# Regularities of Typical Sigma Index on Caterpillar Trees of Pendent Vertices

Jasem Hamoud, Duaa Abdullah

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Jasem Hamoud, Duaa Abdullah. Regularities of Typical Sigma Index on Caterpillar Trees of Pendent Vertices. 2025. hal-04979186

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# Regularities of Typical Sigma Index on Caterpillar Trees of Pendent Vertices

Jasem Hamoud¹* and Duaa Abdullah²†

¹,²*Discrete Mathematics, Moscow Institute of Physics and Technology (National Research University), kerchenskya, Moscow, 117303, Moscow, Russia.

*Corresponding author(s). E-mail(s): khamud@phystech.edu; Contributing authors: abdulla.d@phystech.edu;

†These authors contributed equally to this work.

## Abstract

In this paper, we present a study of topological indices according to caterpillar trees through Albertson index $irr(G) = \sum_{u \in E(G)} |d_G(u) - d_G(v)|$ and the sigma index and total sigma index $\sigma_t(G) = \sum_{1 \le i < j \le n} (d_i - d_j)^2$, which analyses the maximum and minimum values for specific graph properties under certain restrictions of caterpillar tree, where albertson index on caterpillar tree define as $irr(C(n, m)) = \sum_{n=3, m=3} (m-2)(10n-1)$. Through the election of these topological indices, we provide efficient models of these indices on caterpillar trees, as it is known that the Albertson index is the basis on which most of the topological indices are built and we have shown this through the close correlation between the Albertson's index and the Sigma index.

**Keywords:** Albertson index, Sigma index, Caterpillar, Bound

**JEL Classification:** C6

**MSC Classification 2010:** 05C05, 90C35, 68R10

## 1 Introduction

Throughout this paper, we consider finite, undirected, and simple graphs. Caterpillar trees, a type of graph in graph theory, were initially studied in a number of works
by Frank Harary and Edward M. Schwenk [14]. In basically, a caterpillar tree is one in which all vertices are inside one edge of a central path. A tree with all its vertices directly connected to or a component of a central path (the “stalk”) is called a caterpillar graph.

The Albertson and Sigma index are beneficial instruments for analysing caterpillar trees in graph theory. Their work not just advances theory, but it also has practical consequences for a variety of academic areas. Denote by $\Delta, \delta$ represent the graph’s minimum and maximum degrees, or nearly respectively( see Lemma 2). Horoldagva, B., Lkhagva, B., Dorjsembe, S., Gutman, I. in [9] introduce for integer $n, t \ge 0$ where $1 < t < n, t$ is irregularity index, then $G$ is a maximally irregular graph of order $n$ and if $|E(G)| \le \lfloor \frac{n^2}{4} \rfloor$, then $|E(G)| \le \lfloor \frac{t(2n-t+1)}{4} \rfloor$, in [9] introduce “Collatz and Sinogowitz” as $CS(G) = \lambda_1 - \frac{2m}{n}$ where $\lambda_1$ is the largest eigenvalue of the adjacency matrix, and consider $G$ be a simple undirected connected graph, then “The variance” $Var(G)$ of the vertex degrees

$$ \mathrm{Var}(G) = \frac{1}{n} \sum_{v \in V(G)} d_G(v)^2 - \left( \frac{1}{n} \sum_{v \in V(G)} d_G(v) \right)^2 $$

Actually, “Albertson Index” was defined in 1997 the irregularity of $G$ as the sum of imbalances of all edges as

$$ \mathrm{irr}(G) = \sum_{\mathrm{ur} \in \mathrm{E}(G)} |d_G(u) - d_G(v)|. $$

Define the “general Albertson irregularity index” of a graph in [12] (Also see [10]) by Lina, Z., Zhoub, T., Miaob, L., such that for any number $p > 0$ then:

$$ \mathrm{irr}_p(G) = \left( \sum_{uv \in E(G)} |d(u) - d(v)|^p \right)^{\frac{1}{p}}. $$

In [13] proved $\mathrm{irr}_T(G) \le \frac{n^2}{4} \mathrm{irr}(G)$. In this case if $G$ is a tree, then $\mathrm{irr}_T(G) \le (n - 2) \mathrm{irr}(G)$, G. H. Fath, Tabar, Gutman, I., R. Nasiri in [1] mention to all trees of order $n$ if $n > 2$ then we have $\mathrm{irr}(T) \ge 2$, the star $S_n$ by $\mathrm{irr}(S_n) = (n - 1)(n - 2)$. So that the authors provide the maximum value of irregularity graph as $\mathrm{irr}_{\max} = n - 2$.

Dorjsembe, S., Buyantogtokh, L., Gutman, I. in [2] mention to $u_0u_1 \cdots u_t$ be a path in $G$, then $imb_G(u_0, u_t) = \sum_{i=0}^{t-1} |d_G(u_i) - d_G(u_{i+1})|$, then $\mathrm{irr}(G) \ge 2 \lfloor \frac{\Delta}{2} \rfloor$ or given by:

$$ \mathrm{irr}(G) \le \frac{1}{48} (6n^2\Delta + 3\Delta^2n - 2\Delta^3 - 4\Delta). $$

If the tree $T$ is spider with $n$ vertices $\mathrm{irr}(T) = \Delta(\Delta - 1)$, Therefore in [3] the maximum degree of stepwise irregular graphs of order $n$ is $\Delta \le \lfloor \frac{n+1}{2} \rfloor$. if $\delta \le \Delta - 4$, then $\Delta \le \frac{n+9}{4}$ but there are another case as $\Delta < \frac{n+6}{4}$ when $\delta > \Delta - 4$ in this case we have $\Delta < \frac{n+2}{4}$ when $n \equiv 2(4)$.
The relevance of the paper stems with the simple reason that both indices are used in extremal graph theory, which analyses the maximum and minimum values for specific graph properties under certain restrictions. These indices help find extremal structures that optimise or limit irregularity in caterpillar trees, which are differentiated by a core "backbone" to which leaves are attached. Within biological graph theory, topological indices like the Albertson and Sigma indices are used to predict molecular properties based on structure representation. Caterpillar trees can represent specific molecular frameworks, making them valuable for anticipating the chemical and physical features of materials.

This paper is organized as follows. In section 2, we provided a set of necessary preliminaries in order to deal with the research, which we thought it was necessary to put, in section 3, which we divided into two sections, in subsection 3.1 we discussed the first anthropological index, the Albertson index on caterpillar trees, and in subsection 3.2 we discussed the second topological index, the Sigma index on caterpillar trees, and this study was developed due to the importance of topological indicators on the graphs of the caterpillar graphs, including the caterpillar trees.

## 2 Preliminaries

In Lemma 2 we show the relationship between maximum and minimum degree according Albertson index, in Theorem 3 we show the number of caterpillar tree where we considre function in [14] as $c(x) = \frac{x^3(1-3x^2)}{(1-2x)(1-2x^2)}$, in Theorem 4 we show $\sigma_t(G)$ for any graph $G$ of order $n \ge 3$

**Proposition 1** Let $G(V, E)$ is a graph of order $n \ge 0$, the relationship between Albertson index and sigma index given by:

$$ \sqrt{\sigma(G)} \le \text{irr}(G) \le \sqrt{m\sigma(G)}. $$

where $m = |E|$.

**Lemma 2** ([5, 6]) Let $G(V, E)$ a graph (simple and connected) where $|V| = n; n \ge 2$ and $|E| = m$, we have:

$$ \text{irr}(G) \le \frac{(\Delta - \delta)}{\sqrt{\Delta \delta}} \sqrt{m \sum_{uv \in E(G)} d_G(u)d_G(v)}. $$

**Theorem 3** ([14]) Let be $n \ge 3$ vertices, the number of caterpillar tree is given according function $c(x)$ by use $p = n + 4$ as:

$$ C_n = 2^{n-4} + 2^{\lfloor n/2 - 2 \rfloor}. $$

We mention by $C(n, m)$ caterpillar tree with $(n, m)$ vertices, and refer to $n$ main vertices i figure 1 we have $X = (x_1, x_2, \dots, x_n)$ where $n = |X|$, and refer to $m$ a set of branches of main vertices, Zhang, P., Wang, X in[7] show "Caterpillar trees" are employed in chemical graph theory for describing molecular structures, and several
topological indices, which include the Albertson index, have been used to study the characteristics of molecules. We show in Figure 1 Caterpillars trees of C(n, 3) as in the following figure.

<figure><img src="image_2.png" /><figcaption>Fig. 1: Example of Caterpillars C(n, 3)</figcaption></figure>

**Theorem 4 ([4])** Let G is a connected graph of order n which n ∈ N, n ≥ 3 vertices it holds

$$ \sigma_t(G) \le \begin{cases} \lfloor \frac{n}{4} \rfloor \cdot \lfloor \frac{3n}{4} \rfloor (n - 1 - \lfloor \frac{n}{4} \rfloor)^2, & \text{if } n \equiv 0 \text{ mod } 4 \text{ or } n \equiv 3 \text{ mod } 4; \\ \lfloor \frac{n}{4} \rfloor \cdot \lfloor \frac{3n}{4} \rfloor (n - 1 - \lfloor \frac{n}{4} \rfloor)^2, & \text{if } n \equiv 1 \text{ mod } 4 \text{ or } n \equiv 2 \text{ mod } 4. \end{cases} $$

# 3 Main Results

In this section, we will discuss the Albertson Index for this specific tree type (caterpillar tree), as well as its relationship to Albertson Index. Particularly its modified version, is a topological index that uses the degrees of a graph's vertices to calculate its irregularity. Albertson index can provide insight into the structural properties and anomalies of caterpillar trees, a form of tree graph. Caterpillar trees are commonly used in a variety of domains, including chemical graph theory and data structure indexing, because of their basic but informative structure. The next sections discuss how the Albertson index is used and what it signifies for caterpillar trees.

## 3.1 Regularities of Albertson index Bound on Caterpillar Trees

In Proposition 5 we show caterpillar tree for sequence $d = (\underbrace{n, n, \dots, n}_{m+1 \text{ times}}, \underbrace{3, 3, \dots}_{m \text{ times}})$, in

Lemma 6 we show the Albertson index for a sequence of three vertices in many position as a special case of caterpillar tree, in Proposition 8 we disclose a new case of caterpillar tree of order (n, 4), also in Proposition 9 new case of caterpillar tree of order (n, 5), in Theorem 10 we show the general case of caterpillar tree of order (n, m) as we show that in Figure 3.

**Proposition 5** For $n \ge 3$, Albertson index of Caterpillar trees given by:

$$ \mathrm{irr}(C(n, 3)) = \sum_{n=3} (12n - 4). $$

*Proof* Let be n, m ≥ 0 a set of integer point given as sequence $x = (x_1, x_2, \dots, x_n)$ and set $y = (y_1, y_2, \dots, y_m)$, depend on definintion of Albertson index we have $\mathrm{irr}(G) = \sum_{uv \in E(G)} |d(u) - d(v)|$
$d(v)$, then we have for vertices of order 3 the sequence $x = (x_1, x_2, x_3)$ and sequence $y = (y_1, y_2, y_3)$ defined as:

$$ \text{irr}(G) = |x_1 - y_1| + |x_1 - y_2| + |x_1 - y_3| + |x_2 - y_1| + \cdots + |x_n - y_1| + |x_n - y_2| + |x_n - y_3| $$
then:

$$ \text{irr}(G) = \sum_{i=1}^{3} \sum_{j=1}^{3} |x_i - y_j| $$

Let be mention to sequence $x = (x_1, x_2, \dots, x_n)$ by $n$ and set $y = (y_1, y_2, \dots, y_m)$ by $m$, then we have in generally: $\text{irr}(C(n, m)) = (m+1)(n-2)+2$ that is hold to:

$$ \begin{aligned} \text{irr}(C(3, 3)) &= 2m^2 + (m+1).m.(n-2) + 2 \\ &= m(2m + mn - 2m + n - 2) + 2 \\ &= m(m+1).n - 2m + 2 \end{aligned} $$

In fact that is clear, in next table, Table1 we show tested of a sequence of vertices $n = (3, 4, \dots, 1000)$ given as random values from which a sample was selected as:

<table><caption>**Table 1:** Some Value of Albertson index on caterpillar tree.</caption><thead><tr><th>n</th><th>irr(C(n, 3))</th><th>n</th><th>irr(C(n, 3))</th><th>n</th><th>irr(C(n, 3))</th></tr></thead><tbody><tr><td>3</td><td>32</td><td>50</td><td>596</td><td>400</td><td>4796</td></tr><tr><td>4</td><td>44</td><td>60</td><td>716</td><td>500</td><td>5996</td></tr><tr><td>5</td><td>56</td><td>70</td><td>836</td><td>600</td><td>7196</td></tr><tr><td>6</td><td>68</td><td>80</td><td>956</td><td>650</td><td>7796</td></tr><tr><td>7</td><td>80</td><td>90</td><td>1076</td><td>700</td><td>8396</td></tr><tr><td>8</td><td>92</td><td>100</td><td>1196</td><td>750</td><td>8996</td></tr><tr><td>9</td><td>104</td><td>110</td><td>1316</td><td>800</td><td>9596</td></tr><tr><td>10</td><td>116</td><td>150</td><td>1796</td><td>850</td><td>10196</td></tr><tr><td>20</td><td>236</td><td>200</td><td>2396</td><td>900</td><td>10796</td></tr><tr><td>30</td><td>356</td><td>250</td><td>2996</td><td>950</td><td>11396</td></tr><tr><td>40</td><td>476</td><td>300</td><td>3596</td><td>1000</td><td>11996</td></tr></tbody></table>

In Figure 1 we show the general case of caterpillar tree of order $(n, 3)$ and in Figure 2 we show that for many value of a sequence of vertices $n = (3, 4, 5, 6)$, $m = 3$, Albertson index in caterpillars tree as:

<figure><img src="image_3.png" /><figcaption>**Fig. 2:** Caterpillars tree with $(n, 3)$ vertices</figcaption></figure>
In Figure 3 we show that the general case of caterpillar tree of order (n, m), we use the discussion of each case separately from the other and at the end we get the sum of the results in these cases and let start with first vertices x₁ we have albertson index as:

$$|x_1 - y_1| + |x_1 - y_2| + \cdots + |x_1 - y_j|$$

where j is last branches of vertices, then we get on: $\sum_{j=1}^{m} |x_1 - y_j|$.

<figure><img src="image_4.png" /><figcaption>Fig. 3: General case of Caterpillar tree</figcaption></figure>

in vertice $x_2$ we get on: $\sum_{j=1}^{m} |x_2 - y_j|$, and in $x_3$ we get on: $\sum_{j=1}^{m} |x_3 - y_j|$ and in vertice $x_{n-1}$ we have: $\sum_{j=1}^{m} |x_{n-1} - y_j|$, and in last vertice $x_n$ we have: $\sum_{j=1}^{m} |x_n - y_j|$, from figure 3 we show that clear. Finally, after discussing the cases, the Albertson index is the sum of all the cases discussed and is put in a general sum as:

$$\mathrm{irr}(T) = \sum_{j=1}^{m} |x_1 - y_j| + \sum_{j=1}^{m} |x_2 - y_j| + \cdots + \sum_{j=1}^{m} |x_n - y_j|.$$

Then we have:

$$\mathrm{irr}(T) = \sum_{i=1}^{n} \sum_{j=1}^{m} |x_i - y_j|.$$

Finally, According to the definition of Albertson index and by giving some cases of vertices, we got the required result, from which we relied on 3 vertices to 1000 vertices and more. □

**Lemma 6** A sequence of positive integers $d_1, d_2, d_3$; $d_1 \ge d_2 \ge d_3$ and defined numbers : $a \ge b \ge c$. so that, “Albertson index” defined as:

$$\mathrm{irr} = \left\{ \begin{array}{l} \mathrm{irr}_{max} = \sum_{a_i \in \{a,b\}} (a_i - 1)^2 + \sum_{i=0}^{3} k_i c^i \\ \\ \mathrm{irr}_{min} = \sum_{a_i \in \{a,c\}} (a_i - 1)^2 + (b-1)(b-2) + (a-c) \end{array} \right.$$

where

$$\begin{cases} k_0 = 0 \\ k_1 = -(ab + 2a + 2b) \\ k_2 = ab + 3(a + b) \\ k_3 = -(a + b + 3) \end{cases}$$
**Proposition 7** Let be $\text{irr}(C(n, 3))$ Albertson index of caterpillar tree of order $(n, 3)$ vertices, then we have:

$$ \text{irr}_{\max}(C(n, 3)) = \text{irr}_{\min}(C(n, 3)) = \text{irr}(C(n, 3)) $$

This implies that we cannot specify $\text{irr}_{\max}$, $\text{irr}_{\min}$ in caterpillar tree $C(n, 3)$, or in other terms, there are no such values but only the general value of the Albertson index $\text{irr}$.

**Proposition 8** Let be $\text{irr}(C(n, 4))$ Albertson index of caterpillar tree of order $(n, 4)$ vertices, then we have:

$$ \text{irr}C(n, 4) = \sum_{n=3} 20n - 6. $$

In fact, we can illustrate this property with the following Figure 4, in which we show a set of values Albertson index for caterpillar trees of order $(n, 4)$ vertices. According the figure as we know $\text{irr}(T) = \sum_{i=1}^{n} \sum_{j=1}^{m} |x_i - y_j|$, then we have: $C(3, 4) = 54, C(4, 4) = 74, C(5, 4) = 94, C(6, 4) = 114, C(7, 4) = 134$, we show that within the figure.

<figure><img src="image_5.png" /><figcaption>Fig. 4: caterpillar tree $C(n, 4)$ for 4 vertices</figcaption></figure>

In fact, Albertson index changes according to the selected case of caterpillar trees according to the heads, so we presented for three heads and then for four heads as a discussion of this case.
**Proposition 9** Let be $\text{irr}(C(n, 5))$ Albertson index of caterpillar tree of order $(n, 5)$ vertices, then we have:

$$ \text{irr}(C(n, 5)) = \sum_{n=3}^{30n-8} $$

**Theorem 10** Let be integer number's $n, m \ge 3$, and $\text{irr}(C(n, m))$ Albertson index of caterpillar tree of order $(n, m)$ vertices, then we have:

$$ \text{irr}(C(n, m)) = \sum_{n=3, m=3}^{m-2}(10n-1). $$

*Proof* Suppose the sequence of vertices $V = (v_1, v_2, \dots, v_i)$ and denote by $n = |V|$, let a sequence of branches of vertices $U = (u_1, u_2, \dots, u_j)$ and denote by $m = |U|$, then we have:

* For vertices $v_1$ we have: $\text{irr}(C(v_1, U)) = \sum_{j=3}^{p} |v_1 - u_j|$ where $3 \le p \le m$. In this case not enough because we should be take 3 vertices (see Figure 3) at least, so that we have for three vertices already $\text{irr}(C(n, 3)) = \sum_{n=3}^{12n-4}$. Then we suppose $X = (v_1, v_2, v_3)$ and $r = |X|$, then we have:

$$ \text{irr}(C(r, 3)) = \sum_{r=1}^{12r-4}. $$

* For $(n, 4)$ vertices (see Figure 4) we have : $\text{irr}(C(n, 4)) = \sum_{n=3}^{20n-6}$, and $\text{irr}(C(n, 5)) = \sum_{n=3}^{30n-8}$.

For $n$ vertices and among all branches of vertices $U = (u_1, u_2, \dots, u_j)$ in all three case's, in [1] mention to all star trees $S_n$ of order $n$, where Albertson index define as $\text{irr}(S_n) = (n-1)(n-2)$. then we have in caterpillar tree get on the function as:

$$ \zeta(m) = \sum_{m=0}^{10(m-2)}; m = |U| $$

we have for constant terms the function $\psi(m)$ where define as sum of branches of vertices $V = (v_1, v_2, \dots, v_i)$ as :

$$ \psi(m) = \sum_{m=0}^{-(m-2)-2} $$

Notice both functions $\zeta, \psi$ have a sequence of branches of vertices, so that we can write it in main function among all branches of vertices $U = (u_1, u_2, \dots, u_j)$ and $V = (v_1, v_2, \dots, v_i)$ as:

$$ \Gamma(m) = \sum_{m=0}^{\zeta(m) + \psi(m)}. $$

Now we can see that the function $\Gamma$ given us Albertson index for caterpillar trees, so that we can write as:

$$ \text{irr}(C(n, m)) = \sum_{n=3}^{(m-2)(10n+1) - (m-2)} $$

Finally, this leads us to the relationship we assumed in the proof according to the basic definition of Albertson index, which has at least 3 vertices, As required. $\square$
**Theorem 11** Let $n, m, r$ integer numbers where $n \le m \le r$, and $C(m, n, r)$ is caterpillar tree of order $(n, m, r)$ vertices, Albertson index of $C(m, n, r)$ where $n \ge 3, m \ge 3, r \ge 3$ given as:

$$ \mathrm{irr}(C(n, m, r)) = \sum_{n=3, r=3} (n^2(r^2 - 2r) + n(n+1)). $$

or in an equivalent variation according as:

$$ \mathrm{irr}(C(m, n, r)) = \sum \frac{13}{18}(n^3 + m^3 + r^3) $$

where $\{n, m, r\} \in \mathbb{N}$.

*Proof* we will discuss both cases and consider the second case as the first in the discussion and then discuss the general case.

* Case 1: $\mathrm{irr}(C(m, n, r)) = \sum \frac{13}{18}(n^3 + m^3 + r^3)$.

Let be suppose the figure of caterpillar tree of order $(n, m, r)$ vertices given in figure 5, and the sequences is: $V = (v_1, v_2, \dots, v_i)$; $n = |V|$, $U = (u_1, u_2, \dots, u_j)$; $m = |U|$ and $R = (r_1, r_2, \dots, r_k)$; $r = |R|$, then we have as we show that in figure as:

<figure><img src="image_6.png" /><figcaption>Fig. 5: Caterpillar tree of order (3, 4, 5)</figcaption></figure>

the condition of $(n, m, r)$ will be $n \ge 3, m \ge (n+1), r \ge (m+1)$, so that for vertices $v_1$, we notice that for caterpillar tree of order (3, 4, 5), we suppose the function $\zeta(a, b, c)$ where define as $\zeta(a, b, c) = a^3 + b^3 + c^3$, then for $(3, 4, 5)$ we have $\zeta(3, 4, 5) = 216$, by using constant $K = \frac{13}{18}$, in this case we have Albertson index of caterpillar tree $C(3, 4, 5)$ is: $\mathrm{irr}(C(3, 4, 5)) = K\zeta(3, 4, 5)$, then we have: $\mathrm{irr}(C(3, 4, 5)) = 156$. Here in this case we have obtained the function $\zeta(a, b, c)$ and the constant $K$ associated with this function, and from it in the general case if we change the values, which obtain similar results that we can express through the following table 2 that shows these values according to the constant that we considered as $K$ as follows: From the table, we can see that the resulting values are different if the initial condition is changed, so we do not get normal values, which is why we emphasise that the condition $n \ge 3, m \ge (n+1), r \ge (m+1)$ must remain consecutive for both $(n, m, r)$.

* Case 2: General Formula

Assume a sequence is $v = (a, b, c)$ where $a \ge 3, b = a + 1, c = b + 1$, or in another formula we can be write $a \ge 3, b = a + 1, c = a + 2$, then we have, for case of vertices $b, c$ we have:

$$ \mathrm{irr}(C(b, c)) = ((c-1)^2 + 1)(b-1) = c^2b - 2cb + 2b - c^2 + 2c - 2. $$
<table><caption>**Table 2:** Value of $(m, n, r)$ with determination condition.</caption><thead><tr><th>Conditions</th><th>(a, b, c)</th><th>ζ(a, b, c)</th><th>K(ζ(a, b, c))</th></tr></thead><tbody><tr><td>n ≥ 3, m ≥ (n + 1), r ≥ (m + 1)</td><td>(3, 4, 5)</td><td>216</td><td>156</td></tr><tr><td>n ≥ 3, m ≥ (n + 1), r ≥ (m + 2)</td><td>(3, 4, 6)</td><td>307</td><td>3991/18</td></tr><tr><td>n ≥ 3, m ≥ (n + 2), r ≥ (m + 1)</td><td>(3, 5, 6)</td><td>368</td><td>4784/18</td></tr><tr><td>n ≥ 3, m ≥ (n + 1), r ≥ (m + 1)</td><td>(4, 5, 6)</td><td>405</td><td>420</td></tr><tr><td>n ≥ 4, m ≥ (n + 1), r ≥ (m + 2)</td><td>(4, 5, 7)</td><td>532</td><td>6916</td></tr><tr><td>n ≥ 4, m ≥ (n + 2), r ≥ (m + 1)</td><td>(4, 6, 7)</td><td>623</td><td>8099/18</td></tr><tr><td>n ≥ 3, m ≥ (n + 1), r ≥ (m + 1)</td><td>(5, 6, 7)</td><td>648</td><td>930</td></tr><tr><td>n ≥ 3, m ≥ (n + 1), r ≥ (m + 1)</td><td>(6, 7, 8)</td><td>1071</td><td>1806</td></tr></tbody></table>

for vertices $a, c$ we have:

$$ \mathrm{irr}(C(b, c)) = a(c^2b - 2cb + 2b - c^2 + 2c - 1) $$

Therefore we have:

$$ \begin{aligned} \mathrm{irr}(C(a, b, c)) &= a(c^2b - 2cb + 2b - c^2 + 2c - 1) \\ &= a(c(c - 2)(b - 1) + 2b - 1) \\ &= a(b - 1)(c^2 - 2c) + ab. \end{aligned} $$

According to our condition $a \ge 3, b = a + 1, c = a + 2$, we have:

$$ \mathrm{irr}(C(a, b, c)) = a^2(c^2 - 2c) + a(a + 1). $$

As desire.

□

**Proposition 12** For among caterpillar tree $C(a, b, c, d)$, where $b \ge 3, a = b+2, c = b+1, d = a$, then Albertson index on caterpillar tree according Figure 6 is:

$$ \mathrm{irr}(C(a, b, c, d)) = \sum_{i=0}^{6} C_i \cdot x_i $$

where the sum can be written as: $\sum_{i=0}^{6} C_i \cdot x_i = C_0 + C_1 a^4 + C_2 a^3 + C_3 a^2 + C_4 a + C_5 a c + C_6 c$, Where: $C_0 = 16, C_1 = 1, C_2 = -2, C_3 = -3, C_4 = 8, C_5 = 4C_6 = -4$.

*Proof* Actually, this type of tree required dealing with both vertices alone, then we combine the result together, so that let be consider $C(a, b, c, d)$ caterpillar tree of order $(a, b, c, d)$ vertices, with term $b \ge 3, a = b + 2, c = b + 1, d = b + 3$, we show that in Figure 6, then we have for vertices $b, c$ in both side as:

$$ \begin{aligned} \mathrm{irr}(C(b, c)) &= 4b(c - 1) + 4(c - b) \\ &= 4bc - 8b + 4c. \end{aligned} $$

for vertices $a, b, c$ we have:

$$ \begin{aligned} \mathrm{irr}(C(a, b, c)) &= a(b(a - 1)^2 + 2b(a - b + 1) + 2(b - 1)(b - 2) + 3(a - b)) \\ &= a(b(a - 1)^2 + 2ab - 7b + 3a + 4) \\ &= ab(a^2 - 6) + a(3a + 4). \end{aligned} $$
<figure><img src="image_7.png" /><figcaption>Fig. 6: Caterpillar tree of special case in 3 vertices</figcaption></figure>

So that, from all vertices, we have the formula is $a^4 - 2a^3 - 3a^2 + 8a + 4ac - 4c + 16$, therefore we write:

$$ \mathrm{irr}(C(a, b, c, d)) = \sum_{i=0}^{6} C_i \cdot x_i $$

Here, $x_i$ represents the variables $a, a^2, a^3, a^4, ac, c$ and the constant term, and $C_i$ are their coefficients:

* For $i = 0, x_0 = 16, C_0 = 1$, for $i = 1, x_1 = a^4, C_1 = 1$;

* For $i = 2, x_2 = a^3, C_2 = -2$, for $i = 3, x_3 = a^2, C_3 = -3$;

* For $i = 4, x_4 = a, C_4 = 8$, for $i = 5, x_5 = ac, C_5 = 4$ for $i = 6, x_6 = c, C_6 = -4$.

As desire.

□

## 3.2 Regularities of Sigma index ($\sigma$ -Index) on Caterpillar Trees

In this subsection, we introduced Sigma index ($\sigma$ -index) of caterpillar tree is a topological index employed in graph theory for calculating the degree differences between the vertices that are adjacent, where $\sigma_t(G) = \sum_{1 \le i < j \le n} (d_i - d_j)^2$ as we mention in Theorem 4 and Proposition 13, also in Proposition 15 we show Sigma index on Caterpillar tree for special case of order $(n, 4)$, and in Proposition 16 we show that for $(n, 5)$ vertices, in Theorem 17 we show the general case of Sigma index on caterpillar tree of order $(n, m)$ as we show that in Figure 3.

**Proposition 13** For $n \ge 3$, the Sigma index of the Caterpillar trees is given by:

$$ \sigma(C(n, 3)) = 48(n - 2) + 56. $$

*Proof* Let $n, m \ge 0$ be a set of integer points given as sequence $n = (x_1, x_2, \dots, x_n)$ and set $m = (y_1, y_2, \dots, y_m)$, as we known Sigma index $\sigma$ given as: $\sigma(C(n, 3)) = \sum_{uv \in E(G)} (d_u - d_v)^2$, then we have:
$$ \sigma(C(n, 3)) = (x_1 - y_1)^2 + (x_1 - y_2)^2 + (x_1 - y_3)^2 + (x_2 - y_1)^2 + \cdots + (x_{n-1} - y_2)^2 + (x_{n-1} - y_3)^2 + (x_n - y_1)^2 + (x_n - y_2)^2 + (x_n - y_3)^2 $$

That is hold to Sigma index of Caterpillar tree by using $\mathcal{T}_n$ where $\mathcal{T}_n$ is a class of trees as:

$$ \sigma(\mathcal{T}_n) = \sum_{i=1}^{n} \sum_{j=1}^{m} (x_i - y_j)^2 $$

$$ \sigma(C(n, 3)) = \sum_{i=1}^{n} \sum_{j=1}^{3} (x_i - y_j)^2 $$

where a sequence $x = (x_1, x_2, \dots, x_n)$ and $y = (y_1, y_2, \dots, y_m)$. (see general case of caterpillar tree in Figure 3). In figure 7 we show that for Sigma index an Albertson index for some value of vertices, in proposition 5 we have $\text{irr}(C(n, 3)) = 12n - 4$, so that from relationship of sigma index $\sigma(C(n, 3)) = \sum_{i=1}^{n} \sum_{j=1}^{3} (x_i - y_j)^2$ we get on $\sigma(C(n, 3)) = 48(n - 2) + 56$. We show that in table 00 for some value of vertices $n = (3, 4, \dots, 100)$ as: In next table (Table 3) we introduced a sample data about sigma index of caterpillar tree in rang [3, 100] as we show that in the table as: In next figure (Figure 7) we show that for some value from the table 3

<table><caption>**Table 3:** Sample data within the range from 3 up to 100 standard Sigma index tests</caption><thead><tr><th>n</th><th>σ(C(n, 3))</th><th>n</th><th>σ(C(n, 3))</th><th>n</th><th>σ(C(n, 3))</th><th>n</th><th>σ(C(n, 3))</th></tr></thead><tbody><tr><td>3</td><td>104</td><td>11</td><td>488</td><td>23</td><td>1064</td><td>33</td><td>1544</td></tr><tr><td>4</td><td>152</td><td>12</td><td>536</td><td>24</td><td>1112</td><td>34</td><td>1592</td></tr><tr><td>5</td><td>200</td><td>13</td><td>584</td><td>25</td><td>1160</td><td>40</td><td>1880</td></tr><tr><td>6</td><td>248</td><td>14</td><td>632</td><td>26</td><td>1208</td><td>50</td><td>2360</td></tr><tr><td>7</td><td>296</td><td>15</td><td>680</td><td>27</td><td>1256</td><td>60</td><td>2840</td></tr><tr><td>8</td><td>344</td><td>16</td><td>728</td><td>28</td><td>1304</td><td>70</td><td>3320</td></tr><tr><td>9</td><td>392</td><td>17</td><td>776</td><td>29</td><td>1352</td><td>80</td><td>3800</td></tr><tr><td>10</td><td>440</td><td>18</td><td>824</td><td>30</td><td>1400</td><td>90</td><td>4280</td></tr><tr><td>19</td><td>872</td><td>20</td><td>920</td><td>31</td><td>1448</td><td>95</td><td>4520</td></tr><tr><td>21</td><td>968</td><td>22</td><td>1016</td><td>32</td><td>1496</td><td>100</td><td>4760</td></tr></tbody></table>

of value Albertson index and Sigma index as:
<figure><img src="image_8.png" /></figure>

$$ \text{irr}(C(3,3)) = 32, \sigma(C(3,3)) = 104 \quad \text{irr}(C(4,3)) = 44, \sigma(C(4,3)) = 152 $$

<figure><img src="image_9.png" /></figure>

$$ \text{irr}(C(5,3)) = 56, \sigma(C(5,3)) = 200 \quad \text{irr}(C(6,3)) = 68, \sigma(C(6,3)) = 296 $$

<caption>Fig. 7: Sigma index and Albertson index of some value</caption>

Finally, As is well known, the Sigma index is related to the Albertson index, this condition is also given by the definition of sigma index. □

**Remark 1** Both cases of index (Albertson, Sigma) have a different results, as we notice where $\text{irr}(C(n, 3)) = \sum_{n=3}(12n - 4)$ and $\sigma(C(n, 3)) = 48(n - 2) + 56$, in fact for make sigma index of order $(n, 3)$ more effective we can computing it as: $\sigma(C(n, 3)) = \sum_{n=3}(48(n - 2) + 56)$.

**Proposition 14** Let be $\sigma(C(n, 3))$ Sigma index of caterpillar tree of order $(n, 3)$ vertices, then we have:

$$ \sigma_{\max}(C(n, 3)) = \sigma_{\min}(C(n, 3)) = \sigma(C(n, 3)) $$

This implies that we have just $\sigma(C(n, 3))$ without determining the maximum ad minimum.

**Proposition 15** Let be $\sigma(C(n, 4))$ Sigma index of caterpillar tree of order $(n, 4)$ vertices, then we have:

$$ \sigma(C(n, 4)) = \sum_{n=3} 100n - 70. $$

from Figure 4 we show that for both sigma index or Albertson index. We show in next proposition for caterpillar tree of order $(n, 5)$ vertices.

**Proposition 16** Let be $\sigma(C(n, 5))$ Sigma index of caterpillar tree of order $(n, 5)$ vertices, then we have:

$$ \sigma(C(n, 5)) = \sum_{n=3} 180n - 108. $$
From the discussions of the special cases in the propositions 13,15, 16, we can
move on to the formulation of the general theorem, which gives us Albertson index in
caterpillar trees, which is the sum of the square of the difference of the base vertices
and the sub-vertices (branches of vertices).

**Theorem 17** For integer number $n \ge 3$, Sigma index for caterpillar tree $\sigma(C(n, m))$ of order $(n, m)$ vertices given by:

$$
\sigma(C(n, m)) = m^3 n + 2m^3 + m^2 n - 6m^2 - mn + 6m - n + 6.
$$

where $n \ge 3$, and $m \ge n$.

*Proof* We have previously discussed a variety of cases:

* Case 1 for $\sigma(C(n, 3))$: in this case we found in proposition 13 Sigma index on caterpillar tree of order $(n, 3)$ as $\sigma(C(n, 3)) = \sum_{n=3} (48(n-2) + 56)$.

* Case 2 for $\sigma(C(n, 4))$: in this case we found in proposition 15 as we show that in Case 1, so that we have: $\sigma(C(n, 4)) = \sum_{n=3} 100n - 70$

* Case 3 as for $\sigma(C(n, 5))$: in this case we found in proposition 16 as $\sigma(C(n, 4)) = \sum_{n=3} 180n - 108$

Notice the coefficient of $n$ is (48, 100, 108), then we have for a sequence of value $m$ as $d = (3, 4, 6, 7, 8, \dots, 20)$, but for sequence at value $m = 5$ is not correct, the coefficient of $n$ with the sequence of $m$ as $d = (10, 20, \dots, 100)$, is correct, so that in fact, these tested values are not enough to give us the full proof as we want it to be, and it must fulfil all the values without excluding any other values, so we assign the dependent an approximation of the tested values according by:

$$
\sigma(C(n, m)) \approx (52m - 108)n - (30m + 34).
$$

Let be a sequence of vertices as $d = (m, \underbrace{m+2, m+2, \dots, m}_{n \text{ times}}, \underbrace{1, 1, \dots}_{n+2 \text{ times}})$, then we have:

$$
\begin{align*}
\sigma(C) &= \sum_{i=3} \sum_{j=3} (d(V_i) - d(U_j))^2 \\
&= 2(m-1)^3 + 8 + n(m-1)(m+1)^2 \\
&= m^3 n + 2m^3 + m^2 n - 6m^2 - mn + 6m - n + 6
\end{align*}
$$

□

Remark 2 for the general case of Albertson index we found: $\text{irr}(C(n, m)) = \sum_{n=3,m=3} (m-2)(10n-1)$, for the general case of Sigma index, we found $\sigma(C(n, m)) = m^3 n + 2m^3 + m^2 n - 6m^2 - mn + 6m - n + 6$. However, we cannot convert them into a sum as in the Albertson's relationship, which is a good distinction that warrants a study of each indicator individually.

**Theorem 18** Let be $n, m, r$ integer numbers where $n \le m \le r$, and $C(m, n, r)$ is caterpillar tree of order $(n, m, r)$ vertices, Sigma index of $C(m, n, r)$ where $n \ge 3, m \ge 3, r \ge 3$ given as:

$$
\sigma(C(n, m, r)) = \sum_{n=3, r=3} \left( n \left( r^2 (r-3)(n+1) + (r^3 - 3r + (3n+1)) \right) \right)
$$
or in an approximate equivalent variation according as

$$ \sigma(C(m, n, r)) \approx \sum \left( 3 \left( \frac{1}{12}(n+m+r)^3 - (n+m+r) \right) \right). $$

where $\{n, m, r\} \in \mathbb{N}$. (See Figure 5).

**Proof** We will need to prove one of the two relationships, so we have chosen the first relationship, while the second relationship can be obtained by imposing a specific dependency and treating it as such, so for the first case we have: assume a sequence is $v = (a, b, c)$ where $a \ge 3, b = a + 1, c = b + 1$, or in another formula we can be write $a \ge 3, b = a + 1, c = a + 2$, then we have, for case of vertices $b, c$ we have:

$$ ((c-1)^3 + 1)b - 1 = (c^3 - 3c^2 + 3c)b + (-c^3 + 3c^2 - 3c) $$

and for all vertices among sequence $v$, we have:

$$ a.((c^3 - 3c^2 + 3c)b + (-c^3 + 3c^2 - 3c) + 1) = a((c^3 - 3c^2 + 3c)b + (-c^3 + 3c^2 - 3c + 1)). $$

Therefore, we can write it as our condition as:

$$ \begin{aligned} \sigma(C(a, b, c)) &= a((c^3 - 3c^2 + 3c)b + (-c^3 + 3c^2 - 3c + 1)) \\ &= a((c^2(c-3) + 3c)b + c^2(c-3) - 3c + 1) \\ &= a(c^2(c-3)b + (3(b-1)c) + (c^3 - 3c + 1)). \end{aligned} $$

As we soppse $a \ge 3, b = a + 1, c = a + 2$, we have:

$$ \begin{aligned} \sigma(C(a, b, c)) &= a(c^2(c-3)(a+1) + (3ac) + (c^3 - 3c + 1)) \\ &= a(c^2(c-3)(a+1) + (c^3 - 3c + (3a+1))). \end{aligned} $$

□

**Theorem 19** Let be $n, m, r$ integer numbers where $n \le m \le r$, and $C(m, n, r)$ is caterpillar tree of order $(n, m, r)$ vertices, Sigma index of $C(m, n, r)$ where $n \ge 3, m \ge 3, r \ge 3$ given as:

$$ \sigma(C(n, m, r)) = \sum_{n=3, r=3} \left( n \left( r^2(r-3)(n+1) + (r^3 - 3r + (3n+1)) \right) \right) $$

or in an approximate equivalent variation according as

$$ \sigma(C(m, n, r)) \approx \sum \left( 3 \left( \frac{1}{12}(n+m+r)^3 - (n+m+r) \right) \right). $$

where $\{n, m, r\} \in \mathbb{N}$. (See Figure 5).

**Proof** We will need to prove one of the two relationships, so we have chosen the first relationship, while the second relationship can be obtained by imposing a specific dependency and treating it as such, so for the first case we have: assume a sequence is $v = (a, b, c)$ where $a \ge 3, b = a + 1, c = b + 1$, or in another formula we can be write $a \ge 3, b = a + 1, c = a + 2$, then we have, for case of vertices $b, c$ we have:

$$ ((c-1)^3 + 1)b - 1 = (c^3 - 3c^2 + 3c)b + (-c^3 + 3c^2 - 3c) $$

and for all vertices among sequence $v$, we have:

$$ a.((c^3 - 3c^2 + 3c)b + (-c^3 + 3c^2 - 3c) + 1) = a((c^3 - 3c^2 + 3c)b + (-c^3 + 3c^2 - 3c + 1)). $$
Therefore, we can write it as our condition as:

$$
\begin{align*}
\sigma(C(a, b, c)) &= a((c^3 - 3c^2 + 3c)b + (-c^3 + 3c^2 - 3c + 1)) \\
&= a((c^2(c - 3) + 3c)b + c^2(c - 3) - 3c + 1) \\
&= a(c^2(c - 3)b + (3(b - 1)c) + (c^3 - 3c + 1)).
\end{align*}
$$

As we suppose $a \ge 3, b = a + 1, c = a + 2$, we have:

$$
\begin{align*}
\sigma(C(a, b, c)) &= a(c^2(c-3)(a+1) + (3ac) + (c^3 - 3c + 1)) \\
&= a(c^2(c-3)(a+1) + (c^3 - 3c + (3a+1))).
\end{align*}
$$

□

4 Conclusion

There are many and varied topological indices and to study them on caterpillar trees,
we need a huge number of research papers that can cover general cases and special
cases. In this paper, we presented a study of two topological indices, Albertson index
and Sigma index on caterpillar trees in the general case indicated in Figure 3, as
mentioned in the introduction, has important applications in chemistry and computer
science, and we have established the main result for triple vertices where irr(C(n, 3)) =
12n - 4 and about sigma index on caterpillar tree we have σ(C(n, 3)) = 48(n - 2) + 56.
Also we show that in both cases maximum and minimum is one case, and we obtained
a set of characteristics that discuss the special cases of them and then included them
in general.

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<figure><img src="image_10.png" /></figure>
<figure><img src="image_11.png" /></figure>