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Mathematics 2019, 7(4), 324; https://doi.org/10.3390/math7040324

Article
The Bounds of Vertex Padmakar–Ivan Index on k-Trees
1
Department of Mathematics and Physics, Texas A&M International University, Laredo, TX 78041, USA
2
Institute of Computing Science and Technology, Guangzhou University, Guangzhou 510006, China
3
School of Mathematics and Physics, Anhui Jianzhu University, Hefei 230601, China
4
Department of Mathematics, The University of Mississippi, University, MS 38677, USA
*
Author to whom correspondence should be addressed.
Received: 26 January 2019 / Accepted: 11 March 2019 / Published: 1 April 2019

## Abstract

:
The Padmakar–Ivan ($P I$) index is a distance-based topological index and a molecular structure descriptor, which is the sum of the number of vertices over all edges $u v$ of a graph such that these vertices are not equidistant from u and v. In this paper, we explore the results of $P I$-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 $P I$-values on some classes of k-trees and compare them, and our results extend and enrich some known conclusions.
Keywords:
extremal values; PI index; k-trees; distance

## 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 $x , y ∈ V ( G )$ 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: $W ( G ) = ∑ { x , y } ⊂ V ( G ) d ( x , y ) .$ Compared to Wiener index, Szeged index was proposed by Gutman [2] in 1994 that, given $x y ∈ E ( G )$, let $n x y ( x )$ be the number of vertices $w ∈ V ( G )$ such that $d ( x , w ) < d ( y , w )$, $S z ( G ) = ∑ x y ∈ E ( G ) n x y ( x ) n x y ( y ) .$ 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 (PIe) index, $P I e ( G ) = ∑ x y ∈ E ( G ) [ n e ( x ) + n e ( y ) ] ,$ 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 $e = x y$. Based on this idea, Khalifeh et al. [4] introduced a new PI index of vertex version that $P I ( G ) = P I v ( G ) = ∑ x y ∈ E ( G ) [ n x y ( x ) + n x y ( y ) ] .$ Note that, in order to obtain a good recursive formulas, we do not consider the vertices $x , y$ for $n x y ( x )$ and $n x y ( y )$. Thus, $n x y ( x ) + n x y ( y ) ≤ n − 2$.
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 $P I$ 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 $P I ( G )$ and the assumption that we do not consider the vertices $x , y$ for $n x y ( x )$ and $n x y ( y )$, one can obtain that every edge of G has the $P I$-value as $n − 2$. Thus, the following observation is obtained.
Obervation 1.
For a bipartite graph G with n vertices and m edges,$P I ( G ) = ( n − 2 ) m$. In particular, if G is a tree, then$P I ( G ) = ( n − 1 ) ( n − 2 )$.
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 $P I$-indices of k-trees. Our main results are as follows: Theorems 1 and 2 give the exact $P I$-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$n = k p + s$vertices, where$p ≥ 0$is an integer and$s ∈ [ 2 , k + 1 ]$, we have
$( i ) P I ( S n k ) = k ( n − k ) ( n − k − 1 ) , ( i i ) P I ( P n k ) = k ( k + 1 ) ( p − 1 ) ( 3 k p + 6 s − 2 k − 4 ) 6 + ( s − 1 ) s ( 3 k − s + 2 ) 3 .$
Theorem 2.
For any k-spiral$T n , c k ∗$with$n ≥ k$vertices, where$c ∈ [ 1 , k − 1 ]$, we have
$P I ( T n , c k ∗ ) = ( n − k ) ( n − k − 1 ) ( 4 k − n + 2 ) 3 i f n ∈ [ k , 2 k − c ] , 3 c ( n − 2 k + c − 1 ) ( n − 2 k + c ) + ( k − c ) ( 2 c 2 + 3 n c − 4 k c + 3 k n − 4 k 2 − 6 k + 3 n − 2 ) 3 i f n ≥ 2 k − c + 1 .$
Theorem 3 proves that k-stars achieve the maximal values of $P I$-values for k-trees, and Theorem 4 shows that k-paths do not arrive the minimal values and certain $P I$-values of k-spirals are less than that of k-paths.
Theorem 3.
For any k-tree$T n k$with$n ≥ k ≥ 1$, we have$P I ( T n k ) ≤ P I ( S n k )$.
Theorem 4.
For any k-spiral$T n , c k ∗$with$n ≥ k ≥ 1$, we have
$( i ) P I ( P n k ) ≥ P I ( T n , c k ∗ ) i f c ∈ [ 1 , k + 1 2 ) , ( i i ) P I ( P n k ) ≤ P I ( T n , c k ∗ ) i f c ∈ [ k + 1 2 , k − 1 ] .$

## 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 $| G |$ or $| V |$ be the cardinality of V. For any $S ⊆ V ( G )$ and $F ⊆ E ( G )$, 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 $w ( G − S ) ≥ 2$. For any $u , v ∈ V ( G )$, $P u v$ is a path connecting u and v, $d ( u , v )$ is the distance between u and v, $N ( v ) = N G ( v ) = { w ∈ V ( G ) , v w ∈ E ( G ) }$ is the neighborhood of v and $N [ v ] = N ( v ) ∪ { v }$. For any integers $a , b$ with $a ≤ b$, the interval $[ a , b ]$ is the set of all integers between a and b including $a , b$. In addition, let $[ a , b ) = [ a , b ] − { b }$ and $( a , b ] = [ a , b ] − { a }$. In particular, $[ a , b ] = ϕ$ for $a > b$. $f ′ ( x )$ is a derivative of any differentiable function $f ( x )$, where x is the variable. $⌊ x ⌋$ is the largest integer that is less than or equal to x; $⌈ x ⌉$ 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$n ≥ k$, 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$n ≥ k$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$n ≥ k$, the k-path, denoted by$P n k$, is defined as follows: starting with a k-clique$G [ { v 1 , v 2 … v k } ]$. For$i ∈ [ k + 1 , n ]$, the vertex$v i$is adjacent to vertices${ v i − 1 , v i − 2 … v i − k }$only.
Definition 3.
For positive integers$n , k$with$n ≥ k$, the k-star, denoted by$S n k$, is defined as follows: Starting with a k-clique$G [ { v 1 , v 2 … v k } ]$and an independent set S with$| S | = n − k$. For$i ∈ [ k + 1 , n ]$, the vertex$v i$is adjacent to vertices${ v 1 , v 2 … v k }$only.
Definition 4.
For positive integers$n , k , c$with$n ≥ k$and$c ∈ [ 1 , k − 1 ]$, let$v 1 , v 2 , … , v n − c$be the simplicial ordering of$P n − c k − c$. The k-spiral, denoted by$T n , c k ∗$, is defined as$P n − c k − c + K c$, which is,$V ( T n , c k ∗ ) = { v 1 , v 2 , … , v n }$and$E ( T n , c k ∗ ) = E ( P n − c k − c ) ∪ E ( K c ) ∪ { v 1 v l , v 2 v l , … , v n − c v l }$, for$l ∈ [ n − c + 1 , n ]$.
Definition 5.
Let$v ∈ V ( T n k )$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$n ≥ k + 2$, and set$S 1 ( K k ) = ϕ , S 1 ( K k + 1 ) = { v }$, 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${ v 1 , v 2 , … v n }$is called a simplicial elimination ordering of the n-vertex graph G.
In order to consider the $P I$-value of any k-tree G, let $G ′ = G ∪ { u }$ be a k-tree obtained by adding a new vertex u to G. For any $v 1 , v 2 ∈ V ( G )$, let $d ( v 1 , v 2 )$ be the distance between $v 1$ and $v 2$ in G, $d ′ ( v 1 , v 2 )$ be the distance between $v 1$ and $v 2$ in $G ′$. Now, we define a function that measures the difference of $P I$-values of any edge relating a vertex from G to $G ′$ as follows: $f : { w ∈ V ( G ′ ) , x y ∈ E ( G ) }$ to ${ 1 , 0 }$ as follows:
$f ( w , x y ) = 0 , if w = u and d ′ ( x , w ) = d ′ ( y , w ) , 0 , if w ∈ V ( G ) and d ( x , w ) − d ′ ( x , w ) = d ( y , w ) − d ′ ( y , w ) , 1 , if o t h e r w i s e .$
Using the construction of k-trees, we can derive the following lemmas. Note that $P I ( x y ) = n x y ( x ) + n x y ( y )$ and $P I ( x y ) ≤ n − 2$.
Lemma 1.
Let$x y$be any edge of a k-tree G with at least$n ≥ k + 1$vertices, then$P I ( x y ) ≤ n − k − 1$.
Proof.
Since every vertex of any k-tree G with at least $k + 1$ vertices must be in some $( k + 1 )$-cliques, which is, $| N ( x ) ∩ N ( y ) | ≥ k − 1$ for any $x y ∈ E ( G )$, we have $P I ( x y ) ≤ n − ( k − 1 ) − 2 = n − k − 1 .$ □
Lemma 2.
Let$x y$be any edge of a k-tree G with n vertices and$G ′ = G ∪ { u }$be a k-tree obtained by adding u to G. If$w ∈ V ( G )$, then$f ( w , x y ) = 0$.
Proof.
By adding u to G, since $G ′$ is a k-tree, we can get that the distance of any pair of vertices of G will increase at most 1, then $f ( w , x y ) ≤ 1$. If $w ∈ V ( G )$, then there exists a shortest path $P x w$ or $P y w$ such that $u ∉ V ( P x w )$ or $V ( P y w )$, that is, $f ( w , x y ) = 0$. □
Lemma 3.
For any k-path G with n vertices, where$n ≥ k + 2$, let$S 1 ( G ) = { v 1 , v n }$and${ v 1 , v 2 , … , v n }$be the simplicial elimination ordering of G, then$d ( v i , v j ) = ⌈ j − i k ⌉$, for$i < j$and$i , j ∈ [ 1 , n ]$. Furthermore, if$n = k p + s$with$p ≥ 1 , s ∈ [ 2 , k + 1 ]$, then
$d ( v , v k p + s ) = p + 1 i f v ∈ { v 1 , v 2 , … , v s − 1 } , p − i i f v ∈ { v k i + s , v k i + s + 1 , … , v k ( i + 1 ) + s − 1 } , i ∈ [ 0 , p − 1 ] .$
Proof.
If $j − i ≤ k$, 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 $j − i ≥ k + 1$, then $P v i v j = v i v i + k v i + 2 k … v i + ( ⌊ j − i k ⌋ − 1 ) k v i + ⌊ j − i k ⌋ k v j$ is one of the shortest paths between $v i$ and $v j$. Thus, $d ( v i , v j ) = ⌈ j − i k ⌉$ and Lemma 3 is proved. □
Lemma 4.
For any k-spiral$T n , c k ∗$with n vertices and$v i , v j ∈ V ( T n , c k ∗ )$for$i < j$,
$d ( v i , v j ) = 1 , i f j − i ≤ k − c , i , j ∈ [ 1 , n − c ] , 1 , i f i o r j ∈ [ n − c + 1 , n ] , 2 , i f j − i ≥ k − c + 1 , i , j ∈ [ 1 , n − c ] .$
Proof.
If $j − i ≤ k − c$ with $i , j ∈ [ 1 , n − c ]$, 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 $j ∈ [ n − c + 1 , n ]$, without loss of generality, say $v i$ such that $i ∈ [ n − c + 1 , n ]$, then $N [ v i ] = V ( T n , c k ∗ )$, that is, $d ( v i , v j ) = 1$; If $j − i ≥ k − c + 1$ with $i , j ∈ [ 1 , n − c ]$, then $v i ∉ N ( v j )$ 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, $P I ( T n k ) = 0$. Thus, all of the theorems are true and we will only consider the case when $n ≥ k + 2$ below.
Proof of Theorem 1.
For $( i )$, let $V ( S n k ) = { u 1 , u 2 , … , u n }$, $G [ { u 1 , … , u k } ]$ be a k-clique and $N ( u l 0 ) = { u 1 , u 2 , … , u k }$ for $l 0 ≥ k + 1$. Just by Definition 3, we can get that for $i , j ∈ [ 1 , k ]$, $N [ u i ] = N [ u j ] = V ( S n k )$, then $P I ( u i u j ) = n u i u j ( u i ) + n u i u j ( u j ) = 0$; for $i ∈ [ 1 , k ]$ and $l 0 ∈ [ k + 1 , n ]$, $| N [ u i ] − N [ u l 0 ] | = n − k − 1$, then $P I ( u i u l ) = n − k − 1$. Thus, we can get $P I ( S n k ) = ∑ i , j ∈ [ 1 , k ] P I ( u i u j ) + ∑ i ∈ [ 1 , k ] , l 0 ∈ [ k + 1 , n ] P I ( u i u l 0 ) = k ( n − k ) ( n − k − 1 )$.
For $( i i )$, we will proceed it by induction on $| P n k | = n ≥ k + 2$. If $n = k + 2$, let ${ v 1 , v 2 , … , v k + 2 }$ be the simplicial elimination ordering of $P k + 2 k$. By Lemma 3, we can get that $P I ( v 1 v i ) = 1 , P I ( v i v i ′ ) = 0$ and $P I ( v i v k + 2 ) = 1$ for $i , i ′ ∈ [ 2 , k + 1 ]$. Thus, $P I ( P k + 2 k ) = ∑ i = 2 k + 1 P I ( v 1 v i ) + ∑ i = 2 k + 1 P I ( v i v k + 2 ) = 2 k$. Assume that Theorem 1 is true for a k-path with at most $k p + s − 1$ vertices, where $p ≥ 1 , 2 ≤ s ≤ k + 1$. Let $P n k$ be a k-path such that $| P n k | = k p + s$, $V ( P n k ) = { v 1 , v 2 , … , v k p + s }$ and ${ v 1 , v 2 , … , v k p + s }$ be the simplicial elimination ordering of $P n k$. Set $P n − 1 k = P n k − { v k p + s }$, then ${ v 1 , v 2 , … , v k p + s − 1 }$ is the simplicial elimination ordering of $P n − 1 k$ and for any edge $v i v j ∈ E ( P n k )$, $d ( v i , v j )$ or $d ′ ( v i , v j )$ is the distance of $v i$ and $v j$ in $P n − 1 k$ or $P n k$, respectively.
Let $α = [ k ( k + 1 ) ( p − 1 ) ( 3 k p + 6 s − 2 k − 4 ) 6 + ( s − 1 ) s ( 3 k − s + 2 ) 3 ] − [ k ( k + 1 ) ( p − 1 ) ( 3 k p + 6 s − 2 k − 10 ) 6 + ( s − 2 ) ( s − 1 ) ( 3 k − s + 3 ) 3 ] = p k 2 + p k − k 2 − 3 k + 2 k s − s 2 + 3 s − 2$. If we can show that by adding $v k p + s$ to $P n − 1 k$, $P I ( P n k ) = P I ( P n − 1 k ) + α$, then Theorem 1 is true.
Set $w = v k p + s$, $A 1 = { v 1 v s , v 1 v s + 1 , … , v 1 v k + 1 } , A 2 = { v 2 v s , … , v 2 v k + 2 } , … , A s − 1 = { v s − 1 v s , … , v s − 1 v k + s − 1 }$ and $B 1 = { v 1 v 2 , v 1 v 3 , … , v 1 v s − 1 } , B 2 = { v 2 v 3 , … , v 2 v s − 1 } , … , B s − 2 = { v s − 2 v s − 1 } , B s − 1 = ϕ$. By Definition 2 and Lemma 3, we have $d ′ ( v 1 , v k p + s ) = p + 1$, $d ′ ( v s , v k p + s ) = p$ and $d ′ ( v 1 , v k p + s ) = p + 1 , d ′ ( v 2 , v k p + s ) = p + 1$, that is, $d ′ ( v 1 , v k p + s ) ≠ d ′ ( v s , v k p + s )$ and $d ′ ( v 1 , v k p + s ) = d ′ ( v 2 , v k p + s )$. Thus, $f ( w , v 1 v s ) = 1$ and $f ( w , v 1 v 2 ) = 0$. Similarly, for any edge $v h 1 v h 2 ∈ ∪ i = 1 s − 1 A i$ with $h 1 < h 2$, we have $d ′ ( v h 1 , v k p + s ) ≠ d ′ ( v h 2 , v k p + s )$, that is, $f ( w , v h 1 v h 2 ) = 1$; For $v h 1 v h 2 ∈ ∪ i = 1 s − 1 B i$ with $h 1 < h 2$, we have $d ′ ( v h 1 , v k p + s ) = d ′ ( v h 2 , v k p + s )$, that is, $f ( w , v h 1 v h 2 ) = 0$. Thus, we can get that
$f ( v k p + s , x y ) = 1 , if x y ∈ ∪ i = 1 s − 1 A i , 0 , if x y ∈ ∪ i = 1 s − 1 B i .$
For $t ∈ [ 0 , p − 2 ]$, set $A k t + s = { v k t + s v k ( t + 1 ) + s }$, $A k t + s + 1 = { v k t + s + 1 v k ( t + 1 ) + s , v k t + s + 1 v k ( t + 1 ) + s + 1 }$, $… , A k ( t + 1 ) + s − 1 = { v k ( t + 1 ) + s − 1 v k ( t + 1 ) + s , v k ( t + 1 ) + s − 1 v k ( t + 1 ) + s + 1 , … , v k ( t + 1 ) + s − 1 v k ( t + 2 ) + s − 1 }$, and $B k t + s = { v k t + s v k t + s + 1 , … , v k t + s v k ( t + 1 ) + s − 1 } , B k t + s + 1 = { v k t + s + 1 v k t + s + 2 , … , v k t + s + 1 v k ( t + 1 ) + s − 1 } , … , B k ( t + 1 ) + s − 2 = { v k ( t + 1 ) + s − 2 v k ( t + 1 ) + s − 1 } , B k ( t + 1 ) + s − 1 = ϕ$. For $t = 0$ and by Lemma 3, we have $d ′ ( v s , v k p + s ) = p$, $d ′ ( v k + s , v k p + s ) = p − 1$ and $d ′ ( v s , v k p + s ) = p , d ′ ( v s + 1 , v k p + s ) = p$, that is, $d ′ ( v s , v k p + s ) ≠ d ′ ( v k + s , v k p + s )$ and $d ′ ( v s , v k p + s ) = d ′ ( v s + 1 , v k p + s )$. Thus, $f ( w , v s v k + s ) = 1$ and $f ( w , v s v s + 1 ) = 0$. similarly, for any edge $v h 1 v h 2 ∈ ∪ i = k t + s k ( t + 1 ) + s − 1 A i$ with $h 1 < h 2$, we have $d ′ ( v h 1 , v k p + s ) ≠ d ′ ( v h 2 , v k p + s )$, that is, $f ( w , v h 1 v h 2 ) = 1$; for $v h 1 v h 2 ∈ ∪ i = k t + s k ( t + 1 ) + s − 1 B i$ with $h 1 < h 2$, we have $d ′ ( v h 1 , v k p + s ) = d ′ ( v h 2 , v k p + s )$, that is, $f ( w , v h 1 v h 2 ) = 0$. Thus, we can get that
$f ( v k p + s , x y ) = 1 , if x y ∈ ∪ i = k t + s k ( t + 1 ) + s − 1 A i , 0 , if x y ∈ ∪ i = k t + s k ( t + 1 ) + s − 1 B i .$
Next, we consider the edges in the $( k + 1 )$-clique $P n k [ N [ v k p + s ] ]$. For any edge $v h 1 v h 2$ with $h 1 , h 2 ∈ [ k ( p − 1 ) + s , k p + s − 1 ]$, we have $d ′ ( v h 1 , v k p + s ) = d ′ ( v h 2 , v k p + s ) = 1$, that is, $f ( w , v h 1 v h 2 ) = 0$. For any edge $v h v k p + s$ with $h ∈ [ k ( p − 1 ) + s , k p ]$, by Lemma 3, we can obtain that $d ′ ( v 1 , v h ) = p , d ′ ( v 1 , v k p + s ) = p + 1$, $d ′ ( v h − k , v h ) = 1 , d ′ ( v h − k , v k p + s ) = 2$ and when $h ≠ k ( p − 1 ) + s$, $d ′ ( v k ( p − 1 ) + s , v h ) = 1 , d ′ ( v k ( p − 1 ) + s , v k p + s ) = 1$, that is, $d ′ ( v 1 , v h ) ≠ d ′ ( v 1 , v k p + s )$, $d ′ ( v h − k , v h ) ≠ d ′ ( v h − k , v k p + s )$ and $d ′ ( v k ( p − 1 ) + s , v h ) = d ′ ( v k ( p − 1 ) + s , v k p + s )$. Similarly, we get that for $j ∈ [ 1 , p − 1 ] , j ′ ∈ [ 1 , p ]$ and $l ≠ h$,
$d ′ ( v l , v h ) ≠ d ′ ( v l , v k p + s ) if l ∈ [ 1 , s − 1 ] ∪ [ h − j k , k ( p − j ) + s − 1 ] , d ′ ( v l , v h ) = d ′ ( v l , v k p + s ) if l ∈ [ k ( p − j ′ ) + s , h − j ′ k + k − 1 ] ∪ [ h + 1 , k p + s − 1 ] .$
Thus, if $v h = v k ( p − 1 ) + s$, then $d ′ ( v l , v k ( p − 1 ) + s ) ≠ d ′ ( v l , v k p + s )$ with $l ∈ [ 1 , s − 1 ] ∪ { ∪ j = 1 p − 1 [ k ( p − 1 ) + s − j k , ( p − j ) k + s − 1 ] } = [ 1 , ( p − 1 ) k + s − 1 ]$ and $d ′ ( v l , v k ( p − 1 ) + s ) = d ′ ( v l , v k p + s )$ with $l ∈ [ ( p − 1 ) k + s + 1 , k p + s ]$, that is, $P I ( v k ( p − 1 ) + s v k p + s ) = ( p − 1 ) k + s − 1$; similarly, we can obtain that $P I ( v k ( p − 1 ) + s + 1 v k p + s ) = ( p − 1 ) ( k − 1 ) + s − 1 ; P I ( v k ( p − 1 ) + s + 2 v k p + s ) = ( p − 1 ) ( k − 2 ) + s − 1 ; … ; P I ( v k p v k p + s ) = ( p − 1 ) s + s − 1 .$
For any edge $v h v k p + s$ with $h ∈ [ k p + 1 , k p + s − 1 ]$, by Lemma 3, we can obtain that $d ′ ( v h − k , v h ) = 1 , d ′ ( v h − k , v k p + s ) = 2$ and $d ′ ( v k ( p − 1 ) + s , v h ) = 1 , d ′ ( v k ( p − 1 ) + s , v k p + s ) = 1$, that is, $d ′ ( v h − k , v h ) ≠ d ′ ( v h − k , v k p + s )$ and $d ′ ( v k ( p − 1 ) + s , v h ) = d ′ ( v k ( p − 1 ) + s , v k p + s )$. Similarly, we get that for $j ″ ∈ [ 1 , p ]$ and $l ≠ h$,
$d ′ ( v l , v h ) ≠ d ′ ( v l , v k p + s ) if l ∈ [ h − j ″ k , k ( p − j ″ ) + s − 1 ] , d ′ ( v l , v h ) = d ′ ( v l , v k p + s ) if l ∈ [ k ( p − j ″ ) + s , h − j ″ k + k − 1 ] ∪ [ h + 1 , k p + s − 1 ] .$
Thus, if $v h = v k p + 1$, then $d ′ ( v l , v k p + 1 ) ≠ d ′ ( v l , v k p + s )$ for $l ∈ ∪ j ″ = 1 p [ k p + 1 − j ″ k , k ( p − j ″ ) + s − 1 ]$ and $d ′ ( v l , v k p + 1 ) = d ′ ( v l , v k p + s )$ with $l ∈ { ∪ j ″ = 1 p [ k ( p − j ″ ) + s , k ( p + 1 − j ″ ) ] } ∪ [ h + 1 , k p + s − 1 ]$, that is, $P I ( v k p + 1 v k p + s ) = ( s − 1 ) p$; similarly, we have $P I ( v k p + 1 v k p + s ) = ( s − 2 ) p ; … ; P I ( v k p + s − 2 v k p + s ) = 2 p ; P I ( v k p + s − 1 v k p + s ) = p$.
Set $w ∈ V ( P n − 1 k )$, if $x y ∈ E ( P n k )$ with x or $y ≠ v k p + s$, by Lemma 2, we have $f ( w , x y ) = 0$. Thus,
$P I ( P n k ) − P I ( P n − 1 k ) = ∑ x y ∈ ∪ i = 1 k ( p − 1 ) + s − 1 ( A i ∪ B i ) f ( w , x y ) + P I ( v k ( p − 1 ) + s v k p + s ) + P I ( v k ( p − 1 ) + s + 1 v k p + s ) + ⋯ + P I ( v k p + s − 1 v k p + s ) = [ ( k + 2 − s ) + ( k + 3 − s ) + ⋯ + k ] + ( 1 + 2 + ⋯ + k ) ( p − 1 ) + [ k ( p − 1 ) + s − 1 ] + [ ( k − 1 ) ( p − 1 ) + s − 1 ] + [ ( k − 2 ) ( p − 1 ) + s − 1 ] + ⋯ + [ s ( p − 1 ) + s − 1 ] + [ ( s − 1 ) p + ( s − 2 ) p + ⋯ + 2 p + p ] = p k 2 + p k − k 2 − 3 k + 2 k s − s 2 + 3 s − 2 = α .$
Thus, $P I ( P n k ) = k ( k + 1 ) ( p − 1 ) ( 3 k p + 6 s − 2 k − 4 ) 6 + ( s − 1 ) s ( 3 k − s + 2 ) 3$, for $| P n k | = k p + s$ and Theorem 1 is proved. □
Proof of Theorem 2.
We will proceed with it by induction on $n ≥ k + 2$. If $n = k + 2$, by Definition 4, we have $T n , c k ∗$ is also a k-path, that is, $P I ( T n , c k ∗ ) = 2 k$. If $n ≥ k + 3$, assume that Theorem 2 is true for the k-spiral with at most $n − 1$ vertices, we will consider $T n , c k ∗$ with n vertices. Let $T n , c k ∗$ be a k-spiral with $V ( T n , c k ∗ ) = V ( T n − 1 , c k ∗ ) ∪ { v }$ and $E ( T n , c k ∗ ) = E ( T n − 1 , c k ∗ ) ∪ { v v n − 1 , v v n − 2 , … , v v n − k }$ such that $v 1 , v 2 , … , v n − c − 1$ is the simplicial ordering of $P n − c − 1 k − c$, where $T n − 1 , c k ∗ = P n − c − 1 k − c + K c$ with $V ( T n − 1 , c k ∗ ) = { v 1 , v 2 , … , v n − 1 }$ and $E ( T n − 1 , c k ∗ ) = E ( P n − c − 1 k − c ) ∪ E ( K c ) ∪ { v 1 v l , v 2 v l , … , v n − c − 1 v l }$ for $l ∈ [ n − c , n − 1 ]$. For any edge $v i v j ∈ E ( T n , c k ∗ )$, $d ( v i , v j )$ or $d ′ ( v i , v j )$ is the distance of $v i$ and $v j$ in $T n − 1 , c k ∗$ or $T n , c k ∗$, respectively.
For $k + 2 ≤ n ≤ 2 k − c$, let $γ = ( n − k ) ( n − k − 1 ) ( 4 k − n + 2 ) 3 − ( n − k − 1 ) ( n − k − 2 ) ( 4 k − n + 3 ) 3 = ( n − k − 1 ) ( 3 k − n + 2 )$. If we can show that by adding v to $T n − 1 , c k ∗$, $P I ( T n , c k ∗ ) = P I ( T n − 1 , c k ∗ ) + γ$, then Theorem 2 is true.
Set $w = v$ and let $l ∈ [ n − c , n − 1 ]$, by Lemma 4, we have $d ′ ( v l , v ) = 1$ and $d ′ ( v i , v ) = 2$ for $i ∈ [ 1 , n − k − 1 ]$, that is, $f ( w , v l v i ) = 1$; $d ′ ( v l , v ) = d ′ ( v i , v ) = 1$ for $i ∈ [ n − k , n − 1 ]$, that is, $f ( w , v l v i ) = 0$. Set $C 1 = { v 1 v 2 , v 1 v 3 , … , v 1 v n − k − 1 } , C 2 = { v 2 v 3 , v 2 v 4 , … , v 2 v n − k − 1 } , … , C n − k − 2 = { v n − k − 2 v n − k − 1 } , C n − k − 1 = ϕ$, $D 1 = { v 1 v n − k , v 1 v n − k + 1 , … , v 1 v k − c + 1 } , D 2 = { v 2 v n − k , v 2 v n − k + 1 , … , v 2 v k − c + 2 } , … , D n − k − 1 = { v n − k − 1 v n − k , v n − k − 1 v n − k + 1 , … , v n − k − 1 v n − c − 1 }$. By Lemma 4, we have $d ′ ( v 1 , v ) = d ′ ( v 2 , v ) = 2$ and $d ′ ( v n − k , v ) = 1$, that is, $f ( w , v 1 v 2 ) = 0$ and $f ( w , v 1 v n − k ) = 1$. Similarly, for $v h 1 v h 2 ∈ ∪ i = 1 n − k − 1 C i$ with $h 1 < h 2$, we have $d ′ ( v h 1 , v ) = d ′ ( v h 2 , v ) = 2$, that is, $f ( w , v h 1 v h 2 ) = 0$; for $v h 1 v h 2 ∈ ∪ i = 1 n − k − 1 D i$ with $h 1 < h 2$, we have $d ′ ( v h 1 , v ) = 2$ and $d ′ ( v h 2 , v ) = 1$, that is, $f ( w , v h 1 v h 2 ) = 1$. Set $C n − k = { v n − k v n − k + 1 , v n − k v n − k + 2 , … , v n − k v n − c − 1 } , C n − k + 1 = { v n − k + 1 v n − k + 2 , v n − k + 1 v n − k + 3 , … , v n − k + 1 v n − c − 1 } , … , C n − c − 2 = { v n − c − 2 v n − c − 1 }$. By Lemma 4, we have $d ′ ( v n − k , v ) = d ′ ( v n − k − 1 , v ) = 1$, that is, $f ( w , v n − k v n − k − 1 ) = 0$. Similarly, for $v h 1 v h 2 ∈ ∪ i = n − k n − c − 2 C i$ with $h 1 < h 2$, we have $d ′ ( v h 1 , v ) = d ′ ( v h 2 , v ) = 1$, that is, $f ( w , v h 1 v h 2 ) = 0$.
Set $E 1 = { v v i , i ∈ [ n − k , n − c − 1 ] }$, by Lemma 4, we have $d ′ ( v i , v ) = 2 , d ′ ( v i , v n − k ) = 1$ for $i ∈ [ 1 , n − k − 1 ]$ and $d ′ ( v j , v ) = d ′ ( v j , v n − k ) = 1$ for $i ∈ [ n − k + 1 , n ]$. Thus, $P I ( v n − k v ) = n − k − 1$. Similarly, $P I ( v n − k + 1 v ) = P I ( v n − k + 2 v ) = … = P I ( v k − c + 1 v ) = n − k − 1$. In addition, by Lemma 4, we have $d ′ ( v i , v ) = 2 , d ′ ( v i , v k − c + 2 ) = 1$ for $i ∈ [ 2 , n − k − 1 ]$, $d ′ ( v 1 , v ) = d ′ ( v 1 , v k − c + 2 ) = 2$ and $d ′ ( v j , v ) = d ( v j , v k − c + 2 ) = 1$ for $j ∈ [ n − k , n ]$. Thus, $P I ( v k − c + 2 v ) = n − k − 2$. Similarly, we have $P I ( v k − c + 3 v ) = n − k − 3 , P I ( v k − c + 4 v ) = n − k − 4 , … , P I ( v n − c − 1 v ) = 1$. Set $E 2 = { v v l , l ∈ [ n − c , n − 1 ] }$, since $N [ v l ] − N [ v ] = n − k − 1$, we have $P I ( v v l ) = n − k − 1$. Set $E 3 = { v i v l , i ∈ [ 1 , n − c − 1 ] , l ∈ [ n − c , n − 1 ] }$, by Lemma 4, we have $d ′ ( v i , v ) = 2$ for $i ∈ [ 1 , n − k − 1 ]$, $d ′ ( v i , v ) = 1$ for $i ∈ [ n − k , n − c − 1 ]$, $d ′ ( v l , v ) = 1$ for $l ∈ [ n − c , n − 1 ]$. Thus, $f ( w , v i v l ) = 1$ for $i ∈ [ 1 , n − k − 1 ]$ and $f ( w , v i v l ) = 0$ for $i ∈ [ n − k , n − c − 1 ]$.
Set $w ∈ V ( T n k ∗ ) − { v }$, if $x y ∈ E ( T n , c k ∗ )$ with x or $y ≠ v$, by Lemma 2, we have $f ( w , x y ) = 0$. Thus,
$P I ( T n k ∗ ) − P I ( T n − 1 k ∗ ) = ∑ x y ∈ ∪ i = 1 n − c − 2 C i f ( w , x y ) + ∑ x y ∈ i = 1 n − k − 1 D i f ( w , x y ) + ∑ x y ∈ E 1 ∪ E 2 P I ( x y ) + ∑ x y ∈ E 3 f ( w , x y ) = 0 + [ ( 2 k − n − c + 2 ) + ( 2 k − n − c + 3 ) + ⋯ + ( k − c ) ] + [ 1 + 2 + ⋯ + ( n − k − 2 ) + ( n − k − 1 ) ( 2 k − n − c + 2 ) ] + c ( n − k − 1 ) + c ( n − k − 1 ) = ( n − k − 1 ) ( 3 k − n + 2 ) = γ ,$
and Theorem 2 is proved.
For $n ≥ 2 k − c + 1$, let $σ = 3 c ( n − 2 k + c − 1 ) ( n − 2 k + c ) + ( k − c ) ( 2 c 2 + 3 n c − 4 k c + 3 k n − 4 k 2 − 6 k + 3 n − 2 ) 3 − 3 c ( n − 2 k + c − 2 ) ( n − 1 − 2 k + c ) + ( k − c ) ( 2 c 2 + 3 ( n − 1 ) c − 4 k c + 3 k ( n − 1 ) − 4 k 2 − 6 k + 3 n − 2 ) 3 = k 2 − 4 k c + c 2 + 2 n c − 3 c + k$. If we can show that by adding v to $T n − 1 , c k ∗$, $P I ( T n , c k ∗ ) = P I ( T n − 1 , c k ∗ ) + σ$, then Theorem 2 is proved.
Set $w = v$, by Lemma 4, we have $d ′ ( v l , v ) = 1$ for $l ∈ [ n − c , n − 1 ]$, $d ′ ( v i , v ) = 2$ for $i ∈ [ 1 , n − k − 1 ]$ and $d ′ ( v j , v ) = 1$ for $j ∈ [ n − k , n − c − 1 ]$. Thus, $f ( w , v l v i ) = 1$ and $f ( w , v l v j ) = 0$. Set $C 1 = { v 1 v 2 , v 1 v 3 , … , v 1 v k − c + 1 } , C 2 = { v 2 v 3 , v 2 v 4 , … , v 2 v k − c + 2 } , … , C n − 2 k + c − 1 = { v n − 2 k + c − 1 v n − 2 k + c , v n − 2 k + c − 1 v n − 2 k + c + 1 , … , v n − 2 k + c − 1 v n − k − 1 } , C n − 2 k + c = { v n − 2 k + c v n − 2 k + s + 1 , v n − 2 k + c v n − 2 k + s + 2 , … , v n − 2 k + c v n − k − 1 } , C n − 2 k + c + 1 = { v n − 2 k + c + 1 v n − 2 k + c + 2 , v n − 2 k + c + 1 v n − 2 k + c + 3 , … , v n − 2 k + c + 1 v n − k − 1 } , . … , C n − k − 1 = ϕ , D n − 2 k + c = { v n − 2 k + c v n − k } , D n − 2 k + c + 1 = { v n − 2 k + c + 1 v n − k , v n − 2 k + c + 1 v n − k + 1 } , … , D n − k − 1 = { v n − k − 1 v n − k , v n − k − 1 v n − k + 1 , … , v n − k − 1 v n − c − 1 }$.
By Lemma 4, we can get that $d ′ ( v 1 , v ) = d ′ ( v 2 , v ) = 2$ and $d ′ ( v n − k , v ) = 1$, that is, $f ( w , v 1 v 2 ) = 0$ and $f ( w , v 1 v n − k ) = 1$. Similarly, for $v h 1 v h 2 ∈ ∪ i = 1 n − k − 1 C i$ with $h 1 < h 2$, we have $d ′ ( v h 1 , v ) = d ′ ( v h 2 , v ) = 2$, that is, $f ( w , v h 1 v h 2 ) = 0$; for $v h 1 v h 2 ∈ ∪ i = n − 2 k + c n − k − 1 D i$ with $h 1 < h 2$, we have $d ′ ( v h 1 , v ) = 2$ and $d ′ ( v h 2 , v ) = 1$, that is, $f ( w , v h 1 v h 2 ) = 1$. Set $C n − k = { v n − k v n − k + 1 , v n − k v n − k + 2 , … , v n − k v n − c − 1 } , C n − k + 1 = { v n − k + 1 v n − k + 2 , v n − k + 1 v n − k + 3 , … , v n − k + 1 v n − c − 1 } , … , C n − c + 2 = { v n − c − 2 v n − c − 1 }$. By Lemma 4, we can get that $d ′ ( v n − k , v ) = d ′ ( v n − k + 1 , v ) = 1$, that is, $f ( w , v n − k v n − k + 1 ) = 0$. Similarly, for $v h 1 v h 2 ∈ ∪ i = n − k n − c − 2 C i$ with $h 1 < h 2$, we have $d ′ ( v h 1 , v ) = d ′ ( v h 2 , v ) = 1$, that is, $f ( w , v h 1 v h 2 ) = 0$.
Set $E 1 = { v v i , i ∈ [ n − k , n − c − 1 ] }$, by Lemma 4, we have $d ′ ( v , v n − k − 1 ) = 2 , d ′ ( v n − c − 1 , v n − k − 1 ) = 1$, $d ′ ( v , v i ) = d ′ ( v n − c − 1 , v i ) = 1$ for $i ∈ [ n − k , n − c − 2 ] ∪ [ n − c , n − 1 ]$ and $d ′ ( v , v j ) = d ( v n − c − 1 , v j ) = 2$ for $j ∈ [ 1 , n − k − 2 ]$. Thus, $P I ( v v n − c − 1 ) = 1$. Similarly, we have $P I ( v v n − c − 2 ) = 2 , P I ( v v n − c − 3 ) = 3 , … , P I ( v v n − k ) = k − c$. Set $E 2 = { v v l , l ∈ [ n − c , n − 1 ] }$, since $N [ v l ] − N [ v ] = n − k − 1$, we have $P I ( v v l ) = n − k − 1$. Set $E 3 = { v i v l , i ∈ [ 1 , n − c − 1 ] , l ∈ [ n − c , n − 1 ] }$, by Lemma 4, we have $d ′ ( v , v i ) = 2 , d ′ ( v , v l ) = 1$ for $i ∈ [ 1 , n − k − 1 ]$ and $d ′ ( v , v i ) = d ′ ( v , v l ) = 1$ for $i ∈ [ n − k , n − c − 1 ]$. Thus, $f ( w , v i v l ) = 1$ for $i ∈ [ 1 , n − k − 1 ]$ and $f ( w , v i v l ) = 0$ for $i ∈ [ n − k , n − c − 1 ]$.
Set $w ∈ V ( T n k ∗ ) − { v }$, if $x y ∈ E ( T n , c k ∗ )$ with x or $y ≠ v$, by Lemma 2, we have $f ( w , x y ) = 0$. Thus,
$P I ( T n k ∗ ) − P I ( T n − 1 k ∗ ) = ∑ x y ∈ ∪ i = 1 n − c − 2 C i f ( w , x y ) + ∑ x y ∈ i = n − 2 k + c n − k − 1 D i f ( w , x y ) + ∑ x y ∈ E 1 ∪ E 2 P I ( x y ) + ∑ x y ∈ E 3 f ( w , x y ) = 0 + [ 1 + 2 + 3 + ⋯ + ( k − c ) ] + [ 1 + 2 + 3 + ⋯ + ( k − c ) ] + c ( n − k − 1 ) + c ( n − k − 1 ) = k 2 − 4 k c + c 2 + 2 n c − 3 c + k = σ ,$
and Theorem 2 is proved. □
Proof of Theorem 3.
For $n ≥ k + 2$, 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, $P I ( T n k ) = 2 k$. If $n ≥ k + 3$, assume that Theorem 3 is true for the k-tree with at most $n − 1$ vertices, let $v ∈ S 1 ( T n k )$ and $T n − 1 k = T n k − v$, by the induction hypothesis, we have $P I ( T n − 1 k ) ≤ P I ( S n − 1 k ) = k ( n − k − 1 ) ( n − k − 2 )$. By adding back v, let $N ( v ) = { x 1 , x 2 , … , x k }$ and $w = v$. Since $T n k [ v , x 1 , x 2 , … , x k ]$ is a $( k + 1 )$-clique, we have $f ( w , x i x j ) = 0$ for $i , j ∈ [ 1 , k ]$. By Lemmas 1 and 2, we can obtain that $P I ( v x i ) ≤ n − k − 1$ with $i ∈ [ 1 , k ]$ and $f ( w , x y ) ≤ 1$ for any edge $x y ∈ E ( T n k ) − E ( T n k [ v , x 1 , x 2 , … , x k ] )$. Next, set $w ∈ V ( T n k ) − { v }$, by Lemma 2, if $x y ∈ E ( T n k )$ with x or $y ≠ v$, we have $f ( w , x y ) = 0$. Since $| E ( T n k ) − E ( T n k [ v , x 1 , x 2 , … , x k ] ) | = k ( n − k − 1 )$, we have
$P I ( T n k ) = P I ( T n − 1 k ) + ∑ x y ∈ E ( T n k − { v x i , i ∈ [ 1 , k ] } ) f ( w , x y ) + ∑ i = 1 k P I ( v x i ) ≤ P I ( S n − k k ) + k ( n − k − 1 ) + k ( n − k − 1 ) = k ( n − k − 1 ) ( n − k − 2 ) + k ( n − k − 1 ) + k ( n − k − 1 ) = k ( n − k ) ( n − k − 1 ) = P I ( S n k ) .$
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 $P I$-value; then, Theorem 4 is obvious; for $k ≥ 2$, if $k + 2 ≤ n ≤ 2 k − c$, let $n = k p + s$ with $p = 1$ and $s = n − k$, by Theorem 1, we have $P I ( P n k ) − P I ( T n , c k ∗ ) = ( s − 1 ) s ( 3 k − s + 2 ) 3 − ( n − k ) ( n − k − 1 ) ( 4 k − n + 2 ) 3 = ( n − k − 1 ) ( n − k ) [ 3 k − ( n − k ) + 2 ] 3 − ( n − k ) ( n − k − 1 ) ( 4 k − n + 2 ) 3 = 0$, and Theorem 4 is true. If $n ≥ 2 k − c + 1$, $p = n − s k$ and by Theorems 1 and 2, define the new functions as follows: for $z ≥ 2 k − c + 1$, $1 ≤ c ≤ k − 1$ and $2 ≤ s ≤ k + 1$,
$g ( z ) = ( k + 1 ) ( z − s − k ) ( 3 z + 3 s − 2 k − 4 ) 6 + s ( s − 1 ) ( 3 k − s + 2 ) 3 , h ( z , c ) = c ( z − 2 k + c − 1 ) ( z − 2 k + c ) + ( k − c ) ( 2 c 2 + 3 z c − 4 k c + 3 k z − 4 k 2 − 6 k + 3 z − 2 ) 3 , l ( z , c ) = g ( z ) − h ( z , c ) = ( k 2 + 1 2 − c ) z 2 + ( − c 2 + 2 c + 4 k c − 11 k 2 6 − 5 k 2 − 2 3 ) z + k s 2 2 − k 2 s 6 − k s 2 + 5 k 3 3 + 5 k 2 3 + s 2 2 + 4 k 3 − s 3 3 − 6 k 2 c + 3 k c 2 − c 3 3 − 4 k c + c 2 − 2 c 3 , l z ( z ) = l z ( z , c ) = ( k + 1 − 2 c ) z − c 2 + 2 c + 4 k c − 11 k 2 6 − 5 k 2 − 2 3 .$
Then, it is enough to determine whether or not $l ( z , c ) ≥ 0$ is true. By some calculations, we can obtain the following claim:
Claim 1.
$z 1 = 2 k − c + 1$,$z 2 = 2 k − c + 2$are the two roots of$l ( z , c ) = 0$with$c ≠ k + 1 2$.
Proof.
For any $c ∈ [ 1 , k − 1 ]$, let $z 1 = 2 k − c + 1$, $z 2 = 2 k − c + 2$, and we have $l ( 2 k − c + 1 , c ) = 0 , l ( 2 k − c + 2 , c ) = 0$. If $c ≠ k + 1 2$, then Claim is true. □
For fixed $c ∈ [ 1 , k + 1 2 )$, that is, $k 2 + 1 2 − c > 0$, then the function of $l ( z , c )$ about z is open up. Since z is an integer and by Fact 2, we have $l ( z , c ) ≥ 0$ for $z ≥ 2 k − c + 1$ and Theorem 4 is true; if $c = k + 1 2$ and $k ≥ 1$, we have $l z ( z ) = 1 − k 2 12 ≤ 0$, that is, $l ( z , k + 1 2 )$ is decreasing about z. By the proof of Fact 2, we have $l ( 2 k − c + 1 , c ) = 0$. For $z ≥ 2 k − c + 1$, we can get that $l ( z , k + 1 2 ) ≤ l ( 2 k − c + 1 , k + 1 2 ) = 0$ and Theorem 4 is true; for fixed $c ∈ ( k + 1 2 , k − 1 ]$, that is, $k 2 + 1 2 − c < 0$, then the function of $l ( z , c )$ about z is open down. Since z is an integer and by Claim, we can obtain that $l ( z , c ) ≤ 0$ for $z ≥ 2 k − c + 1$ and this finishes the proof of Theorem 4. □

## 4. Conclusions

We can see that the k-stars attain the maximal values of $P I$-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 $P I$-values of k-spirals is even smaller than that of k-paths. Meanwhile, not all $P I$-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 $P I$-value?

## Author Contributions

S.W. contributes for supervision, methodology, validation, project administration and formal analysing. S.W., Z.S., J.-B.L., B.W. contribute for resources, investigation some computations and wrote the initial draft of the paper which were investigated and approved by S.W. and B.W. wrote the final draft.

## Funding

This research was funded by Natural Science Foundation of Guangdong Province under grant 2018A0303130115.

## Conflicts of Interest

The authors declare no conflict of interest.

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