# On Level Energy and Level Characteristic Polynomial of Rooted Trees

Audace A.V. Dossou-Olory, Muhammed F. Killik, Elena V. Konstantinova, Bünyamin Şahin

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Audace A.V. Dossou-Olory, Muhammed F. Killik, Elena V. Konstantinova, Bünyamin Şahin. On Level Energy and Level Characteristic Polynomial of Rooted Trees. 2023. [hal-04199617](https://hal.science/hal-04199617v1)

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# On Level Energy and Level Characteristic Polynomial of Rooted Trees

September 7, 2023

**Audace A. V. Dossou-Olory<fnref n="1" />**

Département d'Hydrologie et Gestion des Ressources en Eau
Institut National de l'Eau, University of Abomey-Calavi

01 BP 4521 Cotonou 01, Bénin

Institut de Mathématiques et de Sciences Physiques, Dangbo, Bénin

Email: audace@aimscac.za

**Muhammed F.Killik**

Department of Mathematics, Faculty of Science
Selcuk University, Konya, Turkey

Email: muhammedfatihkillik@gmail.com

**Elena V. Konstantinova**

Novosibirsk State University, Pirogova str. 2
Sobolev Institute of Mathematics, Ak. Koptyug av. 4
Novosibirsk, 630090, Russia

Email: e.konsta@math.nsc.ru

**Bünyamin Şahin**

Department of Mathematics, Faculty of Science
Selcuk University, Konya, Turkey

Email: bunyamin.sahin@selcuk.edu.tr

## Abstract

Building from the level index, a Wiener-like topological index proposed by Balaji and Mahmoud, we define level matrix and study level energy and level characteristic polynomial of rooted trees. We establish some relations between level matrix and the usual distance matrix. Moreover, we determine various lower and upper bounds on the level energy and calculate the level energy in specific tree families. We also give an explicit expression of the level characteristic polynomial of so-called rooted double stars and rooted binary caterpillars. Finally, we propose and provide some evidence to a conjecture that the rooted path maximises the level energy among all trees with a given number of vertices.

<footnote n="1">¹Corresponding Author</footnote>
# Keywords

Level index, Level characteristic polynomial, Level energy, Distance energy, Distance matrix, Rooted trees.

**2020 Mathematics Subject Classification:** Primary 05C05, 05C50; Secondary 05C12, 05C35.

# 1 Introduction

The first distance based topological index was introduced by Wiener in 1947 [14]. Distances between the vertices of graphs can be showed with a matrix which is called distance matrix. Moreover, the sum of absolute values of the eigenvalues of distance matrix is known as distance energy. Distance energy was introduced by Indulal, Gutman and Vijaykumar, who computed distance energy of graphs having diameter two [8], and several bounds were obtained in [4, 7, 9, 10] thereafter. More details about distance spectra of graphs can be found in the survey [1, 5].

Level index, a Wiener-like topological index, was proposed by Balaji and Mahmoud in 2017 for rooted trees [2]. The authors introduced the level index for statistical investigations, and used it as a measure of disparity/balance within a rooted tree.

In this paper, we build from the level index by defining level matrix and studying level energy and level characteristic polynomial of rooted trees. We also obtain some relations between the level matrix and the distance matrix of rooted trees. Moreover, we establish bounds on the level energy and calculate the level energy of some classes of rooted trees. Finally, we compute the level characteristic polynomial of so-called rooted binary caterpillars, also known as binary Gutman trees or binary benzenoid trees in chemical graph theory [3].

In Section 2 we introduce some preliminaries about distance matrix of connected graphs and level index of rooted trees. Thereafter, we build the level matrix from the level index and provide basic illustrations. The rest of the paper is devoted to the study of the level energy and level characteristic polynomial. In Subsection 3.1 we establish various bounds on the level energy of rooted trees, while we examine the level characteristic polynomial in specific tree classes such as all rooted versions of stars, so-called rooted double stars and rooted binary caterpillars in Subsection 3.2. Finally, we furnish a conjecture that the rooted path has the maximum level energy among all trees, given the number of vertices, and conclude with some remarks.

# 2 Preliminaries

We only consider simple, connected and undirected graphs. A graph $G$ consists of a vertex set $V(G)$ and an edge set $E(G)$. The notation $d(u, v)$ is used to show the distance between two vertices $u$ and $v$ in a graph.
**Definition 2.1** ([8]). Let $G$ be a connected graph and let its vertices be labelled as $v_1, v_2, ..., v_n$. The distance matrix of $G$ is defined as the square matrix $D = D(G) = [d_{ij}]$ where $d_{ij}$ is the distance between vertices $v_i$ and $v_j$ in $G$.

**Definition 2.2** ([8]). The eigenvalues of the distance matrix $D(G)$ are denoted by $\lambda_1, \lambda_2, ..., \lambda_n$ and are called the $D$-eigenvalues of $G$.

Since the distance matrix is symmetric, its eigenvalues are real and can be ordered as $\lambda_1 \ge \lambda_2 \ge \dots \ge \lambda_n$.

**Definition 2.3** ([8]). The distance energy $E_D = E_D(G)$ of a graph $G$ is defined as

$$E_D(G) = \sum_{i=1}^{n} |\lambda_i|.$$

**Definition 2.4.** Let $T$ be a rooted tree and let its vertices be labelled as $v_1, v_2, ..., v_n$. The level of $v \in V(T)$ is the distance from the root of $T$ to $v$. The level matrix of $T$ is defined as the square matrix $L = L(T) = [l_{ij}]$ where $l_{ij}$ is the absolute value of the levels' difference of vertices $v_i$ and $v_j$ in $T$.

Some basic properties of the level matrix are immediate: it is clearly always a symmetric matrix, and the diagonal entries are all equal to 0.

**Definition 2.5** ([2]). The level index of a rooted tree $T$, denoted by $LI(T)$, is given by:

$$LI(T) = \sum_{1 \le i < j \le n} |l_i(T) - l_j(T)|,$$

where $l_i(T)$ shows the level of the vertex $v_i$ in $T$.

The level of a most distant vertex of $T$ is called maximum level and we denote it by $l_{max}$. We can define the level energy and the level characteristic polynomial as follows.

**Definition 2.6.** The level energy $E_L = E_L(T)$ of a rooted tree $T$ is defined as

$$E_L(T) = \sum_{i=1}^{n} |\lambda_i|,$$

where $\lambda_1, \lambda_2, ..., \lambda_n$ are the $L$-eigenvalues of $T$.

For example, consider the tree shown in Figure 1 whose root is the black vertex. The level matrix of $T$ is given as follows:

$$L(T) = \begin{pmatrix} 0 & 1 & 1 & 2 & 2 & 2 \\ 1 & 0 & 0 & 1 & 1 & 1 \\ 1 & 0 & 0 & 1 & 1 & 1 \\ 2 & 1 & 1 & 0 & 0 & 0 \\ 2 & 1 & 1 & 0 & 0 & 0 \\ 2 & 1 & 1 & 0 & 0 & 0 \end{pmatrix},$$
and the level index of $T$ is computed by

$$LI(T) = \frac{1}{2} \sum_{i=1}^{6} \sum_{j=1}^{6} l_{ij} = 14.$$

<figure><img src="image_2.png" /><figcaption>Figure 1: A rooted tree $T$.</figcaption></figure>

Before stating the main theorems of the paper, we also have to report on some important results about distance matrix, distance energy and determinant of block matrices.

**Lemma 2.1 ([8]).** Let $G$ be a connected $n$-vertex graph and let $\lambda_1, \lambda_2, \dots, \lambda_n$ be its $D$-eigenvalues. Then

$$\sum_{i=1}^{n} \lambda_{i} = 0 \quad \text{and} \quad \sum_{i=1}^{n} \lambda_{i}^{2} = 2 \sum_{1 \le i < j \le n} d_{ij}^{2}.$$

**Lemma 2.2 ([8]).** Let $G$ be a connected $n$-vertex graph and $\Delta$ be the absolute value of the determinant of the distance matrix $D(G)$. Then

$$\sqrt{2 \sum_{1 \le i < j \le n} d_{ij}^2 + n(n-1)\Delta_n^2} \le E_D(G) \le \sqrt{2n \sum_{1 \le i < j \le n} d_{ij}^2}.$$

**Lemma 2.3 ([9]).** If $G$ is a connected graph with $n$ vertices, then

$$\sqrt{n(n-1)} \le E_D(G).$$

**Lemma 2.4 ([12]).** Let $A, B, C, D$ be square matrices of the same order, and

$M = \begin{pmatrix} A & B \\ C & D \end{pmatrix}$ be a block matrix such that $CD = DC$. Then

$$\det(M) = \det(AD - BC)$$

holds.

# 3 Main Results

In this section we give several properties of the level energy and compute the level characteristic polynomial of stars, double stars and binary caterpillars. We also formulate a conjecture on a rooted tree with a prescribed number of vertices that maximises the level energy. To begin with, we first establish further intermediate results that are crucial to proving our main theorems.
By a rooted path, we mean a path whose root is one of the endvertices. The n-vertex rooted path is denoted by $P_n$.

**Lemma 3.1.** Let $T$ be a rooted tree but not a rooted path. Then

$$det(L(T)) = 0.$$

*Proof.* If $T$ is a rooted path, then there is only one vertex on each level. Therefore, differences between the levels can be computed as distances between the vertices. Thus, we get

$$L(P_n) = D(P_n).$$

If $T$ is a rooted tree different from a rooted path, then there are two vertices with the same level. This means that two rows of the matrix $L(T)$ are identical and we obtain that $det(L(T)) = 0$. □

The Wiener index of a connected graph $G$ is the sum of distances between all unordered pairs of vertices of $G$. Lemma 3.2 below shows, in particular, that the level index of a tree never exceeds its Wiener index and that the two indices coincide only for the rooted path.

**Lemma 3.2.** Let $T$ be a rooted tree. The following relation is attained between the entries $l_{ij}$ and $d_{ij}$ of $L(T)$ and $D(T)$:

$$l_{ij} \leq d_{ij}.$$

Equality holds if and only if vertices $v_i$ and $v_j$ are on the same path from the root of $T$. In particular, $LI(T) < W(T)$ for $T \neq P_n$.

*Proof.* If vertex $v_i$ and vertex $v_j$ are on the same path from the root of $T$, then $l_{ij} = |l_i(T) - l_j(T)|$ is precisely the distance between $v_i$ and $v_j$.

If $v_i$ and $v_j$ are not on the same path from the root of $T$, then let $u$ be the last vertex on the common subpath from the root (possibly, $u$ can coincide with the root of $T$): in this case, we have

$$l_i(T) - l_j(T) = d(v_i, u) - d(v_j, u) \quad \text{and} \quad d_{ij} = d(v_i, u) + d(v_j, u).$$

Therefore, we get $l_{ij} = |l_i(T) - l_j(T)| < d_{ij}$ since none of the vertices $v_i$ and $v_j$ coincides with $u$. Now it is clear that $LI(T) \leq W(T)$ with equality only for the rooted path. □

**Lemma 3.3.** Let $T$ be a rooted tree. The following relation is attained between entries $l_{ij}$ and $d_{ij}$ of $L(T)$ and $D(T)$:

$$d_{ij} \leq l_i + l_j.$$

*Proof.* If $v_i$ and $v_j$ are in different branches/arms of $T$, then $d_{ij}$ is computed by sum of the distances from the root to the vertices $v_i$ and $v_j$, i.e. $d_{ij} = l_i + l_j$. If $v_i$ and $v_j$ are in the same branch of $T$, it is clear that $d_{ij} \leq l_i + l_j$. □
By tr(M) we mean the trace of a square matrix M. We obtain an analogue of Lemma 2.1 for the level matrix.

**Lemma 3.4.** Let T be a rooted tree and let $\lambda_1, \lambda_2, \dots, \lambda_n$ be the L-eigenvalues of T. Then

$$ \sum_{i=1}^{n} \lambda_i = 0 \quad \text{and} \quad \sum_{i=1}^{n} \lambda_i^2 = 2 \sum_{1 \le i < j \le n} l_{ij}^2. $$

*Proof.* We have

$$ \sum_{i=1}^{n} \lambda_i = \mathrm{tr}(L(T)) = \sum_{i=1}^{n} l_{ii} = 0. $$

A $(i, i)$-component of $L(T)^2$ equals

$$ \sum_{j=1}^{n} l_{ij} l_{ji} = \sum_{j=1}^{n} l_{ij}^2. $$

Thus, we get

$$ \sum_{i=1}^{n} \lambda_i^2 = \mathrm{tr}(L(T)^2) = \sum_{i=1}^{n} \sum_{j=1}^{n} l_{ij}^2 = 2 \sum_{1 \le i < j \le n} l_{ij}^2. $$

□

## 3.1 Bounds on the level energy

In what follows, we consistently assume $n > 1$. Our first main theorem on the level energy states as follows.

**Theorem 3.1.** Let T be an n-vertex rooted tree and $\Delta$ be the absolute value of the determinant of the level matrix $L(T)$. We have

$$ \sqrt{2 \sum_{1 \le i < j \le n} l_{ij}^2 + n(n-1)\Delta^{\frac{2}{n}}} \le E_L(T) \le \sqrt{2n \sum_{1 \le i < j \le n} l_{ij}^2}. $$

In particular,

$$ E_L(P_n) \ge \sqrt{2 \sum_{1 \le i < j \le n} l_{ij}^2 + n(n-1)((n-1)2^{n-2})^{\frac{2}{n}}}. $$

holds.

*Proof.* The upper bound can be established by Cauchy-Schwartz inequality together with Lemma 3.4:

$$ \left(\sum_{i=1}^{n} a_i b_i\right)^2 \le \left(\sum_{i=1}^{n} a_i^2\right) \left(\sum_{i=1}^{n} b_i^2\right). $$
Put $a_i = 1$ and $b_i = |\lambda_i|$ in this inequality to obtain

$$ \left( \sum_{i=1}^{n} |\lambda_i| \right)^2 \le n \sum_{i=1}^{n} \lambda_i^2, $$

which is equivalent to

$$ E_L(T)^2 \le 2n \sum_{1 \le i < j \le n} l_{ij}^2. $$

The lower bound on the level energy is computed as follows:

$$ E_L(T)^2 = \left( \sum_{i=1}^{n} |\lambda_i| \right)^2 \ge \sum_{i=1}^{n} \lambda_i^2 = 2 \sum_{1 \le i < j \le n} l_{ij}^2. $$

If $T$ is not a rooted path, then $\Delta = 0$ (see Lemma 3.1) and we are done. If $T$ is a rooted path, then $T = P_n$ and $L(P_n) = D(P_n)$ (see the proof of Lemma 3.1). Thus, $E_L(P_n) = E_D(P_n)$. Furthermore, a result by Graham and Pollack [6] implies $\Delta = (n-1)2^{n-2}$. By virtue of Lemma 2.2, we conclude that

$$ E_L(P_n) = E_D(P_n) \ge \sqrt{2 \sum_{1 \le i < j \le n} l_{ij}^2 + n(n-1)((n-1)2^{n-2})^{\frac{2}{n}}}. $$

□

To the best of our knowledge, no one knows a neat formula for $E_D(P_n)$, although $E_D(P_n) \approx 0.69482n^2 - 0.7964$ seems to hold [1].

Our next result shows an inequality between the level energy and the distance energy.

**Theorem 3.2.** Let $T$ be a rooted tree with $n$ vertices and with maximum level $l_{max}$. We have

$$ E_L(T) \le l_{max} \sqrt{nE_D(T)}. $$

*Proof.* We know an upper bound on level energy from Theorem 3.1 as

$$ E_L(T) \le \sqrt{2n \sum_{1 \le i < j \le n} l_{ij}^2}. $$

The difference between the levels of any two vertices of $T$ satisfies

$$ l_{ij} = |l_i - l_j| \le l_{max} $$

with equality if and only if one of the vertices $v_i$ and $v_j$ is the root and another is the most distant from the root. Thus, we get

$$ \sqrt{2n \sum_{1 \le i < j \le n} l_{ij}^2} \le \sqrt{2n \sum_{1 \le i < j \le n} l_{max}^2} \le l_{max} \sqrt{2n \frac{n(n-1)}{2}} \le l_{max} \sqrt{nE_D(T)} $$

since $\sqrt{n(n-1)} \le E_D(T)$ by Lemma 2.3. □
In the next theorem, we derive another upper bound on the level energy of a rooted tree with $n$ vertices as well as of the rooted path with $n$ vertices.

**Theorem 3.3.** Let $T$ be a rooted tree on $n$ vertices. We have

$$E_L(T) \le \sqrt{2n \sum_{1 \le i < j \le n} d_{ij}^2} \quad \text{and} \quad E_L(P_n) \le \sqrt{\frac{n^5 - n^3}{6}}.$$

If equality holds, then $T = P_n$.

*Proof.* Using Theorem 3.1 and Lemma 3.2, we obtain

$$E_L(T) \le \sqrt{2n \sum_{1 \le i < j \le n} l_{ij}^2} \le \sqrt{2n \sum_{1 \le i < j \le n} d_{ij}^2}$$

with the second equality holding only if $L(T) = D(T)$, i.e. if $T = P_n$. For the rooted path $P_n$, we have

$$\sum_{1 \le i < j \le n} d_{ij}^2 = \sum_{i=1}^{n} i(n-i)^2 = n^2 \sum_{i=1}^{n-1} i - 2n \sum_{i=1}^{n-1} i^2 + \sum_{i=1}^{n-1} i^3 \\ = n^2 \frac{n(n-1)}{2} - 2n \frac{n(n-1)(2n-1)}{6} + \frac{n^2(n-1)^2}{4} = \frac{n^4 - n^2}{12}.$$

It follows that

$$E_L(P_n) \le \sqrt{\frac{n^5 - n^3}{6}}.$$

<figure><img src="image_3.png" /></figure>

It is known that the path maximises the distance energy among all trees with a given number of vertices. Since $E_L(P_n) = E_D(P_n)$, we can formulate the following.

**Conjecture 3.1.** If $T$ is a rooted tree with $n$ vertices, then

$$E_L(T) \le E_L(P_n).$$

## 3.2 Level characteristic polynomials of some rooted trees

The identity matrix of order $n$ is denoted by $I_n$. By a rooted star, we mean a star whose root is the central vertex. The $n$-vertex rooted star is denoted by $S_n$. We show in Figure 2 the rooted star with 6 vertices.

**Theorem 3.4.** The level energy of the rooted star $S_n$ is given by:

$$E_L(S_n) = 2\sqrt{n-1}.$$
<figure><img src="image_4.png" /><figcaption>Figure 2: The rooted star $S_6$.</figcaption></figure>

**Proof.** We obtain the level matrix and the characteristic matrix of the rooted star $S_n$ as follows:

$$L(S_n) = \begin{pmatrix} 0 & 1 & 1 & \cdots & 1 \\ 1 & 0 & 0 & \cdots & 0 \\ 1 & 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & 0 & 0 & \cdots & 0 \end{pmatrix}, \quad \det(\lambda I_n - L(S_n)) = \begin{vmatrix} \lambda & -1 & -1 & \cdots & -1 \\ -1 & \lambda & 0 & \cdots & 0 \\ -1 & 0 & \lambda & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ -1 & 0 & 0 & \cdots & \lambda \end{vmatrix}$$

Since $L(S_n)$ has rank 2, then there are only two nonzero eigenvalues. Moreover, we note that $L(S_n)$ is the same as the adjacency matrix of $S_n$. Therefore,

$$\varphi(\lambda) = \lambda^{n-2}(\lambda^2 - (n-1)) \quad \text{and} \quad \lambda_{1,2} = \pm\sqrt{n-1},$$

implying that $E_L(S_n) = 2\sqrt{n-1}$. $\square$

<figure><img src="image_5.png" /></figure>

Denote by $R_n$ the n-vertex star rooted at one of its non-central vertex, see Figure 3.

<figure><img src="image_6.png" /><figcaption>Figure 3: The star $R_6$.</figcaption></figure>

**Theorem 3.5.** The level characteristic polynomial of $R_n$ is given by:

$$\varphi(\lambda) = \lambda^{n-3}(\lambda^3 + (-5n + 9)\lambda - 4n + 8).$$

**Proof.** We obtain the level matrix and the level characteristic matrix of $R_n$ as follows:

$$L(S) = \begin{pmatrix} 0 & 1 & 2 & \cdots & 2 \\ 1 & 0 & 1 & \cdots & 1 \\ 2 & 1 & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 2 & 1 & 0 & \cdots & 0 \end{pmatrix} \quad \text{and} \quad \det(\lambda I_n - L(S)) = \begin{vmatrix} \lambda & -1 & -2 & \cdots & -2 \\ -1 & \lambda & -1 & \cdots & -1 \\ -2 & -1 & \lambda & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ -2 & -1 & 0 & \cdots & \lambda \end{vmatrix}$$
We can compute the determinant of the characteristic matrix of $L(S)$ by adding minus two times of second row to the first row, giving us

$$ \det(\lambda I_n - L(R_n)) = \det \begin{pmatrix} \lambda + 2 & -2\lambda - 1 & 0 & \cdots & 0 \\ -1 & \lambda & -1 & \cdots & -1 \\ -2 & -1 & \lambda & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ -2 & -1 & 0 & \cdots & \lambda \end{pmatrix}. $$

Next, we compute the determinant with respect to the first row:

$$ \det(\lambda I_n - L(R_n)) = (\lambda + 2)\det \begin{pmatrix} \lambda & -1 & -1 & \cdots & -1 \\ -1 & \lambda & 0 & \cdots & 0 \\ -1 & 0 & \lambda & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ -1 & 0 & 0 & \cdots & \lambda \end{pmatrix} \\ + (2\lambda + 1)\det \begin{pmatrix} -1 & -1 & -1 & \cdots & -1 \\ -2 & \lambda & 0 & \cdots & 0 \\ -2 & 0 & \lambda & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ -2 & 0 & 0 & \cdots & \lambda \end{pmatrix}. $$

We notice that the matrix in the first term equals the characteristic matrix of $L(S_{n-1})$. Morover we can subtract the second row from all the remaining rows in the matrix of second term. This yields

$$ \det(\lambda I_n - L(R_n)) = (\lambda + 2)\det(\lambda I_n - L(S_{n-1})) \\ + (2\lambda + 1)\det \begin{pmatrix} -1 & -1 & -1 & \cdots & -1 \\ -2 & \lambda & 0 & \cdots & 0 \\ 0 & -\lambda & \lambda & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & -\lambda & 0 & \cdots & \lambda \end{pmatrix}. $$

It follows that

$$ \varphi(\lambda) = (\lambda + 2)\lambda^{n-3}(\lambda^2 - (n-2)) + (2\lambda + 1)(-\lambda^{n-2} - 2(n-2)\lambda^{n-3}) \\ = \lambda^{n-3}(\lambda^3 + (-5n + 9)\lambda - 4n + 8). $$

□

By double rooted star, we mean the tree $DS_n$ rooted at vertex $v_1$ and depicted in Figure 4.
<figure><img src="image_7.png" /><figcaption>Figure 4: The double rooted star $DS_n$.</figcaption></figure>

**Theorem 3.6.** The level characteristic polynomial of $DS_n$ is given by:

$$ \varphi(\lambda) = \lambda^{2n-3}(\lambda^3 - (n^2 + 4n - 1)\lambda - 4n^2 + 4n). $$

*Proof.* The level matrix of $DS_n$ can be given in the following block form:

<table><thead><tr><th></th><th><i>v</i><sub>1</sub></th><th><i>v</i><sub>2</sub></th><th>...</th><th><i>v</i><sub>n</sub></th><th><i>u</i><sub>1</sub></th><th><i>u</i><sub>2</sub></th><th>...</th><th><i>u</i><sub>n</sub></th></tr></thead><tbody><tr><th><i>v</i><sub>1</sub></th><td>0</td><td>1</td><td>...</td><td>1</td><td>2</td><td>2</td><td>...</td><td>2</td></tr><tr><th><i>v</i><sub>2</sub></th><td>1</td><td>0</td><td>...</td><td>0</td><td>1</td><td>1</td><td>...</td><td>1</td></tr><tr><th>...</th><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td></tr><tr><th><i>v</i><sub>n</sub></th><td>1</td><td>0</td><td>...</td><td>0</td><td>1</td><td>1</td><td>...</td><td>1</td></tr><tr><th><i>u</i><sub>1</sub></th><td>2</td><td>1</td><td>...</td><td>1</td><td>0</td><td>...</td><td>...</td><td>0</td></tr><tr><th><i>u</i><sub>2</sub></th><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td></tr><tr><th>...</th><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td></tr><tr><th><i>u</i><sub>n</sub></th><td>2</td><td>1</td><td>...</td><td>1</td><td>0</td><td>...</td><td>...</td><td>0</td></tr></tbody></table>

Thus, the level characteristic polynomial of $DS_n$ is the determinant of the following matrix in block form:

<table><tbody><tr><th>λ</th><td>-1</td><td>-1</td><td>...</td><td>-1</td><td>-1</td><td>-2</td><td>...</td><td>...</td><td>-2</td></tr><tr><th>-1</th><td>λ</td><td>0</td><td>...</td><td>0</td><td>0</td><td>-1</td><td>...</td><td>...</td><td>-1</td></tr><tr><th>...</th><td>0</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td></tr><tr><th>...</th><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td></tr><tr><th>-1</th><td>0</td><td>...</td><td>...</td><td>λ</td><td>0</td><td>...</td><td>...</td><td>...</td><td>...</td></tr><tr><th>-1</th><td>0</td><td>0</td><td>...</td><td>0</td><td>λ</td><td>-1</td><td>...</td><td>...</td><td>-1</td></tr><tr><th>-2</th><td>-1</td><td>-1</td><td>...</td><td>-1</td><td>-1</td><td>λ</td><td>0</td><td>...</td><td>0</td></tr><tr><th>...</th><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>0</td><td>...</td><td>...</td><td>...</td></tr><tr><th>...</th><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td></tr><tr><th>...</th><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td></tr><tr><th>-2</th><td>-1</td><td>-1</td><td>...</td><td>-1</td><td>-1</td><td>0</td><td>...</td><td>...</td><td>0</td></tr></tbody></table>$$ \therefore = \begin{array}{|c|c|} \hline A & B \\ \hline B^T & \lambda I_n \\ \hline \end{array} $$

We apply Lemma 2.4 to obtain

$$ \det(\lambda I_{2n} - L(DS_n)) = \det(\lambda A - BB^T). $$
On the other hand, we compute that

$$BB^T = \begin{pmatrix} 4n & 2n & \cdots & 2n \\ 2n & n & \cdots & n \\ \vdots & \vdots & \ddots & \vdots \\ 2n & n & \cdots & n \end{pmatrix}$$

and

$$\lambda A - BB^T = \begin{pmatrix} \lambda^2 - 4n & -\lambda - 2n & \cdots & \cdots & \cdots & -\lambda - 2n \\ -\lambda - 2n & \lambda^2 - n & -n & \cdots & \cdots & -n \\ \vdots & -n & \lambda^2 - n & -n & \cdots & -n \\ \vdots & \vdots & -n & \ddots & \ddots & \vdots \\ \vdots & \vdots & \vdots & \ddots & \ddots & \vdots \\ \vdots & \vdots & \vdots & \ddots & \ddots & -n \\ -\lambda - 2n & -n & -n & \cdots & \cdots & -n & \lambda^2 - n \end{pmatrix}$$

In order to compute the determinant of this matrix, we perform some elementary row and column operations. First, we add all the remaining rows to the first row to obtain the following matrix:

$$\begin{pmatrix} \lambda^2 + (1-n)\lambda - 2n^2 - 2n & \lambda^2 - \lambda - n^2 - n & \cdots & \cdots & \cdots & \lambda^2 - \lambda - n^2 - n \\ -\lambda - 2n & \lambda^2 - n & -n & \cdots & \cdots & -n \\ \vdots & -n & \lambda^2 - n & -n & \cdots & -n \\ \vdots & \vdots & -n & \ddots & \ddots & \vdots \\ \vdots & \vdots & \vdots & \ddots & \ddots & \vdots \\ \vdots & \vdots & \vdots & \ddots & \ddots & -n \\ -\lambda - 2n & -n & -n & \cdots & \cdots & -n & \lambda^2 - n \end{pmatrix}$$

Next, we subtract the second column from every other column to obtain the equivalent matrix:

$$\begin{pmatrix} (2-n)\lambda - n^2 - n & \lambda^2 - \lambda - n^2 - n & 0 & \cdots & \cdots & 0 \\ -\lambda^2 - \lambda - n & \lambda^2 - n & -\lambda^2 & \cdots & \cdots & -\lambda^2 \\ -\lambda - n & -n & \lambda^2 & 0 & \cdots & 0 \\ \vdots & \vdots & 0 & \ddots & \ddots & \vdots \\ \vdots & \vdots & \vdots & \ddots & \ddots & \vdots \\ \vdots & \vdots & \vdots & \ddots & \ddots & 0 \\ -\lambda - n & -n & 0 & \cdots & 0 & \lambda^2 \end{pmatrix}$$
Next, we expand the determinant with respect to the first row. This gives us:

$$ ((2-n)\lambda - n^2 - n)\det \begin{pmatrix} \lambda^2 - n & -\lambda^2 & \cdots & \cdots & -\lambda^2 \\ -n & \lambda^2 & 0 & \cdots & 0 \\ \vdots & 0 & \ddots & \ddots & \vdots \\ \vdots & \vdots & \ddots & \ddots & \vdots \\ \vdots & \vdots & \ddots & \ddots & 0 \\ -n & 0 & \cdots & \cdots & \lambda^2 \end{pmatrix} \\ - (\lambda^2 - \lambda - n^2 - n)\det \begin{pmatrix} -\lambda^2 - \lambda - n & -\lambda^2 & \cdots & \cdots & -\lambda^2 \\ -\lambda - n & \lambda^2 & 0 & \cdots & 0 \\ \vdots & 0 & \ddots & \ddots & \vdots \\ \vdots & \vdots & \ddots & \ddots & \vdots \\ \vdots & \vdots & \ddots & \ddots & 0 \\ -\lambda - n & 0 & \cdots & \cdots & \lambda^2 \end{pmatrix}, $$

where each submatrix is of order $n-1$. Finally, for each of these submatrices, we add all the remaining rows to the first:

$$ ((2-n)\lambda - n^2 - n)\det \begin{pmatrix} \lambda^2 - n(n-1) & 0 & \cdots & \cdots & 0 \\ -n & \lambda^2 & 0 & \cdots & 0 \\ \vdots & 0 & \ddots & \ddots & \vdots \\ \vdots & \vdots & \ddots & \ddots & \vdots \\ \vdots & \vdots & \ddots & \ddots & 0 \\ -n & 0 & \cdots & \cdots & \lambda^2 \end{pmatrix} \\ - (\lambda^2 - \lambda - n^2 - n)\det \begin{pmatrix} -\lambda^2 - (n-1)(\lambda+n) & 0 & \cdots & \cdots & 0 \\ -\lambda - n & \lambda^2 & 0 & \cdots & 0 \\ \vdots & 0 & \ddots & \ddots & \vdots \\ \vdots & \vdots & \ddots & \ddots & \vdots \\ \vdots & \vdots & \ddots & \ddots & 0 \\ -\lambda - n & 0 & \cdots & \cdots & \lambda^2 \end{pmatrix}. $$

Since these two matrices are triangular, we arrive at:

$$ \begin{aligned} \det(\lambda I_{2n} - L(DS_n)) &= ((2-n)\lambda - n^2 - n)\lambda^{2(n-2)}(\lambda^2 - n(n-1)) \\ &\quad - (\lambda^2 - \lambda - n^2 - n)\lambda^{2(n-2)}(-\lambda^2 - (n-1)(\lambda+n)) \\ &= \lambda^{2n-3}(\lambda^3 - (n^2 + 4n - 1)\lambda - 4n^2 + 4n). \end{aligned} $$

This completes the proof of the theorem. $\square$

<figure><img src="image_8.png" /></figure>
We move our attention to another particular class of rooted trees to which the Gini index was applied in a broader sense, see [2].

Define $T_m$ to be the rooted tree depicted in Figure 5, whose root is $v_1$. This tree belongs to the family of so-called rooted binary caterpillars [3].

<figure><img src="image_9.png" /><figcaption>Figure 5: The rooted binary caterpillar $T_m$.</figcaption></figure>

**Theorem 3.7.** For $m > 2$, the characteristic polynomial of the rooted tree $T_m$ is given by

$$ \varphi(\lambda) = (2\lambda)^{m-1} \left( \lambda \cdot C_m(\lambda/2) + C_{m+1}(\lambda/2) \right) $$

with

$$ C_n(y) = y^n - \sum_{k=2}^{n} 2^{k-2}(k-1) \frac{n^2(n^2-1)(n^2-2^2)\dots(n^2-(k-1)^2)}{k^2(k^2-1)(k^2-2^2)\dots(k^2-(k-1)^2)} y^{n-k} $$

*Proof.* According to Lemma 3.1, we assume that $\lambda \neq 0$.

For every $j \in \{2, 3, \dots, m\}$, there are precisely two vertices on the same level, namely $v_j$ and $v_{j+m}$. The subtree induced by vertices $v_1, v_2, \dots, v_{m+1}$ is a path rooted at $v_1$. Thus the level matrix of $T$ has the following block decomposition:

<table><thead><tr><th></th><th>$v_1$</th><th>$v_2$</th><th>...</th><th>$v_m$</th><th>$v_{m+1}$</th><th>$v_{m+2}$</th><th>$v_{m+3}$</th><th>...</th><th>$v_{2m}$</th></tr></thead><tbody><tr><th>$v_1$</th><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr><tr><th>$v_2$</th><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr><tr><th>$\vdots$</th><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr><tr><th>$v_m$</th><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr><tr><th>$v_{m+1}$</th><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr><tr><th>$v_{m+2}$</th><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr><tr><th>$v_{m+3}$</th><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr><tr><th>$\vdots$</th><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr><tr><th>$v_{2m}$</th><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr></tbody></table>

Denote the rows of this matrix (as well as for the identity matrix $I_{2m}$) by $R_1, R_2, \dots, R_{2m}$ in this order, starting from the first to the last. Then $R_j$ and $R_{j+m}$ are identical rows of $L(T_m)$ for any $j \in \{2, 3, \dots, m\}$.
Now we subtract $R_j$ from $R_{j+m}$ in both $L(T_m)$ and $I_{2m}$; so the rows

$R_{m+2}, R_{m+3}, \dots, R_{2m}$

in $L(T_m)$ all change to zero rows, and the corresponding rows in $I_{2m}$ become

$$
\begin{array}{ccccccccccc}
0 & -1 & 0 & \cdots & 0 & 1 & 0 & 0 & \cdots & 0, \\
& & \uparrow & & & & & & & \\
& & \multicolumn{6}{c}{\text{Pos. } m+2} & & \\
0 & 0 & -1 & 0 & 0 & \cdots & 1 & 0 & \cdots & 0, \\
& & & \uparrow & & & & & & \\
& & & \multicolumn{6}{c}{\text{Pos. } m+3} & & \\
& & & \vdots & & & & & & \\
0 & 0 & \cdots & 0 & -1 & 0 & 0 & \cdots & 0 & 1, \\
& & & \uparrow & & & & & & \\
& & & \multicolumn{6}{c}{\text{Pos. } m} & \uparrow & \\
& & & & & & & & & \multicolumn{2}{c}{\text{Pos. } 2m}
\end{array}
$$

respectively. Thus, we have

$$
\det(xI_{2m} - L(T)) = \det \left( \begin{array}{c|c} xI_{m+1} - D(P_{m+1}) & -B \\ \hline 0 & 0 \\ 0 & -x \\ \vdots & \vdots \\ 0 & 0 \end{array} \right) = \det \left( \begin{array}{c|c} xI_{m+1} - D(P_{m+1}) & -B \\ \hline 0 & 0 \\ 0 & -x \\ \vdots & \vdots \\ 0 & 0 \end{array} \right).
$$

Denote the columns of $L(T_m)$ (as well as for $I_{2m}$) by $C_1, C_2, \dots, C_{2m}$ in this order, starting from the first to the last. Then $C_j$ and $C_{j+m}$ are identical rows of $L(T_m)$ for any $j \in \{2, 3, \dots, m\}$. Now we subtract $C_j$ from $C_{j+m}$ in the above matrix to obtain the following matrix:

$$
\left( \begin{array}{c|c} xI_{m+1} - D(P_{m+1}) & \begin{array}{cccc|c} 0 & 0 & 0 & \cdots & 0 \\ -x & 0 & 0 & \cdots & 0 \\ 0 & -x & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & \cdots & 0 & -x & 0 \\ 0 & 0 & \cdots & 0 & -x \\ 0 & 0 & \cdots & 0 & 0 \end{array} \\ \hline 2xI_{m-1} & \end{array} \right),
$$
which implies that $\det(xI_{2m}-L(T_m))$ is also the determinant of the above block matrix. On the other hand, the product

$$ \begin{array}{c} \begin{array}{ccccc} 1 & 2 & 3 & \cdots & m-1 \end{array} \\ \begin{array}{c} 1 \\ 2 \\ 3 \\ \vdots \\ m \\ m+1 \end{array} \begin{pmatrix} 0 & 0 & 0 & \cdots & 0 \\ -x & 0 & 0 & \cdots & 0 \\ 0 & -x & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & \cdots & 0 & -x & 0 \\ 0 & 0 & \cdots & 0 & -x \\ 0 & 0 & \cdots & 0 & 0 \end{pmatrix} \times \begin{pmatrix} 0 & -x & 0 & \cdots & 0 \\ 0 & 0 & -x & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & \cdots & 0 & -x & 0 \end{pmatrix} \\ (m+1) \times (m-1) \text{ size} \qquad (m-1) \times (m+1) \text{ size} \end{array} $$

yields the square matrix

$$ \begin{pmatrix} 0 & 0 & 0 & 0 & \cdots & 0 \\ 0 & x^2 & 0 & 0 & \cdots & 0 \\ 0 & 0 & x^2 & 0 & \cdots & 0 \\ \vdots & \vdots & \ddots & \ddots & \ddots & \vdots \\ 0 & 0 & \cdots & 0 & x^2 & 0 \\ 0 & 0 & 0 & 0 & \cdots & 0 \end{pmatrix} $$

of order $m+1$. It is known (Schur Complement formula) that

$$ \det \left( \frac{E_1}{E_3} \middle| \frac{E_2}{E_4} \right) = \det(E_4) \det(E_1 - E_2 E_4^{-1} E_3), $$

provided that $E_4$ is invertible. Applying this formula for $x \neq 0$, we obtain:

$$ \det(xI_{2m} - L(T_m)) = (2x)^{m-1}\det(xI_{m+1} - D(P_{m+1}) - \frac{1}{2}xE), $$

where $E$ is the $(m+1) \times (m+1)$ matrix defined by

$$ E = \begin{pmatrix} 0 & 0 & 0 & 0 & \cdots & 0 \\ 0 & 1 & 0 & 0 & \cdots & 0 \\ 0 & 0 & 1 & 0 & \cdots & 0 \\ \vdots & \vdots & \ddots & \ddots & \ddots & \vdots \\ 0 & 0 & \cdots & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & \cdots & 0 \end{pmatrix}. $$

Furthermore, we can write

$$ I_{m+1} - \frac{1}{2}E = \frac{1}{2}I_{m+1} + \frac{1}{2} \begin{pmatrix} 1 & 0 & \cdots & 0 \\ 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 0 \\ 0 & \cdots & 0 & 1 \end{pmatrix} := \frac{1}{2}(I_{m+1} + F), $$
so that

$$xI_{m+1} - D(P_{m+1}) - \frac{1}{2}xE = \frac{1}{2}xI_{m+1} - D(P_{m+1}) + \frac{1}{2}xF,$$

where

$$F = \begin{pmatrix} 1 & 0 & \cdots & 0 \\ 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 0 \\ 0 & \cdots & 0 & 1 \end{pmatrix} = e_1 e_1^T + e_{m+1} e_{m+1}^T$$

with

$$e_1 = \begin{pmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{pmatrix} \quad \text{and} \quad e_{m+1} = \begin{pmatrix} 0 \\ \vdots \\ 0 \\ 1 \end{pmatrix}.$$

Moreover, the matrix determinant lemma gives

$$\det(G + uv^T) = (1 + v^T G^{-1} u) \det(G),$$

provided that G is an invertible matrix. A generalisation (see [13]) of this formula to the noninvertible matrices states that

$$\det(G + uv^T) = \det(G) + v^T \operatorname{adj}(G) u,$$

where adj(G) is the adjugate (transpose of cofactor matrix) of G. In what follows, we use this formula twice.

By setting

$$G_{m+1} = \frac{1}{2}xI_{m+1} - D(P_{m+1}) + \frac{1}{2}xe_{m+1}e_{m+1}^T,$$

and $v^T = \frac{1}{2}xe_1^T$, $u = e_1$, we establish that

$$\begin{aligned} \det(xI_{2m} - L(T_m)) &= (2x)^{m-1}\det(G_{m+1} + uv^T) \\ &= (2x)^{m-1}(\det(G_{m+1}) + v^T \operatorname{adj}(G_{m+1})u). \end{aligned}$$

However, $\det(G_{m+1})u$ is just the first column of the matrix $\det(G_{m+1})$. Consequently, $v^T \operatorname{adj}(G_{m+1})u = \frac{1}{2}xg_{m+1}$ with $g_{m+1}$ being the entry in first row and first column of $\det(G_{m+1})$. It follows that

$$\det(xI_{2m} - L(T_m)) = (2x)^{m-1}(\det(G_{m+1}) + \frac{1}{2}xg_{m+1}). \quad (1)$$

We can get a similar expression for $\det(G_{m+1})$ by setting

$$H_{m+1} = \frac{1}{2}xI_{m+1} - D(P_{m+1}), \quad v^T = \frac{1}{2}xe_{m+1}^T, \quad u = e_{m+1}.$$
This gives us

$$ \det(G_{m+1}) = \det(H_{m+1} + uv^T) = \det(H_{m+1}) + v^T \mathrm{adj}(H_{m+1})u. $$

On the other hand, $v^T \mathrm{adj}(H_{m+1})u = \frac{1}{2}xh_{m+1}$ with $h_{m+1}$ representing the entry in last row and last column of $\mathrm{adj}(H_{m+1})$. It follows that

$$ \det(G_{m+1}) = \det(H_{m+1}) + \frac{1}{2}xh_{m+1}. \qquad (2) $$

We remark that $H_{m+1}$ is the characteristic matrix of the distance matrix of the path $P_{m+1}$ evaluated at $\frac{1}{2}x$. Now we combine equations (1) and (2) to obtain:

$$ \begin{aligned} C(L(T_m); x) &= \det(xI_{2m} - L(T_m)) = (2x)^{m-1}(\det(G_{m+1}) + \frac{1}{2}xg_{m+1}) \\ &= (2x)^{m-1}(\det(H_{m+1}) + \frac{1}{2}xh_{m+1} + \frac{1}{2}xg_{m+1}). \end{aligned} $$

for all $x \neq 0$, where $C(M; y)$ denotes the characteristic polynomial of a matrix $M$ evaluated at $y$. Fortunately, Hosoya, Murakami and Gotoh [7] computed that

$$ \det(H_n) = C(D(P_n); y) = y^n - \sum_{k=2}^{n} 2^{k-2} (k-1) \frac{n^2(n^2-1)(n^2-2^2)\dots(n^2-(k-1)^2)}{k^2(k^2-1)(k^2-2^2)\dots(k^2-(k-1)^2)} y^{n-k}. $$

Furthermore, we note that the $(m+1, m+1)$-cofactor of $H_{m+1}$ is also the determinant of the matrix $H_m$, i.e. $h_{m+1} = \det(H_m)$. Similarly, the $(1, 1)$-cofactor of $G_{m+1}$ is also the determinant of the matrix $\frac{1}{2}xI_m - D(P_m) = H_m$, not that of the matrix $G_m$. Thus $g_{m+1} = \det(H_m)$. Putting everything together, we arrive at

$$ \begin{aligned} C(L(T_m); x) &= (2x)^{m-1}(\det(H_{m+1}) + \frac{1}{2}xh_{m+1} + \frac{1}{2}xg_{m+1}) \\ &= (2x)^{m-1}(\det(H_{m+1}) + \frac{1}{2}x\det(H_m) + \frac{1}{2}x\det(H_m)) \\ &= (2x)^{m-1}\left(x \cdot C(D(P_m); \frac{1}{2}x) + C(D(P_{m+1}); \frac{1}{2}x)\right). \end{aligned} $$

This completes the proof of the theorem. $\square$

<figure><img src="image_10.png" /></figure>

# 4 Conclusion

There are many open problems concerning the level matrix [2]. Some of them are studied in this paper. Naturally, the extremal trees in the set of rooted trees with a given number of vertices deserve to be determined. We have conjectured that the rooted path $P_n$ maximises the level energy, which is an analogue of the distance energy result among $n$-vertex trees. We know that the level energy of
$P_n$ coincides with its distance energy. Ruzieh and Powers [11] provided in 1990 formulas for all the eigenvalues of the distance matrix of paths. However, these formulas are implicit and can only be approximated.

As for the case of Wiener index, it is not difficult to see that the rooted star (resp. rooted path) uniquely minimises (resp. maximises) the level index among all trees with a prescribed number of vertices. In this paper, we have shown that the Wiener index furnishes a sharp upper bound for the level index. It is natural to ask whether there is a similar lower bound that uses other tree invariants.

Given that we have established the level characteristic of the rooted binary caterpillar as a function of the distance characteristic polynomial of paths, we wonder whether a similar explicit formula can be obtained for the general case where the caterpillar is formed by attaching the central vertex of $S_n$ to every vertex of a path, see [2].

On the other hand, in chemistry the electrons of atoms are located on the orbits with respect to their energy levels. Energy levels are related to atomic orbital theory. Therefore, it can be more suitable to find relations between energy levels of electrons and level energy of rooted trees.

## Acknowledgements

Bünyamin Şahin is supported by Selcuk University Scientific Research Project with reference number 23705001. Elena V. Konstantinova was supported by the Mathematical Center in Akademgorodok, under agreement N. 075-15-2022-281 with the Ministry of Science and High Education of the Russian Federation. The work was started in the framework of international cooperation with the department of Mathematics, Faculty of Science, Selcuk University, Turkey, during her visit on May 8–18, 2023.

### CRediT authorship contribution statement

**Audace A. V. Dossou-Olory:** Conceptualization, Investigation, Methodology, Writing - original draft, Writing - review & editing. **Muhammed F.Killik:** Investigation, Methodology, Visualization. **Elena V. Konstantinova:** Supervision, Writing - review & editing **Bünyamin Şahin:** Conceptualization, Project administration.

### Conflict of interest statement

The authors declare that there is no conflict of interest.
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