On trees with a given number of segments and their maximum general $Z$-type index

1.   Introduction

Molecular descriptors are the numerical values that describe certain characteristics of molecules. These numbers are essential to the creation of a number of mathematical chemistry models, such as the QSAR/QSPR (quantitative structure activity/property relationship) models, which help forecast the physicochemical properties and biological activities of new compounds. One specific class of these descriptors is the class of topological indices; such descriptors depend on the graph of the structure of the compound under consideration. There are various such indices in the literature on mathematical chemistry, most of which are studied in relation to chemical graphs. The general $Z$-type index–a topological index introduced recently by Chen and Lin ^[5]–is the subject of this study.

To avoid trivialities, we are explicitly only taking connected graphs into account in this study. The notation $E(G)$ is used to represent the edge set of a graph $G$, while $V(G)$ is used to represent $G$'s vertex set. To express the degree of a vertex $u \in V (G)$, we use $d_{u}(G)$ (or just $d_u$, where there is no possibility of misunderstanding regarding the considred graph). As stated in ^[22], the sum-connectivity index $\chi_{-\frac{1}{2}}$ of a graph $G$ is the sum of the values $(d_{ {u}} + d_{v})^{-1/2}$ across all of $G$'s edges $uv$.

In ^[21], the index $\chi_{-\frac{1}{2}}$ was generalized under the term "general sum-connectivity index" by substituting $\alpha$ for "$-1/2$, " that is, $\chi_{ \alpha}$, where $\alpha$ is a real number that differs from zero. The relationship between a graph's $\chi_{ \alpha}$ and its line graph was examined by Chen ^[4]. Milovanović et al. ^[12] examined a number of inequalities for $\chi_{ \alpha}$. The extremum values of $\chi_{ \alpha}$ for trees with a certain maximum degree were examined by Swartz and Vetrík ^[16]. For graphs with a fixed cyclomatic number, Ali et al. ^[3] addressed a problem about extremum values of $\chi_{ \alpha}$. Zhong and Qian ^[20] investigated a problem involving trees with a given matching number and minimum $\chi_{ \alpha}$. By fixing the diameter and girth of unicyclic networks, Vetrík ^[19] investigated this index. Further information about certain extremal results involving $\chi_{ \alpha}$ can be found in ^[17,18]. The indices $\chi_{1}, \chi_{2}$, and $2\chi_{-1}$ are equivalent to the well-known first Zagreb index ^[15], hyper Zagreb index ^[10], and harmonic index ^[7], respectively.

The platt index ^[8,14] is defined as

The general platt index ^[2] is defined as

Additional detail on the general Platt index can be found in ^[1,3].

The general $Z$-type index ^[5] is defined as

where $\beta$ is a nonzero real number different from $0$. Certainly, $Z_{\alpha, \beta}$ generalizes all the abovementioned indices, namely, the sum-connectivity index, general sum-connectivity index (and hence the first Zagreb index, hyper Zagreb index, and harmonic index), Platt index, general Platt index (and hence the reformulated first Zagreb index ^[11]).

A tree is said to be a chemical tree if the degree of each vertex is less than $5$. A graph with $n$ vertices is called an $n$-order graph. A segment in a tree $T$ is defined ^[6,9] as a non-trivial path of $T$, indicated by $P: x_{1} x_{2} \ldots x_{r}$ such that $d_{x_{1}}(T), d_{x_{r}}(T) \not\in \{2\}$, and $d_{x_{i}}(T) = 2$ whenever $2 \leq i \leq r - 1$; in addition, if $\min\{d_{x_{1}}(T), d_{x_{r}}(T)\} = 1$ and $\max\{d_{x_{1}}(T), d_{x_{r}}(T)\}\ge3$ then $P$ is called a pendent path of $T$; however, if $\min\{d_{x_{1}}(T), d_{x_{r}}(T)\}\ge3$ and $\max\{d_{x_{1}}(T), d_{x_{r}}(T)\}\ge3$ then $P$ is called an internal path of $T$.

In the present paper, our main aim is to study the general $Z$-type index (and hence the abovementioned particular cases of this index, including the general Platt index) of $n$-order trees and chemical trees with a given number of segments. Particularly, the main goal of the present paper is to study the greatest value of $Z_{\alpha, \, \beta}$ of fixed-order chemical trees with a fixed number of segments for $1 < \alpha \le 3$ and $\beta\le2$. Similar results for general trees are also established for $\alpha > 1$ and $\beta\le2$.

2.   Main results

A vertex $u$ in a tree $T$ with $d_u(T) = 1$ or $d_u(T) > 2$ is called a pendent vertex of $T$ or a branching vertex of $T$, respectively. A star-like tree is a tree containing only one branching vertex. To avoid trivialities, throughout this section, we consider trees containing not less than three segments.

Define $\Theta_{i, j}(T): = |\{uv\in E(T): d_u(T) = i, \; d_u(T) = j\}|$ and $m_i(T): = |\{u\in V(T): d_u(T) = i\}|$. For $2\le j \le 4$ and $n\ge3$, we have the following system of equations for an $n$-order chemical tree $T$:

For $u\in V(T)$, we define $N_{T} (u) = \{u'\in V(T):u'u\in E(T)\}$. Let $^{1}\mathcal{T}_{n, r}$ denote the set of $n$-order trees having $r$ segments, provided that the chain of inequalities $3\le r\le n-1$ holds. We denote by $^{1}T_{max}$ a tree having the greatest value of $Z_{\alpha, \, \beta}$ in the set $^{1}\mathcal{T}_{n, r}$ for $\beta\le2$ and $\alpha > 1$.

Lemma 2.1. If $\Theta_{1, k}(^{1}T_{max})\ne0$ for some $k$ with $k\ge3$, then every vertex of degree $2$ (if exists) in $^{1}T_{max}$ has a pendent neighbor.

Proof. We chose $t_1, t_2\in V(^{1}T_{\text{max}})$ in such a way that ${t}_{1}{t}_{2}\in E(^{1}T_{max})$, $d_{ {t}_{2}}(^{1}T_{max})\geq 3$ and $d_{ {t}_{1}}(^{1}T_{max}) = 1$. Also, we assume contrarily that $x, x_1, x_2\in V(^{1}T_{max})$, such that $x_{1}x, x_{2}x\in E(^{1}T_{max})$, $d_x(^{1}T_{max}) = 2$ and $d_{x_{i}}(^{1}T_{max}) \geq 2$ for $i = 1, 2$. If $T^{*}$ denotes the tree formed from $\; ^{1} T_{max}$ by dropping the edges ${t}_{1} {t}_{2}, xx_{1}, xx_{2}$ and adding the edges ${t}_{1}x, {t}_{2}x, x_{1} x_{2}$, then certainly $T^{*} \in\; ^{1} \mathcal{T}_{ n, r}$. In the following, we assume $d_{p}(^{1} T_{max}) = d_p$ for every $p\in V(^{1} T_{max}) = V(T^*)$. The, we have

Let us define a function $f$ of real variables $a, b, c,$ with fixed real numbers $\alpha$ and $\beta$ as

where $\alpha > 1$, $\beta\le2$, $a\ge3$, $b\ge2$ and $c\ge2$. We note that the function $f$ is strictly decreasing in each of its variables $a, b, c.$ Hence, (2.3) gives

a contradiction to the definition of the tree $^{1}T_{max}$. □

Lemma 2.2. The tree $^{1}T_{max}$ must be a star-like tree.

Proof. Since $^{1}T_{max}$ has at least three segments, its number of branching vertices must be nonzero. We chose a vertex $t' \in V(^{1}T_{max})$ of maximum degree. Contrarily, assume that the number of branching vertices of the tree $^{1}T_{max}$ is at least $2$. Among all the branching vertices of $^{1}T_{max}$ different from $t'$, we pick a branching vertex $t \in V(^{1}T_{max})\setminus \{t'\}$ such that the vertices $t'$ and $t$ have the minimum distance between them. two distinct branching vertices. Let $d_{ {t}^{'}}(^{1}T_{max}) = \tau'$ and $d_{t}(^{1}T_{max}) = \tau$. Furthermore, we assume that $N_{^{1}T_{max}}(t) = \{v_l, t_1, \dots, t_{\tau-1}\}$ and $N_{^{1}T_{max}}(t') = \{y, t_1', \dots, t_{\tau'-1}'\}$, where the vertices $v_l$ and $y$ lie on the unique path connecting $t$ and $t'$. Let $T^{*}$ be the tree obtained from $^{1} T_{max}$ by removing $t_1t, t_2t, \dots, t_{\tau-1}t$ and adding $t_1t', t_2t', \dots, t_{\tau-1}t'\}.$ In the remaining proof, we assume that $d_\gamma = d_\gamma(^{1} T_{max})$ for every $\gamma\in V(^{1} T_{max}) = V(T^{*})$.

Case 1: ${t}^{'}{t}\notin E(^{1} T_{max})$.

We note in the present case that $d_y = d_{v_l} = 2$ and hence

We note that there are two real numbers $\tau_1$ and $\tau_2$ satisfying the inequalities $3-\beta < \tau_1 < \tau-\beta+2$ and $\tau'-\beta+2 < \tau_2 < \tau'+\tau-\beta+1$ such that

As $\tau'\ge \tau$, we have $\tau_1 < \tau_2$ and hence from (2.4) and (2.5), we arrive at $Z_{\alpha, \, \beta} (^{1} T_{max})- Z_{\alpha, \, \beta} (T^{*}) < 0$, a contradiction.

Case 2: ${t}^{'}{t}\in E(^{1} T_{max})$.

In the present case, we obtain

again a contradiction.

In both possible cases, we arrive at a contradiction. Therefore, the tree $^{1}T_{max}$ contains exactly one branching vertex. □

Theorem 1. If the chain of inequalities $\lceil (n-1)/2 \rceil \le r \le n-1$ holds then the tree $^{1} T_{max}$ is star-like and $\Theta_{2, 2}(^{1} T_{max}) = 0$.

Proof. Lemma $2.2$ confirms that the tree $^{1} T_{max}$ is star-like. Hence, the number of segments $r$ of $^{1} T_{max}$ is equal to its maximum degree. Suppose, contrarily, that $\Theta_{2, 2}(^{1} T_{max}) > 0$. Then, Lemma $2.1$ confirms that $\Theta_{r, 1}(^{1} T_{max}) = 0$. Consequently, we have $2r = \Theta_{2, r}(^{1} T_{max})+\Theta_{1, 2}(^{1} T_{max})\le |E(^{1} T_{max})|-1 = n-2,$ a contradiction. □

Theorem 2.2. If the chain of inequalities $3\le r \le \lfloor (n-1)/2 \rfloor$ holds then the tree $^{1} T_{max}$ is star-like and $\Theta_{r, 1}(^{1} T_{max}) = 0$.

Proof. Lemma $2.2$ confirms that the tree $^{1} T_{max}$ is star-like. Hence, the number of segments $r$ of $^{1} T_{max}$ is equal to its maximum degree. Suppose, contrarily, that $\Theta_{r, 1}(^{1} T_{max}) > 0$. Then, Lemma $2.1$ confirms that $\Theta_{2, 2}(^{1} T_{max}) = 0$. So, we have $2r > \Theta_{1, 2}(^{1} T_{max})+r = |E(^{1} T_{max})| = n-1,$ a contradiction. □

In the rest of this section, we focus on chemical trees. Denote by $\mathcal{T}_{n, r}$ the set ofall $n$-order chemical trees possessing $r$ segments such that $3\le r\le n-1$. For $r\in\{3, 4\}$, the tree(s) maximizing $Z_{\alpha, \, \beta}$ in the set $\mathcal{T}_{n, r}$ for $\beta\le2$ and $\alpha > 1$ can be obtained directly by utilizing Theorem 2.1 and/or Theorem 2.2. Hence, in the remaining part of the present section, we assume that $5\le r\le n-1$. Denote by $T_{max}$ a tree with the greatest value of $Z_{\alpha, \, \beta}$ in the set $\mathcal{T}_{n, r}$ with the conditions $1 < \alpha\le 3$, $\beta\le2$ and $5 \le r \le n-1$.

Lemma 2.3. It holds that $m_3(T_{max})\le2$.

Proof. Contrarily, assume that $m_3(T_{max})\ge3$. We pick the vertices $x, y, z\in V(T_{max})$ of degree $3$ in such a way that if all these three vertices lie on one path then the vertex $y$ must lie on the path connecting $x$ and $z$. Let $P : (x = )u_{1} u_{2} \ldots u_{l}( = z)$ be the path connecting $z$ and $x$ in $T_{max}$. Let $N_z(T_{max}) = \{z_1, z_2, z_3\}$, where $z_3 = u_{l-1}$. Certainly, $x, y, z,$ must be pairwise nonadjacent when these vertices do not lie on one path. Let $T^{'}$ denote the tree constructed from $T_{max}$ by deleting the edges $zz_{1}, zz_{2}$ and inserting $xz_{1}, yz_{2}$. In the rest of the proof, we take $d_\gamma(T_{max}) = d_\gamma$ for $\gamma \in V(T^{'}) = V(T_{max})$. We may assume, without loss of generality, that $d_{z_2}\le d_{z_1}$.

Case 1: The vertices $x, y, z,$ are pairwise nonadjacent.

In this case, we obtain

We note that each of the vertices $x, y, z,$ may have at most two pendent neighbors; if $x, y, z,$ lie on one path then $y$ may have at most one pendent neighbor. Thus, Equation (2.6) implies that

We note that there exist two real numbers $a_1$ and $a_2$ which satisfy the inequalities $6-\beta < a_1 < 7-\beta$ and $4-\beta < a_2 < 5-\beta$ such that

Since

we have

and hence

which confirms that the right-hand side of Eq (2.8) is negative; thus, (2.7) yields $Z_{\alpha, \, \beta} (T_{max}) < Z_{\alpha, \, \beta} (T^{'})$, a contradiction.

Case 2: Vertices $x, y, z,$ lie on one path, and only one of $x, z,$ is a neighbor of $y$.

We suppose, without loss of generality, that $xy\not\in E(T_{max})$ and $yz \in E(T_{max})$. Then $y = z_3$. Thus, we obtain

a contradiction.

Case 3: Vertices $x, y, z,$ lie on one path provided that $xy\in E(T_{max})$ and $yz \in E(T_{max})$.

In this case, we obtain

Thus, we arrive at $Z_{\alpha, \, \beta} (T_{max}) < Z_{\alpha, \, \beta} (T^{'})$, a contradiction. □

Lemma 2.4. Every internal path of the tree $T_{max}$ has length $1$.

Proof. Assume to the contrary that $P : {t}_1 \ldots {t}_k$ is an internal path of length $k-1\ge2$ in $T_{max}$. Let $x\in V(T_{max})$ be a pendent vertex, and let $t$ be its unique neighbor. Let $T^{'}$ be the tree deduced from $T_{max}$ by dropping ${t} x, {t}_1 {t}_2, {t}_{k-1} {t}_k$ and adding ${t}_1 {t}_k, {t}_2 x, {t}_{k-1} {t}$. In the following, we use $d_\gamma$ to represent the degree of a vertex $\gamma$ in $T_{max}$. It is clear that $T^{'} \in \mathcal{T}_{ n, r}$. On the other hand, we have

a contradiction. □

Lemma 2.5. If $\Theta_{1, s}(T_{max})\ne0$ for some $s$ with $s\geq 3$, then $T_{max}$ contains no pendent path of length larger than $2$.

Proof. Let $t_1, t_2\in V(T_{max})$ be two adjacent vertices such that $d_{ {t}_{1}}(T_{max}) = 1$ and $d_{ {t}_{2}}(T_{max})\geq 3$. Contrarily, we assume that $P:x_1\dots x_k$ is a pendent path having length $k\ge3$ in $T_{max}$, where $d_{x_{k}}(T_{max}) = 1$ and $d_{x_{1}}(T_{max}) \geq 3$. Let $T^{*}$ be the tree formed from $T_{max}$ by deleting ${t}_{1} {t}_{2}, xx_{1}, xx_{2}$ and adding ${t}_{1}x, {t}_{2}x, x_{1} x_{2}$. Clearly, $T^{*} \in\; \mathcal{T}_{ n, r}.$ But,

which is a contradiction. □

Lemma 2.6. If $\Theta_{1, 4}(T_{max}) \neq 0$ then $\Theta_{2, 3}(T_{max}) = 0$.

Proof. We assume to the contrary that $\Theta_{2, 3}(T_{max})\ne0$ as well as $\Theta_{1, 4}(T_{max}) \neq 0$. We consider four vertices ${t}_{2}, t_3, t_4, t_5$ of $T_{max}$ such that $t_4t_5, t_2t_3\in E(T_{max})$ and $(d_{ {t}_{2}}(T_{max}), d_{t_3}(T_{max}), d_{t_4}(T_{max}), d_{t_5}(T_{max})) = (2, 3, 4, 1)$. Let $N_{T_{max}}({t}_{2}) = \{ {t}_{1}, {t}_{3}\}$. Then Lemmas 2.4 and 2.5 confirm that $d_{ {t}_{1}}(T_{max}) = 1$. Let $T^{'}$ be the graph generated from $T_{max}$ by removing ${t}_{1} {t}_{2}, {t}_{2} {t}_{3}, {t}_{4} {t}_{5}$ and inserting ${t}_{1} {t}_{3}, {t}_{2} {t}_{4}, {t}_{2} {t}_{5}.$ Clearly, we have $T^{'} \in \mathcal{T}_{ n, r}$. So, we have

a contradiction. □

Lemma 2.7. If $t_j \in V(T_{max})$ is a vertex of degree three, then $t_j$ has at most one branching neighbor.

Proof. Contrarily, suppose that ${t}_{j-1}, {t}_{ j+1}\in V(T_{max})$ are any two branching vertices adjacent to ${t}_{j}$. Let $P : {t}_{1} {t}_{2} \ldots {t}_{ j-1} {t}_{j} {t}_{j+1}\ldots {t}_{l}$ be the largest path in $T_{\text{max}}$ containing the aforementioned branching vertices. Because of Lemma $2.3$, the path $P$ contains not more than two vertices having degree $3$ (in $T_{\text{max}}$) including ${t}_{j}$. In the case when the path $P$ has two vertices having degree $3$ (in $T_{\text{max}}$) then without loss of generality, we suppose that ${t}_{i} \in V(P)$ is a vertex with degree $3$ (in $T_{\text{max}}$) for some ${i}$ with $1\leq {i} < {j}$. Thus, there is a vertex ${t}_{k} \in V(P)$ of degree $4$ for some $k$ with ${j}+1 \leq k \leq l-1$, which has only one branching neighbor. So, $d_{{t}_{k+1}}(T_{\text{max}}) = 1\;$or$\; 2,$ $d_{{t}_{{j}-1}}(T_{\text{max}}) = 3\;$or$\; 4$, and $d_{{t}_{{j}+1}}(T_{\text{max}}) = 4.$ If $T^{*}$ is the tree constructed from $T_{max}$ by dropping ${t}_{ {j}-1} {t}_{ {j}}, {t}_{ {j}} {t}_{ {j}+1}, {t}_{ k} {t}_{k+1}$ and inserting ${t}_{ {j}-1} {t}_{ {j}+1}, {t}_{k} {t}_{ {j}}, {t}_{ {j}} {t}_{k+1}$, then $T^{*} \in \mathcal{T}_{n, r}$ and

which is a contradiction, where $d_{{t}_{{j}-1}} = d_{{t}_{{j}-1}}(T_{\text{max}})$ and $d_{{t}_{k+1}} = d_{{t}_{k+1}}(T_{\text{max}})$. □

From Lemmas 2.4 and 2.7, next result follows.

Corollary 1. The induced subgraph of $T_{\mathit{\text{max}}}$ formed by its vertices of degree $4$ is a tree.

For every $\ell\in \{2, 3, 4\}$, denote by $^{\ell} \mathcal{T}_{n, r}$ the set of those $n$-order chemical trees having $r$ segments that has/have $\ell-2$ vertex/vertices of degree $3$; so, because of ^[13], it holds that $r \equiv 2\ell \pmod{3}$.

Lemma 2.8. ^[13] The degree sequence of $T\in\mathcal{T}_{n, r}$ having no more than $2$ vertices of degree $3$, with $5 \le r \le n-1$, is as follow:

Theorem 3. If $T\in \mathcal{T}_{n, r}$ with $7\le r < n < \frac{5r+7}{3}$ and $r \equiv 1\! \pmod{3}$, then

Proof. We assume that $^{2} T_{\text{max}}$ is a tree having the maximum value of $Z_{\alpha, \, \beta}$ over the set $\mathcal{T}_{n, r}$ provided that $7\le r < n < \frac{5r+7}{3}$ and $r \equiv 1\! \pmod{3}$. Lemmas 2.3 and 2.8 confirm that $DS_{2}$ is the degree sequence of the tree $^{2} T_{\text{max}}$. Thus, $\Theta_{3, k}(^{2} T_{\text{max}}) = 0$ for every $k\in\{1, 2, 3, 4\}$. The condition $r\geq 7$ confirms that ${m}_4(^{2} T_{\text{max}}) \geq 2$. Now, because of Corollary 1, it holds that $\Theta_{4, 4}(^{2} T_{\text{max}}) = {m}_4 -1$ and so $\Theta_{4, 4}(^{2} T_{\text{max}}) = \frac{ r-4}{3}$. Also, the constraint $n < \frac{5r+7}{3}$ implies that ${m}_1 > {m}_2$; hence, Lemmas $2.4$ and $2.5$ confirm that $\Theta_{2, 2}(^{2} T_{\text{max}}) = 0$. Finally, Eq (2.1) gives

Hence, we calculate $Z_{\alpha, \, \beta} (^{2} T_{max})$, which is the same as the right-hand side of the desired inequality. □

Theorem 4. If $T\in \mathcal{T}_{n, r}$ with $7\le r$ and $n \geq \frac{5r+7}{3}$ and $r \equiv 1\! \pmod{3}$, then

Proof. We assume that $^{2} T_{\text{max}}$ is a tree having the maximum value of $Z_{\alpha, \, \beta}$ over the set $\mathcal{T}_{n, r}$ under the given constraints. Lemmas 2.3 and 2.8 confirm that $DS_{2}$ is the degree sequence of the tree $^{2} T_{\text{max}}$. Thus, $\Theta_{3, k}(^{2} T_{\text{max}}) = 0$ for every $k\in\{1, 2, 3, 4\}$. Now, by using Corollary 1, we obtain $\Theta_{4, 4}(^{2} T_{\text{max}}) = \frac{ r-4}{3}$. Lemmas $2.4$ and $2.5$ confirm that $\Theta_{1, 4}(^{2} T_{\text{max}}) = 0$. Finally, by utilizing Eq (2.1), we obtain

Hence, we calculate $Z_{\alpha, \, \beta} (^{2} T_{max})$, which is the same as the right-hand side of the desired inequality. □

Theorem 5. If $T\in \mathcal{T}_{n, r}$ with $6\le r < n < \frac{5r+3}{3}$ and $r \equiv 0\! \pmod{3}$, then

Proof. We assume that $^{3} T_{\text{max}}$ is a tree having the maximum value of $Z_{\alpha, \, \beta}$ over the set $\mathcal{T}_{n, r}$ under the given constraints. Lemmas 2.3 and 2.8 confirm that $DS_{3}$ is the degree sequence of $^{3} T_{\text{max}}$; hence, $\Theta_{3, 3}(^{3} T_{\text{max}}) = 0$. By using Corollary 1, we have $\Theta_{4, 4}(^{3} T_{\text{max}}) = \frac{ r}{3} -2.$ Lemmas $2.4$ and $2.7$ confirm that $\Theta_{3, 4}(^{3} T_{\text{max}}) = 1$. Also, note that $\Theta_{2, 2}(^{3} T_{\text{max}}) = 0$ and $\Theta_{1, 4}(^{3} T_{\text{max}}) \neq 0$; so, by utilizing Lemma $2.6$, we obtain $\Theta_{2, 3}(^{3} T_{\text{max}}) = 0$. Finally, by using Eq (2.1), we obtain $\Theta_{1, 4}(^{3} T_{\text{max}}) = \frac{5 r}{3}- n, \; \Theta_{1, 2}(^{3} T_{\text{max}}) = \Theta_{2, 4}(^{3} T_{\text{max}}) = n- r-1, \Theta_{3, 1}(^{3} T_{\text{max}}) = 2.$ Hence, we calculate $Z_{\alpha, \, \beta} (^{3} T_{max})$, which is the same as the right-hand side of the desired inequality. □

Theorem 6. If $T\in \mathcal{T}_{n, r}$ with $6\le r < n = \frac{5r+3}{3}$ and $r \equiv 0\! \pmod{3}$, then

Proof. We assume that $^{3} T_{\text{max}}$ is a tree having the maximum value of $Z_{\alpha, \, \beta}$ over the set $\mathcal{T}_{n, r}$ under the given constraints. Lemmas 2.3 and 2.8 confirm that $DS_{3}$ is the degree sequence of $^{3} T_{\text{max}}$; hence, $\Theta_{3, 3} = 0$. Now, by using Corollary 1, Lemmas $2.4$ and $2.7$, and Eq (2.1), we obtain

Hence, we calculate $Z_{\alpha, \, \beta} (^{3} T_{max})$, which is the same as the right-hand side of the desired inequality. □

Theorem 7. If $T\in \mathcal{T}_{n, r}$ with $6\le r$ and $r \equiv 0\! \pmod{3}$ and $n > \frac{5r+3}{3}$, then

Proof. We assume that $^{3} T_{\text{max}}$ is a tree having the maximum value of $Z_{\alpha, \, \beta}$ over the set $\mathcal{T}_{n, r}$ under the given constraints. Lemmas 2.3 and 2.8 confirm that $DS_{3}$ is the degree sequence of $^{3} T_{\text{max}}$; hence, $\Theta_{3, 3}(^{3} T_{\text{max}}) = 0$. Now, by using Corollary 1, Lemmas $2.4$, $2.5$ and $2.7$, and Eq (2.1), we obtain

Hence, we calculate $Z_{\alpha, \, \beta} (^{3} T_{max})$, which is the same as the right-hand side of the desired inequality. □

Theorem 8. If $T\in \mathcal{T}_{n, r}$ with $8\le r < n < \frac{5r-4}{3}$ and $r \equiv 2\! \pmod{3}$, then

Proof. We assume that $^{4} T_{\text{max}}$ is a tree having the maximum value of $Z_{\alpha, \, \beta}$ over the set $\mathcal{T}_{n, r}$ under the given constraints. Lemmas 2.3 and 2.8 confirm that $DS_{4}$ is the degree sequence of $^{4} T_{\text{max}}$. By keeping in mind the given constraints, using Corollary 1, Lemmas $2.4$, $2.6$ and $2.7$, and Eq (2.1), we obtain

Hence, we calculate $Z_{\alpha, \, \beta} (^{4} T_{max})$, which is the same as the right-hand side of the desired inequality. □

Theorem 9. If $T\in \mathcal{T}_{n, r}$ with $8\le r$ and $r \equiv 2\! \pmod{3}$ and $\frac{5 r-4}{3} \leq n \leq \frac{5 r+2}{3}$, then

Proof. We assume that $^{4} T_{\text{max}}$ is a tree having the maximum value of $Z_{\alpha, \, \beta}$ over the set $\mathcal{T}_{n, r}$ under the given constraints. Lemmas 2.3 and 2.8 confirm that $DS_{4}$ is the degree sequence of $^{4} T_{\text{max}}$. By keeping in mind the given constraints, using Corollary 1, Lemmas $2.4$ and $2.7$, and Eq (2.1), we obtain

Hence, we calculate $Z_{\alpha, \, \beta} (^{4} T_{max})$, which is the same as the right-hand side of the desired inequality. □

Theorem 10. If $T\in \mathcal{T}_{n, r}$ with $8\le r$ and $r \equiv 2\! \pmod{3}$ and $n > \frac{5 r+2}{3}$, then

Proof. We assume that $^{4} T_{\text{max}}$ is a tree having the maximum value of $Z_{\alpha, \, \beta}$ over the set $\mathcal{T}_{n, r}$ under the given constraints. Lemmas 2.3 and 2.8 confirm that $DS_{4}$ is the degree sequence of $^{4} T_{\text{max}}$. By keeping in mind the given constraints, using Corollary 1, Lemmas $2.4$, $2.5$ and $2.7$, and Eq (2.1), we obtain

Hence, we calculate $Z_{\alpha, \, \beta} (^{4} T_{max})$, which is the same as the right-hand side of the desired inequality. □

3.   Conclusions

We have characterized graphs attaining the greatest value of $Z_{\alpha, \, \beta}$ in the set of all fixed-order trees with a fixed number of segments for $\alpha > 1$ and $\beta\le2$ (see Theorems 1 and 2). We have also found the largest value of $Z_{\alpha, \, \beta}$ trees belonging to the aforementioned set of trees for $1 < \alpha \le 3$ and $\beta\le2$ (see Theorems 3–10). The obtained results also hold for the general Platt index $Pl_{\alpha}$ because $Z_{\alpha, \, \beta}$ is a generalized version of $Pl_{\alpha}$.

All the results proved in this paper hold for $\alpha > 1$, except Lemma 2.3 (particularly, the desired inequality in its Case 1). It seems to be interesting to prove this lemma (particularly, the desired inequality in its Case 1) for $\alpha > 3$ and $\beta\le2$.

Author contributions

Hicham Saber: Writing-review & editing, Funding acquisition; Zahid Raza: Writing-original draft, Writing-review & editing; Abdalaziz M. Alanazi: Writing-original draft, Writing-review & editing; Adel A. Attiya: Writing-review & editing, Funding acquisition; Akbar Ali: Writing-review & editing, Funding acquisition. All authors have read and approved the final version of the manuscript for publication.

Use of Generative-AI tools declaration

The authors declare that they have not used Artificial Intelligence (AI) tools in the creation of this article.

Acknowledgments

This work is supported by the Scientific Research Deanship, University of Ha'il, Saudi Arabia, through project number RG-24 059.

Conflict of interest

Authors have no conflict of interests to declare.