CCO Commun. Comb. Optim.

Communications in Combinatorics and Optimization Vol. 3 No., 018 pp.179-194 DOI: 10.049/CCO.018.685.109 CCO Commun. Comb. Optim. Leap Zagreb Indices of Trees and Unicyclic Graphs Zehui Shao 1, Ivan Gutman, Zepeng Li 3, Shaohui Wang 4, Pu Wu 1 1 Institute of Computing Science and Technology, Guangzhou University Guangzhou 510006, China zshao@gzhu.edu.cn, puwu1997@16.com Faculty of Science, University of Kragujevac, 34000 Kragujevac, Serbia gutman@kg.ac.rs 3 School of Information Science and Engineering, Lanzhou University Lanzhou 730000, China lizp@lzu.edu.cn 4 Department of Mathematics, Savannah State University, Savannah, GA 31404, USA shaohuiwang@yahoo.com Received: 1 June 018; Accepted: 14 August 018 Published Online: 16 August 018 Communicated by Seyed Mahmoud Sheikholeslami Abstract: By d(v G) and d (v G) are denoted the number of first and second neighbors of the vertex v of the graph G. The first, second, and third leap Zagreb indices of G are defined as LM 1 (G) = v V (G) d (v G), LM (G) = uv E(G) d (u G) d (v G), and LM 3 (G) = v V (G) d(v G) d (v G), respectively. In this paper, we generalize the results of Naji et al. [Commun. Combin. Optim. (017), 99 117], pertaining to trees and unicyclic graphs. In addition, we determine upper and lower bounds on these leap Zagreb indices and characterize the extremal graphs. Keywords: Leap Zagreb index, Zagreb index, degree (of vertex) AMS Subject classification: 05C10, 05C35, 05C90 1. Introduction In this paper, we only consider graphs without multiple edges or loops. The degree of a vertex v in a considered graph is denoted by d(v). A double star S r,s is a tree with exactly two vertices u and v that are not leaves, such that (d(u), d(v)) = (r +1, s+1). Corresponding Author c 018 Azarbaijan Shahid Madani University. All rights reserved.

180 Leap Zagreb Indices of Trees and Unicyclic Graphs The triangular unicyclic graph T r,s,t is a unicyclic graph containing a triangle uvw with vertices u, v, and w, such that (d(u), d(v), d(w)) = (r +, s +, t + ). If uvw is the triangle of an n-vertex triangular unicyclic graph G, and if G uvw = P n 3, then G will be denoted T P n. By S n + e is denoted the graph obtained by connecting with a new edge e two leaf vertices of the star S n. A graph is called a {C 3, C 4 }-free graph if it contains neither C 3 nor C 4. In the interdisciplinary area where chemistry, physics and mathematics meet, molecular graph based structure descriptors, usually referred to as topological indices, are of significant importance [3, 4, 11]. Among the most important such structure descriptors are the classical two Zagreb indices [6, 7], M 1 (G) = u V (G) d(u G) and M (G) = uv E(G) d(u G) d(v G), where G is a graph whose vertex set is V (G) and whose edge set is E(G). By d(u G) is denoted the degree (= number of first neighbors) of the vertex v V (G) and by uv the edge between the vertices u and v. For details of the theory of Zagreb indices see the recent survey [] and the references cited therein. Motivated by the success of Zagreb indices, and following an earlier work [10], Naji et al. [9] introduced the concept of leap Zagreb indices, based on the second degrees of vertices. The k-degree of the vertex v V (G), denoted by d k (v G) or d k (v), is the number of vertices of G whose distance to v is equal to k. Evidently, d 1 (v G) = d(v G). The first, second, and third leap Zagreb indices of a graph G, proposed in [9], are defined respectively as LM 1 (G) = LM (G) = LM 3 (G) = v V (G) uv E(G) v V (G) d (v G) d (u G) d (v G) d(v G) d (v G). In fact, a quantity identical to the third leap Zagreb index LM 3 (G) of a graph G appeared in a paper published in the 1970 [7], but did not attract any attention. The same was the case with a paper from 008 [1]. Quite recently, independently of [9], Ali and Trinajstić re-invented this leap Zagreb index and named it modified first Zagreb connection indices [1]. In [9], the leap Zagreb indices of some graph families and graph joins were determined, as well as of triangle and quadrangle free graphs. In a later work [8], the leap Zagreb indices of graph operations were studied. Leap Zagreb indices are considered in a recent survey [5].

Z. Shao, I. Gutman, Z. Li, S. Wang, P. Wu 181 Motivated by these researches, in the present paper we establish sharp upper and lower bounds on the leap Zagreb indices of trees and unicyclic graphs and characterize the extremal graphs.. Preliminaries The following functions and definitions will be used throughout the paper. Let g(n) = (n 3) (n+6) 9, n 0 (mod 3) (n 3) (n+6) 9 11 9, n 1 (mod 3) (n 3) (n+6) 9 7 9, n (mod 3). It can be seen that g(n) g(n 1) n n 1 3. (1) For an edge e = uv E(G), we define LM (e G) = d (u G) d (v G), and LM (e G) is written as LM (e) when no confusion can arise. Proposition 1. [9] Let P n and S n be the path and star on n vertices. For n 4, (i) LM 1(S n) = (n 1)(n ) and LM 1(P n) = 4(n 3), (ii) LM (S n) = 0 and LM (P n) = 4n 14, (iii) LM 3(S n) = (n 1)(n ) and LM 3(P n) = 4n 10. The following results are immediate and their proofs are omitted. Lemma 1. Let r, s be positive integers with r+s+ = n. Then LM 1(S r,s) = r 3 +s 3 +r +s and LM (S r,s) = rs(1 + r + s). (i) If n is even, then LM (S r,s) (n 1)( n 1). The equality holds if and only if r = s. (ii) If n is odd, then LM (S r,s) 1 (n 3)(n 4 1). The equality holds if and only if r s = 1. Proof. By the definitions of LM 1 and LM, it is clear that LM 1 (S r,s ) = r 3 + s 3 + r + s and LM (S r,s ) = rs(1 + r + s). Then LM (S r,s ) = rs(1 + r + s) = r(n r)(n 1). Therefore, (i) If n is even, then LM (S r,s ) = r(n r)(n 1) ( ) n r+r (n 1) = (n 1)( n 1). The equality holds if and only if r = n r, i.e., r = s = n. (ii) If n is odd, we consider the function h(r) = r(n r)(n 1) and obtain that h(r) = r(n r)(n 1) 1 4 (n 3)(n 1), and the equality holds if and only if r s = 1.

18 Leap Zagreb Indices of Trees and Unicyclic Graphs Lemma. Let n 6. Then LM 1(T P n) = 4n 10, LM 1(S n + e) = (n 3)(n n ). Lemma 3. Let n 6 with r+s+t+3 = n. Then LM (T P n) = 4n 13 and LM (T r,s,t) = (r + s)(r + t) + (r + s)(s + t) + (r + t)(s + t) + r(r + 1)(s + t) + s(s + 1)(r + t) + t(t + 1)(r + s). Lemma 4. Let x y z be non-negative integers such that x + y + z + 3 = n and f(x, y, z) = (x + y)(x + z) + (x + y)(y + z) + (x + z)(y + z) + x(x + 1)(y + z) + y(y + 1)(x + z) + z(z + 1)(x + y). Then f(x, y, z) g(n). Moreover, (i) if n 0 (mod 3), then the equality holds if and only if x = y = z. (ii) If n 1 (mod 3), then the equality holds if and only if x = y = z 1. (iii) If n (mod 3), then the equality holds if and only if x = y 1 = z 1. Lemma 5. Let n 6 with r +s+t+3 = n. Then LM 3(T P n) = 4n 8 and LM 3(T r,s,t) = (n 3)(n + ). Lemma 6. Let G be a {C 3, C 4}-free graph. Then for any v V (G), d (v) = (d(u) 1) u N(v) where N(v) is the set of first neighbors of the vertex v. Recall that Lemma 6 was earlier communicated by Naji et al., [9]. 3. Main results In this section, we obtain bounds on the leap Zagreb indices of trees and unicyclic graphs. 3.1. Extremal trees on leap Zagreb indices Theorem 1. Let T be an n-vertex tree with n 4. Then 4(n 3) LM 1(T ) (n 1)(n ) with the left equality if and only if T = P n and the right equality if and only if T = S n. Proof. Since n 4, we have diam(t ). If diam(t ) =, then T = S n. By Proposition 1, we know that the result is true. If diam(t ) = 3, then T = S r,s, where r and s are two positive integers such that r + s + = n. Then by Lemma 1, LM 1 (T ) = r 3 + s 3 + r + s.

Z. Shao, I. Gutman, Z. Li, S. Wang, P. Wu 183 Since r, s 1, we have LM 1 (T ) (r + s ) (r + s) = (n ) 4(n 3). If n = 4, then T = P 4 and LM 1 (T ) = (n ) = 4(n 3). Thus, the result is true. If n 5, then LM 1 (T ) = (n ) > 4(n 3). On the other hand, LM 1 (T ) = r (r + 1) + s (s + 1) = (n s ) (n s + 1) + s (s + 1) = (n 1)(n ) + s (3n 4) s(3n 10n + 8). Since s < n, we have s(3n 4) (3n 10n+8) < (n )(3n 4) (3n 10n+8) = 0. Thus, LM 1 (T ) < (n 1)(n ). Assume now that diam(t ) 4 and proceed by induction on n. Let P = x 1 x... x diam(t ) be a longest path of T. Then x 1 is a leaf of T. Let T = T x 1. Then we have d (v T ) = d (v T ) 1 if v N(x T ), and d (v T ) = d (v T ) otherwise. By Lemma 6, we have d (x 1 T ) = d(x T ) 1. () Thus, LM 1 (T ) = LM 1 (T ) v N(x T ) = LM 1 (T ) + d (x 1 T ) + d (v T ) + = LM 1 (T ) + (d(x T ) 1) + v N(x T ) v N(x T ) v N(x T ) = LM 1 (T ) + (d(x T ) 1)(d(x T ) ) + d (v T ) (d (v T ) d (v T ) ) (d (v T ) 1) (Applying Eq. ()) v N(x T ) d (v T ). By induction, 4(n 4) LM 1 (T ) (n )(n 3), with the left equality if and only if T = P n 1 and the right equality if and only if T = S n 1. Moreover, by Lemma 6 we have d (v T ) = d(x T ) 1 for each v N(x T )\{x 3 } and d(x T ) d (x 3 T ) n 3 since diam(t ) 4. Note that d(x T ) n 3. Therefore, LM 1 (T ) = LM 1 (T ) + (d(x T ) 1)(d(x T ) ) + v N(x T ) 4(n 4) + d (x 3 T ) 4(n 4) + d(x T ) 4(n 3) d (v T ) with equality if and only if LM 1 (T ) = 4(n 4) and d(x T ) =, which yields T = P n.

184 Leap Zagreb Indices of Trees and Unicyclic Graphs On the other hand, LM 1 (T ) = LM 1 (T ) + (d(x T ) 1)(d(x T ) ) + d (v T ) v N(x T ) LM 1 (T ) + 3(d(x T ) 1)(d(x T ) ) + d (x 3 T ) (n )(n 3) + 3(n 4)(n 5) + (n 3) < (n )(n 3) + (n 3)(3n 10) < (n 1)(n ). By Proposition 1, we have Corollary 1. diam(t ). Let T be an n-vertex tree with n 3. Then LM (T ) = 0 if and only if Proof. If T = S n, then by Proposition 1, part (ii), LM (T ) = 0. Suppose that LM (T ) = 0. Then diam(t ). Otherwise, T would contain a path P = x 1 x x 3 x 4, which would imply LM (T ) d (x T )d (x 3 T ) 1, a contradiction. Therefore, T = S n. We use E(x G) to denote the set of edges incident with x in G. Lemma 7. For any tree T with diam(t ) 4, there exists a tree T with V (T ) = V (T ) and diam(t ) = diam(t ) 1, such that LM (T ) > LM (T ). Proof. Choose a longest path P = x 1 x... x t in T and let N(x T ) = {x 1, x 3, y 1,..., y r }. Then t 5 since diam(t ) 4. We construct a new tree T 1 as follows: V (T 1 ) = V (T ), E(T 1 ) = E(T ) {x 4 x 1, x 4 y 1,..., x 4 y r } \ {x x 1, x y 1,..., x y r }. It can be checked that d (v T 1 ) d (v T ) for any vertex v V (T 1 ) and d (x 5 T 1 ) > d (x 5 T ). Thus, LM (T 1 ) > LM (T ). If diam(t 1 ) = diam(t ) 1, then T 1 is a desired graph and we set T = T 1. Otherwise, diam(t 1 ) = diam(t ). By repeating the procedure we obtain a tree T m with V (T m ) = V (T ), diam(t m ) = diam(t ) 1 and LM (T m ) > LM (T ). Then T m is the desired graph and we set T = T m. Theorem. For any n-vertex tree with n 5 vertices and diam(t ) 3, it holds LM (P n) LM (T ) LM (S r,s), where r, s are two positive integers such that r+s+ = n and r s {0, 1}.

Z. Shao, I. Gutman, Z. Li, S. Wang, P. Wu 185 Proof. The proof is by induction on n. If n = 5, then the result is trivial. Therefore, assume that n 6. Let P = x 1 x... x diam(t ) be a longest path of T and T = T (N(x T )\{x 3 }). Then d (v T ) = d (v T ) d(x T )+1 if v = x 3 and d (v T ) = d (v T ) otherwise. Thus, LM (T ) = LM (T ) d (u T ) d (v T ) + d (u T ) d (v T ) = LM (T ) + + uv E(x 3 T ) uv E(x T )\{x x 3} = LM (T ) + + v N(x 3 T ) uv E(x 3 T ) uv E(x T ) E(x 3 T ) (d (u T ) d (v T ) d (u T ) d (v T )) d (u T ) d (v T ) v N(x T )\{x 3} d (x T ) d (v T ) (d (x 3 T ) d (v T ) d (x 3 T ) d (v T )) = LM (T ) + d (x T )(d(x T ) 1) + (d(x T ) 1) v N(x 3 T ) d (v T ). By induction, we have 4n 18 = LM (P n 1 ) LM (T ) LM (S r,s ), with the left equality if and only if T = P n d(x T )+1 and the right equality if and only if T = S r,s, where r, s are two positive integers such that r +s + = n d(x T )+1 and r s {0, 1}. Since LM (T ) > 0, by Theorem 1, diam(t ) 3. Then d(x T ) and d(x 3 T ). If d(x T ) 3, then LM (T ) LM (T ) + 4d (x T ) + d (v T ) 4n 18 + 4(d(x 3 T ) 1) If d(x 3 T ) 3, then v N(x 3 T ) + d(x 3 T )(d(x 3 T ) 1) 4n 10 > 4n 14. LM (T ) LM (T ) + d (x T ) + v N(x 3 T ) + d(x 3 T )(d(x 3 T ) 1) 4n 10 > 4n 14. d (v T ) 4n 18 + (d(x 3 T ) 1) Assume now that d(x T ) = d(x 3 T ) =. Since n 6, we have diam(t ) 4. If diam(t ) = 4, then T = S 1,n 4 and by Lemma 1, LM (T ) = (n 4)(n ). Thus, LM (T ) = LM (T ) + d (x T )(d(x T ) 1) + (d(x T ) 1) d (v T ) = (n 4)(n ) + 1 + > 4n 14. v N(x 3 T )

186 Leap Zagreb Indices of Trees and Unicyclic Graphs If diam(t ) 5, then d (x 4 T ) and v N(x 3 T ) d (v T ) 3. Thus, LM (T ) = LM (T ) + d (x T )(d(x T ) 1) + (d(x T ) 1) d (v T ) 4n 18 + 1 + 3 = 4n 14, v N(x 3 T ) with the equality if and only if T = P n d(x T )+1 and v N(x 3 T ) d (v T ) = 3. Note that d(x T ) =, from which it follows T = P n. Now we show the right inequality. If diam(t ) = 3, then T is a double star. By Lemma 1, we know that the inequality is true. Now we assume that diam(t ) = p 4. By Lemma 7, there exists a sequence of n-vertex trees T 1, T,..., T p 3 such that diam(t i ) = diam(t ) i and LM (T i 1 ) < LM (T i ), where i = 1,,..., p 3 and T 0 = T. Hence, LM (T ) < LM (T p 3 ). Note that diam(t p 3 ) = diam(t ) (p 3) = 3, that is, T p 3 is a double star. By Lemma 1, LM (T p 3 ) LM (S r,s ), where r, s are two positive integers such that r+s+ = n and r s {0, 1}. Hence, LM (T ) < LM (S r,s ), which completes the proof. Theorem 3. Let T be an n-vertex tree with n 5 vertices. Then (n 5) LM 3(T ) (n 1)(n ). The left equality holds if and only if T = P n whereas the right equality holds if and only if T is a star or a double star. Proof. The proof is by induction on n and is fully analogous to that of Theorem. We omit the details. 3.. Extremal unicyclic graphs on leap Zagreb indices Theorem 4. Let G be an n-vertex unicyclic graph with n 6. Then 4n 10 LM 1(G) (n 3)(n n ) with the left equality if and only if G = T P n and the right equality if and only if G = S n + e. Proof. The proof is by induction on n. If n = 6, then the result is trivial. Now we assume that n 7. Let P = x 1 x... x t be a longest path of G from a leaf to the cycle. If t = 0, then G = C n and LM 1 (G) = 4n < (n 3)(n n ). So the result is true. Assume now that t 1 and let G = G x 1. Then d (v G) = d (v G ) + 1 if

Z. Shao, I. Gutman, Z. Li, S. Wang, P. Wu 187 v N(x G ) and d (v G ) = d (v G) otherwise. Thus, LM 1 (G) = LM 1 (G ) v N(x G ) = LM 1 (G ) + d (x 1 G) + d (v G ) + = LM 1 (G ) + (d(x G) 1) + v N(x G ) v N(x T ) v N(x G ) = LM 1 (G ) + (d(x G) 1)(d(x G) ) + d (v G) (d (v G) d (v G ) ) (d (v G) 1) v N(x G ) d (v G). By induction, we have 4n 14 LM 1 (G ) (n 4)(n 4n + 1), with the left equality if and only if G = T P n 1 and the right equality if and only if G = S n 1 + e. If d(x G) 3 or d (x 3 G), then LM 1 (G) LM 1 (G ) + 4 = 4n 10, with the equality if and only if G = T P n 1, and d(x G) = 3 and d (x 3 G) 1 or d(x G) and d (x 3 G) =. Since n 7, d (x 3 G). So the equality holds if and only if G = T P n 1, d(x G) =, and d (x 3 G) =, which implies that G = T P n. If d(x G) and d (x 3 G) 1, then d(x G) = and d (x 3 G) = 1. Thus, N[x 3 G] = V (G) \ {x 1 } and G = S n 1 + e. It follows hat LM 1 (G) LM 1 (G ) + = (n 4)(n 4n + 1) + > 4n 10. On the other hand, if d(x G) n, then LM 1 (G) LM 1 (G ) + (n 3)(n 4) + (n 3) (n 4)(n 4n + 1) + (n 3)(3n 10) = (n 3)(n n ) (6n 0) < (n 3)(n n ). If d(x G) n 1, then G = S n + e and LM 1 (G) = (n 3)(n n ). Theorem 5. Let G be an n-vertex unicyclic graph with at least eight vertices. Then LM (T P n) < LM (G) < LM (T r,s,t) where r, s, and t are positive integers, such that r s t, r + s + t + 3 = n, and r t 1. Proof. We first prove the left inequality. The proof is by induction on n. If n = 8, then the result can be verified by computer search. Therefore, we assume that n 9. Note that if G = T P n, then LM (G) = 4n 13. Now we show that there exists no unicyclic graph G having minimum LM (G), but G = T Pn. Otherwise, let G be such a graph for which LM (G ) LM (T P n ) = 4n 13. (3)

188 Leap Zagreb Indices of Trees and Unicyclic Graphs Note that LM (C n ) = 4n > 4n 13, implying that G is not a cycle. Therefore, there exists a vertex v with degree one in G. Let vuw be a path in G with d(w G ) 1. Case 1. If d(u G ) 3, then d (v G ). If d (u G ), then LM (uv G ) 4. Let H = G v. Then LM (G ) LM (H) + LM (uv G ) LM (H) + 4. By induction, LM (H) 4(n 1) 13 and so we have LM (G ) 4n 13. Due to LM (G ) being minimum, we can deduce H = T P n 1. Therefore, G is the graph obtained by adding a vertex u and an edge uv, where v is a vertex with degree one in T P n 1. It is clear that LM (G ) is minimum if and only if G = T P n. If d (u G ) = 1, then d (v G ). Then d (x H) = d (x G ) 1 for any x N(u) \ {v}. Therefore, LM (G ) LM (H) + LM (uv G ) + e E(u G )\{uv} LM (e) 4(n 1) 13 + + d (u)(d(u) 1) 4n 13. e E(u H)\{uv} LM (e) Since LM (G ) is minimum, we have H = T P n 1. Therefore, G is the graph obtained by adding a vertex u and an edge uv, where v is a vertex with degree one in T P n 1. Clearly, LM (G ) is minimum if and only if G = T P n. If d (u) = 0, then G is the graph obtained by adding an edge to the star. Clearly, LM (G ) > 4n 13, a contradiction with Eq. (3). Case. If d(u) =, then d (u) = d(w) 1 and d (v) = 1. First, we show that d (w). Otherwise, N[w] {v} = V (G ). Since n 9, there is a vertex x N(w) with d(x) = 1. Then we get the left inequality as desired by the proof of Case 1. Now we have d (w) and so d (y) 1, for each y V (G ). Therefore, LM (G) LM (H) + LM (uv G ) + LM (uw G ) + LM (e) LM (e) LM (uw H) e (E(w G )\{uw}) e (E(w H)\{uw}) 4(n 1) 13 + d (u) + d (u) d (w) + d (v) v N(w)\{u} d (u)(d (w) 1) 4(n 1) 13 + d(w) 1 + d(w) 1 + d (v). v N(w)\{u} If there is a vertex q N(w) {u} with d (q) or d(w) 3, then LM (G) 4(n 1) 13 + d(w) 1 + d(w) 1 + d (v) v N(w)\{u} 4(n 1) 13 + d(w) 1 + d(w) 1 + 4n 13.

Z. Shao, I. Gutman, Z. Li, S. Wang, P. Wu 189 We now only consider the condition d(w) = and d (x) = for any x N(w) {u}. Then all vertices except u, v, x are adjacent to x. Since n 9, there is a leaf adjacent to x and d(x) 3. Then we get the inequality by the proof of Case 1. Now we prove the right inequality. Let G be a unicyclic graph that maximizes LM (G ) and C = v 1 v v k be the unique cycle of G. Then Claim 1. For any u V (G ), d(u, C) 1. Proof. Suppose to the contrary that there exists a vertex u such that d(u, C) = p 1. Assume that u 1 u u p is a shortest path from u 1 to C where u 1 = u and u p = v i for some i = 1,,..., k. If p 4, then we construct a graph G such that V (G ) = V (G ) and E(G ) = E(G ) {u p 3 x x L(u p 1 )} \ {u p 1 x x L(u p 1 )}. Then LM (G ) > LM (G ), contradicting with the assumption that G has maximum LM (G ). If p = 3, then we construct a graph G such that V (G ) = V (G ) and E(G ) = E(G ) {v i 1 x x L(u p 1 )} \ {u p 1 x x L(u p 1 )}. Then LM (G ) > LM (G ), contradicting with the assumption that G has maximum LM (G ). For the upper bound, the proof is by induction on n. If n = 8, we can verify the result by computer search. Now we consider the case n 9 and suppose that for any unicyclic graph G of order n 1, LM (G ) LM (T r,s,t ), (4) where r s t, r + s + t + 3 = n 1 and r t 1. It can be verified that LM (C n ) < LM (T r,s,t ), where r s t, r + s + t + 3 = n and r t 1. Therefore, G is not a cycle. Therefore, each vertex not in C of degree one and has a neighbor in C. If k = 3,then by Lemmas 4 and 5, the right inequality holds. If k = 4, then G is a graph whose removal of all leaf vertices results in C 4. Then by analysis a function with three variables, we can obtain the desired result. If k = 5, then G is a graph whose removal of all leaf vertices results in C 5. Then by analysis a function with four variables, we can obtain the desired result. Now we consider the case k 6. Let v i be a vertex in C with minimum degree. Construct a graph G such that V (G ) = V (G ) \ {v i 1 } and E(G ) = E(G )

190 Leap Zagreb Indices of Trees and Unicyclic Graphs {v i v i } {v i x x L(v i 1 )} \ {v i 1 x x L(v i 1 )}. Let A 1 = {v i 1 u u L(v i 1 G )}, B 1 = {v i 1 v i }, C 1 = {v i u u L(v i G )}, D 1 = E(G ) \ (A 1 B 1 {v i 1 v i } C 1 E 1 ), E 1 = {v i v i+1 }, A = {v i u u L(v i 1 G )}, B = {v i v i }, C = {v i u u L(v i G )}, D = E(G ) \ (A B E 1 ). We define a one-to-one mapping h : E(G ) E(G ) such that h(v p v q ) = v p v q if v p v q D 1, h(v i 1 u) = v i u if u L(v i 1 G ), i.e., v i 1 u A 1, h(v i 1 v i ) = v i v i, i.e., v i 1 u B 1. Let 1 = d (v i G )d (v i 1 G ) d (v i G )d (v i G ). Then by the choice of v i and Lemma 6 we have 1 = (d(v i 3 G ) + d(v i 1 G ) )(d(v i G ) + d(v i G ) ) (d(v i 3 G ) + d(v i G ) )(d(v i G ) + d(v i+1 G ) ). Since d(v j G ) = d(v j G ) for any j {i 3, i, i + 1}, d(v i G ) 0, and d(v i G ) = d(v i G ) + d(v i 1 G ), we have 1 0, i.e., for any e B 1 LM (e G ) LM (h(e) G ). (5) For any vertex u L(v i 1 ) in G, d (u G )d (v i 1 G ) = (d(v i G ) 1)d (v i G ) = (d(v i G ) 1)(d(v i G ) + d(v i G ) ), and d (u G )d (v i G ) = (d(v i G ) 1)d (v i G ) = (d(v i G ) 1)(d(v i G ) + d(v i+1 G ) ).

Z. Shao, I. Gutman, Z. Li, S. Wang, P. Wu 191 Since d(v j G ) = d(v j G ) for any j {i 1, i + }, d (u G )d (v i 1 G ) d (u G )d (v i G ). Then for any e A 1, we have LM (e G ) LM (h(e) G ). Now for any vertex u L(v i ) in G, we have d (u G )d (v i G ) = (d(v i G ) 1)d (v i G ) = (d(v i G ) 1)(d(v i 1 G ) + d(v i+1 G ) ), d (u G )d (v i G ) = (d(v i G ) 1)d (v i G ) = (d(v i G ) + d(v i 1 ) 3)(d(v i G ) + d(v i+1 G ) ). Let = d (u G )d (v i G ) d (u G )d (v i G ). Then = (d(v i G ) 1)(d(v i G ) d(v i 1 G )) + (d(v i 1 G ) )(d(v i G ) + d(v i+1 G ) ) = (d(v i 1 G ) )(d(v i+1 G ) 1 d(v i G ) + 1) + (d(v i G ) )(d(v i G ) 1) + (d(v i 1 G ) )(d(v i G ) 1) 0. Then for any e C 1, we have LM (e G ) LM (h(e) G ). Now for e E 1, we have d (v i G )d (v i+1 G ) = (d(v i+1 G ) + d(v i 1 G ) )(d(v i G ) + d(v i+ G ) ), and d (v i G )d (v i+1 G ) = (d(v i+1 G ) + d(v i G ) )(d(v i G ) + d(v i+ G ) ) = (d(v i+1 G ) + d(v i G ) )(d(v i G ) + d(v i 1 G ) + d(v i+ G ) ). Let 3 = d (v i G )d (v i+1 G ) d (v i G )d (v i+1 G ). Then 3 = (d(v i+1 G ) )(d(v i 1 G ) ) + (d(v i G ) d(v i 1 G ))(d(v i G ) + d(v i+1 G ) ) d(v i G )(d(v i 1 G ) ) (d(v i G ) d(v i 1 G ))(d(v i G ) + d(v i+1 G ) ).

19 Leap Zagreb Indices of Trees and Unicyclic Graphs If d(v i G ) d(v i 1 G ) < 0, then (d(v i G ) d(v i 1 G ))(d(v i G ) + d(v i+1 G ) ) < 0. If d(v i G ) d(v i 1 G ) > 0, then 3 (d(v i G ) d(v i 1 G ))(d(v i G ) + d(v i+1 G ) ) ( d(vi 1 G ) d(v i G ) + d(v i G ) + d(v i+1 G ) ( d(vi G ) + d(v i+1 G ) ( ) ) n. ) As above, for e E 1, ( ) n LM (e G ) LM (e G ). Now we have LM (v i v i 1 G ) = (d(v i 1 G ) + d(v i+1 G ) )(d(v i G ) + d(v i G ) ) ( d(vi 1 G ) + d(v i+1 G ) + d(v i G ) + d(v i G ) 4 ( ) n. ) Finally, it is clear that for any e D 1 LM (e G ) LM (h(e) G ). (6)

Z. Shao, I. Gutman, Z. Li, S. Wang, P. Wu 193 Since E(G ) = A 1 B 1 D 1 {v i 1 v i }, together with Eqs. (5) (6) we obtain LM (G ) = LM (e G ) + LM (e G ) + LM (e G ) e A 1 e B 1 e D 1 LM (e G ) + LM (v i v i+1 G ) + LM (v i v i 1 G ) e C 1 + LM (e G ) + LM (e G ) + LM (e G ) e A e B e D LM (e G ) + LM (v i v i+1 G ) + LM (v i v i+1 G ) e C + LM (v i v i+1 G ) + LM (v i v i 1 G ) ( ) ( n n LM (G ) + + ) < g(n 1) + n n 1 3 g(n), (apply induction assumption Eq. (4)) contradicting to the assumption made on the graph G. By this, the proof of Theorem 5 is completed. In a manner analogous to the proof of Theorem 5 we can establish the following: Theorem 6. Let G be an n-vertex unicyclic graph with n 6. Then 4n 8 LM 3(G) (n 3)(n + ), where r, s, t are positive integers such that r + s + t + 3 = n. The left equality holds if and only if G = T P n and the right equality holds if and only if G = T r,s,t for r, s, t 0. References [1] A. Ali and N. Trinajstić, A novel/old modification of the first zagreb index, Mol. Inf. 37 (018), no. 6/7, Article ID: 1800008. [] B. Borovicanin, K.C. Das, B. Furtula, and I. Gutman, Bounds for Zagreb indices, MATCH Commun. Math. Comput. Chem 78 (017), no. 1, 17 100. [3] J.C. Dearden, The Use of Topological Indices in QSAR and QSPR Modeling, Advances in QSAR Modeling, Springer, Cham, 017, pp. 57 88. [4] J. Devillers and A.T. Balaban (Eds.), Topological Indices and Related Descriptors in QSAR and QSPAR, Gordon & Breach, Amsterdam, 000.

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