\sqrt{635248119}}{36395}\right)t_n-(\frac{3012-\sqrt{635248119}}{36395})}{(\frac{11435617471+\mathrm{453,720}\sqrt{635248119}}{1151329}{)}^{n-i}{t}_n-1}}.$$

Therefore,

$$z_1={\displaystyle \frac{(\frac{11435617471+\mathrm{453,720}\sqrt{635248119}}{1151329}{)}^{n-1}\left(\frac{3012+\sqrt{635248119}}{36395}\right)t_n-(\frac{3012-\sqrt{635248119}}{36395})}{(\frac{11435617471+\mathrm{453,720}\sqrt{635248119}}{1151329}{)}^{n-1}{t}_n-1}}.$$

If $z_n=1$, we obtain the following:

$$\begin{array}{ll}{t}_n& ={\displaystyle \frac{1-\frac{3012-\sqrt{635248119}}{36395}}{1-\frac{3012+\sqrt{635248119}}{36395}}}={\displaystyle \frac{33383+\sqrt{635248119}}{33383+\sqrt{635248119}}}\\ & ={\displaystyle \frac{3485404+133\sqrt{635248119}}{954535}}.\end{array}$$

Thus, we obtain the following:

$$z_1={\displaystyle \frac{{\left(\frac{11435617471+453720\sqrt{635248119}}{1151329}\right)}^{n-1}(\frac{131015223+5360\sqrt{635248119}}{47917657})-\frac{3012-\sqrt{635248119}}{36395}}{{\left(\frac{11435617471+453720\sqrt{635248119}}{1151329}\right)}^{n-1}(\frac{3485404+133\sqrt{635248119}}{954535})-1}}.$$

(19)

Utilizing the expression $z_{n-1}={\displaystyle \frac{84656z_i+51615}{109185z_i+66584}}$ and designating ${\alpha }_n$ and ${\beta }_n$ as the coefficients of $84656z_i+51615$ and $\mathrm{109,185}$ $z_i+66584$, we obtain the following:

$$\begin{gathered} 109185z_n+66584={\alpha }_0(84656z_n+51615)+{\beta }_0(109185z_n+66584), \\ 109185z_{n-1}+66584={\displaystyle \frac{{\alpha }_1(84656z_n+51615)+{\beta }_1(109185z_n+66584)}{{\alpha }_0(84656z_n+51615)+{\beta }_0(109185z_n+66584)}}, \\ 109185z_{n-2}+66584={\displaystyle \frac{{\alpha }_2(84656z_n+51615)+{\beta }_2(109185z_n+66584)}{{\alpha }_1(84656z_n+51615)+{\beta }_1(109185z_n+66584)}}, \\ ⋮ \\ 109185z_{n-i}+66584={\displaystyle \frac{{\alpha }_i\left(84656z_n+51615\right)+{\beta }_i(109185z_n+66584)}{{\alpha }_{i-1}(84656z_n+51615)+{\beta }_{i-1}(109185z_n+66584)}}. \end{gathered}$$

(20)

$$\begin{gathered} 109185z_{n-(i+1)}+66584={\displaystyle \frac{{\alpha }_{i+1}(84656z_n+51615)+{\beta }_{i+1}(109185z_n+66584)}{{\alpha }_i(84656z_n+51615)+{\beta }_i(109185z_n+66584)}}. \\ ⋮ \\ 109185z_2+66584={\displaystyle \frac{{\alpha }_{n-2}(84656z_n+51615)+{\beta }_{n-2}(109185z_n+66584)}{{\alpha }_{n-3}(84656z_n+51615)+{\beta }_{n-3}(109185z_n+66584)}} \end{gathered}$$

(21)

Thus, we obtain the following:

$$\tau \left(F_{TPA}^{\left(n\right)}\right)=3\times 34336^{n-1}{z}_1^2\left[{\alpha }_{n-2}\right(84656z_n+51615)+{\beta }_{n-2}(109185z_n+66584){]}^2$$

(22)

Utilizing Equations (20) and (21) and the expression $z_{n-1}={\displaystyle \frac{84656z_i+51615}{109185z_i+66584}}$, we obtain the following:

$${\alpha }_{i+1}=151240{\alpha }_i-1151329{\alpha }_{i-1};{\beta }_{i+1}=151240{\beta }_i-1151329{\beta }_{i-1}$$

(23)

Its characteristic equation is ${\rho }^2-151240\rho +1151329=0$, with the following roots:

$${\rho }_1=75620+3\sqrt{635248119}\mathrm{and}{\rho }_2=75620-3\sqrt{635248119}.$$

Equation (23) has general solutions that are ${\alpha }_i=a_1{\rho }_1^i+a_2{\rho }_2^i;{\beta }_i=b_1{\rho }_1^i+b_2{\rho }_2^i$.

Using the initial conditions ${\alpha }_0=0,{\beta }_0=1$ and ${\alpha }_1=109185,{\beta }_1=66584$, we obtain the following:

$$\begin{array}{l}{\alpha }_i={\displaystyle \frac{36395\sqrt{635248119}}{1270496238}}(75620+3\sqrt{635248119}{)}^i-{\displaystyle \frac{36395\sqrt{635248119}}{1270496238}}(75620-3\sqrt{635248119}{)}^i;\\ {\beta }_i=({\displaystyle \frac{635248119-3012\sqrt{635248119}}{1270496238}})(75620+3\sqrt{635248119}{)}^i+\\ ({\displaystyle \frac{635248119+3012\sqrt{635248119}}{1270496238}})(75620-3\sqrt{635248119}{)}^i.\end{array}$$

(24)

If $z_n=1$, then $F_{TPA}^{\left(n\right)}$ has no electrically equivalent transformation. By entering Equation (24) into Equation (23), we obtain the following:

$$\begin{array}{ll}\tau \left(F_{TPA}^{\left(n\right)}\right)=& 3\times {\left(34336\right)}^{n-1}{z}_1^2\times \\ & {\left(\begin{array}{c}\left({\displaystyle \frac{889696626522+17650067\sqrt{635248119}}{10123476}}\right){(75620+3\sqrt{635248119})}^{n-2}+\\ \left({\displaystyle \frac{889696626522-17650067\sqrt{635248119}}{10123476}}\right){(75620-3\sqrt{635248119})}^{n-2}\end{array}\right)}^2,n\ge 2.\end{array}$$

(25)

When $n=1$, Equation (25) is satisfied, since $\tau \left(F_{TPA}^{\left(1\right)}\right)=3$. Consequently, the number of spanning trees in the graph family $F_{TPA}^{\left(n\right)}$ generated by triangular prism graph $TPA$ is determined by the following:

$$\begin{array}{ll}\tau \left(F_{TPA}^{\left(n\right)}\right)=& 3\times {\left(34336\right)}^{n-1}{z}_1^2\times \\ & {\left(\begin{array}{c}\left({\displaystyle \frac{889696626522+17650067\sqrt{635248119}}{10123476}}\right){(75620+3\sqrt{635248119})}^{n-2}+\\ \left({\displaystyle \frac{889696626522-17650067\sqrt{635248119}}{10123476}}\right){(75620-3\sqrt{635248119})}^{n-2}\end{array}\right)}^2,n\ge 1.\end{array}$$

(26)

Equation (19) can be inserted into Equation (26), yielding the desired outcome. □

2.3. Complexity of the Graph Family Generated Triangular Prism Graph, $TPP$

The graph family generated by the triangular prism graph $TPP$ and denoted by $F_{TPP}^{\left(n\right)}$, is a recursive definition using the graphs $F_{TPP}^{\left(2\right)}$ and $F_{TPP}^{\left(1\right)}$ (triangle or $K_3$): a replica of $F_{TPP}^{\left(2\right)}$ is used in place of the middle triangle of $F_{TPP}^{\left(2\right)}$ to create the graph $F_{TPP}^{\left(3\right)}$. The middle triangle in the graph $F_{TPP}^{(n-1)}$ is typically swapped out for $F_{TPP}^{\left(2\right)}$ to make $F_{TPP}^{\left(n\right)}$, as shown in Figure 6. $\left|V\left(F_{TPP}^{\left(n\right)}\right)\right|=27n-24$ and $\left|E\left(F_{TPP}^{\left(n\right)}\right)\right|=54n-51,n=1,2,\dots$ are the total vertices and edges of $F_{TPP}^{\left(n\right)}$, respectively. According to this architecture in the large $n$ limit, the average degree of $F_{TPP}^{\left(n\right)}$ is $4$.

Theorem 3.

For $n\ge 1$, the number of spanning trees in the graph family, $F_{TPP}^{\left(n\right)},$ generated by the triangular prism graph $TPP$ is given by the following:

$$3\times {\left(8208\right)}^{n-1}\times z_1^2\times {\left(\lambda {{\rho }_1}^{n-2}+\mu {{\rho }_2}^{n-2}\right)}^2$$

where

$$\begin{array}{l}{z}_1={\displaystyle \frac{{\left(\frac{175701841+13257\sqrt{175655633}}{46208}\right)}^{n-1}\left(\frac{50015497+3863\sqrt{175655633}}{24281582}\right)-\frac{6455-\sqrt{175655633}}{25688}}{{\left(\frac{175701841+13257\sqrt{175655633}}{46208}\right)}^{n-1}\left(\frac{272781961+19233\sqrt{175655633}}{97126328}\right)-1}},\\ {\rho }_1={\displaystyle \frac{13257+\sqrt{175655633}}{2}},{\rho }_2={\displaystyle \frac{13257-\sqrt{175655633}}{2}},\\ \lambda ={\displaystyle \frac{5707051516170+\mathrm{430,627,514}\sqrt{175655633}}{702622532}},\mu ={\displaystyle \frac{5707051516170-430627514\sqrt{175655633}}{702622532}}.\end{array}$$

Proof. 

The electrically equivalent transformation is used to convert the graph $F_{TPP}^{\left(i\right)}$ to the graph $F_{TPP}^{(i-1)}$. The process of transforming the graph $F_{TPP}^{\left(2\right)}$ into the graph $F_{TPP}^{\left(1\right)}$ is shown in Figure 7.

These thirteen transformations are combined to produce the following:

$$\begin{array}{l}\tau \left(F_{TPP}^{\left(n\right)}\right)={\displaystyle \frac{1}{9^4{z}_2}}\times 4^9\times {\left(3z_2+1\right)}^3\times {\left({\displaystyle \frac{9}{4}}\right)}^3\times \left({\displaystyle \frac{9z_2}{3z_2+1}}\right)\times {\left({\displaystyle \frac{4}{5}}\right)}^3\times {\displaystyle \frac{1}{9^4}}\times {\displaystyle \frac{\left(3z_2+1\right)}{\left(4z_2+1\right)}}\times {\left({\displaystyle \frac{19}{4}}\right)}^9\times {\left({\displaystyle \frac{15z_2+4}{3z_2+1}}\right)}^3\times {\left({\displaystyle \frac{45}{19}}\right)}^3\times \\ {\displaystyle \frac{9\left(4z_2+1\right)}{\left(15z_2+4\right)}}\times {\displaystyle \frac{\left(285z_2+76\right)}{9\left(436z_2+115\right)}}\times {\left({\displaystyle \frac{2028z_2+537}{285z_2+76}}\right)}^3\times {\displaystyle \frac{144\left(436z_2+115\right)}{19\left(676z_2+179\right)}}\tau \left(F_{TPP}^{\left(1\right)}\right).\end{array}$$

Thus, we have the following:

$$\tau \left(F_{TPP}^{\left(2\right)}\right)=8208(12844z_2+3401{)}^2\tau (F_{TPP}^{\left(1\right)}).$$

(27)

Furthermore,

$$\begin{array}{ll}\tau \left(F_{TPP}^{\left(n\right)}\right)& ={\displaystyle \prod _{i=2}^n}8208(12844z_i+3401{)}^2\tau (F_{TPP}^{\left(1\right)})\\ & =3\times 8208^{n-1}{z}_1^2[{\prod }_{i=2}^n(12844z_i+3401){]}^2.\end{array}$$

(28)

where $z_{i-1}={\displaystyle \frac{9856z_i+2608}{12844z_i+3401}},i=2,3,\dots ,n.$

Its characteristic equation is $12844{\omega }^2-6455\omega -2608=0$, with the following roots:

$${\omega }_1={\displaystyle \frac{6455-\sqrt{175655633}}{25688}}\mathrm{and}{\omega }_2={\displaystyle \frac{6455+\sqrt{175655633}}{25688}}.$$

These two roots can be subtracted from both sides of $z_{i-1}={\displaystyle \frac{9856z_i+2608}{12844z_i+3401}}$ to obtain the following:

$$\begin{array}{ll}{z}_{i-1}-{\omega }_1& =z_{i-1}-{\displaystyle \frac{6455-\sqrt{175655633}}{25688}}={\displaystyle \frac{9856z_i+2608}{12844z_i+3401}}-{\displaystyle \frac{6455-\sqrt{175655633}}{25688}}\\ & =(13257+\sqrt{175655633})\times {\displaystyle \frac{z_i-\frac{6455-\sqrt{175655633}}{25688}}{2(12844z_i+3401)}}.\end{array}$$

(29)

$$\begin{array}{ll}{z}_{i-1}-{\omega }_1& =z_{i-1}-{\displaystyle \frac{6455+\sqrt{175655633}}{25688}}={\displaystyle \frac{9856z_i+2608}{12844z_i+3401}}-{\displaystyle \frac{6455+\sqrt{175655633}}{25688}}\\ & =(13257-\sqrt{175655633})\times {\displaystyle \frac{z_i-\frac{6455+\sqrt{175655633}}{25688}}{2(12844z_i+3401)}}.\end{array}$$

(30)

Let $t_i={\displaystyle \frac{z_i-\frac{6455-\sqrt{175655633}}{25688}}{z_i-\frac{6455+\sqrt{175655633}}{25688}}}$. Then, $t_{i-1}={\displaystyle \frac{z_{i-1}-\frac{6455-\sqrt{175655633}}{25688}}{z_{i-1}-\frac{6455+\sqrt{175655633}}{25688}}}{t}_i$.

Thus, using Equations (29) and (30), we obtain the following:

$$t_{i-1}=\left({\displaystyle \frac{13257+\sqrt{175655633}}{13257+\sqrt{175655633}}}\right)t_i\mathrm{or}{t}_{i-1}=({\displaystyle \frac{175701841+13257\sqrt{175655633}}{46208}})t_i.$$

Therefore, $t_i=({\displaystyle \frac{175701841+13257\sqrt{175655633}}{46208}}{)}^{n-i}{t}_n$.

From the expression $t_i={\displaystyle \frac{z_i-\frac{6455-\sqrt{175655633}}{25688}}{z_i-\frac{6455+\sqrt{175655633}}{25688}}}$, we have the following:

$$t_i\left(z_i-{\displaystyle \frac{6455+\sqrt{175655633}}{25688}}\right)=z_i-{\displaystyle \frac{6455-\sqrt{175655633}}{25688}},$$

Then,

$$z_i={\displaystyle \frac{(\frac{175701841+13257\sqrt{175655633}}{46208}{)}^{n-i}\left(\frac{6455+\sqrt{175655633}}{25688}\right)t_n-(\frac{6455-\sqrt{175655633}}{25688})}{(\frac{175701841+13257\sqrt{175655633}}{46208}{)}^{n-i}{t}_n-1}}.$$

Thus,

$$z_1={\displaystyle \frac{(\frac{175701841+13257\sqrt{175655633}}{46208}{)}^{n-1}\left(\frac{6455+\sqrt{175655633}}{25688}\right)t_n-(\frac{6455-\sqrt{175655633}}{25688})}{(\frac{175701841+13257\sqrt{175655633}}{46208}{)}^{n-1}{t}_n-1}}.$$

If $z_n=1$, we obtain the following:

$$\begin{array}{ll}{t}_n& ={\displaystyle \frac{1-\frac{6455-\sqrt{175655633}}{25688}}{1-\frac{6455+\sqrt{175655633}}{25688}}}={\displaystyle \frac{19233+\sqrt{175655633}}{19233-\sqrt{175655633}}}\\ & ={\displaystyle \frac{272781961+19233\sqrt{175655633}}{97126328}}.\end{array}$$

Thus, we obtain the following:

$$z_1={\displaystyle \frac{{\left(\frac{175701841+13257\sqrt{175655633}}{46208}\right)}^{n-1}\left(\frac{50015497+3863\sqrt{175655633}}{24281582}\right)-\frac{6455-\sqrt{175655633}}{25688}}{{\left(\frac{\mathrm{175,701,841}+13257\sqrt{175655633}}{46208}\right)}^{n-1}\left(\frac{272781961+\mathrm{19,233}\sqrt{175655633}}{97126328}\right)-1}}.$$

(31)

Using expression $z_{n-1}={\displaystyle \frac{9856z_n+2608}{12844z_n+3401}}$ and designating ${\alpha }_n$ and ${\beta }_n$ as the coefficients of $9856z_n+2608$ and $12844$ $z_n+3401$, respectively, we obtain the following:

$$\begin{gathered} 12844z_n+3401={\alpha }_0(9856z_n+2608)+{\beta }_0(12844z_n+3401), \\ 12844z_{n-1}+3401={\displaystyle \frac{{\alpha }_1(9856z_n+2608)+{\beta }_1(12844z_n+3401)}{{\alpha }_0(9856z_n+2608)+{\beta }_0(12844z_n+3401)}}, \\ 12844z_{n-2}+3401={\displaystyle \frac{{\alpha }_2(9856z_n+2608)+{\beta }_2(12844z_n+3401)}{{\alpha }_1(9856z_n+2608)+{\beta }_1(12844z_n+3401)}}, \\ ⋮ \\ 12844z_{n-i}+3401={\displaystyle \frac{{\alpha }_i(9856z_n+2608)+{\beta }_i(12844z_n+3401)}{{\alpha }_{i-1}(9856z_n+2608)+{\beta }_{i-1}(12844z_n+3401)}}, \end{gathered}$$

(32)

$$\begin{gathered} 12844z_{n-(i+1)}+3401={\displaystyle \frac{{\alpha }_{i+1}(9856z_n+2608)+{\beta }_{i+1}(12844z_n+3401)}{{\alpha }_i(9856z_n+2608)+{\beta }_i(12844z_n+3401)}}, \\ ⋮ \\ 12844z_2+3401={\displaystyle \frac{{\alpha }_{n-2}(9856z_n+2608)+{\beta }_{n-2}(12844z_n+3401)}{{\alpha }_{n-3}(9856z_n+2608)+{\beta }_{n-3}(12844z_n+3401)}}, \end{gathered}$$

(33)

Thus, we obtain the following:

$$\tau \left(F_{TPP}^{\left(n\right)}\right)=3\times 8208^{n-1}{z}_1^2[{\alpha }_{n-2}\left(9856z_n+2608\right)+{\beta }_{n-2}\left(12844z_n+3401\right){]}^2.$$

(34)

Using Equations (32) and (33) and the expression $z_{n-1}={\displaystyle \frac{9856z_n+2608}{12844z_n+3401}}$, we obtain the following:

$${\alpha }_{i+1}=13257{\alpha }_i-23104{\alpha }_{i-1};{\beta }_{i+1}=13257{\beta }_i-23104{\beta }_{i-1}$$

(35)

The characteristic equation of Equation (35) is ${\rho }^2-13257\rho +23104=0$, with the following roots:

$${\rho }_1={\displaystyle \frac{13257+\sqrt{175655633}}{2}}\mathrm{and}{\rho }_1={\displaystyle \frac{13257-\sqrt{175655633}}{2}}.$$

The general solutions of Equation (35) are ${\alpha }_i=a_1{\rho }_1^i+a_2{\rho }_2^i;{\beta }_i=b_1{\rho }_1^i+b_2{\rho }_2^i$.

Using the initial conditions ${\alpha }_0=0,{\beta }_0=1$ and ${\alpha }_1=11886,{\beta }_1=6538$ yields the following:

$$\begin{array}{l}{\alpha }_i={\displaystyle \frac{12844\sqrt{175655633}}{175655633}}({\displaystyle \frac{13257+\sqrt{175655633}}{2}}{)}^i-{\displaystyle \frac{12844\sqrt{175655633}}{175655633}}({\displaystyle \frac{13257-\sqrt{175655633}}{2}}{)}^i;\\ {\beta }_i=\left({\displaystyle \frac{175655633-6455\sqrt{175655633}}{351311266}}\right)({\displaystyle \frac{13257+\sqrt{175655633}}{2}}{)}^i+\\ ({\displaystyle \frac{175655633+6455\sqrt{175655633}}{351311266}})({\displaystyle \frac{13257-\sqrt{175655633}}{2}}{)}^i\end{array}$$

(36)

$F_{TPP}^{\left(n\right)}$ is devoid of any electrically equivalent transformation if $z_n=1$. When Equation (36) is entered into Equation (34), we obtain the following:

$$\begin{array}{l}\tau \left(F_{TPP}^{\left(n\right)}\right)=3\times {\left(8208\right)}^{n-1}{z}_1^2\times \\ {\left(\begin{array}{c}({\displaystyle \frac{5707051516170+\mathrm{430,627,514}\sqrt{175655633}}{702622532}}){({\displaystyle \frac{13257+\sqrt{175655633}}{2}})}^{n-2}+\\ ({\displaystyle \frac{5707051516170-\mathrm{430,627,514}\sqrt{175655633}}{702622532}}){({\displaystyle \frac{13257-\sqrt{175655633}}{2}})}^{n-2})\end{array}\right)}^2,n\ge 2.\end{array}$$

(37)

When $n=1$, $\tau \left(F_{TPP}^{\left(1\right)}\right)=3$, which is in accordance with Equation (37). Consequently, the number of spanning trees in the graph family $F_{TPP}^{\left(n\right)}$ generated by the triangular prism graph $TPP$ is determined by the following:

$$\begin{array}{l}\tau \left(F_{TPP}^{\left(n\right)}\right)=3\times {\left(8208\right)}^{n-1}{z}_1^2\times \\ {\left(\begin{array}{c}({\displaystyle \frac{5707051516170+\mathrm{430,627,514}\sqrt{175655633}}{702622532}}){({\displaystyle \frac{13257+\sqrt{175655633}}{2}})}^{n-2}+\\ ({\displaystyle \frac{5707051516170-\mathrm{430,627,514}\sqrt{175655633}}{702622532}}){({\displaystyle \frac{13257-\sqrt{175655633}}{2}})}^{n-2})\end{array}\right)}^2,n\ge 1.\end{array}$$

(38)

The required outcome is obtained by inserting Equation (31) into Equation (38). □

3. Numerical Results

The next three

Table 1. A portion of the spanning tree count for the graph family $F_{TPS}^{\left(n\right)}$.

$\mathit{n}$	$\mathit{\tau }\left({\mathit{F}}_{\mathit{T}\mathit{P}\mathit{S}}^{\left(\mathit{n}\right)}\right)$
$1$	$3$
$2$	$\mathrm{420,486,266,188}$
$3$	$\mathrm{748,560,566,754,331,739,008,2048}$
$4$	$\mathrm{133,268,897,595,653,249,738,377,310,163,141,918,72}$
$5$	$\mathrm{237,263,357,436,206,999,590,763,513,646,322,104,788,693,884,600,32}$
6	$\mathrm{422,408,392,338,505,175,964,704,222,179,621,861,895,981,7319,158,436,886,426,4192}$

,

Table 2. A portion of the spanning tree count for the graph family $F_{TPA}^{\left(n\right)}$.

$\mathit{n}$	$\mathit{\tau }\left({\mathit{F}}_{\mathit{T}\mathit{P}\mathit{A}}^{\left(\mathit{n}\right)}\right)$
$1$	$3$
$2$	$\mathrm{191,283,645,870,6528}$
$3$	$\mathrm{150,214,672,271,637,848,651,427,658,4448}$
$4$	$\mathrm{117,964,591,658,857,128,830,285,398,804,632,873,538,063,5648}$
$5$	$\mathrm{926,383,866,671,804,364,969,499,356,668,773,371,680,557,376,821,138,390,253,568}$
6	$\mathrm{727,495,476,703,404,696,640,317,472,197,243,686,452,715,337,530,441,487,461,790,067,365,739,757,568}$

and

Table 3. A portion of the spanning tree count for the graph family $F_{TPP}^{\left(n\right)}$.

$\mathit{n}$	$\mathit{\tau }\left({\mathit{F}}_{\mathit{T}\mathit{P}\mathit{P}}^{\left(\mathit{n}\right)}\right)$
$1$	$3$
$2$	$\mathrm{382,537,031,2704}$
$3$	$\mathrm{551,670,656,868,243,420,394,2912}$
$4$	$\mathrm{795,597,731,203,827,982,255,126,904,571,848,2944}$
$5$	$\mathrm{114,737,976,980,586,703,081,300,587,184,353,759,994,835,600,670,72}$
6	$\mathrm{165,470,599,591,600,523,634,016,106,128,093,489,953,795,990,759,650,502,440,386,56}$

display the values of the number of spanning trees in the three graph families $F_{TPS}^{\left(n\right)}$, $F_{TPA}^{\left(n\right)}$, and $F_{TPP}^{\left(n\right)}$.

4. Spanning Tree Entropy

After obtaining precise formulas for the number of spanning trees in each of the three graph families,$F_{TPS}^{\left(n\right)}$, $F_{TPA}^{\left(n\right)}$, and $F_{TPP}^{\left(n\right)},$ we can calculate the spanning tree entropy $Z$, a finite number and an intriguing metric that defines the network topology. This is described in [12] as follows: Consider graph $G$,

$$Z(G)=\underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{\mathit{ln}\tau \left(G\right)}{\left|V\left(G\right)\right|}}.$$

(39)

$$\begin{gathered} Z\left(F_{TPS}^{\left(n\right)}\right)={\displaystyle \frac{1}{27}}\left(\mathrm{l}\mathrm{n}\left[1904\right]+2\mathrm{l}\mathrm{n}\left[15293+\sqrt{233649273}\right]\right)=1.04, \\ Z\left(F_{TPA}^{\left(n\right)}\right)={\displaystyle \frac{1}{27}}\left(\mathrm{l}\mathrm{n}\left[34336\right]+2\mathrm{l}\mathrm{n}\left[75620+3\sqrt{635248119}\right]\right)=1.27, \\ Z\left(F_{TPP}^{\left(n\right)}\right)={\displaystyle \frac{1}{27}}\left(\mathrm{l}\mathrm{n}\left[2052\right]+2\mathrm{l}\mathrm{n}\left[13257+\sqrt{175655633}\right]\right)=1.04. \end{gathered}$$

From these results, we observe the following: graph families $F_{TPS}^{\left(n\right)}$ and $F_{TPP}^{\left(n\right)}$, which have the same average degree $4$, have the same entropy (1.040), while the graph family $F_{TPA}^{\left(n\right)}$ has more edges and average degrees than the graph families $F_{TPS}^{\left(n\right)}$ and $F_{TPP}^{\left(n\right)}$; its entropy is higher (1.27).

Additionally, the apollonian graph [13], which has an average degree of 5 (entropy 1.354), has a higher entropy than all graph families $F_{TPS}^{\left(n\right)}$, $F_{TPA}^{\left(n\right)}$, and $F_{TPP}^{\left(n\right)}$. In addition, the entropy of the fractal scale free lattice [14], which has the entropy $1.04$ and average degree 4, is the same entropy of the graph families $F_{TPS}^{\left(n\right)}$ and $F_{TPP}^{\left(n\right)},$ while the entropies of the graph families $F_{TPS}^{\left(n\right)}$ and $F_{TPP}^{\left(n\right)}$ are smaller than the entropy of the two-dimensional Sierpinski gasket [15], which has an entropy $1.166$ of the same average degree 4. Finally, the entropy of graph families $F_{TPS}^{\left(n\right)}$ and $F_{TPP}^{\left(n\right)}$ is higher than the entropy of graph sequences $\mathrm{J}{\mathrm{S}}_n$ and $R_1\mathrm{J}{\mathrm{S}}_n$, constructed on the Johnson skeleton graph, which have the same mean score of 4 but an entropy value of 1.02 [11].

5. Conclusions

In this study, we calculated the number of spanning trees for three new and huge families of graphs that resulted from three new triangular prism graphs using the electrically equivalent transformation approach. The number of spanning trees for such families of graphs cannot be determined using Kirchhoff’s determinant method or the laborious computation of Laplace spectra. Thus, this is where this method’s strength is found. Additionally, we investigated the entropy of these graph families and how it related to the entropy of other graph families with average degrees that were identical or nearly equal.