# Note on Sharp Bounds for $\sigma$-Irregularity in Caterpillar Trees with Prescribed Degree Sequences

Jasem Hamoud

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Jasem Hamoud. Note on Sharp Bounds for $\sigma$-Irregularity in Caterpillar Trees with Prescribed Degree Sequences. 2026. hal-05441010

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Note on Sharp Bounds for σ-Irregularity in Caterpillar Trees with
Prescribed Degree Sequences

Jasem Hamoud*

Moscow Institute of Physics and Technology, Russia, 141701, Moscow, Institutskii per. 9, Dolgoprudnyi
ORCID:0009-0002-0192-3627

Abstract

Determining sharp bounds for degree-based irregularity indices is a central problem in mathematical chemistry and extremal graph theory, with applications in QSAR/QSPR studies for predicting molecular properties. This paper examines the Albertson index and the Sigma index in trees, focusing on caterpillar trees $\mathbb{C}$ with given degree sequences $\mathcal{D} = (d_1, d_2, \dots, d_n)$. We derive sharp lower and upper bounds for the Sigma index, leveraging its relation to the Albertson index. Lower bounds characterize stable, low- irregularity structures, while upper bounds quantify maximal irregularity in highly asymmetric graphs; the quadratic Sigma index proves more sensitive in extreme cases. A fundamental result establishes that the degree sequence fully determines irregularity values in caterpillar trees, yielding closed-form expressions, precise extremal bounds, and characterizations of extremal trees. These results deepen the theoretical understanding of graph irregularity and improve predictive modeling of chemical structures.

**Keywords:** Irregularity index, Caterpillar tree, Degree Sequences, QSAR/QSPR.

# 1 Introduction

Let $G = (V, E)$ be a simple graph, where $V = V(G)$ is the vertex set known as $V(G) = \{v_1, v_2, \dots, v_n\}$, $|V(G)| = n$ and $E = E(G)$ is the edge set known as $E(G) = \{e_1, e_2, \dots, e_m\}$, $|E(G)| = m$. For two vertices $u, v$ of $G$ are adjacent, or neighbours, if $uv \in E(G)$. Let $d_G(u)$ be the degree of vertex $u \in V(G)$ where it is the number of neighbors of $u$ in $G$. If $d_G(v_1) = d_G(v_2) = \dots = d_G(v_n)$, the graph $G$ said regular, otherwise it is said to be irregular. Let $\text{dis}_G(u, v)$ be the distance of pair vertices $(u, v) \in V(G)$. An irregularity measure [16] of $G$ well-known is the Albertson index [1] denoted by $\text{irr}$ and it defined as

$$ \text{irr}(G) = \sum_{uv \in E(G)} |d_G(u) - d_G(v)| \qquad (1) $$

This measure is not the only possible( see extremely irregular trees [7]). In fact, R. Criado et al. [3] emphasize that irregularity index is

$$ \text{irr}(G) = \frac{1}{n} \sum_{i=1}^{n} \left| d_i - \frac{2m}{n} \right| $$

The extended of Albertson index so-called σ- irregularity [8] and it known as a quadratic irregularity measure defined as

$$ \sigma(G) = \sum_{uv \in E(G)} (d_G(u) - d_G(v))^2 \qquad (2) $$

Zagreb indices appear in most books on topological indices and their applications in QSPR and QSAR [16, 14, 4]. As many authors employing Zagreb indices and other topological indices draw on concepts and terminology

<footnote>*Corresponding author: jasem1994hamoud@gmail.com</footnote>
from chemical graph theory [15]. Molecular graphs are obtained by representing atoms as vertices and bonds as edges [2]. The first(second, resp.) Zagreb index [8, 13] of graph G are defined as

$$M_1(G) = \sum_{v \in V(G)} d_G(v)^2, \quad M_2(G) = \sum_{uv \in E(G)} d_G(u) \cdot d_G(v). \qquad (3)$$

The *Forgotten index* $F(G)$ [8] of graph G is defined as

$$F(G) = \sum_{v \in V(G)} d_G(v)^3. \qquad (4)$$

To determine the unicyclic graphs on $n$ vertices with the minimum value of the max-min degree index [12], it should be focus on investigate the extremal graphs with respect to this index among all chemical unicyclic graphs and, separately, among all chemical trees. A caterpillar tree is a tree in which all vertices lie within distance 1 of a central path subgraph. Equivalently, there exists a path, called the spine or stalk, such that every other vertex is adjacent to it (i.e., pendant leaves attached to the path) [6].

## 2 Preliminaries

Let $T = \mathbb{C}(n, m)$ be a tree contains $n$ vertices and $m$ pendant vertices. Then, $\mathbb{C}(n, m)$ is a caterpillar tree. Let $\mathbb{C}(n, 3)$ be a caterpillar tree illustrations by Figure 1. The degree of $v_1$ denote by $d_1$, the degree of $v_2$ denote by $d_2$, ... and the degree of $v_n$ denote by $d_n$.

<figure><img src="image_2.png" /><figcaption>Figure 1: Caterpillar tree $\mathbb{C}(n, 3)$.</figcaption></figure>

Furthermore, we study the extremal caterpillar tree [10]. In next figure 2, the top configuration the minimal value. In the bottom the maximal value.

<figure><img src="image_3.png" /><figcaption>Figure 2: Extremal caterpillars corresponding to the degree sequence $\mathcal{D} = (6^3, 5^4, 4^5, 3^2, 1^{34})$ are considered.</figcaption></figure>

**Proposition 2.1** ([8]). Let G be a graph, according to (3) and (4) the $\sigma$- irregularity satisfy

$$\sigma(G) = F(G) - 2M_2(G). \qquad (5)$$

A new concept of the irregularity among Albertson index had presented in caterpillar tree for a degree sequence $\mathcal{D} = (d_1, \dots, d_n)$,

$$\mathrm{irr}_{\mathcal{D}}(\mathbb{C}) = (d_n - 1)^2 + (d_1 - 1)^2 + \sum_{i=2}^{n-1} (d_i - 1)(d_i - 2) + \sum_{i=1}^{n-1} |d_i - d_{i+1}|. \qquad (6)$$

Note that $\mathrm{irr}_{\mathcal{D}}(\mathbb{C}) \ge \mathrm{irr}(G)$, during this study of the Sigma index, we deviate from the traditional definition (2) and adopt the formula derived from (6) Theorem 2.2, which is expressed in terms of the degree sequence $\mathcal{D}$ (furthermore see [9]).
**Theorem 2.2.** Let $\mathbb{C}$ be a caterpillar tree on $n + np$ vertices whose spine is the path $v_1v_2\dots v_n$. Let $d_n \ge d_{n-1} \ge \dots \ge d_1 \ge 2$ be given integers and $p > 1$ a given nonnegative integer such that each spine vertex $v_k$ ($k = 1, \dots, n$) has exactly $p$ pendant leaves attached. The degree of $v_k$ in $\mathbb{C}$ is then $d_{\mathbb{C}}(v_k) = d_k + p$. The Sigma index of $\mathbb{C}$ is

$$ \sigma(C) = \sum_{i=1}^{n-1} (d_{i+1} - d_i)^2 + p \sum_{k=1}^{n} (d_k + p - 1)^2. \qquad (7) $$

Also, during Theorem 2.2 we find that $\sigma_{\mathcal{D}}(\mathbb{C}) \ge \sigma(T)$.

## 2.1 Statement of Problem

This work discuss the main question: What is the relationship between Albertson index and $\sigma$- irregularity over $\mathcal{D}$? This relationship had established many bounds for each index. To determine the sharp bound for each index we should be find the sharp bound for both indices as we discussed by Proposition 3.1 and emphasize that by Corollary 3.2.

## 3 Main Result

The first setp by observing the relationship over $\sigma$- irregularity had presented by Proposition 3.1 and emphasize that by Corollary 3.2.

**Proposition 3.1.** Let $\mathbb{C}$ be a caterpillar tree with degree sequence $\mathcal{D}$ and $\sigma_{\mathcal{D}}(\mathbb{C})$ be the $\sigma$- irregularity over $\mathcal{D}$. Then,

$$ \mathrm{irr}(T) \le \mathrm{irr}_{\mathcal{D}}(\mathbb{C}) \le \sigma(T). \qquad (8) $$

*Proof.* Assume $\mathcal{D} = (d_1, \dots, d_n)$ be a degree sequence such that $d_n \ge \dots \ge d_1$. Let $x, y$ be a two vertex of $V(T)$ where $d_T(x) \ne d_T(y)$. Then, if there are a path $\mathcal{P} = xx_1\dots y_1y$. Then, according to [5] satisfy $d_T(x) \ge d_T(y)$ while $d_T(x_1) < d_T(y_1)$. Let $\lambda_{xy}$ be an two edges $xy_1, yx_1$ and $\beta_{xy}$ be an two edges $xx_1$ and $y_1y$. Assume $T'$ be the tree obtained from $T$ by omitted $\beta_{xy}$ and add $\lambda_{xy}$. Then,

$$ \sigma(T) - \sigma(T') = 2[(d_T(x) - d_T(y))(d_T(x_1) - d_T(y_1))]. \qquad (9) $$

Note that, according to (9) the relationship yields $\sigma(T) - \sigma(T') > 0$.

Now, we need to prove the relationship (8) in both side. Thus,

**Case 1.** In this case, we should be prove the left hand from the relationship (8). Thus, according to (9) we noticed that $\mathrm{irr}(T) - \mathrm{irr}(T') > 0$. From another hand, the Albertson index over $\mathcal{D}$ satisfy $\mathrm{irr}_{\mathcal{D}}(\mathbb{C}) - \mathrm{irr}_{\mathcal{D}}(\mathbb{C}') > 0$. Therefore,

$$ \begin{aligned} \mathrm{irr}_{\mathcal{D}}(\mathbb{C}) - \mathrm{irr}(T) &= (d_n - 1)^2 + (d_1 - 1)^2 + \sum_{i=2}^{n-1} (d_i - 1)(d_i - 2) \\ &\ge \sum_{i=2}^{n-1} (d_i - 2)^2 \\ &> 0. \end{aligned} $$

**Case 2.** In this case, we should be prove the right hand from the relationship (8). Thus, the $\sigma$- irregularity over $\mathcal{D}$ satisfy $\sigma_{\mathcal{D}}(\mathbb{C}) - \sigma_{\mathcal{D}}(\mathbb{C}') > 0$. Therefore,

$$ \begin{aligned} \sigma(T) - \mathrm{irr}_{\mathcal{D}}(\mathbb{C}) &= \sum_{uv \in E(G)} (d_G(u) - d_G(v))^2 - \left( (d_n - 1)^2 + (d_1 - 1)^2 + \sum_{i=2}^{n-1} (d_i - 1)(d_i - 2) + \sum_{i=1}^{n-1} |d_i - d_{i+1}| \right) \\ &\ge \sum_{uv \in E(G)} (d_G(u) - d_G(v))^2 - \left( \sum_{i=2}^{n-1} (d_i - 2)^2 + \mathrm{irr}(T) \right) \\ &> 0. \end{aligned} $$

□
Due to the importance of caterpillar trees in establishing these bounds, both indices attain their respective sharp bounds. By correlating the two indices, we obtain the following result directly.

**Corollary 3.2.** Let $\mathbb{C}$ be a caterpillar tree with degree sequence $\mathcal{D}$ and $\sigma_{\mathcal{D}}(\mathbb{C})$ be the $\sigma$-irregularity over $\mathcal{D}$. Then, according to Proposition 3.1,

$$ \mathrm{irr}_{\mathcal{D}}(\mathbb{C}) \le \sigma(T) \le \sigma_{\mathcal{D}}(\mathbb{C}). \qquad (10) $$

Actually, the proof of Corollary 3.2 immediately holds by considering $\mathrm{irr}_{\mathcal{D}}(\mathbb{C}) \le \sigma_{\mathcal{D}}(\mathbb{C})$. Then, the relationship (10) satisfied with both side. The following proposition verifies the aforementioned relationship. We leverage this property to establish the connection between the Albertson index and the Sigma index via the upper bound on their minimum values.

**Proposition 3.3.** Let $\mathbb{C}$ be a caterpillar tree of order $n$ with degree sequence $\mathcal{D} = (d_1, d_2, \dots, d_n)$. Then, the upper bound of the minimum $\sigma$-irregularity satisfies

$$ \sigma_{\mathrm{min},\mathcal{D}}(\mathbb{C}) \le \mathrm{irr}_{\mathrm{min},\mathcal{D}}(\mathbb{C}) + 2(d_{n-1} + d_n) + \sum_{i=1}^{n-2} |d_i - d_{i+2}|, \qquad (11) $$

where $\mathrm{irr}_{\mathrm{min},\mathcal{D}}(\mathbb{C})$ is the minimum value of Albertson index over $\mathcal{D}$.

*Proof.* Let $\mathcal{D} = (d_1, d_2, \dots, d_n)$ be a degree sequence where $d_n \ge d_{n-1} \ge \dots \ge d_1$, then according to (6) the Albertson index satisfy $\mathrm{irr}_{\mathrm{min},\mathcal{D}}(\mathbb{C}) \le \mathrm{irr}_{\mathcal{D}}(\mathbb{C}) \le \mathrm{irr}_{\mathrm{max},\mathcal{D}}(\mathbb{C})$. Then,

$$ \mathrm{irr}_{\mathrm{min},\mathcal{D}}(\mathbb{C}) \le \sum_{i=2}^{n-1} (d_i - 1)(d_i - 2)^2 + \sum_{i=1}^{n-1} |d_i - d_{i+1}|. \qquad (12) $$

Thus, according to Corollary 3.2 we find that $\mathrm{irr}_{\mathcal{D}}(\mathbb{C}) \le \sigma(T)$. Then, for the degrees $d_n$ and $d_{n-1}$ it implies that $(d_n - 1)^2 + (d_{n-1} - 1)^2 \le (d_1 + d_2 - 1)^2$ and $(d_{n-2} - 1)^2 + (d_{n-1} - 1)^2 \le (d_2 + d_3 - 1)^2$. Thus, according to (12) we find that

$$ \sum_{i=2}^{n-1} (d_i - 1)(d_i - 2)^2 + \sum_{i=1}^{n-1} |d_i - d_{i+1}| \le \sum_{i=2}^{n-1} (d_i + 2)(d_i - 1)^2 + \sum_{i=2}^{n-1} (d_i - d_{i+1})^2 \qquad (13) $$

Hence,

$$ \mathrm{irr}_{\mathrm{min},\mathcal{D}}(\mathbb{C}) \le \sum_{i=2}^{n-1} (d_i - 1)(d_i - 2)^2 + \sum_{i=2}^{n-1} (d_i - d_{i+1})^2 \qquad (14) $$

Now, from (13) and (14) it established that for $\sigma$-irregularity where it satisfies

$$ \sigma_{\mathrm{min},\mathcal{D}}(\mathbb{C}) \le \sum_{i=2}^{n-1} (d_i - 1)(d_i - 2)^2 + \sum_{i=1}^{n-1} |d_i - d_{i+1}| + \sum_{i=2}^{n-1} (d_i + 2)(d_i - 1)^2 + \sum_{i=2}^{n-1} (d_i - d_{i+1})^2 \qquad (15) $$

Therefore, from (12)–(15) the relationship (11) holds. $\square$

Determining precise bounds for topological indices remains a central and highly active topic in mathematical chemistry and graph theory. When only loose (non-sharp) upper or lower bounds are available, predictions may suffer from large error margins, rendering the model practically ineffective. In contrast, sharp bounds ensure the narrowest possible interval containing the index of any graph, leading to significantly more accurate predictions and improved correlation coefficients in QSAR/QSPR analyses (see [11]).
# 4 The Sharp Lower Bound of σ-irregularity

In this section, the sharp lower bounds establish precise structural limits, refine predictions, and guide molecular optimization in mathematical chemistry. We presented among Lemma 4.1 the sharp bound of σ-irregularity by considering the relationship between σ-irregularity and Albertson index.

**Lemma 4.1.** Let $\mathbb{C}$ be a caterpillar tree with degree sequence $\mathcal{D} = (d_1, d_2, \dots, d_n)$, the relationship between σ-irregularity and Albertson index satisfy

$$ \sigma_{\mathcal{D}}(\mathbb{C}) \geq \text{irr}_{\mathcal{D}}(\mathbb{C}) + \sum_{i=2}^{n-1} [(d_i - 1)(d_i - 2) - (d_i + 2)(d_i - 1)^2] + \sum_{i=1}^{n-1} (d_i - 1)(d_{i+2} - 1)^2. \quad (16) $$

*Proof.* Recall both topological indices -the Albertson index among (6) and the σ-irregularity among Theorem 2.2- over degree sequence $\mathcal{D} = (d_1, d_2, \dots, d_n)$. Then, by considering Corollary 3.2 holds $\text{irr}_{\mathcal{D}}(\mathbb{C}) \leq \sigma(T) \leq \sigma_{\mathcal{D}}(\mathbb{C})$. Thus, let $uv \in E(T)$ where $u \neq v$, the σ-irregularity satisfy

$$ \begin{aligned} \sigma_{\mathcal{D}}(\mathbb{C}) - \sigma(T) &= \sum_{i \in \{1, n\}} (d_i + 1)(d_i - 1)^2 + \sum_{i=2}^{n-1} (d_i + 2)(d_i - 1)^2 + \\ & \quad + \sum_{i=2}^{n-1} (d_i - d_{i+1})^2 + 2n - 2 - \sum_{uv \in E(G)} (d_G(u) - d_G(v))^2 \\ &= \sum_{i \in \{1, n\}} (d_i + 1)(d_i - 1)^2 + \sum_{i=2}^{n-1} (d_i + 2)(d_i - 1)^2 + 2n - 2 \\ & \geq d_G(u)^2 + d_G(v)^2 + 2d_G(u)d_G(v) |d_G(u) - d_G(v)| + \sum_{i=2}^{n-1} (d_i + 2)(d_i - 1) + 2n - 2 \\ & \geq d_G(u)^2 + d_G(v)^2 + \sum_{i=2}^{n-1} (d_i + 2)(d_i - 1)^3 + 2n - 2. \end{aligned} $$

Hence, it satisfies $\sigma(T) \geq \sqrt{\sigma_{\mathcal{D}}(\mathbb{C}) + 2n - 2}$. Thus, according to (6) holds

$$ \sigma_{\mathcal{D}}(\mathbb{C}) - \text{irr}_{\mathcal{D}}(\mathbb{C}) = 2(d_1 + d_n) + 2n - 4 \quad (17) $$

Therefore, by considering both relationship $\sigma_{\mathcal{D}}(\mathbb{C}) - \sigma(T)$ and $\sigma_{\mathcal{D}}(\mathbb{C}) - \text{irr}_{\mathcal{D}}(\mathbb{C})$, for determine the sharp bound among σ-irregularity it satisfies $\sigma_{\mathcal{D}}(\mathbb{C}) > \sigma(T) + \text{irr}_{\mathcal{D}}(\mathbb{C})$. Now, consider there are a vertex $w \in V(T)$ where $d_T(u) \leq d_T(w) \leq d_T(v)$. If $uw$ is neighbour by $wv$. Then,

$$ \sigma_{\mathcal{D}}(\mathbb{C}) \geq \text{irr}_{\mathcal{D}}(\mathbb{C}) + 2(d_T(u) + d_T(v)) + (d_T(u) - d_T(v))^2 + (d_T(w) - d_T(v))^2. \quad (18) $$

These vertices, which exhibit distinct degrees, clearly demonstrate the correlation between the Albertson Index and the significant values realised by the Sigma Index, which is undeniably a fundamental target centred on determining the most extreme bounds of this index, whether upper or lower. If the degrees $d_T(u)^2 + d_T(v)^2 + 2d_T(w)^2 \leq \text{irr}_{\mathcal{D}}(\mathbb{C}) \leq \sqrt{\sigma_{\mathcal{D}}(\mathbb{C}) + 2n - 2}$ and $\text{irr}_{\mathcal{D}}(\mathbb{C}) > 2(d_1 + d_n) + 2n - 4$. It implies that according to (17) the relationship

$$ \sigma_{\mathcal{D}}(\mathbb{C}) \geq \sum_{i=1}^{n-1} (d_i - 1)(d_{i+2} - 1)^2 + 2(d_1 + d_n) + 2n - 4. \quad (19) $$

Therefore, from (19) yields the relationship $\sigma_{\mathcal{D}}(\mathbb{C}) \geq \text{irr}_{\mathcal{D}}(\mathbb{C}) + 2n - 4$. It is crucial to identify the specific tree class to target for effective molecular of sharp bound. Thus, the sharp upper bound of σ-irregularity is

$$ \sigma_{\mathcal{D}}(\mathbb{C}) \geq \text{irr}_{\mathcal{D}}(\mathbb{C}) + \sum_{i=1}^{n-1} (d_i - 1)(d_{i+2} - 1)^2 + 2n - 4. \quad (20) $$

Consequently, according to relationships (17)-(20) we find that the sharp bound of σ-irregularity is (16) holds. □
For example, let $\mathcal{D}_1 = (4, 8, 12, 14)$, $\mathcal{D}_2 = (5, 10, 15, 18)$ and $\mathcal{D}_3 = (7, 13, 19, 23)$ be a three degree sequences. Then, we find that $\text{irr}_{\mathcal{D}_1}(\mathbb{C}) = 340$, $\sigma_{\mathcal{D}_1}(\mathbb{C}) = 4806$, $\text{irr}_{\mathcal{D}_2}(\mathbb{C}) = 572$, $\sigma_{\mathcal{D}_2}(\mathbb{C}) = 9956$ and $\text{irr}_{\mathcal{D}_3}(\mathbb{C}) = 974$, $\sigma_{\mathcal{D}_3}(\mathbb{C}) = 20962$. Then, the relationship between those cases implies that $\text{irr}(\mathcal{D}_2 - \mathcal{D}_1) < \text{irr}(\mathcal{D}_3 - \mathcal{D}_2)$ and $\sigma(\mathcal{D}_2 - \mathcal{D}_1) < \sigma(\mathcal{D}_3 - \mathcal{D}_2)$. This results emphasize that Lemma 4.1.

We present the lower bound for the $\sigma$-irregularity in the following theorem. This bound depends on several factors, including whether $\mathcal{D} = (d_1, d_2, \dots, d_n)$ is increasing, decreasing, or has randomly positioned elements, which enables us to derive corresponding lower or upper bounds based on $\mathcal{D} = (d_1, d_2, \dots, d_n)$ arrangement.

**Theorem 4.2.** Let $\mathbb{C}$ be a caterpillar tree with degree sequence $\mathcal{D} = (d_1, d_2, \dots, d_n)$ where $d_n \ge d_{n-1} \ge \dots d_2 \ge d_1$ and maximum degree $\Delta$. Then, the $\sigma$-irregularity satisfy

$$ \sigma_{\mathcal{D}}(\mathbb{C}) \ge (d_n - 1)^2 + (d_1 - 1)^2 + \frac{1}{n} \sum_{i=1}^{n-1} d_i (d_{i+1} - 1)^2 + \Delta (\Delta - 1)^2 \quad (21) $$

*Proof.* Recall the degree sequence $\mathcal{D} = (d_1, d_2, \dots, d_n)$ where $d_n \ge d_{n-1} \ge \dots d_2 \ge d_1$ and maximum degree $\Delta$. If the tree $T$ contains only one internal vertex, then $n \le \Delta + 1$. Therefore, the condition $n \ge \Delta + 2$ ensures that $T$ has at least two distinct internal vertices. Let $u$ and $v$ be two vertex where $u \ne v$ with the greatest distance between them in $T$, where $d_T(u) \ge d_T(v)$. Then, for $n$ vertices and $m$ edge satisfy $\Delta (\Delta - 1)^2 > nm$. Then

$$ \sigma_{\mathcal{D}}(\mathbb{C}) \ge d_T(u)^2 + d_T(v)^2 + \frac{1}{n} \sum_{i=1}^{n-1} d_i (d_{i+1} - 1)^2 + \Delta (\Delta - 1)^2 - 2d_T(u)d_T(v). \quad (22) $$

Therefore, it implies that $\sigma_{\mathcal{D}}(\mathbb{C}) > d_T(u)^2 + d_T(v)^2 + 2n - 2$, by considering the maximum and minimum degrees $\delta$ and $\Delta$, respectively. Then, according to (22) satisfied with $\delta \Delta \sqrt{(\Delta - 1)^3} \le nm$. Thus, if $T \cong K_{\delta, \Delta}$ holds when $m < \sqrt{(\delta \Delta)/(\Delta - 1)^3}$. Then, the lower bound of the $\sigma$-irregularity satisfy

$$ \sigma_{\mathcal{D}}(\mathbb{C}) \ge \frac{1}{n} \sum_{i=1}^{n-1} d_i (d_{i+1} - 1)^2 + \delta \Delta \sqrt{(\Delta - 1)^3}. \quad (23) $$

Thus, from both equations (22) and (23) we noticed that $\sigma_{\mathcal{D}}(\mathbb{C}) \ge \sigma(T) + d_T(u)^2 + d_T(v)^2 + 2n - 2$. This bound determines the behaviour of $\sigma$-irregularity over $\mathcal{D}$. Hence, it satisfies

$$ \sigma_{\mathcal{D}}(\mathbb{C}) \ge \frac{1}{n} \sum_{i=1}^{n-1} (d_{i+1} - d_i)^3 + \delta \Delta \sqrt{(\Delta - 1)^3} + 2n - 2. \quad (24) $$

Since the term of degree sequence $\mathcal{D}$ satisfy $(d_n - 1)^2 + (d_1 - 1)^2 \ge 2n - 2$. Hence, by considering the relationship $\sigma_{\mathcal{D}}(\mathbb{C}) \ge \sqrt{(\Delta - 1)^3} + nm + 2n - 2$ for terms $d_1$ and $d_n$ with order $d_n \ge d_{n-1} \ge \dots d_2 \ge d_1$ where $\max(d_n - 1)^2, (d_1 - 1)^2 \ge d_1 + d_n$ and $\sigma_{\mathcal{D}}(\mathbb{C}) > \text{irr}_{\mathcal{D}}(\mathbb{C}) + (d_n - 1)^2 + (d_1 - 1)^2 + 2n - 2$. Then,

$$ \sigma_{\mathcal{D}}(\mathbb{C}) \ge (d_n - 1)^2 + (d_1 - 1)^2 + \frac{1}{n} \sum_{i=1}^{n-1} (d_{i+1} - d_i)^3 + \delta \Delta \sqrt{(\Delta - 1)^3} \quad (25) $$

Finally, from (22)–(25) we find that (21) holds $\square$

All the results obtained to study the impact of the behavior of both topological indices are mutually interrelated. In fact, by means of Theorem 4.2, this interrelation has been clarified through equations (22)–(25).

**Theorem 4.3.** Let $\mathbb{C}$ be a caterpillar tree with degree sequence $\mathcal{D} = (d_1, d_2, \dots, d_n)$ where $d_n \ge d_{n-1} \ge \dots d_2 \ge d_1$. Consider $\alpha = \sqrt{n} - 4$ be an integer. Then, the $\sigma$-irregularity satisfy

$$ \sigma_{\mathcal{D}}(\mathbb{C}) \ge \sum_{i=1}^{n-1} (d_i + 1)^\alpha \sqrt{(d_{i+1} - 1)^\alpha} - 2(d_1 + d_n)^\alpha + 2n - 2. \quad (26) $$

*Proof.* Recall the degree sequence $\mathcal{D} = (d_1, d_2, \dots, d_n)$ where $d_n \ge d_{n-1} \ge \dots d_2 \ge d_1$. Assume $\alpha$ be an integer where $2 \le \alpha \le \sqrt{n}$. Then, according to the relationship of $\sigma$-irregularity over degree sequence $\mathcal{D} = (d_1, d_2, \dots, d_n)$ we should be prove that

$$ \sigma_{\mathcal{D}}(\mathbb{C}) \ge 2(d_1 + d_n)^\alpha + 2n - 2. \quad (27) $$
**Case 1:** By using induction mathematics, for $\alpha = 2$ satisfy $\sigma_{\mathcal{D}}(\mathbb{C}) \ge 2(d_1 + d_n)^2 + 2n - 2$. Then, according Theorem 4.2 it follows the relationship $\sigma_{\mathcal{D}}(\mathbb{C}) \ge (d_n - 1)^2 + (d_1 - 1)^2 + 2n - 2$. Thus (27) is true for $\alpha = 2$. Hence, assume it true for $\alpha$ and we should be prove it true for $\alpha + 1$. Thus,

$$2(d_1 + d_n)^{\alpha+1} - 2(d_1 + d_n)^{\alpha} = 2(d_1 + d_n)^{\alpha}(d_1 + d_n - 1). \qquad (28)$$

Therefore, the term $d_1+d_n \le n$, it implies that $\sigma_{\mathcal{D}}(\mathbb{C}) \ge 2n(d_1+d_n)^{\alpha}$. Thus according to (28) we find that (27) is true for $\alpha + 1$.

**Case 2:** In this case, by considering the relationship (27) among **Case 1** we should be to prove that

$$\sigma_{\mathcal{D}}(\mathbb{C}) \le \sum_{i=1}^{n-1} (d_i + 1)^{\alpha} \sqrt{(d_{i+1} - 1)^{\alpha}}. \qquad (29)$$

Actully, for $\alpha = 2$ we noticed that $(d_1+1)^2(d_2-1)^2 \le (d_2+1)^2(d_3-1)^2$ and $(d_2+1)^2(d_3-1)^2 \le (d_3+1)^2(d_4-1)^2$. Then, $\sigma_{\mathcal{D}}(\mathbb{C}) \le (d_1+1)^2(d_2-1)^2 + (d_2+1)^2(d_3-1)^2 + \dots + (d_{n-1}+1)^2(d_n-1)^2$. Thus (29) is true for $\alpha = 2$. Hence, assume it true for $\alpha$ and we should be prove it true for $\alpha + 1$. Thus, the relationship that

$$\sum_{i=1}^{n-1} (d_i + 1)^{\alpha+1} \sqrt{(d_{i+1} - 1)^{\alpha+1}} - \sum_{i=1}^{n-1} (d_i + 1)^{\alpha} \sqrt{(d_{i+1} - 1)^{\alpha}} \qquad (30)$$

yields

$$\sum_{i=1}^{n-1} (d_i + 1)^{\alpha} \sqrt{(d_{i+1} - 1)^{\alpha}} ((d_i + 1)\sqrt{d_{i+1} - 1} - 1).$$

Therefore, the relationship (30) telescope that

$$\sum_{i=1}^{n-1} ((d_i + 1)\sqrt{d_{i+1} - 1} - 1) \ge 3n - 2. \qquad (31)$$

Thus, we noticed that from (30) and (31) the relationship (29) holds.

**Case 3:** In this case, we observe that the difference among both relationships (27) and (29) satisfied with (26). Thus, we find that

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

Therefore, from **Case 1**, **Case 2** and **Case 3** we find that (26) holds. □

Enhancing the results obtained in Theorem 4.2 through Theorem 4.3 is illustrated by Figure 3, where we demonstrate how the behaviour of $\sigma$-irregularity on $n = 100$ vertices is affected by random values obtained for $\alpha$ in order to link the previous relations to each other. This correlation strengthens the study we presented for $\mathcal{D} = (d_1, d_2, \dots, d_n)$ according to condition $d_n \ge d_{n-1} \ge \dots d_2 \ge d_1$.
<figure><img src="image_4.png" /><figcaption>Figure 3: The impact of α value (5.3 and 3) on σ-irregularity on n = 100 vertices.</figcaption></figure>

**Theorem 4.4.** Let $\mathbb{C}$ be a caterpillar tree with degree sequence $\mathcal{D} = (d_1, d_2, \dots, d_n)$ where $d_n \le d_{n-1} \le \dots d_2 \le d_1$. Then, the σ-irregularity satisfy

$$ \sigma_{\mathcal{D}}(\mathbb{C}) \ge (d_1 - 1)^2 + (d_n - 1)^2 + \sum_{i=1}^{n} d_i^2 + \left\lfloor \frac{2n^2}{3} \right\rfloor + \frac{1}{2}(n^2 + nm - 2). \quad (32) $$

*Proof.* Assume $T$ be a tree of order $n$ with degree sequence $\mathcal{D} = (d_1, d_2, \dots, d_n)$ where $d_n \le d_{n-1} \le \dots d_2 \le d_1$. Then, for vertices set $V(T) = \{v_1, v_2, \dots, v_n\}$ we define $N(v) = \{v_1, v_2, \dots, v_k\}$ all neighbours of $v$ and $n \le k$. It follows that $\sigma$-irregularity over $\mathcal{D} = (d_1, d_2, \dots, d_n)$ satisfy $\sigma_{\mathcal{D}}(\mathbb{C}) \le (n+m)^2$. Thus we noticed that $(d_1 - 1)^2 + (d_n - 1)^2 < n^2$ and $(d_1 - 1)^2 + (d_n - 1)^2 < nm$. Thus,

$$ \sigma_{\mathcal{D}}(\mathbb{C}) \ge (d_1 - 1)^2 + (d_n - 1)^2 + \sum_{i=1}^{n} d_i^2 + nm. \quad (33) $$

Let $u$ and $v$ be two vertex where $u \ne v$ with the greatest distance between them where $d_T(u) \ge d_T(v)$ and $d_T(u) \ne d_1$ or $d_T(v) \ne d_n$, $d_T(v) \ne d_1$ or $d_T(v) \ne d_n$. Then, for the maximum degree $\Delta$ in $T$ satisfied with $\sigma_{\mathcal{D}}(\mathbb{C})$ by the relationship

$$ \sigma_{\mathcal{D}}(\mathbb{C}) \ge (d_1 - 1)^2 + (d_n - 1)^2 + \sum_{i=1}^{n} d_i^2 + (\Delta + 4)^2 + d_T(u)^2 + d_T(v)^2. \quad (34) $$

From both equations (33) and (34) we find that $\sigma_{\mathcal{D}}(\mathbb{C}) \ge (d_1 - 1)^2 + (d_n - 1)^2 + n^2 + nm - 2$. Therefore, the relationship between terms in (32) as

$$ \frac{1}{2}(n^2 + nm - 2) - \left\lfloor \frac{2n^2}{3} \right\rfloor \le (d_1 - 1)^2 + (d_n - 1)^2. $$

Thus, when $n > \Delta$ and $2n^2 > \Delta(\Delta - 1)^2$ it follows $n^2 + nm - 2 > \Delta(\Delta - 1)^2$. Then, for $\sigma_{\mathcal{D}}(\mathbb{C})$ we find that

$$ \sigma_{\mathcal{D}}(\mathbb{C}) \ge \left\lfloor \frac{2n^2}{3} \right\rfloor + \sqrt{\frac{1}{2}(n^2 + nm - 2) + \Delta(\Delta - 1)^2}. \quad (35) $$

Thus, from (33)-(35) we find that (32) holds. $\square$

Through the discussion we presented in Theorem 4.4 for $\mathcal{D} = (d_1, d_2, \dots, d_n)$ given according to the condition $d_n \le d_{n-1} \le \dots d_2 \le d_1$, it became clear to us that this order is related in determining the upper bounds that were discussed through relationship (32). For example, let $\mathcal{D} = (d_1, d_2, \dots, d_n)$ be the degree sequence given among Table 1, where we show the variation of effect of these degrees on behaviour of σ-irregularity.
<table><caption>Table 1: The impact of $\mathcal{D} = (d_1, d_2, \dots, d_n)$.</caption><thead><tr><th>$$ \mathcal{D} = (d_1, d_2, \dots, d_n) $$</th><th>$$ \sigma(T) $$</th><th>$$ \text{Theorm 4.2} $$</th><th>$$ \text{Theorm 4.3} $$</th><th>$$ \text{Theorm 4.4} $$</th></tr></thead><tbody><tr><td>(3,5,7,9,11,13,15,17,19,21,23,25,27,29,31,33)</td><td>165142</td><td>36239</td><td>145566</td><td>30111</td></tr><tr><td>(4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34)</td><td>185536</td><td>39725.9</td><td>164035</td><td>30111</td></tr><tr><td>(5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35)</td><td>207750</td><td>43432</td><td>184198</td><td>30111</td></tr><tr><td>(6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36)</td><td>231880</td><td>47367</td><td>206145</td><td>30111</td></tr><tr><td>(7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37)</td><td>258022</td><td>51536</td><td>229966</td><td>30111</td></tr><tr><td>(8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38)</td><td>286272</td><td>55947.3</td><td>255751</td><td>30111</td></tr><tr><td>(9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39)</td><td>316726</td><td>60606.4</td><td>283590</td><td>28699</td></tr><tr><td>(11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41)</td><td>384630</td><td>70697.4</td><td>345790</td><td>30491</td></tr><tr><td>(12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42)</td><td>422272</td><td>76143</td><td>380331</td><td>31441</td></tr><tr><td>(13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43)</td><td>462502</td><td>81864.8</td><td>417286</td><td>32427</td></tr></tbody></table>

As immediately result from Theorem 4.4 we presented Corollary 4.5 where we established the terms $(d_1 - 1)^3 + (d_n - 1)^3$ and $\Delta(\Delta - 1)^2$ where it satisfied with $(d_1 - 1)^3 + (d_n - 1)^3 + \Delta(\Delta - 1)^2 \le 2n^2$. Figure 4 shows the strongest growth of Sigma index trend with sharp bounds among Table 1.

<figure><img src="image_5.png" /><figcaption>Figure 4: Comparison of bounds in Sigma index across sequences with different starting elements where it depends on the value of first element $d_1$ in $\mathcal{D} = (d_1, d_2, \dots, d_n)$.</figcaption></figure>

**Corollary 4.5.** For any tree of order $n$, the $\sigma$-irregularity over $\mathcal{D} = (d_1, d_2, \dots, d_n)$ where $d_n \le d_{n-1} \le \dots d_2 \le d_1$ satisfy

$$ \sigma_{\mathcal{D}}(\mathbb{C}) \ge (d_1 - 1)^3 + (d_n - 1)^3 + \sum_{i=1}^{n} d_i^2 + \Delta(\Delta - 1)^2. \quad (36) $$

**Theorem 4.6.** Let $\mathbb{C}$ be a caterpillar tree with $n$ vertices and maximum degree $\Delta$, let $a$ and $b$ be two integers where $a+b < n$ and $a.b \ge \Delta^2$. The $\sigma$-irregularity over degree sequence $\mathcal{D} = (d_1, d_2, \dots, d_n)$ where $d_n \le d_{n-1} \le \dots d_2 \le d_1$ satisfy
... $d_2 \le d_1$ satisfy with an integer $\alpha = \sqrt{n} - 4$ as

$$ \sigma_{\mathcal{D}}(\mathbb{C}) \ge \sqrt{\sum_{i=1}^{n} (d_i - 1)^\alpha} + \sqrt{\sum_{i=2}^{n-1} d_i^\alpha} + \sum_{i=1}^{n} d_i^2 - a d_1^2 - b d_n^2 + 2n^2 - 2 \quad (37) $$

*Proof.* Assume $T$ be a tree provide with degree sequence $\mathcal{D} = (d_1, d_2, \dots, d_n)$ where $d_n \le d_{n-1} \le \dots d_2 \le d_1$. Suppose there are $a$ and $b$ two integers where $a+b < n$ and $a*b \ge \Delta^2$ where $\Delta$ is the maximum degree of $T$ and $\alpha = \sqrt{n} - 4$. Then, in order to make it easier to deal with the term given according to relationship (37), we divide it into terms, the concrescence of which we will prove sequentially as follows. For the first term, we observe from Theorem (4.6) that the following relationship is achieved (see Figure 5):

$$ \sqrt{\sum_{i=1}^{n} (d_i - 1)^\alpha} < \sigma_{\mathcal{D}}(\mathbb{C}) < \sum_{i=1}^{n} (d_i - 1)^\alpha. \quad (38) $$

Similarly, for the second term, it satisfies with

$$ \sqrt{\sum_{i=1}^{n} d_i^\alpha} < \sigma_{\mathcal{D}}(\mathbb{C}) < \sum_{i=1}^{n} d_i^\alpha. \quad (39) $$

Thus, for the terms $a d_1^2 - b d_n^2 + 2n^2 - 2$, we noticed that according to our condition $a+b < n$ and $a*b \ge \Delta^2$ it holds the relationship $a d_1^2 - b d_n^2 \ge \Delta(\Delta-1)^2$, also by considering $2n^2 > \Delta(\Delta-1)^2$ we find that $2n^2 > a d_1^2 - b d_n^2$. Thus, the $\sigma$-irregularity over degree sequence $\mathcal{D} = (d_1, d_2, \dots, d_n)$ satisfy

$$ \sigma_{\mathcal{D}}(\mathbb{C}) \le \sum_{i=1}^{n} d_i^2 + a d_1^2 + b d_n^2 + 2n^2 - 2 \quad (40) $$

Hence, from (39) and (40) we noticed that

$$ \sigma_{\mathcal{D}}(\mathbb{C}) \ge \sum_{i=1}^{n} d_i^\alpha - \left( \sum_{i=1}^{n} d_i^2 + a d_1^2 + b d_n^2 + 2n^2 - 2 \right). \quad (41) $$

Finally, from (38)–(41) we find that (37) holds. $\square$

To demonstrate the difference among both equations (38) and (39) we presented the Figure 5 where the blue line is bounds of $\alpha$.

<figure><img src="image_6.png" /><figcaption>Figure 5: The impact of $\alpha$ value with terms $\alpha = \sqrt{n} - 4$ on $\sigma$-irregularity with $n = 100$ vertices.</figcaption></figure>

As a consequence of Corollary 4.5, we obtain a direct result from Theorem 4.6, stated in Corollary 4.7, which relates these bounds to the values of $\alpha$.
**Corollary 4.7.** For any tree of order $n$, the $\sigma$-irregularity over $\mathcal{D} = (d_1, d_2, \dots, d_n)$ where $d_n \le d_{n-1} \le \dots d_2 \le d_1$ satisfy

$$ \sigma_{\mathcal{D}}(\mathbb{C}) \ge \sqrt{\sum_{i=1}^{n} (d_i - 1)^{\alpha}} + \sqrt{\sum_{i=2}^{n-1} d_i^{\alpha}} - \left( (d_1 - 1)^3 + (d_n - 1)^3 + \sum_{i=1}^{n} d_i^2 \right). \quad (42) $$

# 5 The Sharp Upper Bound of $\sigma$-irregularity

In this section, we complete the study of bounds through section 4 where we presented the sharp lower bound of $\sigma$-irregularity. Theorem 5.1 presented the lower bound of $\sigma_{\mathcal{D}}(\mathbb{C})$ according to $d_1$ and $d_n$ with the constant term $2n^2 - 2$.

**Theorem 5.1.** Let $\mathbb{C}$ be a caterpillar tree with degree sequence $\mathcal{D} = (d_1, d_2, \dots, d_n)$ where $d_n \ge d_{n-1} \ge \dots d_2 \ge d_1$. Then, the lower bound of $\sigma$-irregularity satisfy

$$ \sigma_{\mathcal{D}}(\mathbb{C}) \le (d_1 - 1)^3 + (d_n - 1)^3 + \sum_{i=1}^{n-1} d_i \cdot (d_{i+1} - d_i)^4 + 2n^2 - 2. \quad (43) $$

*Proof.* Recall $\mathcal{D} = (d_1, d_2, \dots, d_n)$ where $d_n \ge d_{n-1} \ge \dots d_2 \ge d_1$ and maximum degree $\Delta$ of tree $T$ with $n$ vertices. Then, the following notion yields

$$ \Delta(n - \Delta) \ge \sum_{i=1}^{n-1} (d_{i+1} - d_i)^4 + \sum_{i=1}^{n-1} (d_{i+1} - d_i)^3. \quad (44) $$

Therefore, for the term $n - \Delta$, $\sigma$-irregularity satisfy $\Delta(n - \Delta) < \sigma_{\mathcal{D}}(\mathbb{C}) < \Delta(n - \Delta)^2$. Then, the lower bound satisfied with (43) holds as follow

$$ \sigma_{\mathcal{D}}(\mathbb{C}) \le \Delta(n - \Delta)^2 + \sum_{i=1}^{n-1} (d_{i+1} - d_i)^4 + \sum_{i=1}^{n-1} (d_{i+1} - d_i)^3. \quad (45) $$

Hence, from (45) we noticed that $(d_1 - 1)^3 + (d_n - 1)^3 < \Delta(n - \Delta)^2$ where $d_n^3 > \Delta(n - \Delta)$. Thus $\sigma_{\mathcal{D}}(\mathbb{C}) \ge (d_1 - 1)^3 + (d_n - 1)^3$ and $\sigma_{\mathcal{D}}(\mathbb{C}) \le (d_1 - 1)^3 + (d_n - 1)^3 + \Delta(n - \Delta)^2$. Then

$$ \sigma_{\mathcal{D}}(\mathbb{C}) \le 3(n - \Delta)^2 + \sum_{i=1}^{n-1} d_i \cdot (d_{i+1} - d_i)^4 + \sum_{i=1}^{n-1} d_i \cdot (d_{i+1} - d_i)^3. \quad (46) $$

Therefore, for $n > \Delta$ we find that $2n^2 - 2 < \sigma_{\mathcal{D}}(\mathbb{C}) < 2n^3$. Thus, it follows the relationship $\sigma_{\mathcal{D}}(\mathbb{C}) \le (d_1 - 1)^3 + (d_n - 1)^3 + 2n^2 + (n - \Delta)^2$. Consequence that from (44), (45) and (46) the lower bound of $\sigma_{\mathcal{D}}(\mathbb{C})$ among (43) holds. $\square$

**Theorem 5.2.** Let $\mathbb{C}$ be a caterpillar tree with degree sequence $\mathcal{D} = (d_1, d_2, \dots, d_n)$ where $d_n \ge d_{n-1} \ge \dots d_2 \ge d_1$. Consider $\alpha = \sqrt{n} - 4$ be an integer. Then, the lower bound of $\sigma$-irregularity satisfy

$$ \sigma_{\mathcal{D}}(\mathbb{C}) \le (d_1 - 1)^3 + (d_n - 1)^3 + \sum_{i=1}^{n-1} (d_i + 1)^{\alpha} \sqrt{(d_{i+1} - 1)^{\alpha}} + 2n^2 - 2. \quad (47) $$

*Proof.* Assume $\mathbb{C}$ be a caterpillar tree with $n$ vertices and $\alpha$ be an integer where $\alpha = \sqrt{n} - 4$ with degree sequence $\mathcal{D} = (d_1, d_2, \dots, d_n)$ where $d_n \ge d_{n-1} \ge \dots d_2 \ge d_1$. According to Theorem 4.3 we find that among the relationship (26) implies that

$$ \sigma_{\mathcal{D}}(\mathbb{C}) \ge \sum_{i=1}^{n-1} (d_i + 1)^{\alpha} \sqrt{(d_{i+1} - 1)^{\alpha}} $$
Now, there are three terms in (47) where, the previous relationship satisfied with the lower bound of σ-irregularity. Thus, the terms $d_1$ and $d_n$ satisfy $\sigma_{\mathcal{G}}(\mathbb{C}) \ge (d_1 - 1)^3 + (d_n - 1)^3 + 2n^2 - 2$. Therefore, we noticed that

$$ \sigma_{\mathcal{G}}(\mathbb{C}) \le (d_1 - 1)^4 + (d_n - 1)^4 + 2n^3 - 2. \quad (48) $$

Hence, for the degree sequence $\mathcal{D} = (d_1, d_2, \dots, d_n)$ where $d_n \ge d_{n-1} \ge \dots d_2 \ge d_1$. If $d_i - d_{i-1} < 3$ for all $i \in \mathbb{N}$, we find that the lower bound is

$$ \sigma_{\mathcal{G}}(\mathbb{C}) \ge (d_1 - 1)^3 + (d_n - 1)^3 + \sum_{i=1}^{n-1} (d_i + 1)^\alpha \sqrt{(d_{i+1} - d_i)^\alpha} + 2n^2 - 2. \quad (49) $$

Thus, from (48) and (49) the value of $\alpha$ establish the behaviour of index. Thus, the difference implies that

$$ \sum_{i=1}^{n-1} (d_i + 1)^\alpha \sqrt{(d_{i+1} - 1)^\alpha} - \sum_{i=1}^{n-1} (d_i + 1)^\alpha \sqrt{(d_{i+1} - d_i)^\alpha} = \sum_{i=1}^{n-1} (d_i + 1)^\alpha \cdot [(d_{i+1} - 1)^\alpha - (d_{i+1} - d_i)^\alpha]. $$

Therefore, we obtained on

$$ \sigma_{\mathcal{G}}(\mathbb{C}) \le \sum_{i=1}^{n-1} (d_i + 1)^\alpha \cdot [(d_{i+1} - 1)^\alpha - (d_{i+1} - d_i)^\alpha]. \quad (50) $$

Determining the lower bound through the previous relationship necessarily yields the logical conclusion regarding 47, considering the interdependence we presented through relationships (48)–(50), thus fulfilling relationship (47). □

# 6 Conclusion

Recent research (2024–2025) has emphasized tight lower bounds for both indices on trees, caterpillars, and bipartite graphs. These bounds often connect to parameters such as the maximum degree $\Delta$, minimum degree $\delta$, or Fibonacci-like sequences. They enhance predictive power in mathematical chemistry by revealing how minimal irregularity influences molecular behavior. While upper bounds for highly irregular graphs (e.g., stars) provide complements, lower bounds prove especially valuable for stable, low-irregularity compounds. Upper bounds on the Sigma and Albertson indices play a key role in understanding and quantifying maximal structural irregularity in graphs, with direct implications for predicting molecular behavior in highly asymmetric compounds. The quadratic nature of the Sigma index often renders its upper bounds more discriminative for extreme cases than those of the linear Albertson index. The degree sequence plays a central role: it fully determines the possible irregularity values in caterpillar graphs and enables precise bounds, closed-form expressions, and extremal characterizations for these indices. This renders it essential for both theoretical extremal graph theory and applications in modeling chemical structures.

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