Maximum General Sum-Connectivity Index of Trees and Unicyclic Graphs with Given Order and Number of Pendant Vertices

Abstract

For $a\in \mathbb{R}$, the general sum-connectivity index of a graph G is defined as ${\chi }_a\left(G\right)={\sum }_{uv\in E\left(G\right)}{[d_G\left(u\right)+d_G\left(v\right)]}^a$, where $E\left(G\right)$ is the set of edges of G and $d_G\left(u\right)$ and $d_G\left(v\right)$ are the degrees of vertices u and v, respectively. For trees and unicyclic graphs with given order and number of pendant vertices, we present upper bounds on the general sum-connectivity index ${\chi }_a$, where $0<a<1$. We also present the trees and unicyclic graphs that attain the maximum general sum-connectivity index for $0<a<1$.

1. Introduction

Topological indices have been used for purposes including chemical documentation, quantitative structure-versus-property/activity relationships (QSPR/QSAR), toxicological hazard assessments, isomer discrimination, drug design and combinatorial library design. They have been used in the process of correlating chemical structures with various characteristics such as boiling points and molar heats of formation. Topological indices provide a convenient method of translating chemical constitutions into numerical values that are applied in analyzing correlations with physical properties.

For a graph G, let $V\left(G\right)$ be its set of vertices and $E\left(G\right)$ be its set of edges. The order of G is its number of vertices. The degree $d_G\left(v\right)$ of a vertex v is the number of vertices adjacent to v in G. A pendant vertex is a vertex of degree 1. A tree is a connected graph that does not contain any cycle, and a unicyclic graph is a connected graph that contains exactly one cycle.

The general sum-connectivity index is among the well-known and extensively studied indices. For a graph G, it was defined by Zhou and Trinajstić [1] as

$${\chi }_a\left(G\right)=\sum _{uv\in E\left(G\right)}{[d_G\left(u\right)+d_G\left(v\right)]}^a,$$

where $a\in \mathbb{R}$. Ali, Zhong and Gutman [2] presented a survey paper on ${\chi }_a$. Two papers on ${\chi }_a$ for trees and unicyclic graphs have been published in 2025 so far [3,4]. Note that in [4], graphs with a given maximum degree were studied, while in this paper, we focus on a different parameter. It is interesting that changing the parameter of a graph yields completely different extremal graphs for the interval $0<a<1$ considered in this paper.

Trees and unicyclic graphs are important families of graphs, since they can represent various chemical structures. We study ${\chi }_a$ for trees and unicyclic graphs with order n and k pendant vertices. For $2\le k\le n-1$, let $B_{n,k}$ denote the tree formed by connecting $k-1$ new vertices to one end vertex of a path of order $n-k+1$; see Figure 1. Note that $B_{n,2}$ and $B_{n,n-1}$ are a path and a star, respectively.

Figure 1. Tree $B_{n,k}$ of order n containing k pendant vertices.

For $1\le k\le n-3$, let $C{S}_{n-k,k}$ be the unicyclic graph of order n that contains a cycle having $n-k$ vertices, where one vertex of that cycle is adjacent to k pendant vertices; see Figure 2.

Figure 2. Unicyclic graph $C{S}_{n-k,k}$ of order n containing k pendant vertices and a cycle with $n-k$ vertices.

Below, we name two problems investigated in this paper.

Problem 1.

For trees with order n and k pendant vertices, find the trees with the minimum and maximum general sum-connectivity index ${\chi }_a$.

Let us present known results on Problem 1.

• Tomescu and Kanwal [5] proved that the tree $B_{n,k}$ has the minimum ${\chi }_a$ for $-1\le a<0$.
• Yao [6] determined the trees having the minimum ${\chi }_a$ for $0<a<1$, where $k\le \lfloor {\displaystyle \frac{n+2}{3}}\rfloor$.
• Albalahi and Ali [7] determined the trees with the minimum ${\chi }_a$ for $a>1$, where $4\le k\le \lfloor {\displaystyle \frac{n+5}{3}}\rfloor$.
• Cui and Zhong [8] presented the trees having the maximum ${\chi }_a$ for $-1\le a<0$, where $k\le \lfloor {\displaystyle \frac{n+2}{3}}\rfloor$.
• Tache and Tomescu [9] determined the trees with the maximum ${\chi }_a$ for $a\ge 1$.

Problem 2.

For unicyclic graphs with order n and k pendant vertices, find the graphs with the minimum and maximum ${\chi }_a$.

Less is known about graphs having extremal ${\chi }_a$ among unicyclic graphs with given n and k.

• Tomescu and Arshad [10] showed that the unicyclic graph $C{S}_{n-k,k}$ has the minimum ${\chi }_a$ for $-1\le a<0$.
• Tache and Tomescu [9] determined the unicyclic graphs with the maximum ${\chi }_a$ for $a\ge 1$.

We solve the parts of Problems 1 and 2 about maximum ${\chi }_a$ for $0<a<1$. For trees and unicyclic graphs with order n and k pendant vertices, we present sharp upper bounds on ${\chi }_a$ for $0<a<1$. We show that $B_{n,k}$ is the extremal tree and $C{S}_{n-k,k}$ is the extremal unicyclic graph for our bounds.

2. Preliminary Results

Let us present five lemmas. Lemmas 1, 3 and 5 are used in the proofs of our main results. Lemma 1 was proved in [11]. We include its proof because it is very short.

Lemma 1.

Let $0<a<1$, $1\le x<y$ and $c>0$. Then,

$${(x+c)}^a-x^a>{(y+c)}^a-y^a.$$

Proof. 

For the function $f\left(x\right)={(x+c)}^a-x^a$, we get $f^{\prime }\left(x\right)=a[{(x+c)}^{a-1}-x^{a-1}]<0$, since ${(x+c)}^{a-1}<x^{a-1}$. In that case, $f\left(x\right)$ is strictly decreasing. □

Lemma 2 is used in the proof of Lemma 3.

Lemma 2.

Let $0<a<1$, $x\ge 0$ and $c\ge 2$. Then,

$$1+{\displaystyle \frac{1-a}{x+c}}>{\left(1+{\displaystyle \frac{1}{x+c}}\right)}^{1-a}.$$

Proof. 

Let $0<a<1$, $x\ge 0$, $c\ge 2$ and

$$f\left(x\right)=1+{\displaystyle \frac{1-a}{x+c}}-{\left(1+{\displaystyle \frac{1}{x+c}}\right)}^{1-a}.$$

Then,

$$\begin{array}{cc}\hfill f^{\prime }\left(x\right)& =-{\displaystyle \frac{1-a}{{(x+c)}^2}}+(1-a){\left(1+{\displaystyle \frac{1}{x+c}}\right)}^{-a}{\displaystyle \frac{1}{{(x+c)}^2}}\hfill \\ \hfill & ={\displaystyle \frac{a-1}{{(x+c)}^2}}\left[1-{\left(1+{\displaystyle \frac{1}{x+c}}\right)}^{-a}\right]\hfill \\ \hfill & <0,\hfill \end{array}$$

since ${\displaystyle \frac{a-1}{{(x+c)}^2}}<0$ and $1-{(1+{\displaystyle \frac{1}{x+c}})}^{-a}>0$ for $0<a<1$, $c\ge 2$ and $x\ge 0$. Hence, $f\left(x\right)$ is strictly decreasing. Since $f\left(x\right)\to 0$, as $x\to \infty$, we get $f\left(x\right)>0$. Thus, the result follows. □

Let us show that two functions are increasing.

Lemma 3.

Let $0<a<1$. Then,

$$f_1\left(x\right)={(x+4)}^a+x{(x+3)}^a-x{(x+2)}^a$$

and

$$f_2\left(x\right)=2{(x+5)}^a+(x-1){(x+4)}^a-x{(x+3)}^a$$

are strictly increasing for $x\ge 0$.

Proof. 

Let $0<a<1$, $x\ge 0$ and $c\in \{2,3\}$. First, we show that $[x+(c+1)+xa]{[x+(c+1)]}^{a-1}>(x+c+xa){(x+c)}^{a-1}$, and then we use this inequality to prove that $f_1^{\prime }\left(x\right)>0$ and $f_2^{\prime }\left(x\right)>0$. We have

$$x+c>x+c+xa-xa-ca-x{a}^2=(1-a)(x+c+xa);$$

thus, ${\displaystyle \frac{1}{x+c+xa}}>{\displaystyle \frac{1-a}{x+c}}$. In combination with Lemma 2, we obtain

$$\begin{array}{cc}\hfill {\displaystyle \frac{x+c+1+xa}{x+c+xa}}& =1+{\displaystyle \frac{1}{x+c+xa}}>1+{\displaystyle \frac{1-a}{x+c}}>{\left(1+{\displaystyle \frac{1}{x+c}}\right)}^{1-a}\hfill \\ \\ ={\left({\displaystyle \frac{x+c+1}{x+c}}\right)}^{1-a}={\left({\displaystyle \frac{x+c}{x+c+1}}\right)}^{a-1}.\hfill \end{array}$$

Hence,

$$[x+(c+1)+xa]{[x+(c+1)]}^{a-1}>(x+c+xa){(x+c)}^{a-1}.$$

(1)

Now, we study the function $f_1\left(x\right)={(x+4)}^a+x{(x+3)}^a-x{(x+2)}^a$. The derivative of $f_1\left(x\right)$ is

$$\begin{array}{cc}\hfill f_1^{\prime }\left(x\right)& =a{(x+4)}^{a-1}+{(x+3)}^a+xa{(x+3)}^{a-1}-{(x+2)}^a-xa{(x+2)}^{a-1}\hfill \\ & =a{(x+4)}^{a-1}+(x+3+xa){(x+3)}^{a-1}-(x+2+xa){(x+2)}^{a-1}.\hfill \end{array}$$

Clearly $a{(x+4)}^{a-1}>0$. Using $c=2$ in (1), we get $(x+3+xa){(x+3)}^{a-1}>(x+2+xa){(x+2)}^{a-1}$. Thus $f_1^{\prime }\left(x\right)>0$, so $f_1\left(x\right)$ is strictly increasing for $x\ge 0$.

Let us study $f_2\left(x\right)=2{(x+5)}^a+(x-1){(x+4)}^a-x{(x+3)}^a$. The derivative of $f_2\left(x\right)$ is

$$\begin{array}{cc}\hfill f_2^{\prime }\left(x\right)& =2a{(x+5)}^{a-1}+{(x+4)}^a+(x-1)a{(x+4)}^{a-1}-{(x+3)}^a-xa{(x+3)}^{a-1}\hfill \\ \hfill & =2a{(x+5)}^{a-1}-a{(x+4)}^{a-1}+(x+4+xa){(x+4)}^{a-1}-(x+3+xa){(x+3)}^{a-1}.\hfill \end{array}$$

Using $c=3$ in (1), we get $(x+4+xa){(x+4)}^{a-1}>(x+3+xa){(x+3)}^{a-1}$. Note that $2a{(x+5)}^{a-1}>a{(x+4)}^{a-1}$, since

$$2>1+{\displaystyle \frac{1}{x+4}}>{\left(1+{\displaystyle \frac{1}{x+4}}\right)}^{1-a}={\left({\displaystyle \frac{x+5}{x+4}}\right)}^{1-a}={\left({\displaystyle \frac{x+4}{x+5}}\right)}^{a-1}.$$

Thus, $f_2^{\prime }\left(x\right)>0$, so $f_2\left(x\right)$ is strictly increasing for $x\ge 0$. □

Lemma 4 is used in the proof of Lemma 5.

Lemma 4.

Let $x,c_i,p_i\in \mathbb{R}$ for $i=1,2,3$, where $c_i\ge 0$, $p_i>0$, $p_i\ne 1$ and $\{c_1,c_2,c_3\}\ne \left\{0\right\}$. Then, the function

$$g\left(x\right)=c_1{p}_1^x+c_2{p}_2^x+c_3{p}_3^x$$

is concave up.

Proof. 

The second derivative

$$g^{″}\left(x\right)=c_1{\left(\ln p_1\right)}^2{p}_1^x+c_2{\left(\ln p_2\right)}^2{p}_2^x+c_3{\left(\ln p_3\right)}^2{p}_3^x>0;$$

thus, $g\left(x\right)$ is concave up. □

The inequality presented in Lemma 5 is used in the proof of Theorem 2.

Lemma 5.

Let $x\in \mathbb{R}$ and $k\in \mathbb{Z}$, where $x>0$ and $k\ge 2$. Then,

$$2{(k+4)}^x+(k-3){(k+3)}^x+(k-1){(k+1)}^x-(2k-2){(k+2)}^x>0.$$

Proof. 

Let $k\ge 3$. We consider the function

$$g\left(x\right)=2{\left({\displaystyle \frac{k+4}{k+2}}\right)}^x+(k-3){\left({\displaystyle \frac{k+3}{k+2}}\right)}^x+(k-1){\left({\displaystyle \frac{k+1}{k+2}}\right)}^x.$$

We have

$$\begin{array}{cc}\hfill g(-1)& ={\displaystyle \frac{2(k+2)}{k+4}}+{\displaystyle \frac{(k-3)(k+2)}{k+3}}+{\displaystyle \frac{(k-1)(k+2)}{k+1}}\hfill \\ \hfill & ={\displaystyle \frac{2k+4}{k+4}}+{\displaystyle \frac{k^2-k-6}{k+3}}+{\displaystyle \frac{k^2+k-2}{k+1}}\hfill \\ \hfill & ={\displaystyle \frac{2(k+4)-4}{k+4}}+{\displaystyle \frac{(k-4)(k+3)+6}{k+3}}+{\displaystyle \frac{k(k+1)-2}{k+1}}\hfill \\ \hfill & =\left(2-{\displaystyle \frac{4}{k+4}}\right)+\left(k-4+{\displaystyle \frac{6}{k+3}}\right)+\left(k-{\displaystyle \frac{2}{k+1}}\right)\hfill \\ \hfill & =2k-2-{\displaystyle \frac{4(k+3)(k+1)}{(k+4)(k+3)(k+1)}}+{\displaystyle \frac{6(k+4)(k+1)}{(k+4)(k+3)(k+1)}}-{\displaystyle \frac{2(k+4)(k+3)}{(k+4)(k+3)(k+1)}}\hfill \\ \hfill & =2k-2-{\displaystyle \frac{12}{(k+4)(k+3)(k+1)}}.\hfill \end{array}$$

In that case, $g(-1)<2k-2$ and $g\left(0\right)=2k-2$. By Lemma 4, $g\left(x\right)$ is concave up; thus, $g\left(x\right)>2k-2$ for $x>0$. Hence,

$$2{\left({\displaystyle \frac{k+4}{k+2}}\right)}^x+(k-3){\left({\displaystyle \frac{k+3}{k+2}}\right)}^x+(k-1){\left({\displaystyle \frac{k+1}{k+2}}\right)}^x-(2k-2)>0,$$

and Lemma 5 for $k\ge 3$ follows.

Let $k=2$. Note that from Lemma 4, it does not follow that $g\left(x\right)$ is concave up for $k=2$; therefore, the proof for $k=2$ cannot be included in the proof for the general case.

We need to prove that $2{(k+4)}^x+(k-3){(k+3)}^x+(k-1){(k+1)}^x-(2k-2){(k+2)}^x=2\left(6^x\right)-5^x+3^x-2\left(4^x\right)>0$. Let us consider the function

$$h\left(x\right)={\left({\displaystyle \frac{3}{2}}\right)}^x+{\left({\displaystyle \frac{3}{4}}\right)}^x.$$

By Lemma 4, $h\left(x\right)$ is concave up. Moreover, $h(-1)=h\left(0\right)=2$; thus, $h\left(x\right)>2$ for $x>0$. This implies that ${\left({\displaystyle \frac{3}{2}}\right)}^x+{\left({\displaystyle \frac{3}{4}}\right)}^x-2>0$, so $6^x+3^x-2\left(4^x\right)>0$ for $x>0$. Consequently,

$$2\left(6^x\right)-5^x+3^x-2\left(4^x\right)=(6^x-5^x)+[6^x+3^x-2\left(4^x\right)]>0$$

for $x>0$, since $6^x-5^x>0$. □

3. Main Results

The path and the star with n vertices are the only trees of order n with 2 and $n-1$ pendant vertices, respectively. Thus, we consider trees of order n with k pendant vertices, where $3\le k\le n-2$, in Theorem 1.

Theorem 1.

Let T be a tree of order n with k pendant vertices, where $3\le k\le n-2$. For $0<a<1$,

$${\chi }_a\left(T\right)\le (k-1){(k+1)}^a+{(k+2)}^a+3^a+(n-k-2)4^a$$

with equality if and only if T is $B_{n,k}$.

Proof. 

From $3\le k\le n-2$, we have $n\ge 5$. We use induction on n to prove Theorem 1.

For $n=5$, we obtain $k=3$, and $B_{5,3}$ is the only tree for this case. Thus, Theorem 1 holds.

Let $n\ge 6$. Assume that Theorem 1 is true for all trees of order $n_1=n-1$ with $k_1$ pendant vertices, where $3\le k_1\le n_1-2$.

Let us denote one pendant vertex of T by $v_1$. Let $v_2$ be the vertex adjacent to $v_1$ in T. We define $T_1$ with $V\left(T_1\right)=V\left(T\right)\setminus \left\{v_1\right\}$ and $E\left(T_1\right)=E\left(T\right)\setminus \left\{v_1{v}_2\right\}$. Clearly, $d_T\left(u\right)=d_{T_1}\left(u\right)$ for every vertex $u\in V\left(T_1\right)\setminus \left\{v_2\right\}$.

Case 1. $d_T\left(v_2\right)=2$.

If $k=n-2$, then T must be $B_{n,n-2}$; thus, Theorem 1 holds. In that case, we need to prove Case 1 for $3\le k\le n-3$.

Since $d_T\left(v_2\right)=2$, we get $d_{T_1}\left(v_2\right)=1$. Let $v_3$ be the vertex different from $v_1$ adjacent to $v_2$ in T. Note that $d_T\left(v_3\right)=d_{T_1}\left(v_3\right)\ge 2$. By Lemma 1, we have ${(d_T\left(v_3\right)+2)}^a-{(d_T\left(v_3\right)+1)}^a<4^a-3^a$ if $d_T\left(v_3\right)>2$. Therefore, for $d_T\left(v_3\right)\ge 2$,

$${\chi }_a\left(T\right)-{\chi }_a\left(T_1\right)=3^a+{(d_T\left(v_3\right)+2)}^a-{(d_T\left(v_3\right)+1)}^a\le 4^a$$

with equality if and only if $d_T\left(v_3\right)=2$.

The tree $T_1$ has order $n_1=n-1$ and $k_1=k$ pendant vertices. Since $3\le k_1\le n_1-2$, we get $3\le k\le n-3$. By the induction hypothesis,

$$\begin{array}{cc}\hfill {\chi }_a\left(T_1\right)& \le (k_1-1){(k_1+1)}^a+{(k_1+2)}^a+3^a+(n_1-k_1-2)4^a\hfill \\ \hfill & =(k-1){(k+1)}^a+{(k+2)}^a+3^a+(n-k-3)4^a,\hfill \end{array}$$

with equality if and only if $T_1$ is $B_{n-1,k}$. Thus,

$$\begin{array}{cc}\hfill {\chi }_a\left(T\right)& \le {\chi }_a\left(T_1\right)+4^a\hfill \\ \hfill & \le (k-1){(k+1)}^a+{(k+2)}^a+3^a+(n-k-3)4^a+4^a\hfill \\ \hfill & =(k-1){(k+1)}^a+{(k+2)}^a+3^a+(n-k-2)4^a\hfill \end{array}$$

with the first equality if and only if $d_T\left(v_3\right)=2$, and with the second equality if and only if $T_1$ is $B_{n-1,k}$, which means that T is $B_{n,k}$.

Case 2. $d_T\left(v_2\right)\ge 3$.

If $k=3$, then $d_T\left(v_2\right)$ must be 3. Every tree T with three pendant vertices such that $d_T\left(v_2\right)=3$ except for $B_{n,3}$ has ${\chi }_a$ equal to $2\left(5^a\right)+(n-5)4^a+2\left(3^a\right)$, which is smaller than ${\chi }_a\left(B_{n,3}\right)=5^a+(n-3)4^a+3^a$, since $5^a+3^a<2\left(4^a\right)$ by Lemma 1. In that case, Theorem 1 holds for $k=3$.

We need to prove Case 2 for $4\le k\le n-2$. Let $d_T\left(v_2\right)=r$. We have $3\le r\le k$ and $d_{T_1}\left(v_2\right)=r-1$. Note that $v_2$ is adjacent to at least one non-pendant vertex, say $v_3$ in T, since T is not a star. In that case, $d_T\left(v_3\right)\ge 2$. Clearly, for $w\in N_T\left(v_2\right)\setminus \{v_1,v_3\}$, we have $d_T\left(w\right)\ge 1$. Thus,

$$\begin{array}{cc}\hfill {\chi }_a\left(T\right)-{\chi }_a\left(T_1\right)& ={[d_T\left(v_1\right)+d_T\left(v_2\right)]}^a+{[d_T\left(v_2\right)+d_T\left(v_3\right)]}^a-{[d_{T_1}\left(v_2\right)+d_{T_1}\left(v_3\right)]}^a\hfill \\ \hfill & +\sum _{w\in N_T\left(v_2\right)\setminus \{v_1,v_3\}}({[d_T\left(v_2\right)+d_T\left(w\right)]}^a-{[d_{T_1}\left(v_2\right)+d_{T_1}\left(w\right)]}^a)\hfill \\ \hfill & ={(1+r)}^a+{[r+d_T\left(v_3\right)]}^a-{[r-1+d_T\left(v_3\right)]}^a\hfill \\ \hfill & +\sum _{w\in N_T\left(v_2\right)\setminus \{v_1,v_3\}}({[r+d_T\left(w\right)]}^a-{[r-1+d_T\left(w\right)]}^a)\hfill \\ \hfill & \le {(r+1)}^a+{(r+2)}^a-{(r-1+2)}^a\hfill \\ \hfill & +\sum _{w\in N_T\left(v_2\right)\setminus \{v_1,v_3\}}[{(r+1)}^a-{(r-1+1)}^a]\hfill \\ \hfill =& {(r+1)}^a+{(r+2)}^a-{(r+1)}^a+(r-2)[{(r+1)}^a-r^a]\hfill \\ \hfill =& {(r+2)}^a+(r-2){(r+1)}^a-(r-2)r^a,\hfill \end{array}$$

since by Lemma 1,

$${[r+d_T\left(v_3\right)]}^a-{[r-1+d_T\left(v_3\right)]}^a<{(r+2)}^a-{(r-1+2)}^a$$

if $d_T\left(v_3\right)>2$ (which means that ${[r+d_T\left(v_3\right)]}^a>{(r+2)}^a$) and

$${[r+d_T\left(w\right)]}^a-{[r-1+d_T\left(w\right)]}^a<{(r+1)}^a-{(r-1+1)}^a$$

if $d_T\left(w\right)>1$ (which means that ${[r+d_T\left(w\right)]}^a>{(r+1)}^a$). Thus, the equality ${\chi }_a\left(T\right)-{\chi }_a\left(T_1\right)={(r+2)}^a+(r-2){(r+1)}^a-(r-2)r^a$ holds if and only if $d_T\left(v_3\right)=2$ and $d_T\left(w\right)=1$ for every $w\in N_T\left(v_2\right)\setminus \{v_1,v_3\}$.

The function $f\left(r\right)={(r+2)}^a+(r-2){(r+1)}^a-(r-2)r^a$ is increasing for $r\ge 2$ by Lemma 3 ($f\left(r\right)$ is obtained from $f_1\left(x\right)$ when replacing x with $r-2$). Since $r\le k$, we obtain

$${(r+2)}^a+(r-2){(r+1)}^a-(r-2)r^a\le {(k+2)}^a+(k-2){(k+1)}^a-(k-2)k^a$$

with equality if and only if $r=k$. It follows that

$${\chi }_a\left(T\right)-{\chi }_a\left(T_1\right)\le {(k+2)}^a+(k-2){(k+1)}^a-(k-2)k^a$$

with equality if and only if T is $B_{n,k}$.

Note that $T_1$ has order $n_1=n-1$ and $k_1=k-1$ pendant vertices, where $3\le k_1\le n_1-2$; thus, $4\le k\le n-2$. By the induction hypothesis,

$$\begin{array}{cc}\hfill {\chi }_a\left(T_1\right)& \le (k_1-1){(k_1+1)}^a+{(k_1+2)}^a+3^a+(n_1-k_1-2)4^a\hfill \\ \hfill & =(k-2)k^a+{(k+1)}^a+3^a+(n-k-2)4^a,\hfill \end{array}$$

with equality if and only if $T_1$ is $B_{n-1,k-1}$. Therefore,

$$\begin{array}{cc}\hfill {\chi }_a\left(T\right)& \le {\chi }_a\left(T_1\right)+{(k+2)}^a+(k-2){(k+1)}^a-(k-2)k^a\hfill \\ \hfill & \le (k-2)k^a+{(k+1)}^a+3^a+(n-k-2)4^a\hfill \\ \hfill & +(k-2){(k+1)}^a+{(k+2)}^a-(k-2)k^a\hfill \\ \hfill & =(k-1){(k+1)}^a+{(k+2)}^a+3^a+(n-k-2)4^a\hfill \end{array}$$

with equality ${\chi }_a\left(T\right)=(k-1){(k+1)}^a+{(k+2)}^a+3^a+(n-k-2)4^a$ if and only if T is $B_{n,k}$. □

There is no unicyclic graph of order n with more than $n-3$ pendant vertices, and the cycle of order n is the only unicyclic graph of order n with zero pendant vertices. Thus, we consider unicyclic graphs of order n with k pendant vertices, where $1\le k\le n-3$, in Theorem 2.

Theorem 2.

Let U be a unicyclic graph of order n with k pendant vertices, where $1\le k\le n-3$. For $0<a<1$,

$${\chi }_a\left(U\right)\le k{(k+3)}^a+2{(k+4)}^a+(n-k-2)4^a$$

with equality if and only if U is $C{S}_{n-k,k}$.

Proof. 

From $1\le k\le n-3$, we have $n\ge 4$. We use induction on n to prove Theorem 1.

For $n=4$, we obtain $k=1$ and $C{S}_{3,1}$ is the only unicyclic graph for this case. In that case, Theorem 2 holds.

Let $n\ge 5$. Assume that Theorem 2 is true for all unicyclic graphs of order $n_1=n-1$ with $k_1$ pendant vertices, where $1\le k_1\le n_1-3$.

Let C be the unique cycle of U. Let $v_1$ be a pendant vertex of U furthest from C. Let $v_2$ be the vertex adjacent to $v_1$ in U. We define $U_1$ with $V\left(U_1\right)=V\left(U\right)\setminus \left\{v_1\right\}$ and $E\left(U_1\right)=E\left(U\right)\setminus \left\{v_1{v}_2\right\}$. Clearly, $d_U\left(u\right)=d_{U_1}\left(u\right)$ for every vertex $u\in V\left(U_1\right)\setminus \left\{v_2\right\}$. We distinguish two cases.

Case 1. Every pendant vertex of U is adjacent to a vertex in C.

If U contains only one pendant vertex, then U is $C{S}_{n-1,1}$, so Theorem 2 holds. In that case, we need to prove Case 1 for $2\le k\le n-3$.

We have $v_2\in V\left(C\right)$, and $v_2$ is adjacent to s pendant vertices, where $1\le s\le k$. Thus, $d_U\left(v_2\right)=s+2$ and $d_{U_1}\left(v_2\right)=s+1$. Let $w_1$ and $w_2$ be the two vertices on C adjacent to $v_2$. We have $d_U\left(w_1\right),d_U\left(w_2\right)\ge 2$. Then,

$$\begin{array}{cc}\hfill {\chi }_a\left(U\right)-{\chi }_a\left(U_1\right)& =\sum _{u\in N_U\left(v_2\right)\setminus \{w_1,w_2\}}{[d_U\left(v_2\right)+d_U\left(u\right)]}^a\hfill \\ \hfill & -\sum _{u\in N_{U_1}\left(v_2\right)\setminus \{w_1,w_2\}}{[d_{U_1}\left(v_2\right)+d_{U_1}\left(u\right)]}^a\hfill \\ \hfill & +{[d_U\left(v_2\right)+d_U\left(w_1\right)]}^a-{[d_{U_1}\left(v_2\right)+d_{U_1}\left(w_1\right)]}^a\hfill \\ \hfill & +{[d_U\left(v_2\right)+d_U\left(w_2\right)]}^a-{[d_{U_1}\left(v_2\right)+d_{U_1}\left(w_2\right)]}^a\hfill \\ \hfill & =s{(s+2+1)}^a-(s-1){(s+1+1)}^a+{[s+2+d_U\left(w_1\right)]}^a\hfill \\ \hfill & -{[s+1+d_U\left(w_1\right)]}^a+{[s+2+d_U\left(w_2\right)]}^a-{[s+1+d_U\left(w_2\right)]}^a\hfill \\ \hfill & \le s{(s+3)}^a-(s-1){(s+2)}^a+{(s+2+2)}^a-{(s+1+2)}^a\hfill \\ \hfill & +{(s+2+2)}^a-{(s+1+2)}^a\hfill \\ \hfill & =2{(s+4)}^a+(s-2){(s+3)}^a-(s-1){(s+2)}^a,\hfill \end{array}$$

since by Lemma 1, for $i=1,2$, we have

$${[s+2+d_U\left(w_i\right)]}^a-{[s+1+d_U\left(w_i\right)]}^a<{(s+2+2)}^a-{(s+1+2)}^a$$

if $d_U\left(w_i\right)>2$. Thus, the equality ${\chi }_a\left(U\right)-{\chi }_a\left(U_1\right)=2{(s+4)}^a+(s-2){(s+3)}^a-(s-1){(s+2)}^a$ holds if and only if $d_U\left(w_1\right)=d_U\left(w_2\right)=2$.

The function $f\left(s\right)=2{(s+4)}^a+(s-2){(s+3)}^a-(s-1){(s+2)}^a$ is increasing for $s\ge 1$ by Lemma 3 ($f\left(s\right)$ is obtained from $f_2\left(x\right)$ when replacing x with $s-1$). Since $s\le k$, we obtain

$$2{(s+4)}^a+(s-2){(s+3)}^a-(s-1){(s+2)}^a\le 2{(k+4)}^a+(k-2){(k+3)}^a-(k-1){(k+2)}^a$$

with equality if and only if $s=k$. Thus,

$${\chi }_a\left(U\right)-{\chi }_a\left(U_1\right)\le 2{(k+4)}^a+(k-2){(k+3)}^a-(k-1){(k+2)}^a$$

with equality if and only if $v_2$ is adjacent to k pendant vertices, which means that U is $C{S}_{n-k,k}$.

Note that $U_1$ has order $n_1=n-1$ and $k_1=k-1$ pendant vertices, where $1\le k_1\le n_1-3$; thus, $2\le k\le n-3$. By the induction hypothesis,

$$\begin{array}{cc}\hfill {\chi }_a\left(U_1\right)& \le k_1{(k_1+3)}^a+2{(k_1+4)}^a+(n_1-k_1-2)4^a\hfill \\ \hfill & =(k-1){(k+2)}^a+2{(k+3)}^a+(n-k-2)4^a,\hfill \end{array}$$

with equality if and only if $U_1$ is $C{S}_{n-1-(k-1),k-1}$. Therefore,

$$\begin{array}{cc}\hfill {\chi }_a\left(U\right)& \le {\chi }_a\left(U_1\right)+2{(k+4)}^a+(k-2){(k+3)}^a-(k-1){(k+2)}^a\hfill \\ \hfill & \le (k-1){(k+2)}^a+2{(k+3)}^a+(n-k-2)4^a\hfill \\ \hfill & +2{(k+4)}^a+(k-2){(k+3)}^a-(k-1){(k+2)}^a\hfill \\ \hfill & =k{(k+3)}^a+2{(k+4)}^a+(n-k-2)4^a\hfill \end{array}$$

with equality ${\chi }_a\left(U\right)=k{(k+3)}^a+2{(k+4)}^a+(n-k-2)4^a$ if and only if U is $C{S}_{n-k,k}$.

Case 2. U contains a pendant vertex adjacent to a vertex not in C.

Let us prove that ${\chi }_a\left(U\right)<k{(k+3)}^a+2{(k+4)}^a+(n-k-2)4^a$. Since $v_1$ is a vertex farthest from C, we have $v_2\notin V\left(C\right)$, so $v_2$ is adjacent to one non-pendant vertex, say $v_3$ in U. We have $d_U\left(v_3\right)=d_{U_1}\left(v_3\right)\ge 2$. Let us consider two subcases $d_U\left(v_2\right)=2$ and $d_U\left(v_2\right)\ge 3$.

Case 2.1. $d_U\left(v_2\right)=2$.

All the vertices of the cycle in U are not pendant and $v_2$ is not pendant; therefore, $k\ne n-3$. In that case, $1\le k\le n-4$.

By Lemma 1, we have ${(d_U\left(v_3\right)+2)}^a-{(d_U\left(v_3\right)+1)}^a<4^a-3^a$ if $d_U\left(v_3\right)>2$. Therefore, for $d_U\left(v_3\right)\ge 2$,

$${\chi }_a\left(U\right)-{\chi }_a\left(U_1\right)=3^a+{(d_U\left(v_3\right)+2)}^a-{(d_U\left(v_3\right)+1)}^a\le 4^a$$

with equality if and only if $d_U\left(v_3\right)=2$.

The graph $U_1$ has order $n_1=n-1$ and $k_1=k$ pendant vertices. Since $1\le k_1\le n_1-3$, we get $1\le k\le n-4$. By the induction hypothesis,

$$\begin{array}{cc}\hfill {\chi }_a\left(U_1\right)& \le k_1{(k_1+3)}^a+2{(k_1+4)}^a+(n_1-k_1-2)4^a\hfill \\ \hfill & =k{(k+3)}^a+2{(k+4)}^a+(n-k-3)4^a\hfill \end{array}$$

with equality if and only if $U_1$ is $C{S}_{n-1-k,k}$. Thus,

$$\begin{array}{cc}\hfill {\chi }_a\left(U\right)\le {\chi }_a\left(U_1\right)+4^a& \le k{(k+3)}^a+2{(k+4)}^a+(n-k-3)4^a+4^a\hfill \\ \hfill & =k{(k+3)}^a+2{(k+4)}^a+(n-k-2)4^a.\hfill \end{array}$$

with the first equality if and only if $d_U\left(v_3\right)=2$ and with the second equality if and only if $U_1$ is $C{S}_{n-1-k,k}$. Both equalities cannot be satisfied at the same time, since $d_U\left(v_3\right)=2$ (in combination with $d_U\left(v_2\right)=2$ and $d_U\left(v_1\right)=1$) implies that U contains a pendant path of length at least 3 and $U_1$ being $C{S}_{n-1-k,k}$ implies that U does not contain a pendant path of length greater than 2. Therefore, we obtain ${\chi }_a\left(U\right)<k{(k+3)}^a+2{(k+4)}^a+(n-k-2)4^a$.

Case 2.2. $d_U\left(v_2\right)\ge 3$.

Then, $k\ge 2$ and $v_2$ is adjacent to s pendant vertices, where $2\le s\le k$. In that case, $d_U\left(v_2\right)=s+1$. Then,

$$\begin{array}{cc}\hfill {\chi }_a\left(U\right)-{\chi }_a\left(U_1\right)& =\sum _{u\in N_U\left(v_2\right)\setminus \left\{v_3\right\}}{[d_U\left(v_2\right)+d_U\left(u\right)]}^a-\sum _{u\in N_{U_1}\left(v_2\right)\setminus \left\{v_3\right\}}{[d_{U_1}\left(v_2\right)+d_{U_1}\left(u\right)]}^a\hfill \\ \hfill & +{[d_U\left(v_2\right)+d_U\left(v_3\right)]}^a-{[d_{U_1}\left(v_2\right)+d_{U_1}\left(v_3\right)]}^a\hfill \\ \hfill & =s{(s+2)}^a-(s-1){(s+1)}^a+{[s+1+d_U\left(v_3\right)]}^a-{[s+d_U\left(v_3\right)]}^a\hfill \\ \hfill & \le s{(s+2)}^a-(s-1){(s+1)}^a+{(s+1+2)}^a-{(s+2)}^a\hfill \\ \hfill & =(s-1){(s+2)}^a-(s-1){(s+1)}^a+{(s+3)}^a,\hfill \end{array}$$

since by Lemma 1,

$${[s+1+d_U\left(v_3\right)]}^a-{[s+d_U\left(v_3\right)]}^a<{(s+1+2)}^a-{(s+2)}^a$$

if $d_U\left(v_3\right)>2$.

The function $f\left(s\right)={(s+3)}^a+(s-1){(s+2)}^a-(s-1){(s+1)}^a$ is increasing for $s\ge 1$ by Lemma 3 ($f\left(s\right)$ is obtained from $f_1\left(x\right)$ when x is replaced with $s-1$). Since $s\le k$, we obtain

$$(s-1){(s+2)}^a-(s-1){(s+1)}^a+{(s+3)}^a\le (k-1){(k+2)}^a-(k-1){(k+1)}^a+{(k+3)}^a.$$

In that case,

$${\chi }_a\left(U\right)-{\chi }_a\left(U_1\right)\le (k-1){(k+2)}^a-(k-1){(k+1)}^a+{(k+3)}^a.$$

Note that $U_1$ has order $n_1=n-1$ and $k_1=k-1$ pendant vertices, where $1\le k_1\le n_1-3$; thus, $2\le k\le n-3$. By the induction hypothesis,

$$\begin{array}{cc}\hfill {\chi }_a\left(U_1\right)& \le k_1{(k_1+3)}^a+2{(k_1+4)}^a+(n_1-k_1-2)4^a\hfill \\ \hfill & =(k-1){(k+2)}^a+2{(k+3)}^a+(n-k-2)4^a.\hfill \end{array}$$

Therefore,

$$\begin{array}{cc}\hfill {\chi }_a\left(U\right)& \le {\chi }_a\left(U_1\right)+(k-1){(k+2)}^a-(k-1){(k+1)}^a+{(k+3)}^a\hfill \\ \hfill & \le (k-1){(k+2)}^a+2{(k+3)}^a+(n-k-2)4^a\hfill \\ \hfill & +(k-1){(k+2)}^a-(k-1){(k+1)}^a+{(k+3)}^a\hfill \\ \hfill & =3{(k+3)}^a+(2k-2){(k+2)}^a-(k-1){(k+1)}^a+(n-k-2)4^a.\hfill \end{array}$$

We obtain

$$\begin{array}{cc}\hfill {\chi }_a\left(C{S}_{n-k,k}\right)-{\chi }_a\left(U\right)& \ge k{(k+3)}^a+2{(k+4)}^a+(n-k-2)4^a\hfill \\ \hfill & -[3{(k+3)}^a+(2k-2){(k+2)}^a-(k-1){(k+1)}^a+(n-k-2)4^a]\hfill \\ \hfill & =(k-3){(k+3)}^a+2{(k+4)}^a+(k-1){(k+1)}^a-(2k-2){(k+2)}^a\hfill \\ \hfill & >0\hfill \end{array}$$

for every $k\ge 2$ by Lemma 5. Hence,

$${\chi }_a\left(U\right)<{\chi }_a\left(C{S}_{n-k,k}\right)=k{(k+3)}^a+2{(k+4)}^a+(n-k-2)4^a.$$

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4. Conclusions

By Theorem 1, for $0<a<1$ and any tree T of order n with k pendant vertices, where $3\le k\le n-2$, we have

$${\chi }_a\left(T\right)\le (k-1){(k+1)}^a+{(k+2)}^a+3^a+(n-k-2)4^a$$

with equality if and only if T is $B_{n,k}$.

Let us describe results presented in Theorem 1 for $n=6$. If $n=6$, then $3\le k\le 4$.

• If $n=6$ and $k=3$, then ${\chi }_a\left(T\right)\le 5^a+3\left(4^a\right)+3^a$ and $B_{6,3}$ is the only tree having the largest ${\chi }_a$, so ${\chi }_a\left(B_{6,3}\right)=5^a+3\left(4^a\right)+3^a$.
• If $n=6$ and $k=4$, then ${\chi }_a\left(T\right)\le 6^a+3\left(5^a\right)+3^a$ and $B_{6,4}$ is the only tree having the largest ${\chi }_a$, so ${\chi }_a\left(B_{6,4}\right)=6^a+3\left(5^a\right)+3^a$.

By Theorem 2, for $0<a<1$ and any unicyclic graph U of order n with k pendant vertices, where $1\le k\le n-3$, we have

$${\chi }_a\left(U\right)\le k{(k+3)}^a+2{(k+4)}^a+(n-k-2)4^a$$

with equality if and only if U is $C{S}_{n-k,k}$.

Let us describe results presented in Theorem 2 for $n=6$. If $n=6$, then $1\le k\le 3$.

• If $n=6$ and $k=1$, then ${\chi }_a\left(U\right)\le 2\left(5^a\right)+4\left(4^a\right)$ and $C{S}_{5,1}$ is the only unicyclic graph having the largest ${\chi }_a$, so ${\chi }_a\left(C{S}_{5,1}\right)=2\left(5^a\right)+4\left(4^a\right)$.
• If $n=6$ and $k=2$, then ${\chi }_a\left(U\right)\le 2\left(6^a\right)+2\left(5^a\right)+2\left(4^a\right)$ and $C{S}_{4,2}$ is the only unicyclic graph having the largest ${\chi }_a$, so ${\chi }_a\left(C{S}_{4,2}\right)=2\left(6^a\right)+2\left(5^a\right)+2\left(4^a\right)$.
• If $n=6$ and $k=3$, then ${\chi }_a\left(U\right)\le 2\left(7^a\right)+3\left(6^a\right)+4^a$ and $C{S}_{3,3}$ is the only unicyclic graph having the largest ${\chi }_a$, so ${\chi }_a\left(C{S}_{3,3}\right)=2\left(7^a\right)+3\left(6^a\right)+4^a$.

In Section 1, we stated Problems 1 and 2. We solved parts of those two problems in Theorems 1 and 2. Let us mention those intervals of a for which there are no results on the maximum and minimum ${\chi }_a$ for trees/unicyclic graphs of a given order and number of pendant vertices.

• Find trees with the minimum and maximum ${\chi }_a$ for $a<-1$.
• Find unicyclic graphs with the minimum ${\chi }_a$ for $a<-1$ and $a>0$.
• Find unicyclic graphs with the maximum ${\chi }_a$ for $a<0$.

These problems remain open for further research.