Study of mean-first-passage time and Kemeny’s constant of a random walk by normalized Laplacian matrices of a penta-chain network

Abstract

The mean first-passage time (MFPT), which refers to the expected time it takes for a system to reach a state $j$ given its current state $i$, that is $t_{ji}$, falls under the fundamental theory of Markov processes. The set of mean first-passage time (MFPT) among the positions of a Markov process expands fundamental assumptions of the system’s kinetics through their relation to the spectrum and eigenvectors of the transition matrix, and the moderation times of the random walker which all are of specific computational position. The explicit and precise computation of MFPT of random walks on networks can typically be highly challenging for networks with more than a few nodes, since they translate the global properties of the random walkers and the network they explore. On the other hand, in a connected network, the Kemeny’s constant (KC) gives the expected time of a random walk from an arbitrary vertex $x$ to reach a randomly chosen vertex $y$. The KC is interpreted as a measure of the connectivity level of a network, indicating how effectively the network is interconnected. The KC is an inspiring and helpful quantifier due to its rich applications, mostly in Markov’s chain. In the literature, there are multiple approaches to represent the complete matrix of MFPT. Among them, one widely used and traditional method is to employ the normalized Laplacian matrix. This study presents a new graph spectrum-based approach to compute the MFPT and KC of random walks on penta-chain network ($^{\prime } \Omega$). By using the decomposition theorem of normalized Laplacian polynomial, we computed the normalized Laplacian matrix for the penta-chain network ($^{\prime } \Omega$). Furthermore, by utilizing the roots and coefficients of the obtained matrices, we derived formulas for both the mean first-passage time (MFPT) and the Kemeny’s constant (KC) for $^{\prime } \Omega$. Finally, we compared the result of MFPT and KC with the number of pentagons.

Appendix

Here, we present detailed proofs of our main results within the given context.

Lemma 6.1

For $1 \le j \le 2m$,

$$ b_{j} = \left\{ {\begin{array}{*{20}l} {\frac{12 + 3\sqrt 6 }{{16}}\left( {\frac{\sqrt 3 + \sqrt 2 }{3}} \right)^{j} + \frac{12 - 3\sqrt 6 }{{16}}\left( {\frac{\sqrt 3 - \sqrt 2 }{3}} \right)^{j} } \hfill & {{\text{if}}\;j\;{\text{is}}\;{\text{even}}} \hfill \\ {\frac{4\sqrt 3 + 3\sqrt 3 }{8}\left( {\frac{\sqrt 3 + \sqrt 2 }{3}} \right)^{j} + \frac{4\sqrt 3 - 3\sqrt 3 }{8}\left( {\frac{\sqrt 3 - \sqrt 2 }{3}} \right)^{j} } \hfill & {{\text{if}}\;j\;{\text{is}}\;{\text{odd}}} \hfill \\ \end{array} } \right.. $$

Proof

Since, it is easy to see that $b_{1} = \frac{1}{2}$, $b_{2} = \frac{4}{3},\;b_{3} = \frac{29}{{18}}$. For $3 \le j \le 2m ,$ expanding ${\text{det}}B_{i}$ by its last row gives

$$ b_{j} = \left\{ {\begin{array}{*{20}l} {b_{j - 1} - \frac{1}{9}b_{j - 2} ,} \hfill & {{\text{if}}\;j\;{\text{is}}\;{\text{even}};} \hfill \\ {\frac{4}{3}q_{j - 1} - \frac{1}{9}q_{j - 2} ,} \hfill & {{\text{if}}\;j\;{\text{is}}\;{\text{odd}}.} \hfill \\ \end{array} } \right. $$

Without loss of generality, we assume that $c_{j} = b_{2j}$ for $1 \le j \le m$, let $d_{j} = b_{2i + 1}$ for $1 \le j \le m - 1$. Furthermore, $c_{1} = \frac{4}{3},d_{1} = \frac{29}{{18}}$ for so,

$$ \left\{ {\begin{array}{*{20}l} {c_{j} = d_{j - 1} - \frac{1}{9}c_{j - 1} ,} \hfill \\ {d_{j} = \frac{4}{3}c_{j} - \frac{1}{9}d_{j - 1} .} \hfill \\ \end{array} } \right. $$

(2)

from the first part in (2), we have $d_{j - 1} = c_{j} + \frac{1}{9}c_{j - 1} .$ Therefore, $d_{j} = c_{j + 1} + \frac{1}{9}c_{j} .$ Replacing $d_{j - 1}$ and $d_{j}$ into the second part in (2) gives $c_{j + 1} = \frac{10}{9}c_{j} - \frac{1}{81}c_{j - 1} ,\;j \ge 2$. By following the same procedure, one can deduce that $d_{j + 1} = \frac{10}{9}d_{j} - \frac{1}{81}d_{j - 1} ,\;j \ge 2$, and $b_{j}$. The expression satisfies the following recurrence relation:

$$ b_{j} = \frac{10}{9}b_{j - 2} - \frac{1}{81}b_{j - 4} ,\;b_{1} = \frac{3}{2},\;b_{2} = \frac{4}{3},\;b_{3} = \frac{29}{{18}}. $$

(3)

Then, the characteristic equation of (3) is $x^{4} = \frac{10}{9}x^{2} - \frac{1}{81}$, the roots of which are $x_{1} = \frac{\sqrt 2 + \sqrt 3 }{3},x_{2} = - \frac{\sqrt 2 + \sqrt 3 }{3}$ and $x_{3} = \frac{\sqrt 2 - \sqrt 3 }{3},x_{4} = - \frac{\sqrt 2 - \sqrt 3 }{3}$. The general solution of (3) is given by

$$ b_{j} = \left( {\frac{\sqrt 2 + \sqrt 3 }{3}} \right)^{j} y_{1} + \left( { - \frac{\sqrt 2 + \sqrt 3 }{3}} \right)^{j} y_{2} + \left( {\frac{\sqrt 2 - \sqrt 3 }{3}} \right)^{j} y_{3} + \left( { - \frac{\sqrt 2 - \sqrt 3 }{3}} \right)^{j} y_{4} $$

(4)

In conjunction with the initial conditions provided in Eq. (4), the system of equations leads to the following results:

$$ \left\{ {\begin{array}{*{20}l} {\left( {\frac{\sqrt 2 + \sqrt 3 }{3}} \right)\left( {y_{1} - y_{2} } \right) + \left( { - \frac{\sqrt 2 - \sqrt 3 }{3}} \right)\left( {y_{3} - y_{4} } \right) = \frac{3}{2},} \hfill \\ {\left( {\frac{\sqrt 2 + \sqrt 3 }{3}} \right)^{2} \left( {y_{1} + y_{2} } \right) + \left( { - \frac{\sqrt 2 - \sqrt 3 }{3}} \right)^{2} \left( {y_{3} + y_{4} } \right) = \frac{4}{3},} \hfill \\ {\left( {\frac{\sqrt 2 + \sqrt 3 }{3}} \right)^{3} \left( {y_{1} - y_{2} } \right) + \left( { - \frac{\sqrt 2 - \sqrt 3 }{3}} \right)^{3} \left( {y_{3} - y_{4} } \right) = \frac{29}{{18}},} \hfill \\ {\left( {\frac{\sqrt 2 + \sqrt 3 }{3}} \right)^{4} \left( {y_{1} + y_{2} } \right) + \left( { - \frac{\sqrt 2 - \sqrt 3 }{3}} \right)^{4} \left( {y_{3} + y_{4} } \right) = \frac{79}{{54}}.} \hfill \\ \end{array} } \right. $$

The unique solution of this system is as

$y_{1} = \frac{12 + 6\sqrt 2 + 8\sqrt 3 + 3\sqrt 6 }{{32}}$, $y_{2} = \frac{12 - 6\sqrt 2 - 8\sqrt 3 + 3\sqrt 6 }{{32}}$, $y_{3} = \frac{12 + 6\sqrt 2 - 8\sqrt 3 - 3\sqrt 6 }{{32}}$, $y_{4} = \frac{12 - 6\sqrt 2 + 8\sqrt 3 - 3\sqrt 6 }{{32}}$.

We get our desired result by substituting $y_{1} ,y_{2} ,y_{3}$ and $y_{4}$ in (4).

By expansion-formula, the $\det \mathcal{H}_{R}$ w.r.t its last row we have:

$$ \det \mathcal{H}_{R} = \frac{3}{2}b_{2m} - \frac{1}{6}b_{2m - 1} , $$

And by Lemma 6.1, we can get this lemma easily.

Lemma 6.2

$\det \mathcal{H}_{R} = \frac{{\left( {24 + 11\sqrt 6 } \right)\left( {\sqrt 3 + \sqrt 2 } \right)^{2m} + \left( {24 - 11\sqrt 6 } \right)\left( {\sqrt 3 - \sqrt 2 } \right)^{2m} }}{{32 \times 3^{2m} }}$.

Lemma 6.3

For diagonal entries

$$\begin{aligned} j & = 1,2, \ldots 2m + 1 ,\; b_{2m} \\ & \quad = \frac{{\left[ {192 + 143\sqrt 6 + \left( {924 + 336\sqrt 6 } \right)n} \right]\left( {\sqrt 2 + \sqrt 3 } \right)^{2m} + \left[ {192 - 143\sqrt 6 + \left( {924 - 336\sqrt 6 } \right)n} \right]\left( { - \sqrt 2 + \sqrt 3 } \right)^{2m} }}{{512 \cdot 3^{2m} }}. \\ \end{aligned}$$

.

Proof

Since $b_{ 2m} = \left( { - 1} \right)^{2m} b_{2m}$, is the sum of all of those minors of $\mathcal{H}_{R}$ which contain $2m$ rows and columns, then one has

$$ b_{2m} = \mathop \sum \limits_{j = 1}^{2m + 1} \det \mathcal{H}_{R} \left| {\left\{ j \right\}} \right| = \mathop \sum \limits_{j = 1}^{2m + 1} \det \left( {\begin{array}{*{20}c} {B_{j - 1} } & 0 \\ 0 & {E_{1} } \\ \end{array} } \right) = \mathop \sum \limits_{j = 1}^{2m + 1} \det B_{j - 1} \det E_{j} . $$

Using the property of $\mathcal{H}_{R}$ we have, $\det E_{j} = \det B_{2m + 1 - j} .$

So, $b_{2m} = 2b_{2m} + \mathop \sum \nolimits_{l = 1}^{m - 1} b_{2l} b_{2m - 2l} + \mathop \sum \nolimits_{p = 1}^{m} b_{2p - 1} b_{2m - 2p + 1}$.

From Lemma 6.1, $2b_{2m} = \frac{12 + 3\sqrt 6 }{{16}}\left( {\frac{\sqrt 2 + \sqrt 3 }{3}} \right)^{2m} + \frac{12 - 3\sqrt 6 }{{16}}\left( {\frac{ - \sqrt 2 + \sqrt 3 }{3}} \right)^{2m}$.

$\mathop \sum \limits_{l = 1}^{m - 1} b_{2l} b_{2m - 2l} = \frac{{\left[ {36\left( {11 + 4\sqrt 6 } \right)n - 3\left( {192 + 23\sqrt 6 } \right)} \right]\left( {\sqrt 3 + \sqrt 2 } \right)^{2m} + \left[ {36\left( {11 - 4\sqrt 6 } \right)n - 3\left( {192 - 23\sqrt 6 } \right)} \right]\left( {\sqrt 3 - \sqrt 2 } \right)^{2m} }}{{512 \cdot 3^{2m} }},$and

$$ \mathop \sum \limits_{p = 1}^{m} b_{2p - 1} b_{2m - 2p + 1} = \frac{{\left[ {12\left( {11 + 4\sqrt 6 } \right)n + 5\sqrt 6 } \right]\left( {\sqrt 3 + \surd 2} \right)^{2m} + \left[ {12\left( {11 - 4\sqrt 6 } \right)n - 5\sqrt 6 } \right]\left( {\sqrt 3 - \surd 2} \right)^{2m} }}{{ 128 \cdot 3^{2m} }} . $$

So, by adding the above three equations we can easily calculate the value of $b_{2m} .$

Hence, we introduce a matrix $N$, where $N$ is a matrix obtained from $\mathcal{H}_{Q}$ have $\left( {1, 3m + 1} \right)$-entry and the $\left( {3m + 1, 1} \right)$-entry by replacing $0$. We provide information for j-th order of principal submatrix, $N_{j}$ get from N's first $j$ rows and matching columns.

Lemma 6.4

$\left( { - 1} \right)^{3m} n_{3m} = \frac{7m + 1}{4}\left( \frac{1}{3} \right)^{2m - 1}$.

Proof

Since, the number $\left( { - 1} \right)^{3m} n_{3m}$ is the sum of all determinants of $\mathcal{H}_{Q}$ obtained by deleting j-th row and column.

$$ \left( { - 1} \right)^{3m} n_{3m} = \mathop \sum \limits_{j = 1}^{3m + 1} \det \mathcal{H}_{Q} \left| {\left\{ j \right\}} \right| $$

We obtain the result of this lemma by dividing it into two axioms.

Axiom 1

$1 \le j \le m$.

In this axiom, we delete the j-th row and corresponding column of all matrices present in $\mathcal{H}_{Q}$, the resulting matrix after applying the elementary operations is

$$ \det \mathcal{H}_{Q} = \left| {\begin{array}{*{20}c} {I_{m - 1} } & {F_{{\left( {m - 1} \right) \times \left( {2m + 1} \right)}} } \\ {F_{{\left( {m - 1} \right) \times \left( {2m + 1} \right)}}^{T} } & {G_{{\left( {2m + 1} \right) \times \left( {2m + 1} \right)}} } \\ \end{array} } \right|. $$

So, we get the following matrix.

$$ \det \mathcal{H}_{Q} \left[ { \left\{ { j } \right\} } \right] = \left| {\begin{array}{*{20}c} \frac{1}{2} & { - \frac{1}{\sqrt 6 }} & 0 & 0 & \cdots & 0 & 0 \\ { - \frac{1}{\sqrt 6 }} & 1 & { - \frac{1}{3}} & 0 & \cdots & 0 & 0 \\ 0 & { - \frac{1}{3}} & \frac{2}{3} & { - \frac{1}{3}} & \cdots & 0 & 0 \\ 0 & 0 & { - \frac{1}{3}} & \frac{2}{3} & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & \frac{2}{3} & { - \frac{1}{\sqrt 6 }} \\ 0 & 0 & 0 & 0 & \cdots & { - \frac{1}{\sqrt 6 }} & \frac{1}{2} \\ \end{array} } \right|_{(2m + 1) \times (2m + 1)} . $$

So, after calculation the $\det \mathcal{H}_{Q} \left[ {\left\{ { j } \right\}} \right] = \frac{1}{4}\left( \frac{1}{3} \right)^{2m - 1}$.

Axiom 2

$m + 1 \le j \le 3m + 1$.

In this axiom, we delete the (j − m)-th row and corresponding column of all matrices present in $\mathcal{H}_{Q}$, the resulting matrix after applying the elementary operations is:

$$ \det \mathcal{H}_{Q} \left[ {\left\{ j \right\}} \right] = \left| {\begin{array}{*{20}c} {I_{m} } & {F_{{\left( m \right) \times \left( {2m} \right)}} } \\ {F_{{\left( m \right) \times \left( {2m} \right)}}^{T} } & {G_{{\left( {2m} \right) \times \left( {2m} \right)}} } \\ \end{array} } \right|. $$

So, we get the following matrices;

By deleting $m + 1$ row and column, we have:

$$ \det \mathcal{H}_{Q} \left[ { \left\{ { j } \right\} } \right] = \left| {\begin{array}{*{20}c} \frac{2}{3} & { - \frac{1}{2}} & 0 & 0 & \cdots & 0 & 0 \\ { - \frac{1}{3}} & \frac{2}{3} & { - \frac{1}{3}} & 0 & \cdots & 0 & 0 \\ 0 & { - \frac{1}{3}} & \frac{2}{3} & { - \frac{1}{3}} & \cdots & 0 & 0 \\ 0 & 0 & { - \frac{1}{3}} & \frac{2}{3} & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & \frac{2}{3} & { - \frac{1}{\sqrt 6 }} \\ 0 & 0 & 0 & 0 & \cdots & { - \frac{1}{\sqrt 6 }} & \frac{1}{2} \\ \end{array} } \right|_{(2m) \times (2m)} . $$

Thus,$\det \mathcal{H}_{Q} \left[ {\left\{ { j } \right\}} \right] = \frac{1}{2}\left( \frac{1}{3} \right)^{2m - 1}$ and similarly, we get the same result by deleting $3m + 1$ row and column.

By deleting $m + 2$ row and column, we have:

$$ \det \mathcal{H}_{Q} \left[ { \left\{ { j } \right\} } \right] = \left| {\begin{array}{*{20}c} \frac{1}{2} & 0 & 0 & 0 & \cdots & 0 & 0 \\ 0 & \frac{2}{3} & { - \frac{1}{3}} & 0 & \cdots & 0 & 0 \\ 0 & { - \frac{1}{3}} & \frac{2}{3} & { - \frac{1}{3}} & \cdots & 0 & 0 \\ 0 & 0 & { - \frac{1}{3}} & \frac{2}{3} & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & \frac{2}{3} & { - \frac{1}{\sqrt 6 }} \\ 0 & 0 & 0 & 0 & \cdots & { - \frac{1}{\sqrt 6 }} & \frac{1}{2} \\ \end{array} } \right|_{(2m) \times (2m)} . $$

Thus, $\det \mathcal{H}_{Q} \left[ {\left\{ { j } \right\}} \right] = \frac{1}{4}\left( \frac{1}{3} \right)^{2m - 2}$ and we get the same result by deleting $m + 3,m + 4,m + 5, \ldots , 3m$ rows and columns.

So, collectively we have;

$$ \det \mathcal{H}_{Q} \left[ { \left\{ { j } \right\} } \right] = \left\{ {\begin{array}{*{20}l} {\frac{1}{2}\left( \frac{1}{3} \right)^{2m - 1} } \hfill & {{\text{if}}\;j = m + 1\;{\text{or}}\;3m + 1} \hfill \\ {\frac{1}{4}\left( \frac{1}{3} \right)^{2m - 2} } \hfill & {{\text{if}}\;m + 2 \le j \le 3m} \hfill \\ \end{array} } \right.. $$

Thus, after adding the result of these two axioms we have,

$$ \begin{aligned} \left( { - 1} \right)^{3m} n_{3m} & = \mathop \sum \limits_{j = 1}^{3m + 1} \det \mathcal{H}_{Q} \left| {\left\{ j \right\}} \right| = \mathop \sum \limits_{j = 1}^{ m } \det \mathcal{H}_{Q} \left| {\left\{ j \right\}} \right| + \mathop \sum \limits_{j = m + 1}^{3m + 1} \det \mathcal{H}_{Q} \left| {\left\{ j \right\}} \right| \\ & = m \cdot \frac{1}{4}\left( \frac{1}{3} \right)^{2m - 1} + 2 \cdot \frac{1}{2}\left( \frac{1}{3} \right)^{2m - 1} + \left( {2m - 1} \right)\frac{1}{4}\left( \frac{1}{3} \right)^{2m - 2} \\ & = \frac{7m + 1}{4}\left( \frac{1}{3} \right)^{2m - 1} . \\ \end{aligned} $$

Thus, the proof of Lemma 6.4 is complete.

Lemma 6.5

$\left( { - 1} \right)^{3m - 1} n_{3m - 1} = \frac{{49m^{3} + 42m^{2} + 2m}}{12}\left( \frac{1}{3} \right)^{2m - 1}$.

Proof

Since, the number $\left( { - 1} \right)^{3m - 1} n_{3m - 1}$ is the sum of all determinants of $\mathcal{H}_{Q} (1 \le j < k \le 3m + 1)$ obtained by deleting j-th and k-th row and corresponding column.

$$ \left( { - 1} \right)^{3m - 1} n_{3m - 1} = \mathop \sum \limits_{1 \le j < k \le 3m + 1} \det \mathcal{H}_{Q} \left| {\left\{ {j,k} \right\}} \right| $$

We obtain the result of this lemma by dividing it into three axioms.

Axiom 3

$1 \le j < k \le m$.

In this axiom, we delete the j-th and k-th row and corresponding column of all matrices present in $\mathcal{H}_{Q}$, and the resulting matrix after applying the elementary operations is

$$ \det \mathcal{H}_{Q} \left[ {[j,k\} } \right] = \left| {\begin{array}{*{20}c} {I_{m - 2} } & {F_{{\left( {m - 2} \right) \times \left( {2m + 1} \right)}} } \\ {F_{{\left( {m - 2} \right) \times \left( {2m + 1} \right)}}^{T} } & {G_{{\left( {2m + 1} \right) \times \left( {2m + 1} \right)}} } \\ \end{array} } \right|. $$

We have the following matrix:

$$ \det \mathcal{H}_{Q} \left[ { \left\{ { j,k } \right\} } \right] = \left| {\begin{array}{*{20}c} \frac{1}{2} & { - \frac{1}{\sqrt 6 }} & 0 & 0 & \cdots & 0 & 0 \\ { - \frac{1}{\sqrt 6 }} & 1 & { - \frac{1}{\sqrt 6 }} & 0 & \cdots & 0 & 0 \\ 0 & { - \frac{1}{3}} & \frac{2}{3} & { - \frac{1}{3}} & \cdots & 0 & 0 \\ 0 & 0 & { - \frac{1}{3}} & 1 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & \frac{2}{3} & { - \frac{1}{\sqrt 6 }} \\ 0 & 0 & 0 & 0 & \cdots & { - \frac{1}{\sqrt 6 }} & \frac{1}{2} \\ \end{array} } \right|_{(2m + 1) \times (2m + 1)} . $$

So, after calculation the $\det \mathcal{H}_{Q} \left[ {\left\{ { j,k } \right\}} \right] = \frac{1}{2}\left( {k - j + 1} \right)\left( \frac{1}{3} \right)^{2m - 1} .$

Axiom 4

$m + 1 \le j < k \le 3m + 1$.

In this axiom, we delete the $\left( {j - m} \right){\text{th}}$ and $\left( {k - m} \right){\text{th}}$ row and corresponding column of all matrices present in $\mathcal{H}_{Q}$, the resulting matrix after applying the elementary operations is

$$ \det \mathcal{H}_{Q} \left[ {\left\{ {j,k} \right\}} \right] = \left| {\begin{array}{*{20}c} {I_{m} } & {F_{{\left( m \right) \times \left( {2m - 1} \right)}} } \\ {F_{{\left( m \right) \times \left( {2m - 1} \right)}}^{T} } & {G_{{\left( {2m - 1} \right) \times \left( {2m - 1} \right)}} } \\ \end{array} } \right|. $$

We have the following matrices:

By deleting $m + 2$ and $m + 3$ row and corresponding column, we have;

$$ \det \mathcal{H}_{Q} \left[ { \left\{ { j,k } \right\} } \right] = \left| {\begin{array}{*{20}c} \frac{1}{2} & 0 & 0 & 0 & \cdots & 0 & 0 \\ 0 & \frac{2}{3} & { - \frac{1}{3}} & 0 & \cdots & 0 & 0 \\ 0 & { - \frac{1}{3}} & 1 & { - \frac{2}{3}} & \cdots & 0 & 0 \\ 0 & 0 & { - \frac{1}{3}} & 1 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & \frac{2}{3} & { - \frac{1}{\sqrt 6 }} \\ 0 & 0 & 0 & 0 & \cdots & { - \frac{1}{\sqrt 6 }} & \frac{1}{2} \\ \end{array} } \right|_{(2m - 1) \times (2m - 1)} . $$

Thus, $\det \mathcal{H}_{Q} \left[ {\left\{ { j,k } \right\}} \right] = \frac{1}{4}\left( {k - j} \right)\left( \frac{1}{3} \right)^{2m - 3}$ and similarly, we get the same result by deleting $m + 3 \le j < k \le 3m$ rows and columns.

By deleting $m + 1$ and $m + 2$ row and corresponding column, we have;

$$ \det \mathcal{H}_{Q} \left[ { \left\{ { j,k } \right\} } \right] = \left| {\begin{array}{*{20}c} \frac{2}{3} & { - \frac{1}{3}} & 0 & 0 & \cdots & 0 & 0 \\ { - \frac{1}{3}} & \frac{2}{3} & { - \frac{1}{3}} & 0 & \cdots & 0 & 0 \\ 0 & { - \frac{1}{3}} & \frac{2}{3} & { - \frac{1}{3}} & \cdots & 0 & 0 \\ 0 & 0 & { - \frac{1}{3}} & \frac{2}{3} & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & \frac{2}{3} & { - \frac{1}{\sqrt 6 }} \\ 0 & 0 & 0 & 0 & \cdots & { - \frac{1}{\sqrt 6 }} & \frac{1}{2} \\ \end{array} } \right|_{(2m - 1) \times (2m - 1)} . $$

Thus,$\det {\mathcal{H}}_{Q} \left[ {\left\{ { j,k } \right\}} \right] = \frac{1}{2}\left( {k - 1} \right)\left( \frac{1}{3} \right)^{2m - 2}$ and similarly, we get the same result by deleting $j = m + 1$ and $m + 3 \le k \le 3m$ rows and columns.

By deleting $m + 2$ and $3m + 1$ row and corresponding column, we have;

$$ \det \mathcal{H}_{Q} \left[ { \left\{ { j,k } \right\} } \right] = \left| {\begin{array}{*{20}c} \frac{1}{2} & 0 & 0 & 0 & \cdots & 0 & 0 \\ 0 & \frac{2}{3} & { - \frac{1}{3}} & 0 & \cdots & 0 & 0 \\ 0 & { - \frac{1}{3}} & \frac{2}{3} & { - \frac{1}{3}} & \cdots & 0 & 0 \\ 0 & 0 & { - \frac{1}{3}} & 1 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & \frac{2}{3} & { - \frac{1}{3}} \\ 0 & 0 & 0 & 0 & \cdots & { - \frac{1}{3}} & \frac{2}{3} \\ \end{array} } \right|_{(2m - 1) \times (2m - 1)} . $$

Thus, $\det \mathcal{H}_{Q} \left[ {\left\{ { j,k } \right\}} \right] = \frac{1}{2}\left( {2m + 1 - j} \right)\left( \frac{1}{3} \right)^{2m - 2}$ and similarly, we get the same result by deleting $k = 3m + 1$ and $m + 3 \le j \le 3m$ rows and columns.

By deleting $m + 1$ and $3m + 1$ row and corresponding column, we have;

$$ \det \mathcal{H}_{Q} \left[ { \left\{ { j,k } \right\} } \right] = \left| {\begin{array}{*{20}c} \frac{2}{3} & { - \frac{1}{3}} & 0 & 0 & \cdots & 0 & 0 \\ { - \frac{1}{3}} & \frac{2}{3} & { - \frac{1}{3}} & 0 & \cdots & 0 & 0 \\ 0 & { - \frac{1}{3}} & \frac{2}{3} & { - \frac{1}{3}} & \cdots & 0 & 0 \\ 0 & 0 & { - \frac{1}{3}} & \frac{2}{3} & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & \frac{2}{3} & { - \frac{1}{3}} \\ 0 & 0 & 0 & 0 & \cdots & { - \frac{1}{3}} & \frac{2}{3} \\ \end{array} } \right|_{(2m - 1) \times (2m - 1)} . $$

Thus,

$$ \det \mathcal{H}_{Q} \left[ {\left\{ { j,k } \right\}} \right] = \left( {2m } \right)\left( \frac{1}{3} \right)^{2m - 1} . $$

So, collectively we have: $\det \mathcal{H}_{Q} \left[ { \left\{ { j } \right\} } \right] = \left\{ {\begin{array}{*{20}l} {\frac{1}{4}\left( {k - j} \right)\left( \frac{1}{3} \right)^{2m - 3} } \hfill & {{\text{if}}\;m + 2 \le j < k \le 3m} \hfill \\ {\frac{1}{2}\left( {2m + 1 - j} \right)\left( \frac{1}{3} \right)^{2m - 2} } \hfill & {{\text{if}}\;j = m + 1 ,m + 2 \le k \le 3m} \hfill \\ {\frac{1}{2}\left( {k - 1} \right)\left( \frac{1}{3} \right)^{2m - 2} } \hfill & {{\text{if}}\;m + 2 \le j \le 3m ,k = 3m + 1} \hfill \\ {\left( {2m} \right)\left( \frac{1}{3} \right)^{2m - 1} } \hfill & {j = m + 1 ,k = 3m + 1} \hfill \\ \end{array} } \right.$.

Axiom 5

$1 \le j \le m ,m + 1 \le k \le 3m + 1.$

In this axiom, we delete the j-th $j - th$ and $\left( {k - m} \right){\text{th}}$ row and corresponding column of all matrices present in $\mathcal{H}_{Q}$, and the resulting matrix after applying the elementary operations is

$$ \det \mathcal{H}_{Q} \left[ {\left\{ {j,k} \right\}} \right] = \left| {\begin{array}{*{20}c} {I_{m - 1} } & {F_{{\left( {m - 1} \right) \times \left( {2m } \right)}} } \\ {F_{{\left( {m - 1} \right) \times \left( {2m } \right)}}^{T} } & {G_{{\left( {2m} \right) \times \left( {2m } \right)}} } \\ \end{array} } \right|. $$

So, we get the following matrices:

By deleting 1 and $m + 1$ row and column, we have:

$$ \det \mathcal{H}_{Q} \left[ { \left\{ { j,k } \right\} } \right] = \left| {\begin{array}{*{20}c} 1 & { - \frac{1}{3}} & 0 & 0 & \cdots & 0 & 0 \\ { - \frac{1}{3}} & \frac{2}{3} & { - \frac{1}{3}} & 0 & \cdots & 0 & 0 \\ 0 & { - \frac{1}{3}} & \frac{2}{3} & { - \frac{1}{3}} & \cdots & 0 & 0 \\ 0 & 0 & { - \frac{1}{3}} & \frac{2}{3} & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & \frac{2}{3} & { - \frac{1}{\sqrt 6 }} \\ 0 & 0 & 0 & 0 & \cdots & { - \frac{1}{\sqrt 6 }} & \frac{2}{3} \\ \end{array} } \right|_{(2m) \times (2m)} . $$

Thus,$\det \mathcal{H}_{Q} \left[ {\left\{ { j,k } \right\}} \right] = j\left( \frac{1}{3} \right)^{2m - 1} .$ Similarly, we get the same result for $2 \le j \le n.$

By deleting 1 and $m + 2$ row and column, we have:

$$ \det \mathcal{H}_{Q} \left[ { \left\{ { j,k } \right\} } \right] = \left| {\begin{array}{*{20}c} \frac{1}{2} & 0 & 0 & 0 & \cdots & 0 & 0 \\ 0 & \frac{2}{3} & { - \frac{1}{3}} & 0 & \cdots & 0 & 0 \\ 0 & { - \frac{1}{3}} & \frac{2}{3} & { - \frac{1}{3}} & \cdots & 0 & 0 \\ 0 & 0 & { - \frac{1}{3}} & \frac{2}{3} & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & \frac{2}{3} & { - \frac{1}{\sqrt 6 }} \\ 0 & 0 & 0 & 0 & \cdots & { - \frac{1}{\sqrt 6 }} & \frac{1}{2} \\ \end{array} } \right|_{(2m) \times (2m)} . $$

Thus, $\det \mathcal{H}_{Q} \left[ {\left\{ { j,k } \right\}} \right] = \frac{1}{4}\left( {\left| {k - m - 2j} \right| + 1} \right)\left( \frac{1}{3} \right)^{2m - 2} .$ Similarly, we get the same result for $2 \le j \le m$ and $m + 2 \le k \le 3m.$

By deleting 1 and $3m + 1$ row and column, we have:

$$ \det \mathcal{H}_{Q} \left[ { \left\{ { j,k } \right\} } \right] = \left| {\begin{array}{*{20}c} \frac{1}{2} & { - \frac{1}{\sqrt 6 }} & 0 & 0 & \cdots & 0 & 0 \\ { - \frac{1}{\sqrt 6 }} & 1 & { - \frac{1}{3}} & 0 & \cdots & 0 & 0 \\ 0 & { - \frac{1}{3}} & \frac{2}{3} & { - \frac{1}{3}} & \cdots & 0 & 0 \\ 0 & 0 & { - \frac{1}{3}} & \frac{2}{3} & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & \frac{2}{3} & { - \frac{1}{3}} \\ 0 & 0 & 0 & 0 & \cdots & { - \frac{1}{3}} & \frac{2}{3} \\ \end{array} } \right|_{(2m) \times (2m)} . $$

Thus, $\det \mathcal{H}_{Q} \left[ {\left\{ { j,k } \right\}} \right] = \frac{1}{4}\left( {m - j + 1} \right)\left( \frac{1}{3} \right)^{2m - 1} .$ Similarly, we get the same result for $2 \le j \le m .$

So, collectively we have;

$$\det \mathcal{H}_{Q} \left[ { \left\{ { j , k} \right\} } \right] = \left\{ {\begin{array}{*{20}l} {j\left( \frac{1}{3} \right)^{2m - 1} } \hfill & {{\text{if}}\;k = m + 1} \hfill \\ {\frac{1}{4}\left( {m - j + 1} \right)\left( \frac{1}{3} \right)^{2m - 1} } \hfill & {{\text{if}}\;k = 3m + 1,\;1 \le j \le m} \hfill \\ {\frac{1}{4}\left( {\left| {k - m - 2j} \right| + 1} \right)\left( \frac{1}{3} \right)^{2m - 2} } \hfill & {{\text{if}}\;1 \le j \le m, m + 2 \le k \le 3m} \hfill \\ \end{array} } \right..$$

Thus, after adding the result of these two axioms we have,

$$ \begin{aligned} & \left( { - 1} \right)^{3m - 1} n_{3m - 1} = \mathop \sum \limits_{1 \le j < k \le m} \det \mathcal{H}_{Q} \left| {\left\{ {j,k} \right\}} \right| + \mathop \sum \limits_{m + 1 \le j < k \le 3m + 1} \det \mathcal{H}_{Q} \left| {\left\{ {j,k} \right\}} \right| + \mathop \sum \limits_{1 \le j \le m,m + 1 \le k \le 3m + 1} \det \mathcal{H}_{Q} \left| {\left\{ {j,k} \right\}} \right| \\ & \quad = \frac{{m^{3} + 3m^{2} - 4m}}{12}\left( \frac{1}{3} \right)^{2m - 1} + \frac{{6m^{3} + 3m^{2} + m}}{6}\left( \frac{1}{3} \right)^{2m - 2} + \frac{{4m^{3} + 7m^{2} }}{4}\left( \frac{1}{3} \right)^{2m - 2} = \frac{{ 49m^{3} + 42m^{2} + 2m}}{12}\left( \frac{1}{3} \right)^{2m - 1} . \\ \end{aligned} $$

Thus, we complete the proof of Lemma 6.5.

The below propositions are a direct consequence of the above lemmas.

Proposition 6.6

Let ${}^{\prime }\Omega$ with $m$ pentagons denotes a linear pentagonal chain. Then, we have

$$ \mathcal{L}\left( {{}^{\prime }\Omega } \right) = \mathop \sum \limits_{j = 2}^{3m + 1} \frac{1}{{\eta_{j} }} + \mathop \sum \limits_{k = 1}^{2m + 1} \frac{1}{{\xi_{k} }} $$

(5)

here, the spectrums of $\mathcal{H}_{Q}$ are $0 = \eta_{1} < \eta_{2} \le \cdots \le \eta_{2m}$ and the spectrums of $\mathcal{H}_{R}$ are $0 < \xi_{1} \le \xi_{2} \le \cdots \le \xi_{2m}$.

In the following propositions, we derived the expressions $\sum\nolimits_{j = 2}^{3m + 1} {\frac{1}{{\eta_{j} }}}$ and $\sum\nolimits_{k = 1}^{2m + 1} {\frac{1}{{\xi_{k} }}} .$

Proposition 6.7

Let the spectra of $\mathcal{H}_{Q}$ are denoted by $0 = \eta_{1} < \eta_{2} \le \cdots \le \eta_{2m} .$ Then, $\sum\nolimits_{j = 2}^{3m + 1} {\frac{1}{{\eta_{j} }}} = \frac{{ m\left( {49m^{2} + 42m + 2} \right)}}{{3 \left( {7m + 1} \right)}}.$

Proof

Let $\varphi \left( {\mathcal{H}_{Q} } \right) = x^{2m} + n_{1} x^{2m - 1} + \cdots + n_{2m - 2} x^{2} + n_{2m - 1} x = x\left( {x^{2m - 1} + n_{1} x^{2m - 2} + \cdots + n_{2m - 2} x + n_{2m - 1} } \right),$ where $n_{2m - 1} \ne 0.$ Then, $\eta_{2} ,\eta_{3} , \ldots ,\eta_{2m}$ are the roots of the following equation

$$ x^{2m - 1} + n_{1} x^{2m - 2} + \cdots + n_{2m - 2} x + n_{2m - 1} = 0. $$

That is to say, $\frac{1}{{\eta_{2} }},{ }\frac{1}{{\eta_{3} }}, \ldots ,\frac{1}{{\eta_{2m} }}$ are the roots of $n_{2m - 1} x^{2m - 1} + n_{2m - 2} x^{2m - 2} + \cdots + n_{1} x + 1 = 0.$

From the Vieta’s Theorem, one has

$$ \mathop \sum \limits_{j = 2}^{3m + 1} \frac{1}{{\eta_{j} }} = - \frac{{n_{3m - 2} }}{{n_{3m} }}. $$

(6)

Putting Lemmas 6.4 and 6.5 into (6) yields $\sum\nolimits_{j = 2}^{3m + 1} {\frac{1}{{\eta_{j} }}} = \frac{{ m\left( {49m^{2} + 42m + 2} \right)}}{{3 \left( {7m + 1} \right)}}$ as desired.

Proposition 6.8

Let the spectrums of $\mathcal{H}_{R}$ are denoted by $\xi_{1} ,\xi_{2} , \ldots ,\xi_{2m}$. Then,

$$ \mathop \sum \limits_{k = 1}^{2m + 1} \frac{1}{{\xi_{k} }} = \frac{{\left[ {192 + 143\sqrt 6 + 84\left( {11 + 4\sqrt 6 } \right)m} \right]\left( {\sqrt 3 + \sqrt 2 } \right)^{2m} + \left[ {192 - 143\sqrt 6 + 84\left( {11 - 4\sqrt 6 } \right)m} \right]\left( {\sqrt 3 - \sqrt 2 } \right)^{2m} }}{{16\left[ {\left( {24 + 11\sqrt 6 } \right)\left( {\sqrt 3 + \sqrt 2 } \right)^{2m} + \left( {24 - 11\sqrt 6 } \right)\left( {\sqrt 3 - \sqrt 2 } \right)^{2m} } \right]}}. $$

Proof

Let $\varphi \left( {\mathcal{H}_{R} } \right) = x^{2m} + b_{1} x^{2m - 1} + \cdots + b_{2n - 1} x + b_{2m} ,$ where $b_{2m} \ne 0$. Then, the roots of the following equation are $\xi_{2} ,\xi_{3} , \ldots ,\xi ._{2m}$.

$$ x^{2m} + b_{1} x^{2m - 1} + \cdots + b_{2m - 1} x + b_{2m} = 0. $$

That is to say, $\frac{1}{{\xi_{1} }},\frac{1}{{\xi_{2} }}, \ldots ,\frac{1}{{\xi_{{2{\text{m}}}} }}$ are the roots of:

$$ b_{2m} x^{2m} + b_{2m - 1} x^{2m - 1} + \cdots + b_{1} x + 1 = 0. $$

Be mindful the Vieta’s Theorem, there are

$$ \mathop \sum \limits_{k = 1}^{2m + 1} \frac{1}{{\xi_{k} }} = \frac{{b_{2m} }}{{b_{2m + 1} }} = \frac{{b_{2m} }}{{\det \mathcal{H}_{R} }}. $$

(7)

Putting Lemmas 6.1, 6.2 and 6.3 into (7) yields the required result.