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1 Hong Kong Baptist University HKBU Institutional Repository Department of Mathematics Journal Articles Department of Mathematics 2003 Extremal k*-cycle resonant hexagonal chains Wai Chee Shiu Hong Kong Baptist University, Peter Che Bor Lam Hong Kong Baptist University, Lian-zhu Zhang Zhangzhou Teachers College This ocument is the authors' final version of the publishe article. Link to publishe article: APA Citation Shiu, W., Lam, P., & Zhang, L. (2003). Extremal k*-cycle resonant hexagonal chains. Journal of Mathematical Chemistry, 33 (1), This Journal Article is brought to you for free an open access by the Department of Mathematics at HKBU Institutional Repository. It has been accepte for inclusion in Department of Mathematics Journal Articles by an authorize aministrator of HKBU Institutional Repository. For more information, please contact

2 Extremal k -Cycle Resonant Hexagonal Chains Wai Chee SHIU, Peter Che Bor LAM Department of Mathematics, Hong Kong Baptist University, 224 Waterloo Roa, Kowloon Tong, Hong Kong Lian-zhu ZHANG Department of Mathematics, Zhangzhou Teachers College, Zhangzhou, Fujian , China Abstract Denote by B n the set of all k -cycle resonant hexagonal chains with n hexagons. For any B n B n, let m(b n ) an i(b n ) be the numbers of matchings (=the Hosoya inex) an the number of inepenent sets (=the Merrifiel-Simmons inex) of B n, respectively. In this paper, we give a characterization of the k -cycle resonant hexagonal chains, an show that for any B n B n, m(h n ) m(b n ) an i(h n ) i(b n ), where H n is the helicene chain. Moreover, equalities hol only if B n = H n. Keywors AMS 2000 MSC : k -cycle resonant hexagonal chain, helicene chain, Hosoya inex, Merrifiel-Simmons inex. : 05C10, 05C70, 05C90 1. Introuction A hexagonal system is a 2-connecte plane graph whose every interior face is boune by a regular hexagon. Hexagonal systems are of great importance for theoretical chemistry because they are the natural graph representation of benzenoi hyrocarbons [7]. A consierable amount of research in mathematical chemistry has been evote to hexagonal systems [7 9]. A hexagonal chain with n hexagons is a hexagonal system consisting of n regular hexagons C 1, C 2,..., C n with the properties that (a) for any k, j with 1 k < j n 1, C k an C j have a common ege if an only if j = k + 1, an (b) each vertex belongs to at most two hexagons. Hexagonal chains are the graph representation of an important subclass of benzenoi molecules, namely of the so calle unbranche cataconense benzenois. A great eal of mathematical an mathematico-chemical results on hexagonal chains were obtaine, (see, for example, [7 10, 12, 14, 22 25]). Let G = (V, E) be a graph with vertex set V (G) an ege set E(G). Let e an x be an ege an a vertex of G respectively. We will enote by G e or G x the graph obtaine from G by removing e or x respectively. Denote by N x the set {y V (G) : xy E(G)} {x}. Let S be a subset of V (G). The subgraph of G inuce by S is enote by G[S], an G[V \ S] is enote by G S. Unefine concepts an notations of graph theory are referre to [1, 3] Two eges of a graph G are sai to be inepenent if they are not incient. A subset M of E(G) is calle a matching of G if any two eges of M are inepenent in G. Denote by m(g) the number of matchings of G. In chemical terminology, m(g) is calle the Hosoya inex. The work was partially supporte by the Hong Kong Baptist University Faculty Research Grant; Founation for University Key Teacher by the Ministry of Eucation. 1

3 Two vertices of a graph G are sai to be inepenent if they are not ajacent. A subset I of V (G) is calle an inepenent set of G if any two vertices of I are inepenent. Denote i(g) the number of inepenent sets of G. In chemical terminology, i(g) is calle the Merrifiel-Simmons inex. Clearly, the Hosoya inex or the Merrifiel-Simmons inex of a graph is larger than that of its proper subgraphs. We enote by B n the set of the hexagonal chains with n hexagons. Let B n B n. We enote by V 3 = V 3 (B n ) the set of the vertices with egree 3 in B n. Thus the subgraph B n [V 3 ] is a acyclic graph. If the subgraph B n [V 3 ] is a matching with n 1 eges, then B n is calle a linear chain an enote by L n. If the subgraph B n [V 3 ] is a path, then B n is calle a zig-zag chain an enote by Z n. If the subgraph B n [V 3 ] is a comb, then B n is calle a helicene chain an enote by H n. Figure 1 (a), (b) an (c) illustrate L n, Z n an H n, respectively, where B n [V 3 ] are inicate by heavy eges. C 1 C 2... q C n-1 C n-1 C n C n C n n-1 C n C 1 C 2... q n c n-1 n p n p n-1 p 0 p 1 p 2... C 1 q 1 C 2 2 q 2 c 2 (a) L n (b) Z n (c) H n Figure 1. Note that the consiere hexagonal chains inclue both geometrically planar (e.g. L n an Z n ) an geometrically non-planar (e.g. H n ) species. It is easy to see that B 1 = {L 1 = Z 1 = H 1 }, B 2 = {L 2 = Z 2 = H 2 } an B 3 = {L 3, Z 3 = H 3 }. Let B n B n an label its hexagons consecutively by C 1, C 2,..., C n. Thus the hexagons C 1 an C n are terminal an for j = 1, 2,..., n 1, the hexagons C j an C j+1 have a common ege. We also enote B n by C 1 C 2 C n. In the topological theory of unbranche cataconense hyrocarbons, mathematical chemists are intereste in investigating extremal hexagonal chains with respect to some topological inices, such as the number of Kekulé structures, Winner inex, Hosoya inex, Merrifiel-Simmons inex, graph eigenvalue an total π-electron energy (the total absolute values of eigenvalues of a graph) etc [2, 5, 6, 9, 10, 13, 15 19, 21 23]. Those topological inices of molecular graphs are of great importance in theoretical chemistry [5, 11, 13]. Among hexagonal chains with extremal properties on topological inices, L n, Z n an H n play important roles. We list some of them in Theorems Theorem 1.1(Gutman [10], Zhang [22]). For any n 1 an any B n B n, if B n is neither L n nor Z n, then m(l n ) < m(b n ) < m(z n ). 2

4 Theorem 1.2 (Gutman [10], Zhang [22]). For any n 1 an any B n B n, if B n is neither L n nor Z n, then i(z n ) < i(b n ) < i(l n ). Theorem 1.3 (Gutman [10], Zhang an Tian [23]). Denote by λ 1 (G) the largest eigenvalue of a graph G. Then for any n 1 an any B n B n, if B n is neither L n nor H n, then λ 1 (L n ) < λ 1 (B n ) < λ 1 (H n ). Theorem 1.4 (Zhang, Li an Wang [24, 25]). Denote by π(g) the total π-electron energy of a molecular graph G. Then for any n 1 an any B n B n, if B n is neither L n nor Z n, then π(l n ) < π(b n ) < π(z n ). Let M be a perfect matching of G. A cycle C in G is an M-alternating cycle if eges of C belongs to M an oes not belong to M alternatively. A number of isjoint cycles in a graph G are mutually resonant if there is a perfect matching M of G such that each cycle is an M-alternating cycle. A connecte graph G with perfect matching is sai to be k-cycle resonant if G contains at least k( 1) isjoint cycles, an any t isjoint cycles in G, 1 t k, are mutually resonant. The concept of k-cycle resonant graph was introuce by Guo an Zhang [4]. It is a generalization of k-coverable hexagonal system inuce by Zheng [26]. A graph G is calle k -cycle resonant if G is k-cycle resonant an k is the maximum number of isjoint cycles in G. Denote by B n the set of all k -cycle resonant hexagonal chains with n hexagons. In this paper, we give a characterization of the k -cycle resonant hexagonal chains, an show that for any B n B n, m(h n ) m(b n ) an i(h n ) i(b n ), where H n is the helicene chain. Moerover, equalities hol only if B n = H n. 2. k -Cycle Resonant Hexagonal Chains Any element B n of B n can be obtaine from an appropriately chosen graph B n 1 B n 1 by attaching to it a new hexagon. Let B be a hexagonal chain, C a hexagon an rs an ege of C. It is easy to see that there are three types of attaching: (i) r a, s b; (ii) r b, s c an (iii) r c, s as shown in Figure 2. We call them α-type, β-type an γ-type attaching, respectively. Following [10], we enote by [B] θ the hexagonal chain obtaine from B by θ-type attaching to it a new hexagon C, where θ {α, β, γ}. Obviously, each B n with n 2 can be written as We set B n = βθ 2 θ 3 θ n 1 for short. [ [ [[L2 ] ] ] θ2, where θ ]θ j {α, β, γ}. 3 θ n 1 3

5 B: a b c r s C a C c b a b c C a C b c [B] α [B] β [B] γ Figure 2. For each j, if θ j = β then B n = L n ; if θ j {α, γ} an θ j θ j+1, then B n = Z n ; an if θ j = α (or γ) then B n = H n. Set γ if θ = α; θ = β if θ = β; α if θ = γ. Obviously, the hexagonal chain B n = βθ 2 θ 3 θ n 1 is isomorphic to the hexagonal chain B n = βθ 2 θ 3 θ n 1. In [4], Guo an Zhang give some necessary an sufficient conitions for a graph to be k-cycle resonant. We mention the following results which will be useful for our results. Theorem 2.1. (Guo an Zhang, [4]) A connecte graph with at least k isjoint cycles is k-cycle resonant if an only if G is bipartite an, for 1 t k an any t isjoint cycles W 1, W 2,..., W t in G, G t j=1 W j contains no component of o orer. Theorem 2.2. (Guo an Zhang, [4]) Every 2-cycle resonant hexagonal system is k -cycle resonant, where k is the maximum number of isjoint cycles in the hexagonal system. By Theorems 2.1 an 2.2, we can show that Theorem 2.3. A hexagonal chain B n (n 3) belongs to B n if an only if B n = C 1 C 2 C n = βθ 2 θ 3 θ n 1, where θ j {α, γ}, 2 j n 1. Proof. First, notice that each hexagonal chain is bipartite. Suppose that B n = βθ 2 θ 3 θ n 1 belongs to B n. If there is some j, 2 j n 1, such that θ j = β, then it is easy to see that B n C j 1 C j+1 contains two components of orer one. This contraicts that B n is k -cycle resonant. Now, suppose that B n = βθ 2 θ 3 θ n 1 is a hexagonal chain with θ j {α, γ}, 2 j n 1. First, we show that B n is 2-cycle resonant. Let C be any cycle of B n. Obviously, B n C 4

6 contains no component of o orer. Let C, C be any two isjoint cycles of B n, an let B n (C) an B n (C ) be the sub-chains of B n whose bounary are C an C, respectively. Assume that B n (C) = C i C i+1 C j an B n (C ) = C k C k+1 C l, 1 i j k 2 l 2 n 2. It is easy to see that in this case, B n C C contains three components of orers 4(i 1), 4(k j 2) + 2 an 4(n l), respectively. Thus, by Theorem 2.1, B n is 2-cycle resonant, an hence B n is k -cycle resonant by Theorem 2.2. By Theorem 2.3, every element B n of Bn can be written as B n = βθ 2 θ 3 θ n 1 with θ j {α, γ}, 2 j n 1. Clearly, H n an Z n are k*-resonant. Denote by K(G) the number of perfect matchings (in chemical terminology, it is calle the number of Kekulé structures) of G. In [10], Gutman pointe out that it is well-known that all fully-angularly annulate hexagonal chains (with a given n) have equal an maximal K-value. Hence, by Theorem 2.3, all k -cycle resonant hexagonal chains have equal an maximal K-value. It is easy to see that the equal an maximal K-values K(B n ), (n = 1, 2,... ), are Fibonacci numbers with the initial values K(B 1 ) = 2 an K(B 2 ) = Extremal Properties of H n Among many properties of m(g) an i(g) ([11, 13]); we mention the following results which will be use later. Claim 3.1. Let G be a graph consisting of two components G 1 an G 2. Then (a) m(g) = m(g 1 )m(g 2 ); (b) i(g) = i(g 1 )i(g 2 ). Claim 3.2. Let G be a graph. (a) Suppose uv E(G). Then m(g) = m(g uv) + m(g u v). (b) Suppose u V (G). Then i(g) = i(g u) + i(g N u ). Claim 3.3. Let G be a graph. For each uv E(G), (a) m(g) m(g u) m(g u v) 0; (b) i(g) i(g u) i(g u v) 0. Moreover, equalities hol only if v is the unique neighbor of u. a n s n-1 b n B n-1 t n-1 c n n 5

7 Figure 3. Let B n = C 1 C 2... C n be a any given hexagonal chain. For each k, 1 k n 1, we set B k = C 1 C 2... C k. We use s k 1, t k 1, a k, b k, c k an k to label the vertices of C k such that s k 1 t k 1 is the common ege of C k 1 an C k, an s k 1 a k, a k b k, b k c k, c k k an k t k 1 are the eges of C k. The case k = n is shown in Figure 3. an By Claims 3.1 an 3.2, we obtain the following recurrences: m(b n ) m(b n a n ) m(b n b n ) m(b n 1 ) m(b n 1 s n 1 ) m(b n c n ) = m(b m(b n n ) n 1 t n 1 ) m(b n a n b n ) m(b n 1 s n 1 t n 1 ) m(b n c n n ) i(b n ) i(b n a n ) i(b n b n ) i(b n a n b n ) i(b n N an ) i(b n N bn ) = i(b n 1 ) i(b n 1 s n 1 ) i(b n 1 t n 1 ) i(b n 1 s n 1 t n 1 ) (1). (2) To emonstrate how to obtain the above relations, we prove the first ientity. By applying Claims 3.1(a) an 3.2(a) repeately we have m(b n ) = m(b n s n 1 a n ) + m(b n s n 1 a n ) = m(b n s n 1 a n t n 1 n ) + m(b n s n 1 a n t n 1 n )+ + m(b n s n 1 a n t n 1 n ) + m(b n s n 1 a n t n 1 n ) = m(b n 1 )m(p 4 ) + m(b n 1 t n 1 )m(p 3 )+ + m(b n 1 s n 1 )m(p 3 ) + m(b n 1 s n 1 t n 1 )m(p 2 ) = 5m(B n 1 ) + 3m(B n 1 t n 1 ) + 3m(B n 1 s n 1 ) + 2m(B n 1 s n 1 t n 1 ), where P m is the path with m vertices. Lemma 3.1. For any B n B n (n 1) an {u, v} = {a n, b n } (see Figure 3), we have (a) m(b n ) + m(b n u) 2m(B n v) m(b n u v) > 0; (b) i(b n u v) + i(b n N u ) 2i(B n N v ) > 0. Proof. Since m(b 1 ) = 18, m(b 1 u) = 8, m(b 1 u v) = 5, i(b 1 u v) = 8 an i(b 1 N u ) = 5 = i(b 1 N v ), the lemma hol when n = 1. So we assume n 2. (a) Suppose that u = a n an v = b n. By (1) we have m(b n ) + m(b n a n ) 2m(B n b n ) m(b n a n b n ) 6

8 = 2m(B n 1 ) m(b n 1 s n 1 ) + 2m(B n 1 t n 1 ). Since B n 1 s n 1 is a proper subgraph of B n 1. Thus m(b n 1 ) > m(b n 1 s n 1 ), an hence m(b n ) + m(b n a n ) 2m(B n b n ) m(b n a n b n ) > 0. Suppose that u = b n an v = a n. By (1) we have m(b n ) + m(b n b n ) 2m(B n a n ) m(b n a n b n ) = m(b n 1 ) + 5m(B n 1 s n 1 ) m(b n 1 t n 1 ) + 3m(B n 1 s n 1 t n 1 ). In orer to prove that m(b n ) + m(b n b n ) 2m(B n a n ) m(b n a n b n ) > 0, it suffices to show that 5m(B n 1 s n 1 ) > m(b n 1 ) an 3m(B n 1 s n 1 t n 1 ) > m(b n 1 t n 1 ). Note that, since B n is a k -cycle resonant hexagonal chain, we must have that either s n 1 = a n 1, t n 1 = b n 1 or s n 1 = c n 1, t n 1 = n 1. Moreover, B n 1 B n 1 If s n 1 = a n 1, t n 1 = b n 1, then by (1), we get that 5m(B n 1 s n 1 ) m(b n 1 ) = 5m(B n 1 a n 1 ) m(b n 1 ) = 5[3m(B n 2 ) + 2m(B n 2 t n 2 )] [5m(B n 2 ) + 3m(B n 2 s n 2 ) + 3m(B n 2 t n 2 ) +2m(B n 2 s n 2 t n 2 )] = 10m(B n 2 ) 3m(B n 2 s n 2 ) +7m(B n 2 t n 2 ) 2m(B n 2 s n 2 t n 2 ). Since B n 2 s n 2 an B n 2 s n 2 t n 2 are the proper subgraphs of B n 2 an B n 2 t n 2, respectively, we can get 5m(B n 1 s n 1 ) m(b n 1 ) > 0 in this case. Similarly, we can show that 5m(B n 1 s n 1 ) > m(b n 1 ) in the case s n 1 = c n 1, t n 1 = n 1, an that 3m(B n 1 s n 1 t n 1 ) > m(b n 1 t n 1 ). an (b) Similar to the proof of (a), by (2) we get i(b n ) + i(b n N an ) 2i(B n N bn ) = i(b n 1 ) + 4i(B n 1 s n 1 ) + 2i(B n 1 s n 1 t n 1 ), i(b n ) + i(b n N bn ) 2i(B n N an ) = 4i(B n 1 ) + 3i(B n 1 t n 1 ) 2i(B n 1 s n 1 ) i(b n 1 s n 1 t n 1 ). Notice that B n 1 s n 1 an B n 1 s n 1 t n 1 are the proper subgraphs of B n 1 an B n 1 s n 1, respectively. Therefore i(b n ) + i(b n N u ) 2i(B n N v ) > 0 for {u, v} = {a n, b n }. Let A an B be two hexagonal chains, where A is obtaine from the hexagonal chain A by attaching a hexagon H. The vertices of H are labele a, b, c,, q an p as shown in Figure 4(a). Let r an s be two ajacent vertices of B of egree two. Now, we enote by B n the hexagonal chain obtaine from A an B by ientifying c an r, an, an s, respectively (Figure 4(b)); an 7

9 by G α the hexagonal chain obtaine from A, B by ientifying a an s, an, b an r, respectively (Figure 4(c)). A p q a b c r s B A a p q s= B b c=r A a=s p q B b=r c (a) A an B (b) G γ (c) G α Figure 4. Lemma 3.2. Let A, B, G γ an G α be the k*-cycle resonant hexagonal chains shown in Figure 4. We have (a) if m(a p) > m(a q), then m(g γ ) > m(g γ); (b) if i(a p) < i(a q), then i(g γ ) < i(g α ). Proof. (a) By Claims 3.1(a) an 3.2(a), we have the followings: m(g γ ) = {m(a) + m(a p)}{m(b) + m(b r)} +{m(a q) + m(a p q)}{m(b s) + m(b r s)} +m(a)m(b) + m(a q)m(b s) an m(g α ) = {m(a) + m(a q)}{m(b) + m(b r)} +{m(a p) + m(a p q)}{m(b s) + m(b r s)} +m(a)m(b) + m(a p)m(b s). Thus m(g γ ) m(g α ) = {m(a p) m(a q)}{m(b) + m(b r) 2m(B s) m(b r s)}. By Lemma 3.1(a), m(b)+m(b r) 2m(B s) m(b r s) > 0. Therefore if m(a p) > m(a q), then m(g γ ) > m(g α ). (b) By Claims 3.1(b) an 3.2(b), we have the followings: i(g γ ) = {2i(A) + i(a p)}i(b r s) + {i(a) + i(a p)}i(b N r ) +{2i(A q) + i(a p q)}i(b N s ) an i(g α ) = {2i(A) + i(a q)}i(b r s) + {i(a) + i(a q)}i(b N r ) +{2i(A p) + i(a p q)}i(b N s ). 8

10 Thus i(g γ ) i(g α ) = {i(a p) i(a q)}{i(b r s) + i(b N r ) 2i(B N s )}. By Lemma 3.1(b), i(b r s) + i(b N r ) 2i(B N s ) > 0. Hence, if i(a p) < i(a q), then i(g γ ) < i(g α ). Let H n = C 1 C 2...C n be a helicene chain. We label the common ege of C 1 an C 2 as p 1 q 1 ; an for each k, 2 k n, we label the vertices of V (C k ) V (C k 1 ) as p k, q k, c k an k such that p k 1 p k, p k q k, q k c k, c k k an k q k 1 are eges in H n (see Fig. 1(c)). In Fig. 3, if let B n = H n, B n 1 = H n 1, s n 1 = p n 1, t n 1 = q n 1, then a n = p n an b n = q n. By (1) an (2) we get m(h n ) m(h n 1 ) m(h n p n ) m(h n q n ) = m(h n 1 p n 1 ) m(h n 1 q n 1 ) m(h n p n q n ) m(h n 1 p n 1 q n 1 ) (3) an an Let i(h n ) i(h n p n ) i(h n q n ) i(h n p n q n ) = i(h n 1 ) i(h n 1 p n 1 ) i(h n 1 q n 1 ) i(h n 1 p n 1 q n 1 ). Φ n = m(h n ) m(h n p n ) m(h n p n q n ) Ψ n = i(h n ) i(h n p n ) i(h n p n q n ). (4) Lemma 3.3. For n 1, we have (a) Φ n is a strictly increasing function of n; (b) Ψ n is a strictly ecreasing function of n. Proof. (a) It is easy to see that Φ 1 = 5 an Φ 2 = 34. By (3), we can get Φ n = 3m(H n 1 p n 1 ) + 2m(H n 1 p n 1 q n 1 ). For n 3, we have Φ n Φ n 1 = 3{m(H n 1 p n 1 ) m(h n 2 p n 2 )} +2{m(H n 1 p n 1 q n 1 ) m(h n 2 p n 2 q n 2 )}. Since H n 2 p n 2 an H n 2 p n 2 q n 2 are the proper subgraphs of H n 1 p n 1 an H n 1 p n 1 q n 1, respectively, m(h n 1 p n 1 ) > m(h n 2 p n 2 } an m(h n 1 p n 1 q n 1 ) > m(h n 2 p n 2 q n 2 ). Therefore Φ n > Φ n 1. 9

11 (b) It is easy to see that i(h 1 ) = 18, i(h 1 p 1 ) = 13 an i(h 1 p 1 q 1 ) = 8. Thus Ψ 1 = 3, Ψ 2 = 10 an Ψ 3 = 190. By (4), we get that Ψ n = 2i(H n 1 ) + 2i(H n 1 p n 1 ) i(h n 1 q n 1 ) + i(h n 1 p n 1 q n 1 ) = 6i(H n 2 p n 2 ) 3i(H n 2 p n 2 q n 2 ). Thus, for n 4, we have that Ψ n Ψ n 1 = 6{i(H n 3 p n 3 ) i(h n 2 p n 2 )} +3{i(H n 3 p n 3 q n 3 ) i(h n 2 p n 2 q n 2 )}. Since H n 3 p n 3 an H n 3 p n 3 q n 3 are the proper subgraphs of H n 2 p n 2 an H n 2 p n 2 q n 2, respectively, we have that i(h n 3 p n 3 ) < i(h n 2 p n 2 ) an i(h n 3 p n 3 q n 3 ) < i(h n 2 p n 2 q n 2 ). Therefore Ψ n < Ψ n 1. Lemma 3.4. Let H n be a helicene chain. Then (a) m(h 1 p 1 ) = m(h 1 q 1 ), an for each n 2, m(h n p n ) > m(h n q n ). (b) i(h 1 p 1 ) = i(h 1 q 1 ), an for each n 2, i(h n p n ) < i(h n q n ). Proof. It is easy to obtain that m(h 1 p 1 ) m(h 1 q 1 ) = 0 an m(h 2 p 2 ) m(h 2 q 2 ) > 0. For n 3, by (3) an (4) we have m(h n p n ) m(h n q n ) = m(h n 1 ) m(h n 1 p n 1 ) m(h n 1 p n 1 q n 1 ) {m(h n 1 p n 1 ) m(h n 1 q n 1 )} = Φ n 1 {m(h n 1 p n 1 ) m(h n 1 q n 1 )} = (Φ n 1 Φ n 2 ) + {m(h n 2 p n 2 ) m(h n 2 q n 2 )} an i(h n p n ) i(h n q n ) = {i(h n 1 ) i(h n 1 p n 1 ) i(h n 1 p n 1 q n 1 )} {i(h n 1 p n 1 ) i(h n 1 q n 1 )} = Ψ n 1 + {i(h n 1 q n 1 ) i(h n 1 p n 1 )} = (Ψ n 1 Ψ n 2 ) + {i(h n 2 p n 2 ) i(h n 2 q n 2 )}. By Lemma 3.3(a), we have Φ n 1 Φ n 2 > 0. Hence we get that for each n 3, m(h n p n ) > m(h n q n ). Similarly, we can obtain i(h 1 p 1 ) i(h 1 q 1 ) = 0 an i(h 2 p 2 ) i(h 2 q 2 ) < 0. By Lemma 3.3(b), we have Ψ n 1 Ψ n 2 < 0. Hence we get that for each n 3, i(h n p n ) < i(h n q n ). Theorem 3.5. For any n 1 an any B n Bn, we have (a) m(h n ) m(b n ) m(z n ); 10

12 (b) i(h n ) i(b n ) i(z n ), with relevant equalities holing only if B n = H n, or only if B n = Z n. Proof. We only nee to verify the first inequalities of (a) an (b) accoring to Theorems 1.1 an 1.2. Let B n B n be the hexagonal chain with the smallest number of matchings, (the largest number of inepenent sets, respectively). By Theorem 2.3, B n B n can be written as B n = βθ 2 θ 3 θ n 1 with θ j {α, γ}, 2 j n 1. Assume, without loss of generality, that θ 2 = α, (otherwise, we consier B n = βθ 2 θ 3...θ n 1 ). Suppose that B n H n. Since B 1 = {H 1}, B 2 = {H 2} an B 3 = {H 3}, we have that n 4. Let θ j be the first element of θ 2, θ 3,..., θ n 1 such that θ j = γ. Thus j 3, an B n = βα...αγθ j+1...θ n 1. Referring to Figure 4, set G γ = B n = βα αγθ j+1 θ n 1, A = βα α = H j 1, p = p j 1 an q = q j 1. Let G α = βα ααθ j+1 θ n 1. By Lemma 3.4(a) (Lemma 3.4(b), respectively), we have m(a p) > m(a q), (i(a p) < i(a q), respectively). By Lemma 3.2(a), (Lemma 3.2(b), respectively), we have m(b n ) > m(g α ), (i(b n ) < i(g α ), respectively), which contraicts the choice of B n. The proof of Theorem 3.5 is complete. References [1] J.A. Bony an U.S.R. Murty, Graph Theory with Applications, Macmillan, Lonon an Elsevier, New York, [2] J. Brunvoll, I. Gutman an S.J. Cyvin, Benzenoi chains with maximum number of Kekulé structures, Z. Phys. Chem. (Leipzig) 270 (1989), [3] J. Clark an D.A. Holton, A First Look at Graph Theory, Worl Scienitific, Singapore, New Jersey, Lonon, Hong Kong, [4] X.F. Guo, an F.J. Zhang, k-cycle resonant graphs, Discrete Math. 135 (1994) [5] I. Gutman, O.E. Polansky, Mathematical Concept in Organic Chemistry, Springer, Berlin, [6] I. Gutman an F.J. Zhang O.E., On the orering of graphs with respect their matching numbers, Discrete Appl. Math. 15 (1986) [7] I. Gutman an S.J. Cyvin (es.), Introuction to the Theory of Benzenoi Hyrocarbons, Springer, Berlin, [8] I. Gutman, I. an S.J. Cyvin (es.), Avances in the Theory of Benzenoi Hyrocarbons, Topics in Current Chemistry, Vol 153, Springer, Berlin, [9] I. Gutman, I. (e.), Avances in the Theory of Benzenoi Hyrocarbons 2, Topics in Current Chemistry, Vol 162, Springer, Berlin, [10] I. Gutman, Extremal hexagonal chains, J. of Math. Chem. 12 (1993)

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