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# Sharp Bounds on the σ-Irregularity of Trees with Given Degree Sequences

Jasem Hamoud

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## To cite this version:

Jasem Hamoud. Sharp Bounds on the σ-Irregularity of Trees with Given Degree Sequences. 2025. hal-05347100

HAL Id: hal-05347100

https://hal.science/hal-05347100v1

Preprint submitted on 4 Nov 2025

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Distributed under a Creative Commons Attribution 4.0 International License
# SHARP BOUNDS ON THE σ-IRREGULARITY OF TREES WITH GIVEN DEGREE SEQUENCES

## JASEM HAMOUD

**ABSTRACT.** A fundamental challenge in the study of topological indices is establishing precise bounds, as these results reveal intrinsic relationships among various indices. In this paper, we investigate the sharp upper and lower bounds of the Sigma index for a given degree sequence $\mathcal{D} = (d_1, d_2, \dots, d_n)$. We consider both cases where the degree sequence is non-increasing, $d_1 \ge d_2 \ge \dots \ge d_n$, and non-decreasing, $d_n \ge \dots \ge d_1$, and determine strict bounds accordingly.

**MSC Classification 2010:** 05C05, 05C12, 05C20, 05C25, 05C35, 05C76, 68R10.

**Keywords:** Trees, Degree sequence, graph, Topological indices, Sigma index, Extremal, Irregularity.

## 1. INTRODUCTION

Throughout this paper. Let $G = (V, E)$ be a simple, connected graph, where $n = |V(G)|$, $m = |E(G)|$ and we denote by $d_T(u)$ the degree of vertex $u \in V(T)$. A sequence $\mathcal{D} = (d_1, d_2, \dots, d_n)$ that represents the vertex degrees of a graph is known as a *graphical sequence*. When the terms of this sequence are sorted in descending order so that $d_1 \ge d_2 \ge \dots \ge d_n$ or $d_1 \le d_2 \le \dots \le d_n$, it is called a *degree sequence*. One of the earliest and most influential measures of graph-in our paper we consider the graph is tree- irregularity is the *Albertson index*, introduced in [1, 10], and defined as:

$$ (1) \qquad \mathrm{irr}(T) = \sum_{uv \in E(T)} |d_T(u) - d_T(v)|. $$

To explore existing literature on bounds of Albertson index see [2, 3, 23]. The topological indices form interconnected structures with each other. Topological indices have been extensively defined and applied across various disciplines to analyze diverse properties of entities including atoms and molecules. Numerous such indices have been introduced and rigorously investigated by researchers in both mathematics and chemistry [11, 17, 18]. One of these topological indices called the *Sigma index* is denoted by $\sigma(T)$ and is defined [11] as

$$ (2) \qquad \sigma(T) = \sum_{uv \in E(T)} (d_T(u) - d_T(v))^2. $$

Many studies indicate that the first explicit study of the Sigma index was presented by Gutman et al. [11], but there was a prior study before it presented by Furtula et al. [13] that defined the $\sigma(T)$ as $IRM(G)$ such that has not been investigated in any detail. Therefore, Gutman's study [11] is considered the explicit foundation of the Sigma index, as it established the most important basics related to this index. Let $\sigma \odot(T)$ be the average of Sigma index for the case $d_1 \ge d_2 \ge \dots \ge d_n$, and the case $d_n \ge \dots \ge d_1$. The *reduced Sigma index* defined as

$$ (3) \qquad \sigma_{\mathrm{red}}(T) = \sum_{uv \in E(T)} [(d_T(u) - 1)^2 + (d_T(v) - 1)^2]. $$

<footnote>Corresponding author: Jasem Hamoud, hamoud.math@gmail.com.</footnote>
This particular expression resembles modified Zagreb-type indices (see Definition 2), which are degree-based or related indices for trees, emphasizing the degrees decreased by one rather than the degrees themselves. The increase Sigma index $\sigma_{inc}(T)$ of a tree T defined as

$$ (4) \qquad \sigma_{inc}(T) = \sum_{uv \in E(T)} [(d_T(u) + 1)^2 + (d_T(v) + 1)^2]. $$

The modified Sigma index $\sigma^*(T)$ of a tree T defined as

$$ (5) \qquad \sigma^*(T) = \sum_{uv \in E(T)} \left[ \frac{1}{(d_T(u) + 1)^2 + (d_T(v) + 1)^2} \right]. $$

Among such families, caterpillar trees—characterized by a linear backbone with attached pendant vertices—have emerged as key models in chemical graph theory [14]. Their tractable structure allows the derivation of explicit analytical expressions. While prior work [15, 16] has examined irregularity bounds in these trees, closed-form expressions for both the Albertson and Sigma indices remain scarce.

The main goal of this paper lies in examining the upper and lower bounds of the Sigma index to determine the strictest bound. According to upper or lower bounds it smells like studying the behavior of the Sigma index, which effectively influences the behavior of other topological indices. There are fundamental challenges that to find the sharp and strictly upper and lower bound of Sigma index. The paper also aims to present a new topological indices derived from the Sigma index that will lay the foundation for future studies to overcome major challenges in the optimal improvement of topological indices.

This paper is organized as follows. Section 3 had discussed the efficiency lower bound on $\sigma(T)$ including the optimal behavior of lower bound on $\sigma(T)$ through subsection 3.1 and behavior lower bound on $\sigma(T)$ through subsection 3.2. Section 4 had presented the efficiency of upper bound on $\sigma(T)$ including the optimal behavior of upper bound on $\sigma(T)$ through subsection 4.1. Throughout section 5, it previews some novelty concepts on the Sigma index, where these concepts are represented by the reduced Sigma index among subsection 5.1, the increasing Sigma index, and the modified Sigma index had presented among subsection 5.2.

## 2. PRELIMINARIES

In this section, we recall the notion of a degree sequence, which is a basic descriptor of a graph's structure and plays a central role in our classification of trees. P. Erdös and T. Gallai [5] had presented graphs with prescribed degrees of vertices where among Definition 1 had presented the concept of degree sequence.

**Definition 1 (Degree Sequence [4, 24]).** Let $G = (V, E)$ be a simple graph, where $V = \{v_1, v_2, \dots, v_n\}$, let $\mathcal{D}(G) = (d_G(v_1), \dots, d_G(v_n))$ be a degree sequence of G where $d_G(v_1) \ge d_G(v_2) \ge \dots \ge d_G(v_n)$. If $\mathcal{D}(G) = (k, k, \dots, k)$, then G is regular of degree k. Otherwise, the graph is irregular.

Consider a tree T with $n_1$ leaves and denote its first Zagreb index by $M_1(T)$ had provided among Definition 2.

**Definition 2 (Zagreb Indices [19, 20, 21, 22]).** Let $G = (V, E)$ be a graph. The first and second Zagreb indices are given by

$$ (6) \qquad M_1(G) = \sum_{v \in V(G)} d_G(v)^2, \quad M_2(G) = \sum_{uv \in E(G)} d_G(u)d_G(v), $$

when $n_1$ is an even number, according to (6) the inequality $M_1(T) \ge 9n_1 - 16$ holds in [12, 25]. Equality occurs precisely when every non-leaf vertex in T has degree 4. The next formula captures
the Albertson index in caterpillar trees based on the degrees of the backbone vertices, and will be instrumental in deriving our closed-form expressions.

**Proposition 2.1 ([6]).** For Caterpillar tree with path vertices with degrees $d_1, d_2, \dots, d_n$, we have:

$$ \mathrm{irr}(G) = (d_n - 1)^2 + (d_1 - 1)^2 + \sum_{i=2}^{n-1} (d_i - 1)(d_i - 2) + \sum_{i=1}^{n-1} |d_i - d_{i+1}|. $$

Throughout this paper, we have not use the relationship of Sigma index in (2) but we will the relationship of Sigma index according to ordered among degree sequence such that $d_1 \ge d_2 \ge \dots \ge d_n$ in Theorem 2.1 and for $d_n \ge \dots \ge d_1$ in Theorem 2.2.

**Theorem 2.1 ([6]).** Let $T$ be a tree of order $n$, and let $\mathcal{D} = (d_1, \dots, d_n)$ be a degree sequence such that $d_1 \ge d_2 \ge \dots \ge d_n$. Then, the Sigma index of the tree $T$ is given by:

$$ \sigma(T) = (d_n - 1)^3 + (d_1 - 1)^3 + \sum_{i=1}^{n-1} (d_i - d_{i+1})^2 + \sum_{i=2}^{n-1} (d_i - 1)^2(d_i - 2). $$

**Theorem 2.2 ([6, 7]).** Let $T$ be a tree of order $n$, and let $\mathcal{D} = (d_1, \dots, d_n)$ be a degree sequence such that $d_n \ge \dots \ge d_1$. Then, the Sigma index of the tree $T$ is given by:

$$ \sigma(T) = \sum_{i \in \{1, n\}} (d_i + 1)(d_i - 1)^2 + \sum_{i=2}^{n-1} (d_i + 2)(d_i - 1)^2 + \sum_{i=2}^{n-1} (d_i - d_{i+1})^2 + 2n - 2. $$

**2.1. Statement of the Problem.** We presented the first study of topological indices — specifically the Albertson and Sigma indices — on trees through [6, 7], and extended the research to include advanced results through [8, 9]. Refine topological indices — specifically the Albertson and Sigma indices — on trees by findings sharply bounds to determine the optimal behavior of the Sigma index, which is crucial for establishing the tightest bounds of this index by examining its upper and lower bounds. Consequently, this research aims to anticipate and establish the bounds related to the Sigma index and to examine the influence of these bounds on both values derived according to the ordering of the included sequence.

## 3. THE EFFICIENCY OF LOWER BOUND

In this section, the lower bound of $\sigma \odot (T)$ had presented among Proposition 3.1 for the case $d_1 \ge d_2 \ge \dots \ge d_n$, and the case $d_n \ge \dots \ge d_1$.

**Proposition 3.1.** Let $T$ be a tree with degree sequence $\mathcal{D} = (d_1, d_2, \dots, d_n)$. Then, the lower bound of $\sigma \odot (T)$ satisfy

$$ (7) \qquad \sigma \odot (T) \ge \sum_{i=1}^{n} d_i^3 - \sum_{i=1}^{n-1} \left[ \frac{d_i}{d_{i+1}} \right] - \left\lfloor \frac{2n^2}{3} \right\rfloor. $$

*Proof.* Let us recall degree sequence $\mathcal{D} = (d_1, d_2, \dots, d_n)$ for $\sigma \odot (T)$ we should be discussed with the case $d_1 \ge d_2 \ge \dots \ge d_n$, and the case $d_n \ge \dots \ge d_1$. Thus,

Case 1: If $\mathcal{D} = (d_1, d_2, \dots, d_n)$ ordered such that $d_1 \ge d_2 \ge \dots \ge d_n$. In this case, according to Theorem 2.1, we noticed that the Sigma index satisfies $\sum_{i=1}^{n} d_i^3 \ge (d_n - 1)^3 + (d_1 - 1)^3$. Then,

$$ (8) \qquad \sigma(T) \ge \sum_{i=1}^{n} d_i^3 - \left( (d_n - 1)^3 + (d_1 - 1)^3 + \sum_{i=1}^{n-1} (d_i - d_{i+1})^2 \right). $$
On the other hand, let $\Delta$ be the maximum degree of vertices, since $\sum_{i=1}^{n-1} \lfloor \frac{d_i}{d_{i+1}} \rfloor \le \Delta$ and $2n/3 \ge \Delta$. Thus, from (8) we have

$$ (d_1 - 1)^3 < \sum_{i=1}^{n-1} \lfloor \frac{d_i}{d_{i+1}} \rfloor < \lfloor \frac{2n^2}{3} \rfloor $$

Therefore, in this case, we find that $\sigma(T) \ge \sigma \odot(T)$. Thus, from (8), we find that (7) is satisfied. Case 2: If $\mathcal{D} = (d_1, d_2, \dots, d_n)$ ordered such that $d_n \ge d_{n-1} \ge \dots \ge d_1$. In this case, according to Theorem 2.2, similarly to the discussion we presented in the previous case for the purpose of avoiding repetition, we find that:

$$ (9) \qquad \sigma(T) \ge \sum_{i=1}^{n} d_i^3 - \sum_{i=2}^{n-1} (d_i + 2)(d_i - 1)^2. $$

Therefore,

$$ (10) \qquad \sigma(T) \ge \sum_{i=1}^{n} d_i^3 - \sum_{i=2}^{n-1} d_i^2 + 2n - 2. $$

Hence, we find that $2n > \Delta$ and according to the term of degree sequence $\mathcal{D}$ we noticed that $\sigma(T) > \sum_{i=2}^{n-1} d_i^2 + 2n - 2$. Thus, we find that $\sigma(T) \le \sigma \odot(T)$. Then, from (9) and (10) the relationship (7) holds. $\square$

Assume $\eta(n)$ be the function discussed the value of $n$ where it defined as $\eta(n) = 2n^2 + n - 2$. Among Proposition 3.2, we observe that through Convention 3.1 for the Sigma index.

**Convention 3.1.** For strictly natural numbers, we have the inequality

$$ (11) \qquad 0 < \sqrt{\eta(n)} \left( \left( \frac{1}{\sqrt{2}} \right)^n + \left( \frac{1}{\sqrt{2}} \right)^m \right) < 10. $$

**Proposition 3.2.** Let $T$ be a tree with degree sequence $\mathcal{D} = (d_1, d_2, \dots, d_n)$ where $d_1 \ge d_2 \ge \dots \ge d_n$. Then, the lower bound of $\sigma(T)$ satisfies

$$ (12) \qquad \sigma(T) \ge \left\lfloor \frac{2(n+m)^2}{3} \right\rfloor - 2 \left( n \sum_{i=1}^{n-1} (d_i - d_{i+1})^2 + \sqrt{m\eta(n)} \right). $$

*Proof.* Assume the degree sequence $\mathcal{D} = (d_1, d_2, \dots, d_n)$ where $d_1 \ge d_2 \ge \dots \ge d_n$, and let $\sigma \odot(T)$ be the average Sigma index for the cases $d_1 \ge d_2 \ge \dots \ge d_n$ and $d_n \ge \dots \ge d_1$. Then, according to Convention 3.1, we find that $2m < \sqrt{m\eta(n)} < m^2$. Thus, we have

$$ (13) \qquad \sigma(T) \ge \left\lfloor \frac{2(n+m)^2}{3} \right\rfloor - \sigma \odot(T) + 2\sqrt{m\eta(n)}. $$

Therefore, let $a, b \in \mathbb{N}$ with $a < b$, satisfying $a < \sigma(T) < b$ and $a < \sigma \odot(T) + 2\sqrt{m\eta(n)} < b$. As a direct consequence, we obtain

$$ \sigma(T) \ge \sigma \odot(T) + 2\sqrt{m\eta(n)}. $$

Then,

$$ (14) \qquad \sigma(T) \ge 2a \left( n \sum_{i=1}^{n-1} (d_i - d_{i+1})^2 + \sqrt{m\eta(n)} \right) + \sigma \odot(T) - 2b\sqrt{m\eta(n)}. $$

Hence, from (13) and (14) we notice that

$$ \sigma(T) \le \left\lfloor \frac{2(n+m)^2}{3} \right\rfloor, $$
and this relationship establishes the lower bound of the Sigma index given by (12). □

Through Figure 1, we denote the Sigma index with order $d_n \ge d_{n-1} \ge \dots \ge d_1$ by $T_1$ and that with order $d_1 \ge d_2 \ge \dots \ge d_n$ by $T_2$. The figure illustrates the optimal impact of the Sigma index behavior by showing how it is directly influenced by the degree sequence when it is either increasing or decreasing. This impact is clearly reflected through $\sigma \odot(T)$ in Propositions 3.1 and 3.2.

<figure><img src="image_4.png" /><figcaption>FIGURE 1. Compare $\sigma(T)$ and $\sigma \odot(T)$.</figcaption></figure>

Study of trees in the case of paths and stars was utilized through Lemma 3.2, where we observe both the upper and lower bounds at both sides of the inequality.

**Lemma 3.2.** Let $T$ be a tree and $\mathcal{D} = (d_1, d_2, \dots, d_n)$ be a degree sequence where $d_1 \ge d_2 \ge \dots \ge d_n$. Then,

$$ (15) \qquad (n-1)\sqrt{\eta(n)} \le \sigma(T) \le (n-1)(2n^2 - n - 2), $$

with equality if and only if $T \cong \mathcal{P}_n$ or $T \cong \mathcal{S}_n$.

*Proof.* Let us consider the degree sequence $\mathcal{D} = (d_1, d_2, \dots, d_n)$, which we assume to be decreasing. We need to prove both sides of the inequality (15). If the path $\mathcal{P}_n$ satisfies the Sigma index condition for $n \ge 3$ as $\sigma(\mathcal{P}_n) = 2$, then we find that the Sigma index yields

$$ \sigma(\mathcal{P}_n) > \sqrt{\eta(n)}. $$

Thus, the left-hand side of the inequality (15) holds. The Sigma index of the star tree is known as $\sigma(\mathcal{S}_n) = (n-1)(n-1)^2$. Since $\eta(n) = 2n^2 + n - 2$, we show that $\sigma(T) > \sqrt{\eta(n)}$. Then,

$$ (16) \qquad \sigma(T) \le 2n^2 + (n-1)(n-m)^2. $$

Therefore, we notice that

$$ 2n^2 + (n-1)(n-m)^2 \le (n-1)(2n^2 - n - 2). $$

Then, according to (16), the right-hand side of the inequality (15) holds. □

Throughout the Lemma 3.3, it involves comparing different bounds, considering how the terms behave asymptotically or approximately (e.g., $2(2n-m)^2 \sim 2m^2$) and also justifies why the degree sequence is considered in the non-increasing order $d_1 \ge \dots \ge d_n$ based on a comparison shown in Figure 1 where one configuration $T_1$ is greater than another $T_2$.
**Lemma 3.3.** Let T be a tree and $\mathcal{D} = (d_1, d_2, \dots, d_n)$ be a degree sequence. Then, the lower bound of Sigma index satisfying

$$ (17) \qquad \sigma(T) \ge \sqrt{\frac{2n}{m-2}} \left( 2m^2 - \sum_{i=1}^{n} d_i^2 - \left[ \frac{1}{2}(d_n - 1)^3 + (d_1 - 1)^3 \right] \right). $$

*Proof.* Assume $\mathcal{D} = (d_1, d_2, \dots, d_n)$ is a degree sequence. According to Convention 3.1, we notice that for $d_1 \ge d_2 \ge \dots \ge d_n$ the following inequality holds:

$$ (18) \qquad 2 \le \frac{d_1 + d_2}{3(d_n + d_{n-1})} \sqrt{\frac{2n}{m-2}} \le 4, $$

where $1 < \sqrt{\frac{2n}{m-2}} < 2$. Therefore, considering the relation (18), the lower bound of the Sigma index is given by

$$ (19) \qquad \sigma(T) \ge 2m^2 + \sqrt{\frac{2n}{m-2}}. $$

Henceforth, we consider the Sigma index with $d_1 \ge d_2 \ge \dots \ge d_n$ instead of $d_n \ge d_{n-1} \ge \dots \ge d_1$ because, as shown in Figure 1, we have $T_1 > T_2$. Thus, we observe that

$$ \begin{aligned} \sigma(T) &\ge 2m^2 - \sum_{i=1}^{n} d_i^2 - \left[ \frac{1}{2}(d_n - 1)^3 + (d_1 - 1)^3 \right] \\ &\ge 2m^2 - \sum_{i=2}^{n-1} d_i^2 - \left[ \frac{1}{2}(d_{n-1} - 1)^3 + (d_2 - 1)^3 \right] + 2n - 2 \\ &\ge 2(2n - m)^2 - \sum_{i=1}^{n+1} d_i^2 - \left[ \frac{1}{2}(d_{n+1} - 1)^3 + (d_1 - 1)^3 \right] + 2 \\ &\ge \sqrt{\frac{2n}{m-2}} \left( 2m^2 - \sum_{i=1}^{n} d_i^2 - \left[ \frac{1}{2}(d_n - 1)^3 + (d_1 - 1)^3 \right] \right) + 2n + 2. \end{aligned} $$

Therefore, from this last lower bound and by considering the term $2n + 2$ as a constant term and the approximation $2(2n - m)^2 \sim 2m^2$, according to (18) and (19), we conclude that (17) holds. This completes the proof. $\square$

**3.1. The Optimal Lower Bound of Sigma Index.** The findings previously discussed in relation to Properties 3.1 and 3.2, along with the theoretical framework that refined these results in Lemmas' 3.2 and 3.3, underscore the critical importance of identifying whether the sequence is increasing or decreasing, as elucidated in Figure 1. Leveraging this established foundation, we aim to present optimal results concerning the lower bound of the sigma index, which poses the primary challenge in achieving the most precise and robust outcome, both methodologically and substantively.

We presented among Theorem 3.4, basic bounds on the Sigma index $\sigma(T)$ satisfied with $\sigma(T) \ge 2n^2$ and $\sigma(T) < \Delta(\Delta - 1)^2$, which imply $\sigma(T) \ge 2n^2 + \Delta(\Delta - 1)^2$.

**Theorem 3.4.** Let T be a tree with n vertices and m edges, let $\mathcal{D} = (d_1, d_2, \dots, d_n)$ be a degree sequence with maximum degree $\Delta$. Then, the lower bound of Sigma index satisfying

$$ (20) \qquad \sigma(T) \ge \frac{2nm\sqrt{2n^2 - nm + 2}}{\Delta(\Delta - 1)^2} + \left\lfloor \frac{2(n + m)^3}{3} \right\rfloor - 2n^2 + \Delta(\Delta - 1)^2 + 2. $$
Proof. Let us consider the decreasing degree sequence $\mathcal{D} = (d_1, d_2, \dots, d_n)$, with $n$ vertices, $m$ edges, and maximum degree $\Delta$. Clearly, $\sigma(T) \ge 2n^2$ and $\sigma(T) < \Delta(\Delta - 1)^2$. Thus, we observe that

$$ (21) \qquad \sigma(T) \ge 2n^2 + \Delta(\Delta - 1)^2. $$

Assume $\mathcal{T}_{n,\Delta}$ is the class of trees where $T \in \mathcal{T}_{n,\Delta}$. Then, let $d_T(v_\ell)$ be the degree of vertex $v_\ell$, with degrees ordered as $d_T(v_1) \le d_T(v_2) \le \dots \le d_T(v_\ell) \le \dots \le d_T(v_{n-1}) \le d_T(v_n)$. It follows that

$$ (22) \qquad \sigma(T) \ge 2d_T(v_\ell)^2 + \Delta(d_T(v_n) - d_T(v_1))^2 + 2n - 2. $$

For $d_T(v_1) \ge 3$, we observe that

$$ \begin{aligned} \sigma(T) &\ge \left(d_T(v_1)^2 + \sum_{i=1}^n d_T(v_i)^3\right)^2 + 2d_T(v_\ell)^2 - \\ &\quad \left(\sum_{i=2}^{n-1} (d_T(v_n) - d_T(v_2))^2 + \sum_{i=2}^{n-1} (d_T(v_{n-1}) - d_T(v_3))^2 + \dots + (d_T(v_3) - d_T(v_2))^2\right) \\ &\ge \left(d_T(v_1)^2 + \sum_{i=1}^n d_T(v_i)^3\right)^2 + 2d_T(v_\ell)^2 - \sum_{i=2}^{n-1} \sum_{j=i+1}^n (d_T(v_j) - d_T(v_i))^2 \\ &\ge \left(d_T(v_1)^2 + \sum_{i=1}^n d_T(v_i)^3\right)^2 + 2d_T(v_\ell)^2 - \sum_{2 \le i \le j \le n} (d_T(v_j) - d_T(v_i))^2 \\ &\ge \left(d_T(v_1)^2 + \sum_{i=1}^n d_T(v_i)^3\right)^2 + 2d_T(v_\ell)^2 - m \sum_{i=2}^n d_T(v_i)^2 - \left(\sum_{i=2}^n d_T(v_i)\right)^2. \end{aligned} $$

Hence, from (22), it follows that $m \le \sqrt{2n^2 - nm + 2} \le n$. Since $\Delta(\Delta - 1)^2 < 2nm \le \sigma(T)$, we deduce that

$$ (23) \qquad \sigma(T) \ge \frac{2nm\sqrt{2n^2 - nm + 2}}{\Delta(\Delta - 1)^2} + \frac{2n^2}{m}. $$

Given that

$$ 2nm\sqrt{2n^2 - nm + 2} \ge \left\lfloor \frac{2(n+m)^3}{3} \right\rfloor, $$

we observe that the lower bound for $\sigma(T)$ from (23) satisfies

$$ (24) \qquad \sigma(T) \ge \left\lfloor \frac{2(n+m)^3}{3} \right\rfloor - \frac{2n^3}{m}. $$

Finally, for $T \in \mathcal{T}_{n,\Delta}$ with $d_T(v_1) \ge 3$, we obtain that

$$ (25) \qquad \sigma(T) \ge d_T(v_1)^2 + \sum_{i=1}^n d_T(v_i)^3 - m \sum_{i=2}^n d_T(v_i)^2. $$

From (21) through (25), the inequality (20) is validated. This completes the proof. $\square$

**Corollary 3.5.** Let $\mathcal{T}_{n,\Delta}$ be the class of trees, for a tree $T \in \mathcal{T}_{n,\Delta}$ according to Theorem 3.4 satisfying

$$ (26) \qquad \sigma(T) \ge \left( \sum_{i=1}^{n} d_T(v_i)^3 \right)^2 + 2(d_T(v_n) + d_T(v_1))^2 - m \sum_{i=2}^{n-1} d_T(v_i)^2 - \left( \sum_{i=2}^{n-1} d_T(v_i) \right)^2. $$
The characteristic results discussed in Theorem 3.6 focus on establishing lower and upper bounds for the Sigma index $\sigma(T)$ of trees $T$ in the class $\mathcal{T}_{n,\Delta}$. These bounds combine to explicitly characterize the strict lower bound of the Sigma index using the term in the original Theorem 3.4, equation (27).

**Theorem 3.6.** Let $T$ be a tree with $n$ vertices and $m$ edges, let $\mathcal{D} = (d_1, d_2, \dots, d_n)$ be a degree sequence with maximum degree $\Delta$. Then, the lower bound of Sigma index satisfying

$$ (27) \quad \sigma(T) \ge \left( d_T(v_1)^2 + \sum_{i=1}^n d_T(v_i)^3 \right)^2 + \sqrt{\frac{5(n+m)^4}{\Delta(\Delta-1)^2}} - \sum_{i=1}^n \left( \frac{d_T(v_1)^2 + 3n!}{2^n(m-1)!} \right) - \Delta(\Delta-n)^2. $$

*Proof.* Recall some information from Theorem 3.4. Let $\mathcal{T}_{n,\Delta}$ be the class of trees such that $T \in \mathcal{T}_{n,\Delta}$. Let $d_T(v_\ell) \ge 3$ be the degree of vertex $v_\ell$, where $d_T(v_1) \le d_T(v_2) \le \dots \le d_T(v_\ell) \le \dots \le d_T(v_{n-1}) \le d_T(v_n)$. Assume the decreasing degree sequence $\mathcal{D} = (d_1, d_2, \dots, d_n)$, with $n$ vertices, $m$ edges, and maximum degree $\Delta$. Then,

$$ (28) \quad \sigma(T) \ge \left( d_T(v_1)^2 + \sum_{i=1}^n d_T(v_i)^3 \right)^2 - m \sqrt{\frac{5(n+m)^4}{\Delta(\Delta-1)^2}} - \left( \sum_{i=2}^n d_T(v_i) \right)^2 + 2d_T(v_\ell), $$

by considering the term

$$ \sqrt{\frac{5(n+m)^4}{\Delta(\Delta-1)^2}} \le \sum_{i=2}^n d_T(v_i)^2 - 2d_T(v_\ell)^2. $$

Therefore, from (28), the Sigma index satisfies $\sigma(T) \le \Delta(\Delta - n)^2$ and $nm \le \Delta(4\Delta - n)^2$. We notice that

$$ (29) \quad \sigma(T) \ge \left( d_T(v_1)^2 + \sum_{i=1}^n d_T(v_i)^3 \right)^2 - \left( \frac{2nm}{3} + \Delta(4\Delta - n)^2 \right). $$

From another perspective, since the condition $d_n \ge d_{n-1} \ge \dots \ge d_1$, which also holds for the reverse order $d_1 \ge d_2 \ge \dots \ge d_n$, we observe that the optimal improvement of the lower bound of the Sigma index must achieve an optimal improvement of these lower bounds. From this, it follows that $5(n+m)^4 \ge \Delta(4\Delta - n)^2$ and

$$ (30) \quad m \sum_{i=2}^n d_T(v_i)^2 + \left( \sum_{i=2}^n d_T(v_i) \right)^2 \le d_T(v_1)^2 + \sum_{i=1}^n d_T(v_i)^3 + 2d_T(v_\ell)^2. $$

Hence, from (29) and (30), we find that

$$ (31) \quad \sigma(T) \le \left( d_T(v_1)^2 + \sum_{i=1}^n d_T(v_i)^3 \right)^2 + \sqrt{\frac{5(n+m)^4}{\Delta(\Delta-1)^2}}. $$

Thus, we need to prove the lower bound of the Sigma index given in (27). According to other terms in (27), we notice that

$$ (32) \quad \sum_{i=1}^n \frac{d_T(v_1)^2 + 2n!}{2^n(m-1)!} \le nd_T(v_1)^2 + \frac{2nm^2 + d_T(v_1)^2}{2^n}. $$

Therefore, we have

$$ (33) \quad \sigma(T) \ge \left( d_T(v_1)^2 + \sum_{i=1}^n d_T(v_i)^3 \right)^2 - nd_T(v_1)^2 + \frac{2nm^2 + d_T(v_1)^2}{2^n}. $$
Thus, these convergent bounds of the Sigma index contribute to strictly obtaining the lower bound. Then, we have

$$ (34) \qquad \sigma(T) \ge \left( d_T(v_1)^2 + \sum_{i=1}^n d_T(v_i)^3 \right)^2 - \left( \sum_{i=1}^n \left( \frac{d_T(v_1)^2 + 3n!}{2^n(m-1)!} \right) + \Delta(\Delta - n)^2 \right). $$

Finally, according to (33) and (34), we notice that the lower bound of the Sigma index strictly holds from the term (27). □

**3.2. Behavior Lower Bound of Sigma Index.** The study of the behavior of any topological index is considered highly important due to several factors, including determining the bounds that are closest to this index. From this perspective, we review through Theorems 3.7 and 3.8 the optimal behavior of the Sigma index according to the given minimum bound. Through Properties 4 and 5, we attempted to present the optimal minimum bound for the Sigma index, which effectively contributes to determining the innovative behavior of the minimum bound of the Sigma index. This is considered the real challenge in understanding the direction of this index and its impact.

**Theorem 3.7.** Let $T$ be a tree with $n$ vertices and $m$ edges, let $\mathcal{D} = (d_1, d_2, \dots, d_n)$ be a degree sequence with maximum degree $\Delta$. Then, according to Theorem 3.6 the lower bound of the Sigma index satisfies

$$ (35) \qquad \sigma(T) \ge \sqrt{\sum_{i=1}^{n} d_i^3 - \sum_{i=2}^{n-1} d_i^2 + \left\lfloor \frac{2n-1}{2} \right\rfloor \left\lfloor \frac{3n-2}{3} \right\rfloor + \frac{1}{2} n \Delta}. $$

**Theorem 3.8.** Let $T$ be a tree with $n$ vertices and $m$ edges, let $\mathcal{D} = (d_1, d_2, \dots, d_n)$ be a degree sequence with maximum degree $\Delta$. Then, the lower bound of the Sigma index satisfies

$$ (36) \qquad \sigma(T) \ge \frac{1}{2}\eta(T) + \frac{2n^4 + n^3 - (m-1)^2 + n - 2}{3m^2 - 2n + 2} + \sqrt{\left\lfloor \frac{2n-1}{2} \right\rfloor \left\lfloor \frac{3n-2}{3} \right\rfloor}. $$

*Proof.* Let $\mathcal{D} = (d_1, d_2, \dots, d_n)$ be a degree sequence with maximum degree $\Delta$. Then, among the functions $\eta(T)$, we find that $\sigma(T) \ge \eta(T)$. Thus, the inequality is valid for $\sigma(T) > \eta(T)/2$. The Sigma index satisfies $\sigma(T) \le 2n^4 + n^3 - (m-1)^2 + n - 2$ and $\sigma(T) \le 3m^2 - 2n + 2$. Thus,

$$ (37) \qquad \sigma(T) \ge \frac{2n^4 + n^3 - (m-1)^2 + n - 2}{3m^2 - 2n + 2}, $$

where this relationship satisfies

$$ \frac{2n^4 + n^3 - (m-1)^2 + n - 2}{3m^2 - 2n + 2} \ll \frac{1}{2}\eta(T). $$

Therefore, according to Theorem (3.7), we notice that

$$ (38) \qquad \left\lfloor \frac{2n-1}{2} \right\rfloor \left\lfloor \frac{3n-2}{3} \right\rfloor \le n(n-\Delta)^2 + 2n + 2. $$

Thus, from (37) and (38) we find that

$$ (39) \qquad \sigma(T) \ge nm + \sqrt{n(n-\Delta)^2 + 2n + 2}. $$

Hence, from (39) we establish that $\sigma(T) \le \frac{1}{2}\eta(T)+nm$, which implies that $nm+\sqrt{n(n-\Delta)^2 + 2n + 2} \ge \eta(T)/2$, and the lower bound of the Sigma index yields

$$ (40) \qquad \sigma(T) \ge \frac{1}{2}\eta(T) + nm - \frac{2n^4 + n^3 - (m-1)^2 + n - 2}{3m^2 - 2n + 2} + \sqrt{\left\lfloor \frac{2n-1}{2} \right\rfloor \left\lfloor \frac{3n-2}{3} \right\rfloor}. $$
Actually, according to (40), we find that for (36) the equivalent relation for the term related to the square root achieves $3m^2 - 2n + 2 \le \sqrt{n(n-1)}$, and here we notice that

$$ (41) \qquad \sigma(T) \ge \frac{1}{2}\eta(T) + \frac{2n^4 + n^3 - (m-1)^2 + n - 2}{3m^2 - 2n + 2} + \sqrt{n(n-1)}. $$

Finally, according to (37)–(41), we find that the lower bound of the Sigma index satisfies $\sigma(T) \ge \sqrt{n(n-1)}$. Thus, the relation (36) holds. $\square$

The following corollaries provide lower bounds for the Sigma index $\sigma(T)$ of a tree $T$, building upon the main results established in Theorems 3.7 and 3.8.

**Corollary 3.9.** For any tree $T$ according to Theorems 3.7 and 3.8 the lower bound of Sigma index satisfy

$$ (42) \qquad \sigma(T) \ge d_1^3 + d_n^3 + \sqrt{n(m-1) + \eta(T)} + \frac{1}{2}n\Delta. $$

This result refines the lower bound for $\sigma(T)$ by directly incorporating boundary degree terms $d_1^3 + d_n^3$, which reflect the influence of extremal degrees on $\sigma(T)$.

**Corollary 3.10.** According to Theorems 3.7, 3.8 and Corollary 3.9, the lower bound of Sigma index satisfy

$$ (43) \qquad \sigma(T) \ge \frac{2n^4 + n^3}{3m^2 - 2n + 2} + \sqrt{n(m-1) + \eta(T)} + \frac{1}{2}n\Delta. $$

## 4. THE EFFICIENCY OF UPPER BOUND

In this section, Lemma 4.1 gives us the sharp upper bound for the Sigma index, which significantly $2n(m-1)^2 \sim 2n^3$.

**Lemma 4.1.** Let $T$ be a tree, and $\mathcal{D} = (d_1, \dots, d_n)$ be a degree sequence where $d_1 \ge d_2 \ge \dots \ge d_n$. Then, the upper bound of Sigma index is $\sigma(T) \le 2n(m-1)^2$.

*Proof.* Assume $\mathcal{D} = (d_1, \dots, d_n)$ be a degree sequence where $d_1 \ge d_2 \ge \dots \ge d_n$ with the maximum degree $\Delta$ of $T$. Then, according to Theorem 3.8 the Sigma index satisfy $\sigma(T) \le 2n^4 + n^3 - (m-1)^2 + n - 2$ and $\sigma(T) \le 3m^2 - 2n + 2$. Thus, the sharp upper bound of Sigma index satisfied with the term $2(n+m)^2$. Therefore, let $k \ge 2$ and $k \in \mathbb{N}$, we need to show that $\sigma(T) \le 2nk + (km-1)^2$. In this case, clearly $\sigma(T) > 2nk$ where we consider $2 \le k \le n - \Delta$. On the other hand, we notice that

$$ (44) \qquad \sigma(T) \le \sqrt{2k + n^2} + (km - 1)^2. $$

Thus, the equation (44) established that $\sqrt{2k + n^2} \le (km - 1)^2$. Thus, the upper bound of Sigma index is $\sigma(T) \le (km - 1)^2$. $\square$

Let $\mathcal{D} = (d_1, \dots, d_n)$ be a degree sequence satisfying $d_1 \le d_2 \le \dots \le d_n$ had employed for Lemma 4.2. According to Lemma 4.1, if the degrees satisfy the reverse ordering $d_1 \ge d_2 \ge \dots \ge d_n$, then the $\sigma(T)$ of the tree $T$ is bounded in Lemma 4.1 by $2n(m-1)^2 \sim 2n^3$.

**Lemma 4.2.** Let $T$ be a tree, and $\mathcal{D} = (d_1, \dots, d_n)$ be a degree sequence where $d_1 \le d_2 \le \dots \le d_n$. Then, the upper bound of Sigma index is

$$ (45) \qquad \sigma(T) \le 2(d_n + d_1)^2 + \sum_{i=1}^{n} d_i(d_i - 1)^2 + 2(n + m)^3. $$
Proof. Assume $\mathcal{D} = (d_1, \dots, d_n)$ is a degree sequence where $d_1 \le d_2 \le \dots \le d_n$. Then, according to Lemma 4.1, we find that when $d_1 \ge d_2 \ge \dots \ge d_n$, the upper bound of the Sigma index is $\sigma(T) \le 2n(m-1)^2$. Thus, we notice that

$$ (46) \qquad 2(d_n + d_1)^2 + \sum_{i=1}^{n} d_i(d_i - 1)^2 \le 2n(m-1)^2. $$

Then, equation (46) implies that

$$ (47) \qquad \sigma(T) \le 2(d_n + d_1)^2 + \sum_{i=1}^{n} d_i(d_i - 1)^2 + 2n(m-1)^2. $$

Therefore, by considering the term $2n(m-1)^2$ satisfying $2n(m-1)^2 \ll 2(n+m)^3$, we find that

$$ \begin{align*} \sigma(T) &\le 2(d_n + d_1)^2 + \sum_{i=1}^{n} d_i(d_i - 1)^2 + 2n(m-1)^2 \\ &\le 2(d_{n-1} + d_2)^2 + d_n^2 + d_1^2 + \sum_{i=2}^{n-1} d_i(d_i - 1)^2 + 2n(m-2)^2 \\ &\le 2(d_{n+1} + d_1)^2 - (d_{n-1} + d_2)^2 + \sum_{i=2}^{n+1} d_i(d_i - 1)^2 + 2n(m+1)^2. \end{align*} $$

Hence, according to the term $2n(m-1)^2 \ll 2(n+m)^3$ and equation (47), the upper bound of the Sigma index satisfies

$$ (48) \qquad \sigma(T) \le 2(d_n + d_1)^2 + 2(n+m)^3. $$

Thus, from (48), we note that (45) holds. $\square$

Let $\mathcal{P}(n)$ be a function of vertices where $\mathcal{P}(n) = n(n+m-\Delta)^2$. Then, we noticed that $\sigma(T) \ge \sqrt{\mathcal{P}(n)}$, where among Theorem 4.3 had presented the upper bound of Sigma index.

**Theorem 4.3.** Let $T$ be a tree, and $\mathcal{D} = (d_1, \dots, d_n)$ be a degree sequence. Then, the upper bound of Sigma index is

$$ (49) \qquad \sigma(T) \le 2(d_n + d_1)^2 + \sum_{i=1}^{n} d_i(d_i - 1)^2 + \frac{1}{2}\eta(T) + \sqrt{\mathcal{P}(n)} $$

Proof. Assume $\mathcal{D} = (d_1, \dots, d_n)$ is a degree sequence with maximum degree $\Delta$. Then, $\sigma(T) \gg \frac{\eta(T)}{2}$ and $\sigma(T) \gg \sqrt{\mathcal{P}(n)}$. Thus, considering $\Delta(\Delta - 1)^2$ as the constant term, we find that

$$ (50) \qquad \sigma(T) \le \frac{1}{2}\eta(T) + \sqrt{\mathcal{P}(n)} + \Delta(\Delta - 1)^2. $$

In fact, to discuss the remaining bounds, we need to specify the degree sequence in the cases when it is decreasing and when it is increasing, which we demonstrate by discussing each case separately as follows.

Case 1: If $d_1 \le d_2 \le \dots \le d_n$, then the Sigma index satisfies $\sigma(T) \gg 2(d_n + d_1)^2$. Thus, we find that

$$ (51) \qquad \sigma(T) \le 2(d_n + d_1)^2 + \sum_{i=1}^{n} d_i(d_i - 1)^2 + \Delta(\Delta - 1)^2. $$
Since $\sigma(T) \ge \sqrt{\mathcal{P}(n)}$, and according to (50) the upper bound of the Sigma index is $\sigma(T) \le \sqrt{\mathcal{P}(n)} + 2(d_n + d_1)^3 + \Delta(\Delta - 1)^2$. Thus, we notice that

$$ (52) \qquad \sigma(T) \le \sqrt{\mathcal{P}(n)} + 2(d_n + d_1)^3 + \sum_{i=1}^{n} (d_i(d_i - 1)^2). $$

Now, let us generate a sequence according to the terms $d_1 \le d_2 \le \dots \le d_n$ as $\mathcal{A}_{se} = (d_n - d_{n-1}, d_{n-1} - d_{n-2}, \dots, d_2 - d_1)$, such that $\mathcal{A}_{se} = (a_1, a_2, \dots, a_m)$. Then, $\sigma_{\mathcal{G}}(T) \ge \sigma_{\mathcal{A}_{se}}(T)$. Thus, according to (52), we notice that

$$ (53) \qquad \sigma(T) \le \sigma_{\mathcal{A}_{se}}(T) + \sqrt{\mathcal{P}(n)} + 2(d_n + d_1)^3. $$

Actually, from (51) to (53), we find that when $d_1 \le d_2 \le \dots \le d_n$, the upper bound of the Sigma index (49) holds.

Case 2: If $d_1 \ge d_2 \ge \dots \ge d_n$, then the Sigma index satisfies $\sigma(T) \ge 2(d_n + d_1)^2 + 2(d_{n-1} + d_2)^2 + d_n^2 + d_1^2$. Thus, we find that

$$ (54) \qquad \sigma(T) \le d_n^2 + d_1^2 + \sum_{i=n}^{2} (d_i - d_{i-1})^2 + \Delta(\Delta - 1)^2. $$

In this case, we can employ the previous discussion from Case 1 to complete this case by referring back to Figure 1, which shows the closeness of the Sigma index between both cases. Here, we note that

$$ (55) \qquad \sigma(T) \le \sum_{i=n}^{2} (d_i - d_{i-1})^2 + \frac{1}{2}\eta(T). $$

Thus, according to (54) and (55), the relationship of bounds among the Sigma index yields

$$ \sum_{i=n}^{2} (d_i - d_{i-1})^2 + \frac{1}{2}\eta(T) \ge 2(d_n + d_1)^2 + 2(d_{n-1} + d_2)^2 + d_n^2 + d_1^2 + \sqrt{\mathcal{P}(n)}. $$

Then, we notice that

$$ (56) \qquad \sigma(T) \le \sum_{i=1}^{n} d_i^3 + \sum_{i=n}^{2} (d_i - d_{i-1})^2 + \frac{1}{2}\eta(T). $$

Therefore, the relationship (56) establishes the upper bound of the Sigma index such that

$$ (57) \qquad \sigma(T) \le \sum_{i=1}^{n} d_i(d_i - 1)^2 + \frac{1}{2}\eta(T) + \sqrt{\mathcal{P}(n)} + \sum_{i=n}^{2} (d_i - d_{i-1})^2. $$

Since $\sum_{i=1}^{n} d_i^3 \ge 2(d_n + d_1)^2$ and $\sum_{i=1}^{n} d_i^3 \ge 2(d_n + d_1)^2 + 2(d_{n-1} + d_2)^2 + d_n^2 + d_1^2$, we notice that the relationship (49) holds. The inequality (49) is confirmed to hold, establishing the validity of these bounds on the Sigma index $\sigma(T)$. $\square$

**4.1. The Optimal Behavior of Upper Bound on Sigma Index.** Throughout the Theorem 4.4, it establishes upper bounds for the Sigma index $\sigma(T)$ of a tree with degree sequence $\mathcal{D}$. It shows that among equation (58) and provides refined inequalities involving parameters $n, m$, and $k$. It confirms that for $2 \le k \le 10$, the bounds hold and relate to the growth of $\sigma(T)$ with respect to $k$ and $(n, m)$.
**Theorem 4.4.** Let $T$ be a tree, and $\mathcal{D} = (d_1, \dots, d_n)$ be a degree sequence. Then, the upper bound of Sigma index is

$$ (58) \qquad \sigma(T) \le \sum_{i=n}^{i=2} (d_i - d_{i-1})^3 + \left\lfloor \frac{2(n+m)}{3} \right\rfloor + \left\lceil \frac{3m^2}{4} \right\rceil + \frac{3n^2 + nm - 2}{2(2m-3)^2}. $$

*Proof.* Assume $\mathcal{D} = (d_1, \dots, d_n)$ is a degree sequence. According to Theorem 4.3, applying equation (56) to the term $\sum_{i=n}^{i=1} (d_i - d_{i-1})^3$, we find that

$$ (59) \qquad \sigma(T) \le \sum_{i=n}^{i=2} (d_i - d_{i-1})^3 + \sum_{i=1}^{n} d_i^3 + \Delta(\Delta - 1)^2. $$

Now, let us discuss the remaining bounds in relation to (58), where the degree sequence is not specified as increasing or decreasing; we will handle it as required. Since $m = n - 1$, we note that $2(n+m) \le \Delta(\Delta-1)^2$ and $3m^2 \le \Delta(\Delta-1)^2$. Thus,

$$ (60) \qquad \sigma(T) \le \left\lfloor \frac{2(n+m)}{3} \right\rfloor + \left\lceil \frac{3m^2}{4} \right\rceil + \Delta(\Delta - 1)^2. $$

Let $k > 1$ be an integer. We emphasize the relationship in (61), where by replacing the term $\eta(T)$ with $\Delta(\Delta - 1)^2$, we need to confirm the inequality for the upper bound of the Sigma index as

$$ (61) \qquad \sigma(T) \le k^2 \left\lfloor \frac{2(n+m)}{3} \right\rfloor + k \left\lceil \frac{3m^2}{4} \right\rceil + 2nk - 2. $$

Thus, we observe that

$$ (62) \qquad k^2 \left\lfloor \frac{2(n+m)}{3} \right\rfloor + k \left\lceil \frac{3m^2}{4} \right\rceil + 2nk - 2 \le \left\lfloor \frac{2(nk)^2}{3} \right\rfloor + \left\lceil \frac{3km^2}{4} \right\rceil + 2(n+m)k. $$

Therefore, according to (62), it follows that for $2 \le k \le 10$, the inequality

$$ (63) \qquad \sigma(T) \le \left\lfloor \frac{2(nk)^2}{3} \right\rfloor + \left\lceil \frac{3km^2}{4} \right\rceil + 2(n+m)k $$

holds. Hence, from (62) and (63), we confirm that the relationship in (61) holds. Moreover, according to (59) and (60), we see that

$$ (64) \qquad \sigma(T) \le \sum_{i=n}^{i=2} (d_i - d_{i-1})^3 + \left\lfloor \frac{2(n+m)^2}{3} \right\rfloor + \left\lceil \frac{3m^2}{4} \right\rceil. $$

Finally, for the last term $3n^2 + nm - 2$, we have $3n^2 + nm - 2 < 2(2m - 3)^2$. Thus, from (64), we conclude that the relationship in (58) holds. $\square$

**Theorem 4.5** establishes an upper bound by the relationship (65) showing the Sigma index's dependence on degree increments and structural constants.

**Theorem 4.5.** Let $T$ be a tree, and $\mathcal{D} = (d_1, \dots, d_n)$ be a degree sequence. Then, the upper bound of Sigma index is

$$ (65) \qquad \sigma(T) \le (d_1 - 1)^3 + (d_n - 1)^3 + \sum_{i=n}^{i=2} d_i (d_i - d_{i-1})^3 + 2n^2 + 3m - 2. $$

*Proof.* Assume $\mathcal{D} = (d_1, \dots, d_n)$ is a degree sequence. According to Theorems 4.3 and 4.4, the lower bound of the Sigma index satisfies

$$ (66) \qquad \sigma(T) \ge (d_1 - 1)^3 + (d_n - 1)^3 + \sum_{i=2}^{n} d_i (d_i - d_{i-1})^3. $$
Now, we need to prove that (65) holds with respect to the term $2n^2 + 3m - 2$ by considering the lower bound (66). Assume the maximum degree is $\Delta$, where $\sigma(T) \gg \Delta(\Delta - 1)^2$. Then, clearly, $\Delta(\Delta - 1)^2 \le 2n^2$. Thus,

$$ (67) \qquad \sigma(T) \le 2n^2 + \Delta(\Delta - 1)^2. $$

Hence, for the terms involving $d_1$ and $d_n$, we have $(d_1 - 1)^3 + (d_n - 1)^3 \le \Delta(\Delta - 1)^2$ and therefore $\sigma(T) \le (d_1 - 1)^3 + (d_n - 1)^3 + 2n^2 + \Delta(\Delta - 1)^2$. Thus,

$$ (68) \qquad \sigma(T) \le 2n^2 + \sum_{i=1}^{n} (d_i - 1)^3. $$

From (67) and (68), we observe

$$ \begin{aligned} \sigma(T) &\le 2n^2 + \sum_{i=1}^{n} d_i (d_i - 1)^3 \\ &\le 2n^2 + \sum_{i=2}^{n-1} d_i (d_i - 1)^3 + d_1^3 + d_n^3 \\ &\le (d_1 - 1)^3 + (d_n - 1)^3 + \sum_{i=2}^{n} d_i (d_i - d_{i-1})^3 + 2n^2. \end{aligned} $$

Accordingly, we note that the optimal behavior of the Sigma index associated with the vertices yields

$$ (69) \qquad \sigma(T) \le \sum_{i=2}^{n} d_i (d_i - d_{i-1})^3 + 2n^2 + 3m - 2. $$

Thus, according to (69) and considering the constant term $(d_1 - 1)^3 + (d_n - 1)^3$, the relationship (65) holds. $\square$

## 5. NOVELTY ON SIGMA INDEX

Through this section, we will review some novelty concepts on the Sigma index, where these concepts are represented by the reduced Sigma index given by (3), the increasing Sigma index, and the modified Sigma index. These indices are currently discussed on caterpillar trees, which we will extend to include different trees and graphs in future studies.

5.1. On Reduced Sigma Index. To emphasize that the reduced Sigma index through this subsection. For example, among Figure 2, we presented the reduced Sigma index according to (3) for the tree with degree sequence $\mathcal{D} = (4, 3, 2, 2, 2, 1, 1, 1, 1, 1)$ is 54.

<figure><img src="image_5.png" /><figcaption>FIGURE 2. Tree with degree sequence $\mathcal{D} = (4, 3, 2, 2, 2, 1, 1, 1, 1, 1)$.</figcaption></figure>

Example 1: Let $T$ be a tree with $n = 20$ vertices given among Figure 3. Through this figure we improvement the concept of reduced Sigma index where $\sigma_{red}(T) = (106, 98, 98, 84, 82, 60)$.
<figure><img src="image_6.png" /><figcaption>FIGURE 3. The reduced Sigma index for n = 20 vertices.</figcaption></figure>

**Proposition 5.1.** Let T be a tree with degree sequence $\mathcal{D} = (d_1, d_2, \dots, d_n)$. Then, the lower bound of $\sigma_{red}(T)$ satisfy

$$ (70) \qquad \sigma_{red}(T) \geq \sum_{i=1}^{n} (d_i - 1)^3 - \sum_{i=1}^{n-1} \left[ \frac{2d_i}{3d_{i+1}} \right] - \left\lfloor \frac{3n^2}{4} \right\rfloor. $$

*Proof.* Immediately according to Proposition 3.1 by considering $\sigma_{red}(T) \geq \sigma \odot (T)$. $\square$

**Proposition 5.2.** Let T be a tree with degree sequence $\mathcal{D} = (d_1, d_2, \dots, d_n)$. Then, the lower bound of $\sigma_{red}(T)$ satisfy

$$ (71) \qquad \sigma_{red}(T) \geq \sum_{i=1}^{n} (d_i - 1)^4 - 2n^2 + 3m - 2. $$

*Proof.* Recall the degree sequence $\mathcal{D} = (d_1, d_2, \dots, d_n)$ with the maximum degree $\Delta$. Then, we noticed that $\sigma_{red}(T) \geq \Delta(\Delta - 1)^2$. Then, according to Theorem 3.6 we have

$$ (72) \qquad \sigma_{red}(T) \geq \left( d_T(v_1)^2 + \sum_{i=1}^{n} d_T(v_i)^3 \right)^2 - \sum_{i=1}^{n} \left( \frac{d_T(v_1)^2 + 3n!}{2^n(m-1)!} \right) - \Delta(\Delta - n)^2. $$

Thus,

$$ \begin{align*} \sigma_{red}(T) &\geq \sum_{i=1}^{n} (d_i - 1)^4 - \sum_{uv \in E(T)} \left[ (d_T(u) - 1)^2 + (d_T(v) - 1)^2 \right] \\ &\geq \sum_{i=1}^{n} (d_i - 1)^2 (d_i^2 - 3d_i + 1) + \Delta(\Delta - 1)^2 \\ &\geq \sum_{i=n}^{n} d_i (d_i - d_{i-1})^3 - 2n^2 + \Delta(\Delta - 1)^2 \\ &\geq \sum_{i=1}^{n} d_i^3 - \sum_{i=2}^{n-1} (d_i - d_{i-1})^3 + (d_1 - 1)^3 + (d_n - 1)^3 - 2n^2 + \Delta(\Delta - 1)^2 \\ &\geq \sum_{i=1}^{n} (d_i - 1)^4 - 2n^2 + 3m - 2. \end{align*} $$

<figure><img src="image_7.png" /></figure>
**Theorem 5.1.** Let $T$ be a tree with degree sequence $\mathcal{D} = (d_1, d_2, \dots, d_n)$. Then, the lower bound of $\sigma_{red}(T)$ satisfy

$$ (73) \qquad \sigma_{red}(T) \ge 2n^2 + \sum_{i=1}^{n} d_i (d_i - 1)^3 - \frac{2nm\sqrt{2n^2 - nm + 2}}{\Delta(\Delta - 1)^2} + \frac{2(n+m)^2}{m}. $$

*Proof.* Since $\sigma(T) \ge 2n^2$ and $\sigma(T) < \Delta(\Delta - 1)^2$. Thus, we find that $\sigma(T) \ge \sigma_{red}(T)$ according to Theorem 3.4 the reduced Sigma index satisfies $\sigma_{red}(T) \ge (d_1 - 1)^3 + (d_n - 1)^3$ and Proposition 5.2,

$$ (74) \qquad \sigma_{red}(T) \le 2n^2 + \Delta(\Delta - 1)^2. $$

Thus, when $m \le \sqrt{2n^2 - nm + 2} \le n$. Then, the term holds the relationship $\Delta(\Delta - 1)^2 < 2nm \le \sigma_{red}(T)$. Thus,

$$ (75) \qquad \sigma_{red}(T) \ge \frac{2nm\sqrt{2n^2 - nm + 2}}{\Delta(\Delta - 1)^2} + \frac{2(n+m)^2}{m}. $$

Therefore, from (74) and (75) we noticed that

$$ (76) \qquad \frac{2nm\sqrt{2n^2 - nm + 2}}{\Delta(\Delta - 1)^2} + \frac{2(n+m)^2}{m} \ge 2n^2 $$

Thus, let $\theta(n)$ be the sharp bound of $\sigma_{red}(T)$ satisfied with (76), since $\Delta \to \infty$, the first term of (76) tends to zero, so the maximum is attained at small $\Delta$. For $2nm \ll (n+m)^2$, the term $\sqrt{2n^2 - nm + 2} \sim \sqrt{n^2 + 2}$, so the first term roughly behaves such that

$$ (77) \qquad \frac{2nm\sqrt{2n^2 - nm + 2}}{\Delta(\Delta - 1)^2} \sim \frac{2n^2\sqrt{n^2 + 2}}{\Delta(\Delta - 1)^2}. $$

Thus, according to (76) and (77) for large $n$ with $2nm \ll (n+m)^2$ we noticed that the sharp bound $\theta(n)$ satisfies

$$ (78) \qquad \frac{2nm\sqrt{2n^2 - nm + 2}}{\Delta(\Delta - 1)^2} + \frac{2(n+m)^2}{m} - 2n^2 \sim n^3 $$

Therefore, according to (78) by considering the relationship $\sigma_{red}(T) \ge (d_1 - 1)^3 + (d_n - 1)^3$, we find that

$$ (79) \qquad \sigma_{red}(T) \ge \sum_{i=1}^{n} d_i (d_i - 1)^3 - n^3. $$

Finally, from (74)–(79) we find that the relationship (73) holds. $\square$

**Example 2:** Let $T$ be a tree with $n = 30$ vertices given among Figure 4. Through this figure we improvement the concept of reduced Sigma index where $\sigma_{red}(T) = (206, 192, 148, 126, 118, 118, \dots)$.
<figure><img src="image_8.png" /><figcaption>FIGURE 4. The reduced Sigma index for n = 30 vertices.</figcaption></figure>

5.2. **Novelty on Sigma Index for Future Works.** Let $\alpha > 1$, according to (4) the general increase Sigma index $\sigma_{inc}^{\alpha}(T)$ of a tree $T$ defined as

$$ (80) \qquad \sigma_{inc}^{\alpha}(T) = \sum_{uv \in E(T)} [(d_T(u) + 1)^{\alpha} + (d_T(v) + 1)^{\alpha}]. $$

According to (5), the general modified Sigma index $\sigma_{\alpha}^{*}(T)$ of a tree $T$ defined as

$$ (81) \qquad \sigma_{\alpha}^{*}(T) = \sum_{uv \in E(T)} \left[ \frac{1}{(d_T(u) + 1)^{\alpha} + (d_T(v) + 1)^{\alpha}} \right]. $$

Parameters of Sigma index had provided among (4) and (80) holds Theorem 5.2 for all caterpillar trees.

**Theorem 5.2.** Let $\mathcal{D} = (d_1, \dots, d_n)$ be a degree sequence with $n \ge 3$, such that $d_n > d_1 > d_{n-2} > \dots > d_3 > d_2 > d_{n-1}$. Then, the caterpillar tree with this ordering attains the maximum value of the increase Sigma index $\sigma_{inc}(T)$ among all caterpillar trees whose path vertices have the degree sequence $\mathcal{D}$.

Also, parameters of Sigma index had provided among (5) and (81) holds Theorem 5.3 the maximum value for all caterpillar trees.

**Theorem 5.3.** Let $\mathcal{D} = (d_1, \dots, d_n)$ be a degree sequence with $n \ge 3$, such that $d_n > d_1 > d_{n-2} > \dots > d_3 > d_2 > d_{n-1}$. Then, the caterpillar tree with this ordering attains the maximum value of the modified Sigma index $\sigma^{*}(T)$ among all caterpillar trees whose path vertices have the degree sequence $\mathcal{D}$.

## 6. CONCLUSION

Through this paper, we examined the lower bound of the Sigma index $\sigma(T)$ for trees characterized by specified degree sequences. It established exact and optimal lower bounds considering both non-increasing and non-decreasing degree arrangements. The central findings, encapsulated in Propositions 3.1 and 3.2 alongside the enhanced inequalities detailed in Lemmas 3.2 and 3.3, emphasize the importance of how the degree sequence ordering influences the precision of the bounds. Building on these results, the developed theorems and corollaries deliver a comprehensive description of the Sigma index's lower bounds by taking into account the maximum degree and other structural aspects of the tree.
Further insights into the asymptotic and minimal behavior of the Sigma index are provided by Theorems 3.7 and 3.8, underscoring key characteristics valuable for ongoing and future studies. Through section 4, Lemma 4.1 gives us the sharp upper bound for the Sigma index, which significantly $2n(m-1)^2 \sim 2n^3$. Theorem 4.5 establishes an upper bound by the relationship

$$ \sigma(T) \le (d_1 - 1)^3 + (d_n - 1)^3 + \sum_{i=n}^{i=2} d_i (d_i - d_{i-1})^3 + 2n^2 + 3m - 2. $$

These contributions refine previous estimates and clarify the role of extreme degree values in determining the Sigma index.

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**JASEM HAMOUD** DEPARTMENT OF DISCRETE MATHEMATICS, MOSCOW INSTITUTE OF PHYSICS AND TECHNOLOGY, DOLGOPRUDNY 141700, RUSSIA.

Email address: hamoud.math@gmail.com