Leap Zagreb Indices of Some Wheel Related Graphs
Shiladhar P., A. M. Naji and N. D. Soner Department of Studies in Mathematics,
University of Mysore, Manasagangotri, Mysuru-570 006, INDIA.
email:shiladharpawar@gmail.com and ndsoner@yahoo.co.in.
(Received on: March 12, 2018)
ABSTRACT
Recently, A. M. Naji et al.10, introduced leap Zagreb indices of a graph based on the second degrees of vertices (number of their second neighbours). The first leap Zagreb index 𝐿𝑀1(𝐺) is equal to the sum of squares of the second degrees of the vertices, the second leap Zagreb index 𝐿𝑀2(𝐺) is equal to the sum of the products of the second degrees of pairs of adjacent vertices of G and the third leap Zagreb index 𝐿𝑀3(𝐺) is equal to the sum of the products of the first degrees with the second degrees of the vertices. In this paper, exact expressions for leap Zagreb indices of wheel 𝑊𝑛, and some related graphs as gear 𝐺𝑛, helm 𝐻𝑛, flower 𝐹𝑙𝑛 and sunflower 𝑆𝑓𝑛 graphs are computed.
2010 Mathematics Subject Classification : 05C07, 05C12, 05C76.
Keywords: Degrees, Second degrees (of vertex), Leap Zagreb indices, Wheel graphs.
1. INTRODUCTION
In this paper, we are concerned only with simple graphs, i.e., finite graphs having no
loops, multiple and directed edges. Let 𝐺 = (𝑉, 𝐸) be such a graph with vertex set 𝑉 (𝐺) and
edges set 𝐸(𝐺). As usual, we denote by 𝑛 = | 𝑉| and 𝑚 = | 𝐸 | to the number of vertices and
edges in a graph 𝐺, respectively. The distance 𝑑
𝐺(𝑢, 𝑣) between any two vertices u and v of a
graph 𝐺 is equal to the length of (number of edges in) a shortest path connecting them. For a
vertex 𝑣 ∈ 𝑉 (𝐺) and a positive integer 𝑘, the open 𝑘-neighborhood of 𝑣 in a graph 𝐺 is
denoted by 𝑁
𝑘(𝑣/𝐺) and is defined as 𝑁
𝑘(𝑣/𝐺) = {𝑢 ∈ 𝑉 (𝐺) ∶ 𝑑
𝐺(𝑢, 𝑣) = 𝑘}. The k-
distance degree of a vertex v in G is denoted by 𝑑
𝑘(𝑣/𝐺) (or simply 𝑑
𝑘(𝑣) if no
misunderstanding) and is defined as the number of k- neighbours of the vertex v in G, i.e.,
𝑑
𝑘(𝑣/𝐺) = |𝑁
𝑘(𝑣/𝐺)|. It is clearly that 𝑑
1(𝑣/𝐺) = 𝑑(𝑣/𝐺) for every 𝑣 ∈ 𝑉(𝐺). For a
vertex v of G, the eccentricity 𝑒(𝑣) = 𝑚𝑎𝑥{𝑑
𝐺(𝑣, 𝑢) ∶ 𝑢 ∈ 𝑉 (𝐺)}. The diameter of G is
𝑑𝑖𝑎𝑚(𝐺) = 𝑚𝑎𝑥{𝑒(𝑣) ∶ 𝑣 ∈ 𝑉 (𝐺)} and the radius of G is 𝑟𝑎𝑑(𝐺) = 𝑚𝑖𝑛{𝑒(𝑣) 𝑣 ∈ 𝑉 (𝐺)}. For any terminology or notation not mention here, we refer to
9.
A topological index of a graph is a graph invariant number calculated from a graph representing a molecule and applicable in chemistry. The Zagreb indices have been introduced, more than forty four years ago, by I. Gutman and Trinajestic
7, in 1972, and elaborated in
6. They are defined as:
𝑀
1( 𝐺) = ∑ 𝑑
12( 𝑣/𝐺)
𝑣∈𝑉( 𝐺)
and
𝑀
2(𝐺) = ∑ 𝑑
1𝑢,𝑣∈𝐸( 𝐺)
( 𝑢/𝐺) 𝑑
1(𝑣/𝐺) .
For properties of the two Zagreb indices see
2,3,6,12,15and the papers cited therein. In recent years, some novel variants of ordinary Zagreb indices have been introduced and studied, such as Zagreb coincides
1,8, multiplicative Zagreb indices
5,13,15, multiplicative sum Zagreb index
4,14, and multiplicative Zagreb coincides
16and etc.
Recently, A. M. Naji et al.
10, have been introduced a new distance-degree-based topological indices conceived depending on the second degrees of vertices (number of their second neighbours), and are so-called leap Zagreb indices of a graph G and are defined as, respectively.
𝐿𝑀
1(𝐺) = ∑ 𝑑
22(𝑣/𝐺)
𝑣∈𝑉(𝐺)
.
𝐿𝑀
2(𝐺) = ∑ 𝑑
2(𝑢/𝐺) 𝑑
2(𝑣/𝐺)
𝑢𝑣∈𝐸(𝐺)
.
𝐿𝑀
3(𝐺) = ∑ 𝑑(𝑢/𝐺) 𝑑
2(𝑣/𝐺)
𝑣∈𝑉(𝐺)
.
For properties of the leap Zagreb indices see
10,11.
In this paper, the explicit formulae for leap Zagreb indices of wheel and some related graph are presented.
2. MAIN RESULT
Definition.2.1. The wheel graph 𝑊
𝑛with 𝑛 + 1 vertices is defined to be the join of 𝐾
1and 𝐶
𝑛, where 𝐾
1is the complete graph with one vertex and 𝐶
𝑛is the cycle graph with n vertices.
Clearly, |V (𝑊
𝑛)| = n + 1 and |E (𝑊
𝑛)| = 2n. The vertex corresponding to 𝐾
1is known as apex
while the vertices corresponding to 𝐶
𝑛are 𝐾
1known as rim vertices.
Figure 1: Wheel graph 𝑾𝒏.
Lemma 2.1. Let 𝑊
𝑛be a wheel graph with n+1 vertices ( as shown in fig. 1 ).Then we have 𝑑(𝑣
0) = 𝑛, 𝑑
2(𝑣
0) = 0.
𝑑(𝑣
𝑖) = 3, 𝑑
2(𝑣
𝑖) = 𝑛 − 3.
Theorem 2.2. Let 𝑊
𝑛, n ≥ 4 be a wheel graph with n+1 vertices. Then 1) 𝐿𝑀
1(𝑊
𝑛) = 𝑛(𝑛
2− 6𝑛 + 9)
2) 𝐿𝑀
2( 𝑊
𝑛) = 𝑛( 𝑛 − 3)
23) 𝐿𝑀
3( 𝑊
𝑛) = 3𝑛(𝑛 − 3).
Proof. Let 𝑊
𝑛, 𝑛 ≥ 4 be a wheel with 𝑛 + 1 vertices and let 𝑣
0,𝑣
1,…, 𝑣
𝑛be a vertices set of 𝑤
𝑛, where 𝑣
0be the apex vertex and 𝑣
1,𝑣
2,…,𝑣
𝑛be the rim vertices of 𝑊
𝑛.
1) By Lemma 2.1., and 𝑓𝑟𝑜𝑚 𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝐿𝑀
1( 𝐺) 𝑤𝑒 ℎ𝑎𝑣𝑒 , 𝐿𝑀
1(𝑊
𝑛) = ∑ 𝑑
22𝑣∈𝑉( 𝑊𝑛)
(𝑣)
= 𝑑
22(𝑣
0) + ∑ 𝑑
22𝑛
𝑖=1
(𝑣
𝑖)
= 0 + ∑
𝑛𝑖=11(𝑛 − 3)
2= 0 + 𝑛(𝑛 − 3)
2= 𝑛(𝑛
2− 6𝑛 + 9)
Therefore 𝐿𝑀
1(𝑊
𝑛) = 𝑛(𝑛
2− 6𝑛 + 9). ∎
2) From figure 1., we have two types of edges 𝑣
0𝑣
𝑖and 𝑣
𝑖𝑣
𝑖+1, for indices being taken modolu 𝑛, and from definition of L𝑀
2(𝐺) we have,
𝐿𝑀
2(𝑊
𝑛) = ∑ 𝑑
2𝑢,𝑣∈𝐸(𝑊𝑛)
(𝑣
0)𝑑
2(𝑣)
= ∑
𝑣0𝑣𝑑
2𝑖∈𝐸(𝑊𝑛)
(𝑣
0)𝑑
2(𝑣
𝑖) + ∑
𝑛𝑣𝑖𝑣𝑑
2𝑖+1∈𝐸(𝑊𝑛)
(𝑣
𝑖)𝑑
2(𝑣
𝑖+1) = 0 + ∑
𝑛𝑖=1(𝑛 − 3)(𝑛 − 3)
= 0 + 𝑛 (𝑛 − 3)
2Therefore, 𝐿𝑀
2(𝑊𝑛)= 𝑛( 𝑛 − 3)
2. ∎ 3) From definition of L𝑀
3(𝐺) we have,
𝐿𝑀
3(𝑊
𝑛) = ∑ 𝑑(𝑣) 𝑑
2(𝑣)
𝑣∈𝑉(𝑊𝑛)
= 𝑑(𝑣
0)𝑑
2(𝑣
0) + ∑
𝑛𝑖=1𝑑 (𝑣
𝑖)𝑑
2(𝑣
𝑖) = 0 + ∑
𝑛𝑖=13(𝑛 − 3 )
= 3𝑛(𝑛 − 3)
Therefore 𝐿𝑀
3(𝑤
𝑛) = 3𝑛(𝑛 − 3). ∎
Definition 2.2. The gear graph 𝐺
𝑛, also sometimes known as a bipartite wheel graph, is a graph obtained from wheel graph 𝑊
𝑛by added a vertex between each pair of adjacent rim vertices. It contains three type of vertices, the vertex of degree n called apex, n vertices of degree three and n vertices of degree two. The gear graph 𝐺
𝑛has 2n+1 vertices and 3n edges.
Figure 2: Gear graph 𝑮𝒏.
Lemma 2.3. Let 𝐺
𝑛be a gear graph with 2𝑛 + 1 vertices ( as shown in fig. 2). Then we have, 𝑑(𝑣
0) = 𝑛 𝑑
2(𝑣
0) = 𝑛
𝑑(𝑣
𝑖) = 3 𝑑
2(𝑣
𝑖) = 𝑛 − 1 𝑑(𝑢
𝑖) = 2 𝑑
2(𝑢
𝑖) = 3 Theorem 2.4. Let 𝐺
𝑛be a gear graph with 2𝑛 + 1 vertices. Then 1) L𝑀
1(𝐺
𝑛) = 𝑛(𝑛
2− 𝑛 + 10).
2) 𝐿𝑀
2(𝐺
𝑛) = 𝑛(𝑛 − 1)(𝑛 + 6).
3) 𝐿𝑀
3(𝐺
𝑛) = 𝑛( 4𝑛 + 3).
Proof. Let 𝐺
𝑛be a gear graph and let 𝑣
0be the apex vertex, 𝑣
𝑛, 𝑣
2, 𝑣
3, … , 𝑣
𝑛be the vertices of 𝐺
𝑛with three degree and 𝑢
1, 𝑢
2, … , 𝑢
𝑛be the vertices of 𝐺
𝑛with two degree.
1) By Lemma 2.3., and from definition of L𝑀
1(𝐺) , we have , 𝐿𝑀
1(𝐺
𝑛) = ∑ 𝑑
22(𝑣)
𝑣∈𝑉(𝐺𝑛)
.
= 𝑑
22(𝑣
0) + ∑
𝑛𝑖=1𝑑
22(𝑣
𝑖) + ∑
𝑛𝑖=1𝑑
22(𝑢
𝑖) = 𝑛
2+ ∑
𝑛𝑖=1(𝑛 − 1)
2+ ∑
𝑛𝑖=13
2= 𝑛
2+ 𝑛(𝑛 − 1)
2+ 3
2𝑛 = 𝑛
2+ 𝑛(𝑛
2− 2𝑛 + 1) + 9𝑛 = 𝑛
2+ 𝑛
3− 2𝑛
2+ 𝑛 + 9𝑛 = 𝑛
3− 𝑛
2+ 10𝑛
= 𝑛( 𝑛
2− 𝑛 + 10).
Therefore 𝐿𝑀
1(𝐺
𝑛) = 𝑛( 𝑛
2− 𝑛 + 10). ∎ 2) From definition of L𝑀
2( 𝐺), we have ,
𝐿𝑀
2(𝐺
𝑛) = ∑
𝑢𝑣∈𝐸( 𝑍𝑛)𝑑
2(𝑢)𝑑
2(𝑣)
= ∑
𝑛𝑖=1𝑑
2(𝑣
0)𝑑
2(𝑣𝑖) + ∑
𝑛𝑖=1𝑑
2( 𝑢
𝑖) 𝑑
2( 𝑣
𝑖) = ∑
𝑛𝑖=1𝑛(𝑛 − 1 ) + ∑
2𝑛𝑖=13(𝑛 − 1)
= 𝑛 . 𝑛 (𝑛 − 1) + 2𝑛. 3(𝑛 − 1) = 𝑛
2(𝑛 − 1) + 6𝑛(𝑛 − 1) = 𝑛
3− 𝑛
2+ 6𝑛
2− 6𝑛 Therefore, L𝑀
2(𝐺
𝑛) = 𝑛(𝑛 − 1)(𝑛+6). ∎ 3) From definition of L𝑀
3( 𝐺), we have,
𝐿𝑀
3(𝐺
𝑛) = ∑ 𝑑(𝑣)𝑑
2(𝑣)
𝑣∈𝑉(𝐺𝑛)
.
= 𝑑(𝑣
0)𝑑
2(𝑣
0) + ∑
𝑛𝑖=1𝑑(𝑣
𝑖) 𝑑
2(𝑣
𝑖) + ∑
𝑛𝑖=1𝑑 (𝑢
𝑖)𝑑
2(𝑢
𝑖) = 𝑛 . 𝑛 + ∑
𝑛𝑖=13 (𝑛 − 1) + ∑
𝑛𝑖=12 (3)
= 𝑛
2+ 3𝑛(𝑛 − 1) + 6𝑛 = 𝑛
2+ 3𝑛
2− 3𝑛 + 6𝑛 = 4𝑛
2+ 3n
Therefore 𝐿𝑀
3(𝐺
𝑛) = 𝑛(4𝑛 + 3). ∎
Definition 2.3. The helm graph 𝐻
𝑛is a graph obtained from wheel graph 𝑊
𝑛by attaching a
pendant edge to each rim vertex. The helm graph contains three types of vertices, the vertex
of degree n called apex, n pendant vertices and n rim vertices of degree four. The helm graph
𝐻
𝑛has 2n + 1 vertices and 3n edges.
Figure 3 : Helm graph 𝑯𝒏
Lemma 2.5. Let 𝐻
𝑛be a helm graph with 2n + 1 vertices( as shown in fig. 3). Then we have, 𝑑(𝑣
0) = 𝑛 𝑑
2(𝑣
0) = 𝑛
𝑑(𝑣
𝑖) = 4 𝑑
2(𝑣
𝑖) = 𝑛 − 1 𝑑(𝑢
𝑖) = 1 𝑑
2(𝑢
𝑖) = 3 Theorem 2.6. Let 𝐻
𝑛be a helm graph with 2n + 1 vertices. Then 1) 𝐿𝑀
1(𝐻
𝑛) = 𝑛( 𝑛
2– 𝑛 + 10)
2) 𝐿𝑀
2(𝐻
𝑛) = 2𝑛( 𝑛
2− 1) 3) 𝐿𝑀
3(𝐻
𝑛) = 𝑛(5𝑛 − 1).
Proof . Let 𝐻
𝑛be a helm graph with 2n + 1 vertices and let 𝑣
0be the apex vertex, 𝑣
1, 𝑣
2, 𝑣
3, … , 𝑣
𝑛be the rim vertices with four degrees and 𝑢
1, 𝑢
2, 𝑢
3, … , 𝑢
𝑛be the pendant vertices.
1) By Lemma. 2.5., and from definition of L𝑀
1( 𝐺), we have,
𝐿𝑀
1(𝐻
𝑛) = ∑ 𝑑
22(𝑣)
𝑣∈𝑉(𝐻𝑛)
.
= 𝑑
22(𝑣
0) + ∑
𝑛𝑖=1𝑑
22(𝑣
𝑖) + ∑
𝑛𝑖=1𝑑
22(𝑢
𝑖) = 𝑛
2+ 𝑛(𝑛 − 1)
2+ 3
2𝑛
= 𝑛
2+ 𝑛( 𝑛
2− 2𝑛 + 1) + 9𝑛 = 𝑛
2+ 𝑛
3− 2𝑛
2+ 𝑛 + 9𝑛 = 𝑛
3− 𝑛
2+ 10𝑛
Therefore L𝑀
1(𝐻
𝑛) = 𝑛( 𝑛
2− 𝑛 + 10). ∎ 2) From definition of L𝑀
2( 𝐺), we have,
𝐿𝑀
2(𝐻
𝑛) = ∑ 𝑑
2(𝑢)
𝑢𝑣∈𝐸( 𝐻𝑛)
𝑑
2(𝑣)
= ∑ 𝑑
2( 𝑣
0)
𝑛
𝑣0,𝑣𝑖∈𝐸(𝐻𝑛)
𝑑
2(𝑣
𝑖) + ∑ 𝑑
2𝑛
𝑣𝑖,𝑢𝑖
(𝑣
𝑖)𝑑
2(𝑢
𝑖) + ∑ 𝑑
2𝑛
𝑣𝑖,𝑣𝑖+1
(𝑣
𝑖)𝑑
2( 𝑣
𝑖+1) = ∑
𝑛𝑖=1𝑛 ( 𝑛 − 1 ) + ∑
𝑛𝑖=1(𝑛 − 1) 3 + ∑
𝑛𝑖=1( 𝑛 − 1 ) (𝑛 − 1) = 𝑛
2( 𝑛 − 1) + 3𝑛(𝑛 − 1) + 𝑛(𝑛 − 1)
2= 𝑛
3− 𝑛
2+ 3𝑛
2− 3𝑛 + 𝑛( 𝑛
2− 2𝑛 + 1) = 𝑛
3− 𝑛
2+ 3𝑛
2− 3𝑛 + 𝑛
3− 2𝑛
2+ 𝑛 = 2𝑛
3− 2𝑛
Therefore L𝑀
2(𝐻
𝑛) = 2𝑛(𝑛
2− 1). ∎
3) From definition of L𝑀
3( 𝐺), we have, 𝐿𝑀
3(𝐻
𝑛) = ∑ 𝑑(𝑣)𝑑
2𝑣∈𝑉(𝐻𝑛)
(𝑣).
= 𝑑(𝑣
0)𝑑
2(𝑣
0) + ∑
𝑛𝑖=1𝑑 (𝑣
𝑖)𝑑
2(𝑣
𝑖) + ∑
𝑛𝑖=1𝑑 (𝑢
𝑖)𝑑
2(𝑣
𝑖) = 𝑛 . 𝑛 + ∑
𝑛𝑖=14 ( n − 1) + ∑
ni=11 ( 3)
= 𝑛
2+ 4𝑛( 𝑛 − 1) + 3𝑛 = 𝑛
2+ 4𝑛
2− 4𝑛 + 3𝑛 = 5𝑛
2− 𝑛
Therefore L𝑀
3(𝐻
𝑛) = 𝑛 (5𝑛 − 1). ∎
Definition 2.4. The flower graph 𝐹𝑙
𝑛is a graph obtained from a helm graph by joining each pendant vertex to the apex of the helm graph. There are three types of vertices, the apex of degree 2n, n vertices of degree four and n vertices of degree two. The flower graph 𝐹𝑙
𝑛has 2n + 1 vertices and 4n edges.
Figure 4 : Flower graph 𝑭𝒍𝒏.
Lemma 2.7., Let 𝐹𝑙
𝑛be a flower graph with 2n + 1 vertices ( as shown in fig. 4). Then we have, 𝑑(𝑣
0) = 2𝑛 , 𝑑
2(𝑣
0) = 0,
𝑑(𝑣
𝑖) = 4 , 𝑑
2(𝑣
𝑖) = 𝑛 − 5, 𝑑(𝑣
𝑖) = 2, 𝑑
2(𝑢
𝑖) = 𝑛 − 2 . Theorem 2.8. Let 𝐹𝑙
𝑛be a flower graph with 2n + 1 vertices. Then 1) L𝑀
1(𝐹𝑙
𝑛) = 𝑛(2𝑛
2− 14𝑛 + 29 )
2) 𝐿𝑀
2(𝐹𝑙
𝑛) = 𝑛(2𝑛
2− 17𝑛 + 35) 3) 𝐿𝑀
3( 𝐹𝑙
𝑛) = 2𝑛( 3𝑛 − 12).
Proof. Let 𝐹𝑙
𝑛be a flower graph with 2n + 1 vertices, 4n edges and let 𝑣
0be the apex vertex, 𝑣
1, 𝑣
2, 𝑣
3, … , 𝑣
𝑛be the rim vertices with four degrees and 𝑢
1, 𝑢
2, 𝑢
3, … , 𝑢
𝑛be the extreme vertices with two degrees. Since 𝑑(𝑣
0) = 2𝑛.
1) By Lemma 2.7., and from definition of L𝑀
1( 𝐺), we have, 𝐿𝑀
1(𝐹𝑙
𝑛) = ∑ 𝑑
22(𝑣).
𝑣∈𝑉(𝐹𝑙𝑛)
= 𝑑
22(𝑣
0) + ∑
𝑛𝑖=1𝑑
22(𝑣
𝑖) + ∑
𝑛𝑖=1𝑑
22(𝑢
𝑖) = 0 + ∑
𝑛𝑖=1( 𝑛 − 5)
2+ ∑
𝑛𝑖=1( 𝑛 − 2)
2= 𝑛( 𝑛 − 5)
2+ 𝑛(𝑛 − 2)
2= 𝑛( 𝑛
2− 10𝑛 + 25) + 𝑛( 𝑛
2− 4𝑛 + 4) = 𝑛
3− 10𝑛
2+ 25𝑛 + 𝑛
3− 4𝑛
2+ 4𝑛 = 2𝑛
3− 14𝑛
2+ 29𝑛
Therefore 𝐿𝑀
1(𝐹𝑙
𝑛) = 𝑛(2𝑛
2− 14𝑛 + 29). ∎
2) From figure 4., we have four types of edges,𝑣
0𝑣
𝑖, 𝑣
0𝑢
𝑖, 𝑣
𝑖𝑢
𝑖𝑎𝑛𝑑 𝑣
𝑖𝑣
𝑖+1, 𝑓𝑜𝑟 𝑖 = 1,2, … , 𝑛 and definition of L𝑀
2( 𝐺), we have,
𝐿𝑀
2(𝐹𝑙
𝑛) = ∑ 𝑑
2𝑢,𝑣 ∈𝐸(𝐹𝑙𝑛)
(𝑢
𝑖)𝑑
2(𝑣
𝑖)
= ∑𝑛 𝑑2𝑖=1 ( 𝑣0)𝑑2(𝑣𝑖) + ∑𝑛 𝑑2
𝑖=1 (𝑣0)𝑑2(𝑢𝑖) + ∑𝑛 𝑑2
𝑖=1 (𝑣𝑖)𝑑2(𝑢𝑖) + ∑𝑛 𝑑2
𝑖=1 (𝑣𝑖)𝑑2(𝑣𝑖+1)
= 0 + 0 + 𝑛( 𝑛 − 5)( 𝑛 − 2) + 𝑛 ( 𝑛 − 5) ( 𝑛 − 5)
= 𝑛(𝑛 − 5)[(𝑛 − 2) + (𝑛 − 5) ] = 𝑛(𝑛 − 5) ( 2𝑛 − 7)
= 2𝑛
3− 7𝑛
2− 10𝑛
2+ 35
Therefore L𝑀
2(𝐹𝑙
𝑛) = 𝑛( 2𝑛
2− 17𝑛 + 35). ∎ 3) Form definition of L𝑀
3( 𝐺) we have,
𝐿𝑀
3(𝐹𝑙
𝑛) = ∑ 𝑑(𝑣)𝑑
2( 𝑣)
𝑣∈𝑉( 𝐹𝑙𝑛)
.
= 𝑑(𝑣
0)𝑑
2( 𝑣
0) + ∑
𝑛𝑖=1𝑑 ( 𝑣
𝑖)𝑑
2( 𝑣
𝑖) + ∑
𝑛𝑖=1𝑑 (𝑢
𝑖)𝑑
2(𝑢
𝑖)
= 0 + ∑
𝑛𝑖=14 (𝑛 − 5) + ∑
𝑛𝑖=12 (𝑛 − 2) = 4𝑛(𝑛 − 5) + 2𝑛( 𝑛 − 2)
= 4𝑛
2− 20𝑛 + 2𝑛
2− 4𝑛 = 6𝑛
2− 24𝑛
Therefore L𝑀
3(𝐹𝑙
𝑛) = 2𝑛( 3𝑛 − 12). ∎
Definition 2.5. The sunflower graph 𝑆𝑓
𝑛is the graph obtained from the flower graph 𝐹𝑙
𝑛by attaching n pendant edges to the apex vertex. 𝑆𝑓
𝑛has four types of vertices, the apex of 3n, n vertices of degree four, n vertices of degree two and n pendant vertices.
Figure 5 : Sunflower graph 𝑺𝒇𝒏.
Lemma 2.9. Let 𝑆𝑓
𝑛be a sunflower graph with 3n + 1 vertices ( as shown in fig. 5).Then we have, 𝑑(𝑣
0) = 3𝑛, 𝑑
2(𝑣
0) = 0,
𝑑(𝑣
𝑖) = 4 , 𝑑
2(𝑣
𝑖) = 3𝑛 − 4, 𝑑(𝑢
𝑖) = 2, 𝑑
2(𝑢
𝑖) = 3𝑛 − 2 , 𝑑(𝑤
𝑖) = 1, 𝑑
2(𝑤
𝑖) = 3𝑛 − 1.
Theorem 2.10. Let 𝑆𝑓
𝑛be a sunflower graph with 3n + 1 vertices. Then 1) L𝑀
1(𝑆𝑓
𝑛) = 3𝑛( 9𝑛
2− 14𝑛 + 7)
2) 𝐿𝑀
2(𝑆𝑓
𝑛) = 6𝑛( 3𝑛
2− 7𝑛 + 4) 3) 𝐿𝑀
3(𝑆𝑓
𝑛) = 21𝑛( 𝑛 − 1).
Proof. Let 𝑆𝑓
𝑛be a sunflower graph with 3n + 1 vertices, 5n edges and let 𝑣
0be the apex vertex, 𝑣
1, 𝑣
2, 𝑣
3, … , 𝑣
𝑛be the rim vertices with four degrees and 𝑢
1, 𝑢
2, 𝑢
3, … , 𝑢
𝑛be the extreme vertices with two degrees. since d(𝑣
0)= 3n.
1) By Lemma 2.9., and from definition of L𝑀
1(𝐺), we have, 𝐿𝑀
1(𝑆𝑓
𝑛) = ∑
𝑣∈𝑉(𝑆𝑓𝑛)𝑑
22( 𝑣).
= ∑
𝑛𝑖=1𝑑
22(𝑣
0) + ∑
𝑛𝑖=1𝑑
22(𝑣
𝑖) + ∑
𝑛𝑖=1𝑑
22(𝑢
𝑖) + ∑
𝑛𝑖=1𝑑
22(𝑤
𝑖)
= 0 + ∑
𝑛𝑖=1( 3𝑛 − 4)
2+ ∑
𝑛𝑖=1( 3𝑛 − 2)
2+ ∑
𝑛𝑖=1( 3𝑛 − 1)
2= 𝑛( 3𝑛 − 4)
2+ 𝑛(3𝑛 − 2)
2+ 𝑛( 3𝑛 − 1)
2= 𝑛( 9𝑛
2− 24𝑛 + 16) + 𝑛( 9𝑛
2− 12𝑛 + 4) + 𝑛(9𝑛
2− 6𝑛 + 1) = 9𝑛
3− 24𝑛
2+ 16𝑛 + 9𝑛
3− 12𝑛
2+ 4𝑛 + 9𝑛
3− 6𝑛
2+ 𝑛 = 27𝑛
3− 42𝑛
2+ 21𝑛
Therefore 𝐿𝑀
1(𝑆𝑓
𝑛) = 3𝑛(9𝑛
2− 14𝑛 + 7). ∎
2) From definition of L𝑀
2(𝐺)we have, 𝐿𝑀
2(𝑆𝑓
𝑛) = ∑ 𝑑
2(𝑢)𝑑
2(𝑣).
𝑢𝑣∈𝐸(𝑆𝑓𝑛)
= ∑ 𝑑
2𝑛
𝑖=1
(𝑣
0)𝑑
2(𝑣
𝑖) + ∑ 𝑑
2𝑛
𝑖=1
(𝑣
0)𝑑
2(𝑢
𝑖) + ∑ 𝑑
2(𝑣
0)
𝑛
𝑖=1
𝑑
2(𝑤
𝑖) +
∑ 𝑑
2(𝑣
𝑖)
𝑣𝑖,𝑣𝑖+1
𝑑
2(𝑣
𝑖+1) + ∑ 𝑑
2(𝑣
𝑖)
𝑛
𝑖=1
𝑑
2(𝑢
𝑖)
= 0 + 0 + 0 + ∑
𝑛𝑖=1( 3𝑛 − 4) ( 3𝑛 − 4) + ∑
𝑛𝑖=1( 3𝑛 − 4) ( 3𝑛 − 2) = 𝑛(3𝑛−4)
2+ 𝑛( 3𝑛 − 4)( 3𝑛 − 2)
= 𝑛 ( 3𝑛 − 4)[ 3𝑛 − 4 + 3𝑛 − 2]
= 𝑛 ( 3𝑛 − 4 ) ( 6𝑛 − 6) = 18𝑛
3− 18𝑛
2− 24𝑛
2+ 24𝑛 = 6𝑛 ( 3𝑛
2− 7𝑛 + 4)
Therefore, L𝑀
2(𝑆𝑓
𝑛) = 6𝑛( 3𝑛
2− 7𝑛 + 4). ∎ 3) From definition of L𝑀
3( 𝐺) and Lemma 2.9, we have,
𝐿𝑀
3(𝑆𝑓
𝑛) = ∑ 𝑑(𝑣)𝑑
2(𝑣).
𝑣∈𝑉(𝑆𝑓𝑛)
= 𝑑(𝑣
0)𝑑
2(𝑣
0) + ∑ 𝑑(𝑣
𝑖)𝑑
2(𝑣
𝑖) + ∑ 𝑑(𝑢
𝑖)𝑑
2(𝑢
𝑖) + ∑ 𝑑(𝑤
𝑖)𝑑
2(𝑤
𝑖)
𝑛
𝑖=1 𝑛
𝑖=1 𝑛
𝑖=1
