Maximum atom-bond sum-connectivity index of n -order trees with ﬁxed number of leaves

Let G be a graph. For an edge e of G , denote by d e the number of edges adjacent to e . The atom-bond sum-connectivity (ABS) index of G is deﬁned as ABS ( G ) = (cid:80) e ∈ E ( G ) (cid:112) 1 − 2( d e + 2) − 1 . A graph of order n is known as an n -order graph. The problem of determining trees possessing the minimum ABS index among all n -order trees with ﬁxed number of leaves has recently been attacked in two preprints independently. This article provides a complete solution to the maximal version of the aforementioned problem


Introduction and statement of the result
Those graph-theoretical terms that are used in the present article, without providing their definitions here, can be found in the books [7,8,12].
A graph invariant is a property of a graph that is preserved by isomorphism [12]. The degree sequence and the size of a graph are examples of graph invariants. In chemical graph theory, those graph invariants that take a single number are called topological indices [20]. According to Ivan Gutman [13], among all topological indices the connectivity index "is the most studied, most often applied, and most popular". The connectivity index was introduced by Milan Randić in [18], where he referred to it as the branching index; this index is nowadays known as the Randić index. The readers interested in getting details about the connectivity index may consult the books [14,15], review articles [16,19], and related papers cited therein.
Because of the success of the connectivity index, many modified versions of this index have been introduced and investigated. The sum-connectivity (SC) index [21] and the atom-bond connectivity (ABC) index [9,10] are among the renowned and well-investigated variants of the connectivity index. Most of the mathematical aspects of the ABC and SC indices can be found in the review articles [1] and [4], respectively.
The present paper is concerned with a recently proposed variant of the ABC index, namely the atom-bond sumconnectivity (ABS) index [2]. The ABS index was defined by utilizing the main concept of the SC and ABC indices. For a graph G, its ABS index is defined by the equation where E(G) denotes the set of edges of G and d y indicates the degree of the vertex y in G (similarly, d z is defined). Certainly, the ABS index is a vertex-degree-based topological index; however, it can also be considered as an edgedegree-based topological index because the number "d y + d z − 2" is called the degree of the edge yz ∈ E(G).
The mathematical study of the ABS index was initiated in the article [2], where some extremal results about the ABS index for (general) graphs and (molecular) trees were reported. Chemical applicability of the ABS index was investigated in [3], where the maximum and minimum values of the ABS index of unicyclic graphs were also found (a connected graph with the same number of edges and vertices is known as a unicyclic graph). Gowtham and Gutman [11] established several inequalities involving the difference between SC and ABS indices. The maximum value of the ABS index of graphs with several fixed parameters was investigated in [6].
A graph of order n is known as an n-order graph. A leaf of a graph is a vertex of degree one; a leaf is also known as a pendent vertex or an end vertex. The problem of determining trees possessing the minimum ABS index among all n-order trees with fixed number of leaves has recently been attacked in [5,17] independently; the present article provides a complete solution to the maximal version of this problem (for the case of general graphs, see [6] ). Now, we state the main result of this article. − 1 Figure 1: The graph S n, mentioned in Theorem 1.1.

Theorem 1.1.
In the class of all n-order trees with leaves, S n, is the unique graph having the maximum ABS index, where 2 ≤ ≤ n − 2 and S n, is the tree obtained from the path graph P n− +1 of order n − + 1 by attaching − 1 leaves to exactly one leaf of P n− +1 (see Figure 1). The mentioned maximum value is

Proof of Theorem 1.1
We prove the theorem by induction on n. For n = 4 one has = 2 and for n = 5 one has ∈ {2, 3}; in any case, the result trivially holds because there is only one tree in each case (and that tree is S n, ). Assume that n ≥ 6 and that the result is true for every tree of order n − 1 with leaves, where 2 ≤ ≤ n − 3. Suppose that T is an n-order tree with leaves with the constraint 2 ≤ ≤ n − 2. It is enough to show that with equality if and only if T isomorphic to the graph S n, . Take uv ∈ E(T ) such that d v = 1 and d u ≥ 2. Let T − v be the tree obtained from T by removing the vertex v. Note that the degree of every vertex, except u, of T − v is the same in both T and T − v. In what follows, d u represents the degree of u in T .
Let u 1 , · · · , u du−1 be all the neighbors of u different from v. Then Since ≤ n − 2, at least one of the vertices u 1 , · · · , u du−1 is non-leaf. Also, the function φ defined by is strictly decreasing in y. Thus, Equation (2) yields with equality if and only if exactly one of the vertices u 1 , · · · , u du−1 is non-leaf provided that this unique non-leaf has the degree 2. Since d u ≤ and the function ψ defined by is strictly increasing for z > 2, from Equation (3) it follows that Observe that the equality in (4) holds if and only if d u = and exactly one of the vertices u 1 , · · · , u du−1 is non-leaf, which has the degree 2. Note that, in the present case, the inequality ≥ 3 holds and the tree T − v has exactly − 1 leaves. Since 2 ≤ − 1 ≤ n − 3, by the inductive hypothesis, we have with equality if and only if T − v isomorphic to the graph S n−1, −1 . Thus, from (4) and (5) the desired inequality (that is, (1)) follows.
Case 2. d u = 2. If = n − 2, then T is isomorphic to the graph S n,n−2 and hence we are done. Thus, in the following, we assume that ≤ n − 3. Let u ∈ V (T ) be the unique neighbor of u different from v. Certainly, d u ≥ 2 because n ≥ 6. It holds that where the right equality holds if and only if d u = 2. Observe that, in the present case, the tree T − v has exactly leaves. Since 2 ≤ ≤ n − 3, by the inductive hypothesis, it holds that Note that the equation holds if and only if T − v is isomorphic to the graph S n−1, and d u = 2, which is equivalent to say that T is isomorphic to the graph S n, .