The Bounds of Vertex Padmakar–Ivan Index on k-Trees

Abstract

The Padmakar–Ivan ($PI$) index is a distance-based topological index and a molecular structure descriptor, which is the sum of the number of vertices over all edges $uv$ of a graph such that these vertices are not equidistant from u and v. In this paper, we explore the results of $PI$-indices from trees to recursively clustered trees, the k-trees. Exact sharp upper bounds of PI indices on k-trees are obtained by the recursive relationships, and the corresponding extremal graphs are given. In addition, we determine the $PI$-values on some classes of k-trees and compare them, and our results extend and enrich some known conclusions.

1. Introduction

Let G be a simple connected non-oriented graph with vertex set $V(G)$ and edge set $E(G)$. The distance $d(x,y)$ between the vertices $x,y\in V(G)$ is the minimum length of the paths between x and y in G. The oldest and most thoroughly examined molecular descriptor is Wiener index or path number [1], which was first considered in trees by Wiener in 1947 as follows: $W(G)={\sum }_{\{x,y\}\subset V(G)}d(x,y).$ Compared to Wiener index, Szeged index was proposed by Gutman [2] in 1994 that, given $xy\in E(G)$, let $n_{xy}(x)$ be the number of vertices $w\in V(G)$ such that $d(x,w)<d(y,w)$, $Sz(G)={\sum }_{xy\in E(G)}{n}_{xy}(x)n_{xy}(y).$ Based on the considerable success of Wiener index and Sz index, Khadikar proposed a new distance-based index [3] to be used in the field of nano-technology, that is edge Padmakar–Ivan (PI_e) index, $P{I}_e(G)={\sum }_{xy\in E(G)}[n_e(x)+n_e(y)],$ where $n_e(x)$ denotes the number of edges which are closer to the vertex x than to the vertex y, and $n_e(y)$ denotes the number of edges which are closer to the vertex y than to the vertex x, respectively.

It is easy to see that the above concept does not count edges equidistant from both ends of the edge $e=xy$. Based on this idea, Khalifeh et al. [4] introduced a new PI index of vertex version that $PI(G)=P{I}_v(G)={\sum }_{xy\in E(G)}[n_{xy}(x)+n_{xy}(y)].$ Note that, in order to obtain a good recursive formulas, we do not consider the vertices $x,y$ for $n_{xy}(x)$ and $n_{xy}(y)$. Thus, $n_{xy}(x)+n_{xy}(y)\le n-2$.

Nowadays, Padmakar–Ivan indices are widely used in Quantitative Structure–Activity Relationship (QSAR) and Quantitative Structure–Property Relationship (QSPR) [5,6], and there are many interesting results [5,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26] between graph theory and chemistry. For instances, Klavžar [27] provided PI-partitions and arbitrary Cartesian product. Pattabiraman and Paulraja [28] presented the formulas for vertex PI indices of the strong product of a graph and the complete multipartite graph. Ilić and Milosavljević [29] established basic properties of weighted vertex PI index and some lower and upper bounds on special graphs. Wang and Wei [30] studied vertex PI index on an extention of trees (cacti). In [31], Das and Gutman obtained a lower bound on the vertex PI index of a connected graph in terms of numbers of vertices, edges, pendent vertices, and clique number. Hoji et al. [32] provided exact formulas for the vertex PI indices of Kronecker product of a connected graph G and a complete graph. Since the tree is a basic class of graphs in mathematics and chemistry, and these results indicate that either the stars or the paths attain the maximal or minimal bounds for particular chemical indices, then a natural question is how about the situations for vertex Padmakar–Ivan index?

Because $PI$ index is a distance-based index and not very easy to calculate, we first consider the bipartite graph G with n vertices. Note that the tree is a subclass of bipartite graphs which have no odd cycles. By the definition of $PI(G)$ and the assumption that we do not consider the vertices $x,y$ for $n_{xy}(x)$ and $n_{xy}(y)$, one can obtain that every edge of G has the $PI$-value as $n-2$. Thus, the following observation is obtained.

Obervation 1.

For a bipartite graph G with n vertices and m edges,$PI(G)=(n-2)m$. In particular, if G is a tree, then$PI(G)=(n-1)(n-2)$.

Next, we will consider the graphs with odd cycles. In particular, the general tree, k-tree, contains a lot of odd cycles. Then, we are going to consider the PI indices of k-trees and figure out whether or not a k-star or a k-path attains the maximal or minimal bound for $PI$-indices of k-trees. Our main results are as follows: Theorems 1 and 2 give the exact $PI$-values of k-stars, k-paths and k-spirals (see Definitions 1–5 below).

Theorem 1.

For any k-star$S_n^k$and k-path$P_n^k$with$n=kp+s$vertices, where$p\ge 0$is an integer and$s\in [2,k+1]$, we have

$$\begin{array}{c}(i)PI(S_n^k)=k(n-k)(n-k-1),\hfill \\ (ii)PI(P_n^k)=\frac{k(k+1)(p-1)(3kp+6s-2k-4)}{6}+\frac{(s-1)s(3k-s+2)}{3}.\hfill \end{array}$$

Theorem 2.

For any k-spiral$T_{n,c}^{k\ast }$with$n\ge k$vertices, where$c\in [1,k-1]$, we have

$$PI(T_{n,c}^{k\ast })=\left\{\begin{array}{ccc}\hfill \frac{(n-k)(n-k-1)(4k-n+2)}{3}& if& n\in [k,2k-c],\hfill \\ \hfill \frac{3c(n-2k+c-1)(n-2k+c)+(k-c)(2c^2+3nc-4kc+3kn-4k^2-6k+3n-2)}{3}& if& n\ge 2k-c+1.\hfill \end{array}\right.$$

Theorem 3 proves that k-stars achieve the maximal values of $PI$-values for k-trees, and Theorem 4 shows that k-paths do not arrive the minimal values and certain $PI$-values of k-spirals are less than that of k-paths.

Theorem 3.

For any k-tree$T_n^k$with$n\ge k\ge 1$, we have$PI(T_n^k)\le PI(S_n^k)$.

Theorem 4.

For any k-spiral$T_{n,c}^{k\ast }$with$n\ge k\ge 1$, we have

$$\begin{array}{cc}(i)& PI(P_n^k)\ge PI(T_{n,c}^{k\ast })ifc\in [1,\frac{k+1}{2}),\hfill \\ (ii)& PI(P_n^k)\le PI(T_{n,c}^{k\ast })ifc\in [\frac{k+1}{2},k-1].\hfill \end{array}$$

2. Preliminary

In this section, we first give some notations and lemmas that are crucial in the following sections. As usual, $G=(V,E)$ is a connected finite simple undirected graph with vertex set $V=V(G)$ and edge set $E=E(G)$. Let $\left|G\right|$ or $\left|V\right|$ be the cardinality of V. For any $S\subseteq V(G)$ and $F\subseteq E(G)$, we use $G[S]$ to denote the subgraph of G induced by S, $G-S$ to denote the subgraph induced by $V(G)-S$ and $G-F$ to denote the subgraph of G obtained by deleting F. $w(G-S)$ is the number of components of $G-S$ and S is a cut set if $w(G-S)\ge 2$. For any $u,v\in V(G)$, $P_{uv}$ is a path connecting u and v, $d(u,v)$ is the distance between u and v, $N(v)=N_G(v)=\{w\in V(G),vw\in E(G)\}$ is the neighborhood of v and $N[v]=N(v)\cup \left\{v\right\}$. For any integers $a,b$ with $a\le b$, the interval $[a,b]$ is the set of all integers between a and b including $a,b$. In addition, let $[a,b)=[a,b]-\left\{b\right\}$ and $(a,b]=[a,b]-\left\{a\right\}$. In particular, $[a,b]=\varphi$ for $a>b$. $f^{\prime }(x)$ is a derivative of any differentiable function $f(x)$, where x is the variable. $\lfloor x\rfloor$ is the largest integer that is less than or equal to x; $\lceil x\rceil$ is the smallest integer that is greater than or equal to x. It is clear that d is from 0 to the diameter of graphs. Other undefined notations are referred to [33].

It is commonly known that a chordal graph G with at least three vertices is a triangulated graph and contains a simplicial vertex, whose neighborhood induces a clique. During recent decades, there are many interesting studies related to chordal graphs. In 1969, Beineke and Pippert [7] gave the definition of k-trees, which is a significant subclass of chordal graphs. Now, we just give some definitions about k-trees below.

Definition 1.

For positive integers$n,k$with$n\ge k$, the k-tree, denoted by$T_n^k$, is defined recursively as follows: The smallest k-tree is the k-clique$K_k$. If G is a k-tree with$n\ge k$vertices and a new vertex v of degree k is added and joined to the vertices of a k-clique in G, then the obtained graph is a k-tree with$n+1$vertices.

Definition 2.

For positive integers$n,k$with$n\ge k$, the k-path, denoted by$P_n^k$, is defined as follows: starting with a k-clique$G[\{v_1,v_2\dots v_k\}]$. For$i\in [k+1,n]$, the vertex$v_i$is adjacent to vertices$\{v_{i-1},v_{i-2}\dots v_{i-k}\}$only.

Definition 3.

For positive integers$n,k$with$n\ge k$, the k-star, denoted by$S_n^k$, is defined as follows: Starting with a k-clique$G[\{v_1,v_2\dots v_k\}]$and an independent set S with$\left|S\right|=n-k$. For$i\in [k+1,n]$, the vertex$v_i$is adjacent to vertices$\{v_1,v_2\dots v_k\}$only.

Definition 4.

For positive integers$n,k,c$with$n\ge k$and$c\in [1,k-1]$, let$v_1,v_2,\dots ,v_{n-c}$be the simplicial ordering of$P_{n-c}^{k-c}$. The k-spiral, denoted by$T_{n,c}^{k\ast }$, is defined as$P_{n-c}^{k-c}+K_c$, which is,$V(T_{n,c}^{k\ast })=\{v_1,v_2,\dots ,v_n\}$and$E(T_{n,c}^{k\ast })=E(P_{n-c}^{k-c})\cup E(K_c)\cup \{v_1{v}_l,v_2{v}_l,\dots ,v_{n-c}{v}_l\}$, for$l\in [n-c+1,n]$.

Definition 5.

Let$v\in V(T_n^k)$be a vertex of degree k whose neighbors form a k-clique of$T_n^k$, then v is called a k-simplicial vertex. Let$S_1(T_n^k)$be the set of all k-simplicial vertices of$T_n^k$, for$n\ge k+2$, and set$S_1(K_k)=\varphi ,S_1(K_{k+1})=\left\{v\right\}$, where v is any vertex of$K_{k+1}$. Let$G_0=G,G_i=G_{i-1}-v_i$, where$v_i$is a k-simplicial vertex of$G_{i-1}$, then$\{v_1,v_2,\dots v_n\}$is called a simplicial elimination ordering of the n-vertex graph G.

In order to consider the $PI$-value of any k-tree G, let $G^{\prime }=G\cup \left\{u\right\}$ be a k-tree obtained by adding a new vertex u to G. For any $v_1,v_2\in V(G)$, let $d(v_1,v_2)$ be the distance between $v_1$ and $v_2$ in G, $d^{\prime }(v_1,v_2)$ be the distance between $v_1$ and $v_2$ in $G^{\prime }$. Now, we define a function that measures the difference of $PI$-values of any edge relating a vertex from G to $G^{\prime }$ as follows: $f:\{w\in V(G^{\prime }),xy\in E(G)\}$ to $\{1,0\}$ as follows:

$$f(w,xy)=\left\{\begin{array}{ccc}\hfill 0,& \mathrm{if}& w=u\mathrm{and}{d}^{\prime }(x,w)=d^{\prime }(y,w),\hfill \\ \hfill 0,& \mathrm{if}& w\in V(G)\mathrm{and}d(x,w)-d^{\prime }(x,w)=d(y,w)-d^{\prime }(y,w),\hfill \\ \hfill 1,& \mathrm{if}& otherwise.\hfill \end{array}\right.$$

Using the construction of k-trees, we can derive the following lemmas. Note that $PI(xy)=n_{xy}(x)+n_{xy}(y)$ and $PI(xy)\le n-2$.

Lemma 1.

Let$xy$be any edge of a k-tree G with at least$n\ge k+1$vertices, then$PI(xy)\le n-k-1$.

Proof. 

Since every vertex of any k-tree G with at least $k+1$ vertices must be in some $(k+1)$-cliques, which is, $|N(x)\cap N(y)|\ge k-1$ for any $xy\in E(G)$, we have $PI(xy)\le n-(k-1)-2=n-k-1.$ □

Lemma 2.

Let$xy$be any edge of a k-tree G with n vertices and$G^{\prime }=G\cup \left\{u\right\}$be a k-tree obtained by adding u to G. If$w\in V(G)$, then$f(w,xy)=0$.

Proof. 

By adding u to G, since $G^{\prime }$ is a k-tree, we can get that the distance of any pair of vertices of G will increase at most 1, then $f(w,xy)\le 1$. If $w\in V(G)$, then there exists a shortest path $P_{xw}$ or $P_{yw}$ such that $u\notin V(P_{xw})$ or $V(P_{yw})$, that is, $f(w,xy)=0$. □

Lemma 3.

For any k-path G with n vertices, where$n\ge k+2$, let$S_1(G)=\{v_1,v_n\}$and$\{v_1,v_2,\dots ,v_n\}$be the simplicial elimination ordering of G, then$d(v_i,v_j)=\lceil \frac{j-i}{k}\rceil$, for$i<j$and$i,j\in [1,n]$. Furthermore, if$n=kp+s$with$p\ge 1,s\in [2,k+1]$, then

$$d(v,v_{kp+s})=\left\{\begin{array}{ccc}\hfill p+1& if& v\in \{v_1,v_2,\dots ,v_{s-1}\},\hfill \\ \hfill p-i& if& v\in \{v_{ki+s},v_{ki+s+1},\dots ,v_{k(i+1)+s-1}\},i\in [0,p-1].\hfill \end{array}\right.$$

Proof. 

If $j-i\le k$, then $v_i,v_j$ must be in the same $(k+1)$-clique of G, and we have $d(v_i,v_j)=1$; if $j-i\ge k+1$, then $P_{v_i{v}_j}=v_i{v}_{i+k}{v}_{i+2k}\dots v_{i+(\lfloor \frac{j-i}{k}\rfloor -1)k}{v}_{i+\lfloor \frac{j-i}{k}\rfloor k}{v}_j$ is one of the shortest paths between $v_i$ and $v_j$. Thus, $d(v_i,v_j)=\lceil \frac{j-i}{k}\rceil$ and Lemma 3 is proved. □

Lemma 4.

For any k-spiral$T_{n,c}^{k\ast }$with n vertices and$v_i,v_j\in V(T_{n,c}^{k\ast })$for$i<j$,

$$d(v_i,v_j)=\left\{\begin{array}{ccc}\hfill 1,& if& j-i\le k-c,i,j\in [1,n-c],\hfill \\ \hfill 1,& if& iorj\in [n-c+1,n],\hfill \\ \hfill 2,& if& j-i\ge k-c+1,i,j\in [1,n-c].\hfill \end{array}\right.$$

Proof. 

If $j-i\le k-c$ with $i,j\in [1,n-c]$, by Definition 4, we can get that $v_i,v_j$ must be in the same $(k+1)$-clique of G and $d(v_i,v_j)=1$; If i or $j\in [n-c+1,n]$, without loss of generality, say $v_i$ such that $i\in [n-c+1,n]$, then $N[v_i]=V(T_{n,c}^{k\ast })$, that is, $d(v_i,v_j)=1$; If $j-i\ge k-c+1$ with $i,j\in [1,n-c]$, then $v_i\notin N(v_j)$ and $P_{v_i{v}_j}=v_i{v}_n{v}_j$ is one of the shortest paths between $v_i$ and $v_j$, that is, $d(v_i,v_j)=2$. Thus, Lemma 4 is proved. □

3. Main Proofs

In this section, we give the proofs of main results by inductions. For a k-tree $T_n^k$, if $n=k$ or $k+1$, then $T_n^k$ is a k or (k + 1)-clique, that is, $PI(T_n^k)=0$. Thus, all of the theorems are true and we will only consider the case when $n\ge k+2$ below.

Proof of Theorem 1.

For $(i)$, let $V(S_n^k)=\{u_1,u_2,\dots ,u_n\}$, $G[\{u_1,\dots ,u_k\}]$ be a k-clique and $N(u_{l_0})=\{u_1,u_2,\dots ,u_k\}$ for $l_0\ge k+1$. Just by Definition 3, we can get that for $i,j\in [1,k]$, $N[u_i]=N[u_j]=V(S_n^k)$, then $PI(u_i{u}_j)=n_{u_i{u}_j}(u_i)+n_{u_i{u}_j}(u_j)=0$; for $i\in [1,k]$ and $l_0\in [k+1,n]$, $|N[u_i]-N[u_{l_0}]|=n-k-1$, then $PI(u_i{u}_l)=n-k-1$. Thus, we can get $PI(S_n^k)={\sum }_{i,j\in [1,k]}PI(u_i{u}_j)+{\sum }_{i\in [1,k],l_0\in [k+1,n]}PI(u_i{u}_{l_0})=k(n-k)(n-k-1)$.

For $(ii)$, we will proceed it by induction on $|P_n^k|=n\ge k+2$. If $n=k+2$, let $\{v_1,v_2,\dots ,v_{k+2}\}$ be the simplicial elimination ordering of $P_{k+2}^k$. By Lemma 3, we can get that $PI(v_1{v}_i)=1,PI(v_i{v}_{i^{\prime }})=0$ and $PI(v_i{v}_{k+2})=1$ for $i,i^{\prime }\in [2,k+1]$. Thus, $PI(P_{k+2}^k)={\sum }_{i=2}^{k+1}PI(v_1{v}_i)+{\sum }_{i=2}^{k+1}PI(v_i{v}_{k+2})=2k$. Assume that Theorem 1 is true for a k-path with at most $kp+s-1$ vertices, where $p\ge 1,2\le s\le k+1$. Let $P_n^k$ be a k-path such that $|P_n^k|=kp+s$, $V(P_n^k)=\{v_1,v_2,\dots ,v_{kp+s}\}$ and $\{v_1,v_2,\dots ,v_{kp+s}\}$ be the simplicial elimination ordering of $P_n^k$. Set $P_{n-1}^k=P_n^k-\left\{v_{kp+s}\right\}$, then $\{v_1,v_2,\dots ,v_{kp+s-1}\}$ is the simplicial elimination ordering of $P_{n-1}^k$ and for any edge $v_i{v}_j\in E(P_n^k)$, $d(v_i,v_j)$ or $d^{\prime }(v_i,v_j)$ is the distance of $v_i$ and $v_j$ in $P_{n-1}^k$ or $P_n^k$, respectively.

Let $\alpha =[\frac{k(k+1)(p-1)(3kp+6s-2k-4)}{6}+\frac{(s-1)s(3k-s+2)}{3}]-[\frac{k(k+1)(p-1)(3kp+6s-2k-10)}{6}+\frac{(s-2)(s-1)(3k-s+3)}{3}]=p{k}^2+pk-k^2-3k+2ks-s^2+3s-2$. If we can show that by adding $v_{kp+s}$ to $P_{n-1}^k$, $PI(P_n^k)=PI(P_{n-1}^k)+\alpha$, then Theorem 1 is true.

Set $w=v_{kp+s}$, $A_1=\{v_1{v}_s,v_1{v}_{s+1},\dots ,v_1{v}_{k+1}\},A_2=\{v_2{v}_s,\dots ,v_2{v}_{k+2}\},\dots ,A_{s-1}=\{v_{s-1}{v}_s,\dots ,v_{s-1}{v}_{k+s-1}\}$ and $B_1=\{v_1{v}_2,v_1{v}_3,\dots ,v_1{v}_{s-1}\},B_2=\{v_2{v}_3,\dots ,v_2{v}_{s-1}\},\dots ,B_{s-2}=\left\{v_{s-2}{v}_{s-1}\right\},B_{s-1}=\varphi$. By Definition 2 and Lemma 3, we have $d^{\prime }(v_1,v_{kp+s})=p+1$, $d^{\prime }(v_s,v_{kp+s})=p$ and $d^{\prime }(v_1,v_{kp+s})=p+1,d^{\prime }(v_2,v_{kp+s})=p+1$, that is, $d^{\prime }(v_1,v_{kp+s})\ne d^{\prime }(v_s,v_{kp+s})$ and $d^{\prime }(v_1,v_{kp+s})=d^{\prime }(v_2,v_{kp+s})$. Thus, $f(w,v_1{v}_s)=1$ and $f(w,v_1{v}_2)=0$. Similarly, for any edge $v_{h_1}{v}_{h_2}\in {\cup }_{i=1}^{s-1}{A}_i$ with $h_1<h_2$, we have $d^{\prime }(v_{h_1},v_{kp+s})\ne d^{\prime }(v_{h_2},v_{kp+s})$, that is, $f(w,v_{h_1}{v}_{h_2})=1$; For $v_{h_1}{v}_{h_2}\in {\cup }_{i=1}^{s-1}{B}_i$ with $h_1<h_2$, we have $d^{\prime }(v_{h_1},v_{kp+s})=d^{\prime }(v_{h_2},v_{kp+s})$, that is, $f(w,v_{h_1}{v}_{h_2})=0$. Thus, we can get that

$$f(v_{kp+s},xy)=\left\{\begin{array}{ccc}\hfill 1,& \mathrm{if}& xy\in {\cup }_{i=1}^{s-1}{A}_i,\hfill \\ \hfill 0,& \mathrm{if}& xy\in {\cup }_{i=1}^{s-1}{B}_i.\hfill \end{array}\right.$$

For $t\in [0,p-2]$, set $A_{kt+s}=\left\{v_{kt+s}{v}_{k(t+1)+s}\right\}$, $A_{kt+s+1}=\{v_{kt+s+1}{v}_{k(t+1)+s},v_{kt+s+1}{v}_{k(t+1)+s+1}\}$, $\dots ,A_{k(t+1)+s-1}=\{v_{k(t+1)+s-1}{v}_{k(t+1)+s},v_{k(t+1)+s-1}{v}_{k(t+1)+s+1},\dots ,v_{k(t+1)+s-1}{v}_{k(t+2)+s-1}\}$, and $B_{kt+s}=\{v_{kt+s}{v}_{kt+s+1},\dots ,v_{kt+s}{v}_{k(t+1)+s-1}\},B_{kt+s+1}=\{v_{kt+s+1}{v}_{kt+s+2},\dots ,v_{kt+s+1}{v}_{k(t+1)+s-1}\},\dots ,B_{k(t+1)+s-2}=\left\{v_{k(t+1)+s-2}{v}_{k(t+1)+s-1}\right\},B_{k(t+1)+s-1}=\varphi$. For $t=0$ and by Lemma 3, we have $d^{\prime }(v_s,v_{kp+s})=p$, $d^{\prime }(v_{k+s},v_{kp+s})=p-1$ and $d^{\prime }(v_s,v_{kp+s})=p,d^{\prime }(v_{s+1},v_{kp+s})=p$, that is, $d^{\prime }(v_s,v_{kp+s})\ne d^{\prime }(v_{k+s},v_{kp+s})$ and $d^{\prime }(v_s,v_{kp+s})=d^{\prime }(v_{s+1},v_{kp+s})$. Thus, $f(w,v_s{v}_{k+s})=1$ and $f(w,v_s{v}_{s+1})=0$. similarly, for any edge $v_{h_1}{v}_{h_2}\in {\cup }_{i=kt+s}^{k(t+1)+s-1}{A}_i$ with $h_1<h_2$, we have $d^{\prime }(v_{h_1},v_{kp+s})\ne d^{\prime }(v_{h_2},v_{kp+s})$, that is, $f(w,v_{h_1}{v}_{h_2})=1$; for $v_{h_1}{v}_{h_2}\in {\cup }_{i=kt+s}^{k(t+1)+s-1}{B}_i$ with $h_1<h_2$, we have $d^{\prime }(v_{h_1},v_{kp+s})=d^{\prime }(v_{h_2},v_{kp+s})$, that is, $f(w,v_{h_1}{v}_{h_2})=0$. Thus, we can get that

$$f(v_{kp+s},xy)=\left\{\begin{array}{ccc}\hfill 1,& \mathrm{if}& xy\in {\cup }_{i=kt+s}^{k(t+1)+s-1}{A}_i,\hfill \\ \hfill 0,& \mathrm{if}& xy\in {\cup }_{i=kt+s}^{k(t+1)+s-1}{B}_i.\hfill \end{array}\right.$$

Next, we consider the edges in the $(k+1)$-clique $P_n^k[N[v_{kp+s}]]$. For any edge $v_{h_1}{v}_{h_2}$ with $h_1,h_2\in [k(p-1)+s,kp+s-1]$, we have $d^{\prime }(v_{h_1},v_{kp+s})=d^{\prime }(v_{h_2},v_{kp+s})=1$, that is, $f(w,v_{h_1}{v}_{h_2})=0$. For any edge $v_h{v}_{kp+s}$ with $h\in [k(p-1)+s,kp]$, by Lemma 3, we can obtain that $d^{\prime }(v_1,v_h)=p,d^{\prime }(v_1,v_{kp+s})=p+1$, $d^{\prime }(v_{h-k},v_h)=1,d^{\prime }(v_{h-k},v_{kp+s})=2$ and when $h\ne k(p-1)+s$, $d^{\prime }(v_{k(p-1)+s},v_h)=1,d^{\prime }(v_{k(p-1)+s},v_{kp+s})=1$, that is, $d^{\prime }(v_1,v_h)\ne d^{\prime }(v_1,v_{kp+s})$, $d^{\prime }(v_{h-k},v_h)\ne d^{\prime }(v_{h-k},v_{kp+s})$ and $d^{\prime }(v_{k(p-1)+s},v_h)=d^{\prime }(v_{k(p-1)+s},v_{kp+s})$. Similarly, we get that for $j\in [1,p-1],j^{\prime }\in [1,p]$ and $l\ne h$,

$$\left\{\begin{array}{ccc}\hfill d^{\prime }(v_l,v_h)\ne d^{\prime }(v_l,v_{kp+s})& \mathrm{if}& l\in [1,s-1]\cup [h-jk,k(p-j)+s-1],\hfill \\ \hfill d^{\prime }(v_l,v_h)=d^{\prime }(v_l,v_{kp+s})& \mathrm{if}& l\in [k(p-j^{\prime })+s,h-j^{\prime }k+k-1]\cup [h+1,kp+s-1].\hfill \end{array}\right.$$

Thus, if $v_h=v_{k(p-1)+s}$, then $d^{\prime }(v_l,v_{k(p-1)+s})\ne d^{\prime }(v_l,v_{kp+s})$ with $l\in [1,s-1]\cup \left\{{\cup }_{j=1}^{p-1}[k(p-1)+s-jk,(p-j)k+s-1]\right\}=[1,(p-1)k+s-1]$ and $d^{\prime }(v_l,v_{k(p-1)+s})=d^{\prime }(v_l,v_{kp+s})$ with $l\in [(p-1)k+s+1,kp+s]$, that is, $PI(v_{k(p-1)+s}{v}_{kp+s})=(p-1)k+s-1$; similarly, we can obtain that $PI(v_{k(p-1)+s+1}{v}_{kp+s})=(p-1)(k-1)+s-1;PI(v_{k(p-1)+s+2}{v}_{kp+s})=(p-1)(k-2)+s-1;\dots ;PI(v_{kp}{v}_{kp+s})=(p-1)s+s-1.$

For any edge $v_h{v}_{kp+s}$ with $h\in [kp+1,kp+s-1]$, by Lemma 3, we can obtain that $d^{\prime }(v_{h-k},v_h)=1,d^{\prime }(v_{h-k},v_{kp+s})=2$ and $d^{\prime }(v_{k(p-1)+s},v_h)=1,d^{\prime }(v_{k(p-1)+s},v_{kp+s})=1$, that is, $d^{\prime }(v_{h-k},v_h)\ne d^{\prime }(v_{h-k},v_{kp+s})$ and $d^{\prime }(v_{k(p-1)+s},v_h)=d^{\prime }(v_{k(p-1)+s},v_{kp+s})$. Similarly, we get that for $j^{″}\in [1,p]$ and $l\ne h$,

$$\left\{\begin{array}{ccc}\hfill d^{\prime }(v_l,v_h)\ne d^{\prime }(v_l,v_{kp+s})& \mathrm{if}& l\in [h-j^{″}k,k(p-j^{″})+s-1],\hfill \\ \hfill d^{\prime }(v_l,v_h)=d^{\prime }(v_l,v_{kp+s})& \mathrm{if}& l\in [k(p-j^{″})+s,h-j^{″}k+k-1]\cup [h+1,kp+s-1].\hfill \end{array}\right.$$

Thus, if $v_h=v_{kp+1}$, then $d^{\prime }(v_l,v_{kp+1})\ne d^{\prime }(v_l,v_{kp+s})$ for $l\in {\cup }_{j^{″}=1}^p[kp+1-j^{″}k,k(p-j^{″})+s-1]$ and $d^{\prime }(v_l,v_{kp+1})=d^{\prime }(v_l,v_{kp+s})$ with $l\in \left\{{\cup }_{j^{″}=1}^p[k(p-j^{″})+s,k(p+1-j^{″})]\right\}\cup [h+1,kp+s-1]$, that is, $PI(v_{kp+1}{v}_{kp+s})=(s-1)p$; similarly, we have $PI(v_{kp+1}{v}_{kp+s})=(s-2)p;\dots ;PI(v_{kp+s-2}{v}_{kp+s})=2p;PI(v_{kp+s-1}{v}_{kp+s})=p$.

Set $w\in V(P_{n-1}^k)$, if $xy\in E(P_n^k)$ with x or $y\ne v_{kp+s}$, by Lemma 2, we have $f(w,xy)=0$. Thus,

$$\begin{array}{ccc}\hfill PI(P_n^k)-PI(P_{n-1}^k)& =& {\sum }_{xy\in {\cup }_{i=1}^{k(p-1)+s-1}(A_i\cup B_i)}f(w,xy)+PI(v_{k(p-1)+s}{v}_{kp+s})\hfill \\ & & +PI(v_{k(p-1)+s+1}{v}_{kp+s})+\cdots +PI(v_{kp+s-1}{v}_{kp+s})\hfill \\ & =& [(k+2-s)+(k+3-s)+\cdots +k]+(1+2+\cdots +k)(p-1)\hfill \\ & & +[k(p-1)+s-1]+[(k-1)(p-1)+s-1]+[(k-2)(p-1)+s\hfill \\ & & -1]+\cdots +[s(p-1)+s-1]+[(s-1)p+(s-2)p+\cdots +2p+p]\hfill \\ & =& p{k}^2+pk-k^2-3k+2ks-s^2+3s-2\hfill \\ & =& \alpha .\hfill \end{array}$$

Thus, $PI(P_n^k)=\frac{k(k+1)(p-1)(3kp+6s-2k-4)}{6}+\frac{(s-1)s(3k-s+2)}{3}$, for $|P_n^k|=kp+s$ and Theorem 1 is proved. □

Proof of Theorem 2.

We will proceed with it by induction on $n\ge k+2$. If $n=k+2$, by Definition 4, we have $T_{n,c}^{k\ast }$ is also a k-path, that is, $PI(T_{n,c}^{k\ast })=2k$. If $n\ge k+3$, assume that Theorem 2 is true for the k-spiral with at most $n-1$ vertices, we will consider $T_{n,c}^{k\ast }$ with n vertices. Let $T_{n,c}^{k\ast }$ be a k-spiral with $V(T_{n,c}^{k\ast })=V(T_{n-1,c}^{k\ast })\cup \left\{v\right\}$ and $E(T_{n,c}^{k\ast })=E(T_{n-1,c}^{k\ast })\cup \{v{v}_{n-1},v{v}_{n-2},\dots ,v{v}_{n-k}\}$ such that $v_1,v_2,\dots ,v_{n-c-1}$ is the simplicial ordering of $P_{n-c-1}^{k-c}$, where $T_{n-1,c}^{k\ast }=P_{n-c-1}^{k-c}+K_c$ with $V(T_{n-1,c}^{k\ast })=\{v_1,v_2,\dots ,v_{n-1}\}$ and $E(T_{n-1,c}^{k\ast })=E(P_{n-c-1}^{k-c})\cup E(K_c)\cup \{v_1{v}_l,v_2{v}_l,\dots ,v_{n-c-1}{v}_l\}$ for $l\in [n-c,n-1]$. For any edge $v_i{v}_j\in E(T_{n,c}^{k\ast })$, $d(v_i,v_j)$ or $d^{\prime }(v_i,v_j)$ is the distance of $v_i$ and $v_j$ in $T_{n-1,c}^{k\ast }$ or $T_{n,c}^{k\ast }$, respectively.

For $k+2\le n\le 2k-c$, let $\gamma =\frac{(n-k)(n-k-1)(4k-n+2)}{3}-\frac{(n-k-1)(n-k-2)(4k-n+3)}{3}=(n-k-1)(3k-n+2)$. If we can show that by adding v to $T_{n-1,c}^{k\ast }$, $PI(T_{n,c}^{k\ast })=PI(T_{n-1,c}^{k\ast })+\gamma$, then Theorem 2 is true.

Set $w=v$ and let $l\in [n-c,n-1]$, by Lemma 4, we have $d^{\prime }(v_l,v)=1$ and $d^{\prime }(v_i,v)=2$ for $i\in [1,n-k-1]$, that is, $f(w,v_l{v}_i)=1$; $d^{\prime }(v_l,v)=d^{\prime }(v_i,v)=1$ for $i\in [n-k,n-1]$, that is, $f(w,v_l{v}_i)=0$. Set $C_1=\{v_1{v}_2,v_1{v}_3,\dots ,v_1{v}_{n-k-1}\},C_2=\{v_2{v}_3,v_2{v}_4,\dots ,v_2{v}_{n-k-1}\},\dots ,C_{n-k-2}=\left\{v_{n-k-2}{v}_{n-k-1}\right\},C_{n-k-1}=\varphi$, $D_1=\{v_1{v}_{n-k},v_1{v}_{n-k+1},\dots ,v_1{v}_{k-c+1}\},D_2=\{v_2{v}_{n-k},v_2{v}_{n-k+1},\dots ,v_2{v}_{k-c+2}\},\dots ,D_{n-k-1}=\{v_{n-k-1}{v}_{n-k},v_{n-k-1}{v}_{n-k+1},\dots ,v_{n-k-1}{v}_{n-c-1}\}$. By Lemma 4, we have $d^{\prime }(v_1,v)=d^{\prime }(v_2,v)=2$ and $d^{\prime }(v_{n-k},v)=1$, that is, $f(w,v_1{v}_2)=0$ and $f(w,v_1{v}_{n-k})=1$. Similarly, for $v_{h_1}{v}_{h_2}\in {\cup }_{i=1}^{n-k-1}{C}_i$ with $h_1<h_2$, we have $d^{\prime }(v_{h_1},v)=d^{\prime }(v_{h_2},v)=2$, that is, $f(w,v_{h_1}{v}_{h_2})=0$; for $v_{h_1}{v}_{h_2}\in {\cup }_{i=1}^{n-k-1}{D}_i$ with $h_1<h_2$, we have $d^{\prime }(v_{h_1},v)=2$ and $d^{\prime }(v_{h_2},v)=1$, that is, $f(w,v_{h_1}{v}_{h_2})=1$. Set $C_{n-k}=\{v_{n-k}{v}_{n-k+1},v_{n-k}{v}_{n-k+2},\dots ,v_{n-k}{v}_{n-c-1}\},C_{n-k+1}=\{v_{n-k+1}{v}_{n-k+2},v_{n-k+1}{v}_{n-k+3},\dots ,v_{n-k+1}{v}_{n-c-1}\},\dots ,C_{n-c-2}=\left\{v_{n-c-2}{v}_{n-c-1}\right\}$. By Lemma 4, we have $d^{\prime }(v_{n-k},v)=d^{\prime }(v_{n-k-1},v)=1$, that is, $f(w,v_{n-k}{v}_{n-k-1})=0$. Similarly, for $v_{h_1}{v}_{h_2}\in {\cup }_{i=n-k}^{n-c-2}{C}_i$ with $h_1<h_2$, we have $d^{\prime }(v_{h_1},v)=d^{\prime }(v_{h_2},v)=1$, that is, $f(w,v_{h_1}{v}_{h_2})=0$.

Set $E_1=\{v{v}_i,i\in [n-k,n-c-1]\}$, by Lemma 4, we have $d^{\prime }(v_i,v)=2,d^{\prime }(v_i,v_{n-k})=1$ for $i\in [1,n-k-1]$ and $d^{\prime }(v_j,v)=d^{\prime }(v_j,v_{n-k})=1$ for $i\in [n-k+1,n]$. Thus, $PI(v_{n-k}v)=n-k-1$. Similarly, $PI(v_{n-k+1}v)=PI(v_{n-k+2}v)=\dots =PI(v_{k-c+1}v)=n-k-1$. In addition, by Lemma 4, we have $d^{\prime }(v_i,v)=2,d^{\prime }(v_i,v_{k-c+2})=1$ for $i\in [2,n-k-1]$, $d^{\prime }(v_1,v)=d^{\prime }(v_1,v_{k-c+2})=2$ and $d^{\prime }(v_j,v)=d(v_j,v_{k-c+2})=1$ for $j\in [n-k,n]$. Thus, $PI(v_{k-c+2}v)=n-k-2$. Similarly, we have $PI(v_{k-c+3}v)=n-k-3,PI(v_{k-c+4}v)=n-k-4,\dots ,PI(v_{n-c-1}v)=1$. Set $E_2=\{v{v}_l,l\in [n-c,n-1]\}$, since $N[v_l]-N[v]=n-k-1$, we have $PI(v{v}_l)=n-k-1$. Set $E_3=\{v_i{v}_l,i\in [1,n-c-1],l\in [n-c,n-1]\}$, by Lemma 4, we have $d^{\prime }(v_i,v)=2$ for $i\in [1,n-k-1]$, $d^{\prime }(v_i,v)=1$ for $i\in [n-k,n-c-1]$, $d^{\prime }(v_l,v)=1$ for $l\in [n-c,n-1]$. Thus, $f(w,v_i{v}_l)=1$ for $i\in [1,n-k-1]$ and $f(w,v_i{v}_l)=0$ for $i\in [n-k,n-c-1]$.

Set $w\in V(T_n^{k\ast })-\left\{v\right\}$, if $xy\in E(T_{n,c}^{k\ast })$ with x or $y\ne v$, by Lemma 2, we have $f(w,xy)=0$. Thus,

$$\begin{array}{ccc}\hfill PI(T_n^{k\ast })-PI(T_{n-1}^{k\ast })& =& {\sum }_{xy\in {\cup }_{i=1}^{n-c-2}{C}_i}f(w,xy)+{\sum }_{xy{\in }_{i=1}^{n-k-1}{D}_i}f(w,xy)+{\sum }_{xy\in E_1\cup E_2}PI(xy)+\hfill \\ & & {\sum }_{xy\in E_3}f(w,xy)\hfill \\ & =& 0+[(2k-n-c+2)+(2k-n-c+3)+\cdots +(k-c)]\hfill \\ & & +[1+2+\cdots +(n-k-2)+(n-k-1)(2k-n-c+2)]\hfill \\ & & +c(n-k-1)+c(n-k-1)\hfill \\ & =& (n-k-1)(3k-n+2)\hfill \\ & =& \gamma ,\hfill \end{array}$$

and Theorem 2 is proved.

For $n\ge 2k-c+1$, let $\sigma =\frac{3c(n-2k+c-1)(n-2k+c)+(k-c)(2c^2+3nc-4kc+3kn-4k^2-6k+3n-2)}{3}-\frac{3c(n-2k+c-2)(n-1-2k+c)+(k-c)(2c^2+3(n-1)c-4kc+3k(n-1)-4k^2-6k+3n-2)}{3}=k^2-4kc+c^2+2nc-3c+k$. If we can show that by adding v to $T_{n-1,c}^{k\ast }$, $PI(T_{n,c}^{k\ast })=PI(T_{n-1,c}^{k\ast })+\sigma$, then Theorem 2 is proved.

Set $w=v$, by Lemma 4, we have $d^{\prime }(v_l,v)=1$ for $l\in [n-c,n-1]$, $d^{\prime }(v_i,v)=2$ for $i\in [1,n-k-1]$ and $d^{\prime }(v_j,v)=1$ for $j\in [n-k,n-c-1]$. Thus, $f(w,v_l{v}_i)=1$ and $f(w,v_l{v}_j)=0$. Set $C_1=\{v_1{v}_2,v_1{v}_3,\dots ,v_1{v}_{k-c+1}\},C_2=\{v_2{v}_3,v_2{v}_4,\dots ,v_2{v}_{k-c+2}\},\dots ,C_{n-2k+c-1}=\{v_{n-2k+c-1}{v}_{n-2k+c},v_{n-2k+c-1}{v}_{n-2k+c+1},\dots ,v_{n-2k+c-1}{v}_{n-k-1}\},C_{n-2k+c}=\{v_{n-2k+c}{v}_{n-2k+s+1},v_{n-2k+c}{v}_{n-2k+s+2},\dots ,v_{n-2k+c}{v}_{n-k-1}\},C_{n-2k+c+1}=\{v_{n-2k+c+1}{v}_{n-2k+c+2},v_{n-2k+c+1}{v}_{n-2k+c+3},\dots ,v_{n-2k+c+1}{v}_{n-k-1}\},.\dots ,C_{n-k-1}=\varphi ,D_{n-2k+c}=\left\{v_{n-2k+c}{v}_{n-k}\right\},D_{n-2k+c+1}=\{v_{n-2k+c+1}{v}_{n-k},v_{n-2k+c+1}{v}_{n-k+1}\},\dots ,D_{n-k-1}=\{v_{n-k-1}{v}_{n-k},v_{n-k-1}{v}_{n-k+1},\dots ,v_{n-k-1}{v}_{n-c-1}\}$.

By Lemma 4, we can get that $d^{\prime }(v_1,v)=d^{\prime }(v_2,v)=2$ and $d^{\prime }(v_{n-k},v)=1$, that is, $f(w,v_1{v}_2)=0$ and $f(w,v_1{v}_{n-k})=1$. Similarly, for $v_{h_1}{v}_{h_2}\in {\cup }_{i=1}^{n-k-1}{C}_i$ with $h_1<h_2$, we have $d^{\prime }(v_{h_1},v)=d^{\prime }(v_{h_2},v)=2$, that is, $f(w,v_{h_1}{v}_{h_2})=0$; for $v_{h_1}{v}_{h_2}\in {\cup }_{i=n-2k+c}^{n-k-1}{D}_i$ with $h_1<h_2$, we have $d^{\prime }(v_{h_1},v)=2$ and $d^{\prime }(v_{h_2},v)=1$, that is, $f(w,v_{h_1}{v}_{h_2})=1$. Set $C_{n-k}=\{v_{n-k}{v}_{n-k+1},v_{n-k}{v}_{n-k+2},\dots ,v_{n-k}{v}_{n-c-1}\},C_{n-k+1}=\{v_{n-k+1}{v}_{n-k+2},v_{n-k+1}{v}_{n-k+3},\dots ,v_{n-k+1}{v}_{n-c-1}\},\dots ,C_{n-c+2}=\left\{v_{n-c-2}{v}_{n-c-1}\right\}$. By Lemma 4, we can get that $d^{\prime }(v_{n-k},v)=d^{\prime }(v_{n-k+1},v)=1$, that is, $f(w,v_{n-k}{v}_{n-k+1})=0$. Similarly, for $v_{h_1}{v}_{h_2}\in {\cup }_{i=n-k}^{n-c-2}{C}_i$ with $h_1<h_2$, we have $d^{\prime }(v_{h_1},v)=d^{\prime }(v_{h_2},v)=1$, that is, $f(w,v_{h_1}{v}_{h_2})=0$.

Set $E_1=\{v{v}_i,i\in [n-k,n-c-1]\}$, by Lemma 4, we have $d^{\prime }(v,v_{n-k-1})=2,d^{\prime }(v_{n-c-1},v_{n-k-1})=1$, $d^{\prime }(v,v_i)=d^{\prime }(v_{n-c-1},v_i)=1$ for $i\in [n-k,n-c-2]\cup [n-c,n-1]$ and $d^{\prime }(v,v_j)=d(v_{n-c-1},v_j)=2$ for $j\in [1,n-k-2]$. Thus, $PI(v{v}_{n-c-1})=1$. Similarly, we have $PI(v{v}_{n-c-2})=2,PI(v{v}_{n-c-3})=3,\dots ,PI(v{v}_{n-k})=k-c$. Set $E_2=\{v{v}_l,l\in [n-c,n-1]\}$, since $N[v_l]-N[v]=n-k-1$, we have $PI(v{v}_l)=n-k-1$. Set $E_3=\{v_i{v}_l,i\in [1,n-c-1],l\in [n-c,n-1]\}$, by Lemma 4, we have $d^{\prime }(v,v_i)=2,d^{\prime }(v,v_l)=1$ for $i\in [1,n-k-1]$ and $d^{\prime }(v,v_i)=d^{\prime }(v,v_l)=1$ for $i\in [n-k,n-c-1]$. Thus, $f(w,v_i{v}_l)=1$ for $i\in [1,n-k-1]$ and $f(w,v_i{v}_l)=0$ for $i\in [n-k,n-c-1]$.

Set $w\in V(T_n^{k\ast })-\left\{v\right\}$, if $xy\in E(T_{n,c}^{k\ast })$ with x or $y\ne v$, by Lemma 2, we have $f(w,xy)=0$. Thus,

$$\begin{array}{ccc}\hfill PI(T_n^{k\ast })-PI(T_{n-1}^{k\ast })& =& {\sum }_{xy\in {\cup }_{i=1}^{n-c-2}{C}_i}f(w,xy)+{\sum }_{xy{\in }_{i=n-2k+c}^{n-k-1}{D}_i}f(w,xy)+{\sum }_{xy\in E_1\cup E_2}PI(xy)\hfill \\ & & +{\sum }_{xy\in E_3}f(w,xy)\hfill \\ & =& 0+[1+2+3+\cdots +(k-c)]+[1+2+3+\cdots +(k-c)]\hfill \\ & & +c(n-k-1)+c(n-k-1)\hfill \\ & =& k^2-4kc+c^2+2nc-3c+k\hfill \\ & =& \sigma ,\hfill \end{array}$$

and Theorem 2 is proved. □

Proof of Theorem 3.

For $n\ge k+2$, we will proceed it by introduction on $|T_n^k|=n$. If $n=k+2$, $T_n^k$ is also a k-path, that is, $PI(T_n^k)=2k$. If $n\ge k+3$, assume that Theorem 3 is true for the k-tree with at most $n-1$ vertices, let $v\in S_1(T_n^k)$ and $T_{n-1}^k=T_n^k-v$, by the induction hypothesis, we have $PI(T_{n-1}^k)\le PI(S_{n-1}^k)=k(n-k-1)(n-k-2)$. By adding back v, let $N(v)=\{x_1,x_2,\dots ,x_k\}$ and $w=v$. Since $T_n^k[v,x_1,x_2,\dots ,x_k]$ is a $(k+1)$-clique, we have $f(w,x_i{x}_j)=0$ for $i,j\in [1,k]$. By Lemmas 1 and 2, we can obtain that $PI(v{x}_i)\le n-k-1$ with $i\in [1,k]$ and $f(w,xy)\le 1$ for any edge $xy\in E(T_n^k)-E(T_n^k[v,x_1,x_2,\dots ,x_k])$. Next, set $w\in V(T_n^k)-\left\{v\right\}$, by Lemma 2, if $xy\in E(T_n^k)$ with x or $y\ne v$, we have $f(w,xy)=0$. Since $|E(T_n^k)-E(T_n^k[v,x_1,x_2,\dots ,x_k])|=k(n-k-1)$, we have

$$\begin{array}{ccc}\hfill PI(T_n^k)& =& PI(T_{n-1}^k)+{\sum }_{xy\in E(T_n^k-\{v{x}_i,i\in [1,k]\})}f(w,xy)+{\sum }_{i=1}^{k}PI(v{x}_i)\hfill \\ & \le & PI(S_{n-k}^k)+k(n-k-1)+k(n-k-1)\hfill \\ & =& k(n-k-1)(n-k-2)+k(n-k-1)+k(n-k-1)\hfill \\ & =& k(n-k)(n-k-1)\hfill \\ & =& PI(S_n^k).\hfill \end{array}$$

Thus, this finishes the proof of Theorem 3. □

Proof of Theorem 4.

For $k=1$ and by Fact 1, we can get that every tree with the same vertices has the same $PI$-value; then, Theorem 4 is obvious; for $k\ge 2$, if $k+2\le n\le 2k-c$, let $n=kp+s$ with $p=1$ and $s=n-k$, by Theorem 1, we have $PI(P_n^k)-PI(T_{n,c}^{k\ast })=\frac{(s-1)s(3k-s+2)}{3}-\frac{(n-k)(n-k-1)(4k-n+2)}{3}=\frac{(n-k-1)(n-k)[3k-(n-k)+2]}{3}-\frac{(n-k)(n-k-1)(4k-n+2)}{3}=0$, and Theorem 4 is true. If $n\ge 2k-c+1$, $p=\frac{n-s}{k}$ and by Theorems 1 and 2, define the new functions as follows: for $z\ge 2k-c+1$, $1\le c\le k-1$ and $2\le s\le k+1$,

$$\begin{array}{ccc}\hfill g(z)& =& \frac{(k+1)(z-s-k)(3z+3s-2k-4)}{6}+\frac{s(s-1)(3k-s+2)}{3},\hfill \\ \hfill h(z,c)& =& c(z-2k+c-1)(z-2k+c)+\frac{(k-c)(2c^2+3zc-4kc+3kz-4k^2-6k+3z-2)}{3},\hfill \\ \hfill l(z,c)& =& g(z)-h(z,c)\hfill \\ & =& (\frac{k}{2}+\frac{1}{2}-c)z^2+(-c^2+2c+4kc-\frac{11k^2}{6}-\frac{5k}{2}-\frac{2}{3})z\hfill \\ & +& \frac{k{s}^2}{2}-\frac{k^{2}s}{6}-\frac{ks}{2}+\frac{5k^3}{3}+\frac{5k^2}{3}+\frac{s^2}{2}+\frac{4k}{3}-\frac{s^3}{3}-6k^{2}c+3k{c}^2-\frac{c^3}{3}-4kc+c^2-\frac{2c}{3},\hfill \\ \hfill l_z(z)& =& l_z(z,c)\hfill \\ & =& (k+1-2c)z-c^2+2c+4kc-\frac{11k^2}{6}-\frac{5k}{2}-\frac{2}{3}.\hfill \end{array}$$

Then, it is enough to determine whether or not $l(z,c)\ge 0$ is true. By some calculations, we can obtain the following claim:

Claim 1.

$z_1=2k-c+1$,$z_2=2k-c+2$are the two roots of$l(z,c)=0$with$c\ne \frac{k+1}{2}$.

Proof. 

For any $c\in [1,k-1]$, let $z_1=2k-c+1$, $z_2=2k-c+2$, and we have $l(2k-c+1,c)=0,l(2k-c+2,c)=0$. If $c\ne \frac{k+1}{2}$, then Claim is true. □

For fixed $c\in [1,\frac{k+1}{2})$, that is, $\frac{k}{2}+\frac{1}{2}-c>0$, then the function of $l(z,c)$ about z is open up. Since z is an integer and by Fact 2, we have $l(z,c)\ge 0$ for $z\ge 2k-c+1$ and Theorem 4 is true; if $c=\frac{k+1}{2}$ and $k\ge 1$, we have $l_z(z)=\frac{1-k^2}{12}\le 0$, that is, $l(z,\frac{k+1}{2})$ is decreasing about z. By the proof of Fact 2, we have $l(2k-c+1,c)=0$. For $z\ge 2k-c+1$, we can get that $l(z,\frac{k+1}{2})\le l(2k-c+1,\frac{k+1}{2})=0$ and Theorem 4 is true; for fixed $c\in (\frac{k+1}{2},k-1]$, that is, $\frac{k}{2}+\frac{1}{2}-c<0$, then the function of $l(z,c)$ about z is open down. Since z is an integer and by Claim, we can obtain that $l(z,c)\le 0$ for $z\ge 2k-c+1$ and this finishes the proof of Theorem 4. □

4. Conclusions

We can see that the k-stars attain the maximal values of $PI$-values for k-trees. One of the guesses is that the k-paths attain the minimal values. Actually, it is not the case and some $PI$-values of k-spirals is even smaller than that of k-paths. Meanwhile, not all $PI$-values of k-spirals are less than the values of all other k-trees. This fact indicates an interesting problem—which type of k-trees will achieve the minimum $PI$-value?