Fixed-Order Chemical Trees with Given Segments and Their Maximum Multiplicative Sum Zagreb Index

: Topological indices are often used to predict the physicochemical properties of molecules. The multiplicative sum Zagreb index is one of the multiplicative versions of the Zagreb indices, which belong to the class of most-examined topological indices. For a graph G with edge set E = { e 1 , e 2 , · · · , e m } , its multiplicative sum Zagreb index is deﬁned as the product of the numbers D ( e 1 ) , D ( e 2 ) , · · · , D ( e m ) , where D ( e i ) is the sum of the degrees of the end vertices of e i . A chemical tree is a tree of maximum degree at most 4. In this research work, graphs possessing the maximum multiplicative sum Zagreb index are determined from the class of chemical trees with a given order and ﬁxed number of segments. The values of the multiplicative sum Zagreb index of the obtained extremal trees are also obtained.


Introduction
A characteristic of a graph that is preserved under graph isomorphism is commonly referred to as a graph invariant [1].In chemical graph theory, graphical invariants that take numerical quantities are usually named topological invariants, or simply, topological indices.Zagreb indices, particularly the first and second Zagreb indices denoted by M 1 and M 2 , respectively, belong to well-examined categories of topological indices.Initially, they appeared in connection with the study molecules [2,3].These indices can be defined as where uv is an edge between the vertices u and v, while d x is the degree of the vertex x.Some information about the chemical applications of M 1 and M 2 can be found in [4,5].These indices have also been the subject of extensive research into their relationship, comparison, and other mathematical properties [6][7][8][9][10][11][12][13][14].Many existing facts of the Zagreb indices can be found in survey papers [15][16][17][18][19].
In 2012, Eliasi et al. [25] introduced a modification of the multiplicative first Zagreb index known as the multiplicative sum Zagreb index [32].The mathematical formulation of the multiplicative sum Zagreb index is as follows: In [25], it was shown that the path has the least Π * 1 -value across all connected graphs with the specified order.The trees achieving the second least Π * 1 -value were also determined in [25].Xu and Das [31] described the extremal trees, unicyclic graphs and bicyclic graphs of a specified order with respect to Π * 1 .Azari and Iranmanesh [33] established bounds on Π * 1 for graph operations.Further mathematical features of Π * 1 can be found in [27,34].
The focus of this work is strictly on the mathematical structure of chemical graphs.(Such graphs have found applications in chemistry; see, for example, the recent article [35]).More precisely, in this study, graphs possessing the greatest Π * 1 -values are determined from the class of chemical trees with a given order and fixed number of segments.The Π * 1 -values of the obtained extremal trees are also obtained.

Preliminaries
In this section, some definitions as well as notations used in this paper are given.Undefined terminology from graph theory can be found in some standard books.The graphs under discussion here are simple, undirected, and finite.The degree of a vertex v is represented by d v .The distance between two vertices u and v is denoted by d(u, v).A tree of maximum degree at most four is called a chemical tree.The notion |A| represents the cardinality of the set A. Consider a non-trivial path then every vertex of P i,j different from vertex v 1 and v k is called an internal vertex of P i,j .The path P i,j is referred to as an internal path if i, j ≥ 3 provided that all internal vertices have degree 2 (if exist).Furthermore, the path P i,j is an external path if one of the two numbers i, j, is equal to 1 and the other has a value greater than 2 provided that all internal vertices have degree 2 (if exist).A branching vertex in a tree is a vertex with a degree greater than 2. If S is an external path or an internal path in a tree G, then S is called a segment of the graph G.We call the path graph P n of order n also a segment of P n .Thus, a path graph has only one segment and there is no graph with exactly two segments.Let G(n, s) denote the class of all chemical trees with exactly n vertices and s segments, where 3 ≤ s ≤ n − 1.Let D(G) be the degree sequence of a graph G. Define For a chemical tree G, we write (for the sake of simplicity) the degree sequence of G as For example, if a chemical tree has the degree sequence (4, 4, 1, 1, 1, 1, 1, 1) then we write it as ((2) 4 , (0) 3 , (0) 2 , (6) 1 ).Let N G (u) be the set of neighbors of the vertex u ∈ G.In a graph G, let E i,j (G) (or simply E i,j , when there is no confusion about G) be the set of edges of G with end vertices of degrees i and j.Certainly, E i,j (G) = E j,i (G).For e ∈ E i,j , define w(e) = i + j.

Main Results
Before proceeding to our main results, we give some crucial lemmas that provide some useful information on obtaining the greatest possible value of the multiplicative sum Zagreb index for trees belonging to the class G(n, s).Lemma 1.For a chemical tree G m ∈ G(n, s) with maximum multiplicative sum Zagreb index, the following statements are true: If there is an internal path of the form P i,j in G m of length 1, then there is no internal path of the form P p,q in G m , with p + q < i + j, having length larger than 1.(d) If a path P i,j , with 6 ≤ i + j ≤ 8, contains an internal vertex of degree four then | E 1,3 |= 0. (e) The graph G m does not possess an internal path having a length of 1 and an internal path having a length larger than 2 simultaneously.(f) If G m has exactly two vertices of degree 3 then G m does not possess simultaneously the internal paths P 3,3 and P 4,4 both of length 1.
Proof.Throughout this proof, whenever the degree notion d α is used, it represents the degree of a vertex α in the graph G m .
(a) If | E 1,2 | = 0 then there must be a vertex, say v 2 , of degree two laying on an external path , 4}, and i ≥ 3. Since s ≥ 5, G m has another branching vertex, say v j (i < j), which forms an internal path v i − v j in the graph G m .Let v i+1 be the neighbor of v i that lies on the path v i − v j (the vertex v i+1 may coincide with v j ).Now, we consider a tree G m 1 that can be found in the class G(n, s) and is obtained from G m by the following operation: There exists a positive real number Θ such that these two graphs satisfy the following: Since (b) Assume that the hypothesis holds but the conclusion does not hold; that is, suppose that the inequality |E 1,2 |> 1 holds.Let v denotes the only branching vertex in G m (as 3 ≤ s ≤ 4) with two distinct external paths having length of at least 2. Now, a new tree G m 1 is constructed from G m using the following operation: This operation emphasizes that G m 1 ∈ G(n, s).There is Θ > 0 such that Since (c) Assume contrarily that G m possesses an internal path of the form P p,q of length larger than 1, such that p + q < i + j.Suppose that P p,q : ) is considered that is obtained using the following operation: There is a number Θ > 0, such that The following possible cases are discussed next: In this case, p + q ≤ 7. Thus, 5 ≤ p + d v 3 ≤ 7 and consequently, Equation (3) gives Case (2): Either i = 3 and j = 4 or i = 4 and j = 3.
In both possible cases, we arrive at a contradiction because of our contrary assumption that G m possess an internal path of the form P p,q of length larger than 1, such that p + q < i + j.
(d) Contrarily, assume that | E 1,3 | = 0.Then, there exists yz ∈ E 1,3 , such that d y = 1 and Without loss of generality, suppose that u k is the only internal vertex on P i,j ; otherwise, we may consider a subpath P i ,j of P i,j containing exactly one internal vertex of degree 4, such that 6 ≤ i + j ≤ 8.The following cases are to be discussed here: Case (1): If u k+1 = u g , then we assume that u k+2 is a neighbor of u k+1 not lying on the path P i,j ; otherwise, we assume that u k+2 is a neighbor of u k+1 lying on the u k+1 − u g path.Whether z lies on P i,j or not, in either of the two cases, we define a new graph as follows: Note that the tree G m 1 belongs to the collection G(n, s).Whether z lies on P i,j or not, in either of the two cases, there exists a positive real number Θ, such that which is a contradiction.
In this case, we consider a tree G m 1 ∈ G(n, s) obtained using the operation described as follows: Whether z lies on P i,j or not, in either of the two cases, there exists a positive real number Θ, such that Note that there are the following three possibilities concerning the degrees of the vertices u k−1 and u k+1 : In every case, Equation (4) gives (e) Although the proof of this part is slightly different from that of part (c), we provide its proof here for the sake of completeness.Assume contrarily that G m simultaneously possesses an internal path P p,q : ) is considered that is obtained using the following operation: There is a number Θ > 0, such that Since 3 ≤ i ≤ 4 and 3 ≤ j ≤ 4, Equation ( 5) provides First, we discuss the case when y 1 = u.Consider a tree G m 1 ∈ G(n, s) that is obtained using the following operation: Then, we obtain ), a contradiction.Next, consider the case when y 1 = u.Then, y 1 has a neighbor, say y 2 , of degree 2 lying on the y 1 − u path.Let y 3 be the neighbor of y 2 lying on the y 2 − u path.The vertices y 3 and u may be the same.By part (e), d y 3 = 4. Now, consider a tree G m 2 ∈ G(n, s) that is obtained using the following operation: , which yields a contradiction again.
Corollary 1.The graph constructed by attaching s − 1 pendent vertices to a single pendent vertex of the path P n−(s−1) possesses uniquely the greatest multiplicative sum Zagreb index in G(n, s) for every n ≥ s + 3 with s ∈ {3, 4}.The mentioned greatest value is 3(s + 2)(s + 1) s−1 4 n−s−2 .
Because of Corollary 1, in the rest of the current section, we focus on the case when 5 ≤ s ≤ n − 1 for the class G(n, s).To prove our next lemma, we need the following existing result: Lemma 2 ([36]).For every graph G ∈ G(n, s), the following statements holds: (a) The equation x 3 = 0 holds if and only if 3 , s = 3t + 2 for some positive integer t.
Lemma 3. Let G m ∈ G(n, s) be a chemical tree with maximum multiplicative sum Zagreb index and t be a positive integer.Then, Proof.Throughout this proof, whenever the degree notion d α is used, it represents the degree of a vertex α in the graph G m .First, we show that x 3 ≤ 2. We suppose on the contrary that the inequality x 3 ≥ 3 holds.Take p, q, r ∈ V(G m ), such that d p = d q = d r = 3.If all these three vertices are on one path, then (without loss of generality) suppose that q lies on the unique p − r path in G m .In either of the two cases, suppose that N G m (r) = {r 1 , r 2 , r 3 } is the set of neighbors of r with the condition that r 1 is located on p − r path (r 1 may coincide with q).Now, a chemical tree G m 1 is obtained in the collection G(n, s) using the following operation: For the case when all the three vertices p, q, r, are on one path, see Figure 1.Note that there is a real number Θ > 0, such that Since 2 ≤ d r 1 ≤ 4, inequality (6) yields a contradiction.Hence, the inequality x 3 ≤ 2 holds, which together with Lemma 2 gives the required result.
We now define three subclasses of G(n, s) as follows when n ≥ 8 and s = 3t + 1 for some integer t ≥ 2: Three examples G 1 , G 2 , and G 3 , one from each of the classes G 1 , G 2 , and G 3 , respectively, are depicted in Figure 2. Theorem 1.Let G m ∈ G(n, s) be a chemical tree with maximum multiplicative sum Zagreb index such that n ≥ 8 and s = 3t + 1 for some integer t ≥ 2.
Proof.By Lemma 3, it holds that x 3 = 0.If x 2 = 0 then G m ∈ G 1 and we are done.If Next, we define five subclasses of G(n, s) as follows when n ≥ 8 and s = 3t for some integer t ≥ 2: For example, the graphs H 2 and H 4 given in Figure 3 belong to H 3 .
For example, the graph H 7 given in Figure 3 belongs to H 4 .
• H 5 = {T ∈ G(n, s) : For example, the graph H 8 given in Figure 3 belongs to H 5 .
Theorem 2. Let G m ∈ G(n, s) be a chemical tree with maximum multiplicative sum Zagreb index such that n ≥ 8 and s = 3t for some integer t ≥ 2. Then G m ∈ ∪ We next define six subclasses of G(n, s) as follows when n ≥ 7 and s = 3t + 2 for some integer t ≥ 1: For example, the graphs I 0 , I 1 , I 2 , I 3 , I 4 , and I 7 given in Figure 4 belong to I 2 .
For example, the graph I 5 depicted in Figure 4 belongs to I 3 .
For example, the graph I 6 given in Figure 4 belongs to I 4 .
For example, the graph I 8 shown in Figure 4 belongs to I 5 .
Theorem 3. Let G m ∈ G(n, s) be a chemical tree with the greatest multiplicative sum Zagreb index such that n ≥ 7 and s = 3t + 2 for some integer t ≥ 1.Then G m ∈ ∪ 6 i=1 I i .
By using the above degree sequence of G m , we have x 2 = 0. Thus, in this case, G m consists of vertices of degree 4 and 1 only.Consequently, we have | E Therefore, Case (2): Note, in this case, that 1 ≤ x 2 ≤ x 4 − 1.By keeping in mind Lemma 1, we obtain Therefore, 1) .
Note, in the present case, that x 2 > x 4 − 1. Bearing in mind Lemma 1, we obtain Therefore, .
Theorem 5. Let G m ∈ G(n, s) be the chemical tree with maximum multiplicative sum Zagreb index such that s = 3 t for some integer t ≥ 2.Then, Proof.By Lemma 3, we have Case (1): n = s + 1.
In the present case, it holds that x 2 = 0. We discuss two possible subcases of the present case as follows: Note, in the present subcase, that s ∈ {6, 9}.Since x 2 = 0 in the consider case, if s = 6 then Equation ( 7) yields D(G m ) = ((1) 4 , (1) 3 , (0) 2 , (5 ) 1 ), and hence n = 7.Thus, G m is the graph constructed by attaching two new pendent vertices to a pendent vertex of the star S 5 .Similarly, if s = 9, then Equation (7) gives , (0) 2 , (7 ) 1 ), and hence n = 10.Thus, by Lemma 1(e), G m is the graph H 1 depicted in Figure 3. Hence, if s ∈ {6, 9}, then Note, in the present subcase, that s ≥ 12. Recall also that x 2 = 0 in the present case.By Lemma 1(d), every neighbor of the unique vertex of degree 3 in G m is of degree 4; for example, see H 2 in Figure 3. Hence, Observe, in the current case, that x 2 ≥ 1.In the following, we discuss two subcases according to whether s ≤ 9 or s > 9.
If n = 11 then, by Lemma 1, G m is the graph H 3 depicted in Figure 3.If n = 12 then, again by Lemma 1, G m is the graph constructed from H 3 by inserting a vertex of degree 2 on its unique internal path of length 1.Thus, all possible non-zero values of | E i,j | are given as follows: Observe, in the current subcase, that s ≥ 12 and x 2 ≥ 1.
Subcase (2.2.1): s ≥ 12 and 4s−9 3 < n ≤ 4s 3 Note that that is, x 4 − 3 < x 2 ≤ x 4 .Hence, by Lemma 1, every internal path of the form P 4,4 (in G m ) has length 2, and its unique vertex of degree 3 is adjacent to 3 − [x 2 − (x 4 − 3)] vertex/vertices of degree 4; for example, consider a graph constructed from the graph H 4 (shown in Figure 3) by inserting a vertex of degree 2 on each of the x 2 − (x 4 − 3) internal path(s) of the form P 3,4 .Hence, the possible non-zero values of | E i,j | of G m are given as follows: Subcase (2.2.2): s ≥ 12 and s + 1 < n ≤ 4s−9 3 .
Observe, in the current subcase, that s ≥ 15.Note that that is, x 2 ≤ x 4 − 3. Hence, by Lemma 1, x 2 internal path(s) of the form P 4,4 in G m has/have length 2; for example, consider a graph constructed from the graph H 2 (shown in Figure 3) by inserting a vertex of degree 2 on each of the x 2 internal path(s) of the form P 4,4 .Hence, the possible non-zero values of | E i,j | are given as follows: Case (3): n > 4s 3 .
In the following, we discuss subcases for the current case.Thus, by Lemma 1, G m is a graph constructed from H 1 (given in Figure 3) by inserting at least one vertex of degree 2 on each of two internal paths of length 1.
(f) Assume contrarily that G m possesses simultaneously the internal paths P 3,3 and P 4,4 both of length 1.Let xy, uv ∈ V(G m ), such that d y = 3 = d x , the distance d(y, u) is minimum, and d v = 4 = d u .Note that y and u lie on the unique x − v path.Let y 1 be the neighbor of y lying on the y − u path.By part (c), the degree of y 1 is 4.

1 Figure 1 .
Figure 1.The transformation applied on the graph G m to obtain a new graph G m 1 in Lemma 3.

G 2 G 3 G 1 Figure 2 .
Figure 2. Three examples G 1 , G 2 , and G 3 , one from each of the classes G 1 , G 2 , and G 3 , respectively.