Leap Zagreb Indices of Some Wheel Related Graphs

Leap Zagreb Indices of Some Wheel Related Graphs

Article  in  Journal of Computer and Mathematical Sciences · March 2018

DOI: 10.29055/jcms/751

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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 𝐺_𝑛, helm 𝐻_𝑛, flower 𝐹𝑙_𝑛 and sunflower 𝑆𝑓_𝑛 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 𝑑₁(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 to⁹.

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 Trinajestic⁷, in 1972, and elaborated in⁶. They are defined as:

𝑀₁( 𝐺) = ∑ 𝑑₁² ( 𝑣/𝐺)

𝑣∈𝑉( 𝐺)

and

𝑀₂(𝐺) = ∑ 𝑑₁

𝑢,𝑣∈𝐸( 𝐺)

( 𝑢/𝐺) 𝑑₁(𝑣/𝐺) .

For properties of the two Zagreb indices see2,3,6,12,15 and 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 coincides¹⁶ and 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.

𝐿𝑀₁(𝐺) = ∑ 𝑑₂² (𝑣/𝐺)

𝑣∈𝑉(𝐺)

.

𝐿𝑀₂(𝐺) = ∑ 𝑑₂(𝑢/𝐺) 𝑑₂(𝑣/𝐺)

𝑢𝑣∈𝐸(𝐺)

.

𝐿𝑀₃(𝐺) = ∑ 𝑑(𝑢/𝐺) 𝑑₂(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_𝑛 with 𝑛 + 1 vertices is defined to be the join of 𝐾₁ and 𝐶_𝑛 , where 𝐾₁ is the complete graph with one vertex and 𝐶_𝑛 is the cycle graph with n vertices.

Clearly, |V (𝑊_𝑛)| = n + 1 and |E (𝑊_𝑛)| = 2n. The vertex corresponding to 𝐾₁ is known as apex while the vertices corresponding to 𝐶_𝑛 are 𝐾₁known as rim vertices.

Figure 1: Wheel graph 𝑾𝒏.

Lemma 2.1. Let 𝑊_𝑛 be a wheel graph with n+1 vertices ( as shown in fig. 1 ).Then we have 𝑑(𝑣₀) = 𝑛, 𝑑₂(𝑣₀) = 0.

𝑑(𝑣_𝑖) = 3, 𝑑₂(𝑣_𝑖) = 𝑛 − 3.

Theorem 2.2. Let 𝑊_𝑛, n ≥ 4 be a wheel graph with n+1 vertices. Then 1) 𝐿𝑀₁(𝑊_𝑛) = 𝑛(𝑛²− 6𝑛 + 9)

2) 𝐿𝑀₂ ( 𝑊_𝑛 ) = 𝑛( 𝑛 − 3)² 3) 𝐿𝑀₃ ( 𝑊_𝑛) = 3𝑛(𝑛 − 3).

Proof. Let 𝑊_𝑛, 𝑛 ≥ 4 be a wheel with 𝑛 + 1 vertices and let 𝑣₀,𝑣_1,…, 𝑣_𝑛 be a vertices set of 𝑤_𝑛, where 𝑣₀ be the apex vertex and 𝑣_1,𝑣_2, …,𝑣_𝑛 be the rim vertices of 𝑊_𝑛.

1) By Lemma 2.1., and 𝑓𝑟𝑜𝑚 𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝐿𝑀₁ ( 𝐺) 𝑤𝑒 ℎ𝑎𝑣𝑒 , 𝐿𝑀₁(𝑊_𝑛) = ∑ 𝑑₂²

𝑣∈𝑉( 𝑊_𝑛)

(𝑣) = 𝑑₂²(𝑣₀) + ∑ 𝑑₂²

𝑛

𝑖=1

(𝑣_𝑖)

= 0 + ∑^𝑛_𝑖=11(𝑛 − 3)² = 0 + 𝑛(𝑛 − 3)² = 𝑛(𝑛²− 6𝑛 + 9)

Therefore 𝐿𝑀₁(𝑊_𝑛) = 𝑛(𝑛²− 6𝑛 + 9). ∎

2) From figure 1., we have two types of edges 𝑣₀ 𝑣_𝑖 and 𝑣_𝑖𝑣_𝑖+1, for indices being taken modolu 𝑛, and from definition of L𝑀₂(𝐺) we have,

𝐿𝑀₂(𝑊_𝑛 ) = ∑ 𝑑₂

𝑢,𝑣∈𝐸(𝑊_𝑛)

(𝑣₀)𝑑₂(𝑣)

= ∑_𝑣0𝑣 𝑑₂

𝑖∈𝐸(𝑊𝑛) (𝑣₀)𝑑₂(𝑣_𝑖) + ∑^𝑛_𝑣𝑖𝑣 𝑑₂

𝑖+1∈𝐸(𝑊𝑛) (𝑣_𝑖)𝑑₂(𝑣_𝑖+1) = 0 + ∑^𝑛_𝑖=1(𝑛 − 3)(𝑛 − 3)

= 0 + 𝑛 (𝑛 − 3)²

Therefore, 𝐿𝑀_2(𝑊𝑛) = 𝑛( 𝑛 − 3)² . ∎ 3) From definition of L𝑀₃(𝐺) we have,

𝐿𝑀3(𝑊𝑛) = ∑ 𝑑(𝑣) 𝑑2(𝑣)

𝑣∈𝑉(𝑊_𝑛)

= 𝑑(𝑣₀)𝑑₂(𝑣₀) + ∑^𝑛_𝑖=1𝑑(𝑣_𝑖)𝑑₂(𝑣_𝑖) = 0 + ∑^𝑛_𝑖=13(𝑛 − 3)

= 3𝑛(𝑛 − 3)

Therefore 𝐿𝑀₃(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_𝑛 by 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_𝑛 has 2n+1 vertices and 3n edges.

Figure 2: Gear graph 𝑮_𝒏.

Lemma 2.3. Let 𝐺_𝑛 be a gear graph with 2𝑛 + 1 vertices ( as shown in fig. 2). Then we have, 𝑑(𝑣₀) = 𝑛 𝑑₂(𝑣₀) = 𝑛

𝑑(𝑣_𝑖) = 3 𝑑₂(𝑣_𝑖) = 𝑛 − 1 𝑑(𝑢_𝑖) = 2 𝑑₂(𝑢_𝑖) = 3 Theorem 2.4. Let 𝐺_𝑛 be a gear graph with 2𝑛 + 1 vertices. Then 1) L𝑀₁(𝐺_𝑛) = 𝑛(𝑛²− 𝑛 + 10).

2) 𝐿𝑀₂ (𝐺_𝑛 ) = 𝑛(𝑛 − 1)(𝑛 + 6).

3) 𝐿𝑀₃ (𝐺_𝑛 ) = 𝑛( 4𝑛 + 3).

Proof. Let 𝐺_𝑛 be a gear graph and let 𝑣₀ be the apex vertex, 𝑣_𝑛, 𝑣₂, 𝑣₃, … , 𝑣_𝑛 be the vertices of 𝐺_𝑛 with three degree and 𝑢₁, 𝑢₂, … , 𝑢_𝑛 be the vertices of 𝐺_𝑛 with two degree.

1) By Lemma 2.3., and from definition of L𝑀₁ (𝐺) , we have , 𝐿𝑀₁(𝐺_𝑛) = ∑ 𝑑₂²(𝑣)

𝑣∈𝑉(𝐺_𝑛)

.

= 𝑑₂²(𝑣0) + ∑^𝑛_𝑖=1𝑑₂²(𝑣𝑖) + ∑^𝑛_𝑖=1𝑑₂²(𝑢𝑖) = 𝑛² + ∑^𝑛_𝑖=1(𝑛− 1)²+ ∑^𝑛_𝑖=13²

= 𝑛²+ 𝑛(𝑛 − 1)²+ 3²𝑛 = 𝑛²+ 𝑛(𝑛²− 2𝑛 + 1) + 9𝑛 = 𝑛²+ 𝑛³− 2𝑛²+ 𝑛 + 9𝑛 = 𝑛³− 𝑛²+ 10𝑛

= 𝑛( 𝑛² − 𝑛 + 10).

Therefore 𝐿𝑀1(𝐺_𝑛) = 𝑛( 𝑛²− 𝑛 + 10). ∎ 2) From definition of L𝑀₂( 𝐺), we have ,

𝐿𝑀₂(𝐺_𝑛) = ∑_{𝑢𝑣∈𝐸( 𝑍𝑛)}𝑑₂(𝑢)𝑑₂(𝑣)

= ∑^𝑛_𝑖=1𝑑₂(𝑣₀)𝑑₂(𝑣𝑖) + ∑^𝑛_𝑖=1𝑑₂( 𝑢_𝑖) 𝑑₂( 𝑣_𝑖) = ∑^𝑛_𝑖=1𝑛(𝑛 − 1) + ∑^2𝑛_𝑖=13(𝑛 − 1)

= 𝑛 . 𝑛 (𝑛 − 1) + 2𝑛. 3(𝑛 − 1) = 𝑛²(𝑛 − 1) + 6𝑛(𝑛 − 1) = 𝑛³ − 𝑛² + 6𝑛²− 6𝑛 Therefore, L𝑀₂(𝐺_𝑛) = 𝑛(𝑛 − 1)(𝑛+6). ∎

3) From definition of L𝑀₃( 𝐺), we have, 𝐿𝑀₃(𝐺_𝑛) = ∑ 𝑑(𝑣)𝑑₂(𝑣)

𝑣∈𝑉(𝐺_𝑛)

.

= 𝑑(𝑣₀ )𝑑₂(𝑣₀) + ∑^𝑛_𝑖=1𝑑(𝑣_𝑖)𝑑₂(𝑣_𝑖) + ∑^𝑛_𝑖=1𝑑(𝑢_𝑖)𝑑₂(𝑢_𝑖) = 𝑛 . 𝑛 + ∑^𝑛_𝑖=13(𝑛 − 1) + ∑^𝑛_𝑖=12(3)

= 𝑛²+ 3𝑛(𝑛 − 1) + 6𝑛 = 𝑛²+ 3𝑛²− 3𝑛 + 6𝑛 = 4𝑛² + 3n

Therefore 𝐿𝑀₃(𝐺_𝑛) = 𝑛(4𝑛 + 3). ∎

Definition 2.3. The helm graph 𝐻_𝑛 is a graph obtained from wheel graph 𝑊_𝑛 by 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_𝑛 has 2n + 1 vertices and 3n edges.

Figure 3 : Helm graph 𝑯_𝒏

Lemma 2.5. Let 𝐻_𝑛 be a helm graph with 2n + 1 vertices( as shown in fig. 3). Then we have, 𝑑(𝑣₀) = 𝑛 𝑑₂(𝑣₀) = 𝑛

𝑑(𝑣_𝑖) = 4 𝑑₂(𝑣_𝑖) = 𝑛 − 1 𝑑(𝑢_𝑖) = 1 𝑑₂(𝑢_𝑖) = 3 Theorem 2.6. Let 𝐻_𝑛 be a helm graph with 2n + 1 vertices. Then 1) 𝐿𝑀₁(𝐻_𝑛) = 𝑛( 𝑛² – 𝑛 + 10)

2) 𝐿𝑀₂ (𝐻_𝑛 ) = 2𝑛( 𝑛²− 1) 3) 𝐿𝑀₃ (𝐻_𝑛 ) = 𝑛(5𝑛 − 1).

Proof . Let 𝐻_𝑛 be a helm graph with 2n + 1 vertices and let 𝑣₀ be the apex vertex, 𝑣₁, 𝑣₂, 𝑣₃, … , 𝑣_𝑛 be the rim vertices with four degrees and 𝑢₁, 𝑢₂, 𝑢₃, … , 𝑢_𝑛 be the pendant vertices.

1) By Lemma. 2.5., and from definition of L𝑀₁( 𝐺), we have,

𝐿𝑀₁(𝐻_𝑛) = ∑ 𝑑₂²(𝑣)

𝑣∈𝑉(𝐻_𝑛)

.

= 𝑑₂²(𝑣₀) + ∑^𝑛_𝑖=1𝑑₂²(𝑣_𝑖) + ∑^𝑛_𝑖=1𝑑₂²(𝑢_𝑖) = 𝑛² + 𝑛(𝑛 − 1)²+ 3²𝑛

= 𝑛²+ 𝑛( 𝑛²− 2𝑛 + 1) + 9𝑛 = 𝑛²+ 𝑛³− 2𝑛²+ 𝑛 + 9𝑛 = 𝑛³− 𝑛²+ 10𝑛

Therefore L𝑀₁(𝐻_𝑛) = 𝑛( 𝑛²− 𝑛 + 10). ∎ 2) From definition of L𝑀₂( 𝐺), we have,

𝐿𝑀₂(𝐻_𝑛) = ∑ 𝑑₂(𝑢)

𝑢𝑣∈𝐸( 𝐻_𝑛)

𝑑₂(𝑣) = ∑ 𝑑₂( 𝑣₀)

𝑛

𝑣_0,𝑣_𝑖∈𝐸(𝐻_𝑛)

𝑑₂(𝑣_𝑖) + ∑ 𝑑₂

𝑛

𝑣_𝑖,𝑢_𝑖

(𝑣_𝑖)𝑑₂(𝑢_𝑖) + ∑ 𝑑₂

𝑛

𝑣_𝑖,𝑣_𝑖+1

(𝑣_𝑖)𝑑₂( 𝑣_𝑖+1) = ∑^𝑛_𝑖=1𝑛( 𝑛 − 1 ) + ∑^𝑛_𝑖=1(𝑛 − 1) 3 + ∑^𝑛_𝑖=1( 𝑛 − 1 )(𝑛 − 1) = 𝑛²( 𝑛 − 1) + 3𝑛(𝑛 − 1) + 𝑛(𝑛 − 1)²

= 𝑛³− 𝑛²+ 3𝑛²− 3𝑛 + 𝑛( 𝑛²− 2𝑛 + 1) = 𝑛³− 𝑛²+ 3𝑛²− 3𝑛 + 𝑛³− 2𝑛²+ 𝑛 = 2𝑛³− 2𝑛

Therefore L𝑀₂(𝐻_𝑛) = 2𝑛(𝑛²− 1). ∎

3) From definition of L𝑀₃( 𝐺), we have, 𝐿𝑀₃(𝐻_𝑛) = ∑ 𝑑(𝑣)𝑑₂

𝑣∈𝑉(𝐻_𝑛)

(𝑣).

= 𝑑(𝑣₀)𝑑₂(𝑣₀) + ∑^𝑛_𝑖=1𝑑(𝑣_𝑖)𝑑₂(𝑣_𝑖) + ∑^𝑛_𝑖=1𝑑(𝑢_𝑖)𝑑₂(𝑣_𝑖) = 𝑛 . 𝑛 + ∑^𝑛_𝑖=14( n − 1) + ∑ⁿ_i=11( 3)

= 𝑛²+ 4𝑛( 𝑛 − 1) + 3𝑛 = 𝑛²+ 4𝑛²− 4𝑛 + 3𝑛 = 5𝑛²− 𝑛

Therefore L𝑀₃(𝐻 _𝑛 ) = 𝑛 (5𝑛 − 1). ∎

Definition 2.4. The flower graph 𝐹𝑙_𝑛 is 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_𝑛 has 2n + 1 vertices and 4n edges.

Figure 4 : Flower graph 𝑭𝒍_𝒏.

Lemma 2.7., Let 𝐹𝑙_𝑛 be a flower graph with 2n + 1 vertices ( as shown in fig. 4). Then we have, 𝑑(𝑣₀) = 2𝑛 , 𝑑₂(𝑣₀) = 0,

𝑑(𝑣_𝑖) = 4 , 𝑑₂(𝑣_𝑖) = 𝑛 − 5, 𝑑(𝑣_𝑖) = 2, 𝑑₂(𝑢_𝑖) = 𝑛 − 2 . Theorem 2.8. Let 𝐹𝑙𝑛 be a flower graph with 2n + 1 vertices. Then 1) L𝑀₁(𝐹𝑙_𝑛) = 𝑛(2𝑛²− 14𝑛 + 29 )

2) 𝐿𝑀₂ (𝐹𝑙_𝑛) = 𝑛(2𝑛²− 17𝑛 + 35) 3) 𝐿𝑀3 ( 𝐹𝑙𝑛) = 2𝑛( 3𝑛 − 12).

Proof. Let 𝐹𝑙_𝑛 be a flower graph with 2n + 1 vertices, 4n edges and let 𝑣₀ be the apex vertex, 𝑣₁, 𝑣₂, 𝑣₃, … , 𝑣_𝑛 be the rim vertices with four degrees and 𝑢₁, 𝑢₂, 𝑢₃, … , 𝑢_𝑛be the extreme vertices with two degrees. Since 𝑑(𝑣₀) = 2𝑛.

1) By Lemma 2.7., and from definition of L𝑀₁( 𝐺), we have, 𝐿𝑀1(𝐹𝑙𝑛) = ∑ 𝑑₂²(𝑣).

𝑣∈𝑉(𝐹𝑙_𝑛)

= 𝑑₂²(𝑣₀) + ∑^𝑛_𝑖=1𝑑₂²(𝑣_𝑖) + ∑^𝑛_𝑖=1𝑑₂²(𝑢_𝑖) = 0 + ∑^𝑛_𝑖=1(𝑛 − 5)²+ ∑^𝑛_𝑖=1(𝑛 − 2)² = 𝑛( 𝑛 − 5)²+ 𝑛(𝑛 − 2)²

= 𝑛( 𝑛²− 10𝑛 + 25) + 𝑛( 𝑛²− 4𝑛 + 4) = 𝑛³− 10𝑛²+ 25𝑛 + 𝑛³− 4𝑛²+ 4𝑛 = 2𝑛³ − 14𝑛² + 29𝑛

Therefore 𝐿𝑀₁(𝐹𝑙_𝑛) = 𝑛(2𝑛²− 14𝑛 + 29). ∎

2) From figure 4., we have four types of edges,𝑣₀𝑣_𝑖, 𝑣₀𝑢_𝑖, 𝑣_𝑖𝑢_𝑖 𝑎𝑛𝑑 𝑣_𝑖𝑣_𝑖+1, 𝑓𝑜𝑟 𝑖 = 1,2, … , 𝑛 and definition of L𝑀₂ ( 𝐺), we have,

𝐿𝑀₂(𝐹𝑙_𝑛) = ∑ 𝑑₂

𝑢,𝑣 ∈𝐸(𝐹𝑙_𝑛)

(𝑢_𝑖)𝑑₂(𝑣_𝑖) = ∑𝑛 𝑑2

𝑖=1 ( 𝑣₀)𝑑₂(𝑣_𝑖) + ∑𝑛 𝑑2

𝑖=1 (𝑣₀)𝑑₂(𝑢_𝑖) + ∑𝑛 𝑑2

𝑖=1 (𝑣_𝑖)𝑑₂(𝑢_𝑖) + ∑𝑛 𝑑2

𝑖=1 (𝑣_𝑖)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)

= 2𝑛³− 7𝑛²− 10𝑛²+ 35

Therefore L𝑀₂(𝐹𝑙_𝑛) = 𝑛( 2𝑛²− 17𝑛 + 35). ∎ 3) Form definition of L𝑀₃ ( 𝐺) we have,

𝐿𝑀3(𝐹𝑙_𝑛) = ∑ 𝑑(𝑣)𝑑2( 𝑣)

𝑣∈𝑉( 𝐹𝑙_𝑛)

.

= 𝑑(𝑣₀)𝑑₂( 𝑣₀) + ∑^𝑛_𝑖=1𝑑( 𝑣_𝑖)𝑑₂( 𝑣_𝑖) + ∑^𝑛_𝑖=1𝑑(𝑢_𝑖)𝑑₂(𝑢_𝑖)

= 0 + ∑^𝑛_𝑖=14(𝑛 − 5) + ∑^𝑛_𝑖=12(𝑛 − 2) = 4𝑛(𝑛 − 5) + 2𝑛( 𝑛 − 2)

= 4𝑛²− 20𝑛 + 2𝑛²− 4𝑛 = 6𝑛²− 24𝑛

Therefore L𝑀₃(𝐹𝑙_𝑛) = 2𝑛( 3𝑛 − 12). ∎

Definition 2.5. The sunflower graph 𝑆𝑓_𝑛 is the graph obtained from the flower graph 𝐹𝑙_𝑛 by attaching n pendant edges to the apex vertex. 𝑆𝑓_𝑛 has 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 𝑺𝒇_𝒏.

Lemma 2.9. Let 𝑆𝑓_𝑛 be a sunflower graph with 3n + 1 vertices ( as shown in fig. 5).Then we have, 𝑑(𝑣₀) = 3𝑛, 𝑑₂(𝑣₀) = 0,

𝑑(𝑣𝑖) = 4 , 𝑑₂(𝑣_𝑖) = 3𝑛 − 4, 𝑑(𝑢_𝑖) = 2, 𝑑₂(𝑢_𝑖) = 3𝑛 − 2 , 𝑑(𝑤_𝑖) = 1, 𝑑₂(𝑤_𝑖) = 3𝑛 − 1.

Theorem 2.10. Let 𝑆𝑓_𝑛 be a sunflower graph with 3n + 1 vertices. Then 1) L𝑀₁(𝑆𝑓_𝑛) = 3𝑛( 9𝑛²− 14𝑛 + 7)

2) 𝐿𝑀₂ (𝑆𝑓_𝑛) = 6𝑛( 3𝑛² − 7𝑛 + 4) 3) 𝐿𝑀₃ (𝑆𝑓_𝑛 ) = 21𝑛( 𝑛 − 1).

Proof. Let 𝑆𝑓_𝑛 be a sunflower graph with 3n + 1 vertices, 5n edges and let 𝑣₀be the apex vertex, 𝑣₁, 𝑣₂, 𝑣₃, … , 𝑣_𝑛 be the rim vertices with four degrees and 𝑢₁, 𝑢₂, 𝑢₃, … , 𝑢_𝑛 be the extreme vertices with two degrees. since d(𝑣₀)= 3n.

1) By Lemma 2.9., and from definition of L𝑀₁(𝐺), we have, 𝐿𝑀₁(𝑆𝑓_𝑛) = ∑_{𝑣∈𝑉(𝑆𝑓𝑛)}𝑑₂²( 𝑣).

= ∑^𝑛_𝑖=1𝑑₂²(𝑣₀) + ∑^𝑛_𝑖=1𝑑₂²(𝑣_𝑖) + ∑^𝑛_𝑖=1𝑑₂²(𝑢_𝑖)+ ∑^𝑛_𝑖=1𝑑₂²(𝑤_𝑖) = 0 + ∑^𝑛_𝑖=1(3𝑛 − 4)²+ ∑^𝑛_𝑖=1( 3𝑛 − 2)²+ ∑^𝑛_𝑖=1( 3𝑛 − 1)²

= 𝑛( 3𝑛 − 4)²+ 𝑛(3𝑛 − 2)²+ 𝑛( 3𝑛 − 1)²

= 𝑛( 9𝑛²− 24𝑛 + 16) + 𝑛( 9𝑛²− 12𝑛 + 4) + 𝑛(9𝑛²− 6𝑛 + 1) = 9𝑛³− 24𝑛²+ 16𝑛 + 9𝑛³− 12𝑛²+ 4𝑛 + 9𝑛³− 6𝑛²+ 𝑛 = 27𝑛³− 42𝑛²+ 21𝑛

Therefore 𝐿𝑀₁(𝑆𝑓_𝑛) = 3𝑛(9𝑛²− 14𝑛 + 7). ∎

2) From definition of L𝑀₂ (𝐺)we have, 𝐿𝑀₂(𝑆𝑓_𝑛) = ∑ 𝑑₂(𝑢)𝑑₂(𝑣).

𝑢𝑣∈𝐸(𝑆𝑓_𝑛)

= ∑ 𝑑₂

𝑛

𝑖=1

(𝑣₀)𝑑₂(𝑣_𝑖) + ∑ 𝑑₂

𝑛

𝑖=1

(𝑣₀)𝑑₂(𝑢_𝑖) + ∑ 𝑑₂(𝑣₀)

𝑛

𝑖=1

𝑑₂(𝑤_𝑖) +

∑ 𝑑₂(𝑣_𝑖)

𝑣_{𝑖,𝑣𝑖+1}

𝑑₂(𝑣_𝑖+1) + ∑ 𝑑₂(𝑣_𝑖)

𝑛

𝑖=1

𝑑₂(𝑢_𝑖)

= 0 + 0 + 0 + ∑^𝑛_𝑖=1(3𝑛 − 4) ( 3𝑛 − 4) + ∑^𝑛_𝑖=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) = 18𝑛³− 18𝑛²− 24𝑛²+ 24𝑛 = 6𝑛 ( 3𝑛²− 7𝑛 + 4)

Therefore, L𝑀₂(𝑆𝑓_𝑛) = 6𝑛( 3𝑛²− 7𝑛 + 4). ∎ 3) From definition of L𝑀₃( 𝐺) and Lemma 2.9, we have,

𝐿𝑀₃(𝑆𝑓_𝑛) = ∑ 𝑑(𝑣)𝑑₂(𝑣).

𝑣∈𝑉(𝑆𝑓_𝑛)

= 𝑑(𝑣₀)𝑑₂(𝑣₀) + ∑ 𝑑(𝑣_𝑖)𝑑₂(𝑣_𝑖) + ∑ 𝑑(𝑢_𝑖)𝑑₂(𝑢_𝑖) + ∑ 𝑑(𝑤_𝑖)𝑑₂(𝑤_𝑖)

𝑛

𝑖=1 𝑛

𝑖=1 𝑛

𝑖=1

= 0 + ∑^𝑛_𝑖=14 ( 3𝑛 − 4)+ ∑^𝑛_𝑖=12 ( 3𝑛 − 2) + ∑^𝑛_𝑖=11 (3𝑛 − 1) = 4𝑛( 3𝑛 − 4) + 2𝑛 ( 3𝑛 − 2) + 𝑛 (3𝑛 − 1)

= 12𝑛²− 16𝑛 + 6𝑛²− 4𝑛 + 3𝑛²− 𝑛 = 21𝑛²− 21𝑛

Therefore 𝐿𝑀₃(𝑆𝑓_𝑛) = 21𝑛( 𝑛 − 1). ∎

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