Extremal hexagonal chains with respect to the coefficients sum of the permanental polynomial

Applied Mathematics and Computation 291 (2016) 30–38

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Extremal hexagonal chains with respect to the coefficients sum of the permanental polynomial Wei Li a, Zhongmei Qin b,∗, Heping Zhang c a

Department of Applied Mathematics, School of Science, Northwestern Polytechnical University, Xi’an, Shaanxi 710072, PR China Center for Combinatorics and LPMC, Nankai University, Tianjin 300071, PR China c School of Mathematics and Statistic, Lanzhou University, Lanzhou, Gansu 730000, PR China b

a r t i c l e

i n f o

a b s t r a c t

MSC: 05C31 05C75

A hexagonal system is a graphical representation of a benzenoid hydrocarbon in theoretical chemistry. A hexagonal chain is a cata-condensed hexagonal system with no branchings. In this paper we consider extremal hexagonal chains with respect to the coefficients sum of the permanental polynomial. We prove that the linear chain attains the minimum value of this sum and the zigzag chain attains the maximum value of this sum.

Keywords: Hexagonal chain Permanental polynomial Coefficients sum

© 2016 Elsevier Inc. All rights reserved.

1. Introduction A hexagonal system is 2-connected plane graph, and each of the inner faces is a unit hexagon. It is the graph representation of the benzenoid hydrocarbon. Harary [6] first realized that hexagonal systems are very attractive objectives in graph-theoretical study and initiated their great importance for mathematical investigations. A hexagonal chain is a hexagonal system with the properties that (a) it has no vertex belonging to three hexagons, and (b) it has no hexagon with more than two adjacent hexagons. The extremal problems on the hexagonal chains attracted much attention, and those extremal hexagonal chains with respect to the number of perfect matchings, Kirchhoff index, largest eigenvalue and energy were investigated [4,14–17]. In the following, we introduce some definitions and notations used in this paper. Let M = (mi j ) be an n × n matrix. The permanent of M is defined as

perM =

 σ

m1σ1 m2σ2 . . . mnσn ,

where the sum ranges over all the permutations σ of {1, 2, . . . , n}. Let G be a graph on n vertices and A(G ) = (ai j )n×n the adjacency matrix of G. Let I denote an identity matrix of order n. The permanental polynomial of G is per(xI − A(G )) [13],  written as π (G, x)= nk=0 bk xn−k . By setting x = 1, we get that π (G, 1) is the coefficients sum of the permanental polynomial of G. It was proved that the permanental coefficients are related to the graph structure [11]:

(−1 )i bi =



2k ( H )

for 1 ≤ i ≤ n,

H

∗

Corresponding author. Tel.: +86 15122627176. E-mail addresses: [email protected] (W. Li), [email protected] (Z. Qin), [email protected] (H. Zhang).

http://dx.doi.org/10.1016/j.amc.2016.06.025 0 096-30 03/© 2016 Elsevier Inc. All rights reserved.

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W. Li et al. / Applied Mathematics and Computation 291 (2016) 30–38

31

a c

...

b

... ... Fig. 1. (a) a linear chain; (b) a zigzag chain; (c) a helix chain.

where the sum ranges over all subgraphs H on i vertices whose components are single edges or cycles, and k(H) is the number of cycles in H. Particularly, for a bipartite graph G, one has b2i+1 = 0 and

b 2i =



m2 ( H )

for 1 ≤ i ≤ 

H

n , 2

(2)

where the sum takes over all the induced subgraph H of G on 2i vertices and m(H) is the number of perfect matchings of H. The permanental polynomials of graphs were first systematically studied in 1980’s [1,11]. The investigation on permanental polynomials of chemical graphs were started by Kasum et al. [7]. Recently, some studies on the permanental polynomials of hexagonal systems, fullerenes and general graphs have appeared [2,3,5,9,10]. Gutman and Cash [5] discussed the relation between the permanental polynomial and the characteristic polynomial of hexagonal system, and established some formulae on these two polynomials. Li and Zhang [9] gave recursive expressions to show the permanental polynomials of various hexagonal chains. For more details we refer to a survey paper [8] in a book [12]. In this paper, we focus on the coefficients sum of the permanental polynomial, and characterize the extremal hexagonal chains with respect to this sum. The organization of this paper is as follows. In Section 2 we introduce the roll-attaching operation of hexagonal chains, which is very helpful to prove the main result. In Section 3 we provide some auxiliary results on the coefficients sums of the permanental polynomials of hexagonal chains, and in Section 4 we prove that the linear chain takes the minimum value of this sum and the zigzag chain takes the maximum value of this sum. 2. The roll-attaching operation of hexagonal chains In this section, we introduce some definitions and operations on the hexagonal chains, which will be needed in the proof of our main result. Denote by Bn the set of hexagonal chains with n hexagons. Each hexagonal chain Bn ∈ Bn (n ≥ 2) can be obtained from a hexagon by adding hexagons gradually. Thus we denote Bn as C1C2 . . . Cn , where Ci is the ith hexagon in Bn . Let V3 = V3 (Bn ) be the set of vertices of Bn with degree 3. Then the graph Bn [V3 ] induced by the edges with end-vertices in V3 is a acyclic graph, and such a graph is signed by thick line as shown in Fig. 1. If Bn [V3 ] is a matching with n − 1 edges, then Bn is a linear chain, denoted by Ln (Fig. 1(a)). If Bn [V3 ] is a path, then Bn is a zigzag chain, denoted by Zn (Fig. 1(b)). If Bn [V3 ] is a comb, then Bn is a helix chain, denoted by Hn (Fig. 1(c)). Thus we can see that B1 = {L1 = Z1 = H1 }, B2 = {L2 = Z2 = H2 } and B3 = {L3 , Z3 = H3 }. Any element Bn ∈ Bn can be obtained from Bn−1 in Bn−1 by attaching a hexagon as follows. Let B be a hexagonal chain. Let C be a hexagon and rs an edge of C. There are three cases for attaching C to B as depicted in Fig. 2: (1) r ≡ a, s ≡ b; (2) r ≡ b, s ≡ c; (3) r ≡ c, s ≡ d, denoted by α -type, β -type and γ -type respectively. Denote by [B]k the hexagonal chain obtained from a hexagonal chain B by attaching a hexagon C through k-type attaching, where k ∈ {α , β , γ } [16]. Each hexagonal chain Bn can be written as Bn = k1 k2 . . . kn , where ki ∈ {α , β , γ } for 1 ≤ i ≤ n. As it is irrelevant to which type the first hexagon is, we set k1 = β . Then Bn = β k2 . . . kn . If ki = β for every i, then Bn = Ln ; if ki ∈ {α , γ } for every i and ki = ki+1 , then Bn = Zn ; if ki = α (resp. γ ) for every i, then Bn = Hn . According to [16], let k¯ be the rolling of k. Then the roll-attaching operation k¯ is

⎧ ⎨α , k¯ = β , ⎩ γ,

if k = γ ; if k = β ; if k = α .

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a b r C

c s d

a

a b

C

a

c

b b

C

d

c

c

C

d

d α − type

β − type

γ − type

Fig. 2. Three cases of attaching a hexagon C to a hexagonal chain.

For a given hexagonal chain Bn = C1C2 . . . Cn = β k2 . . . kn , we call B = CiCi+1 . . . Cn a hexagonal subchain of Bn , denoted by B = ki ki+1 . . . kn (where k1 = β when i = 1). Let B = k¯ i k¯ i+1 . . . k¯ n . Then B is the rolling of B. Similarly, Bn = β¯ k¯ 2 k¯ 3 . . . k¯ n is the rolling of Bn . It is obvious that B and B are isomorphism. Suppose Bn = C1C2 . . . Cn = β k2 k3 . . . kn is a hexagonal chain. Let B = C1C2 . . . Ci and B

= Ci+1Ci+2 . . . Cn . Denote by ri si the common edge of Ci and Ci+1 . Then Bn can be regarded as a hexagonal chain obtained from B by attaching B through ki type attaching. Label the vertices of V (Ci ) − V (Ci−1 ) by a, b, c, d in a clockwise direction. We can see that if ki = α , then ri ≡ a, si ≡ b; if ki = β , then ri ≡ b, si ≡ c; if ki = γ , then ri ≡ c, si ≡ d. See Fig. 2. For a given hexagonal chain Bn = β k2 . . . ki−1 ki ki+1 . . . kn and its subchain B

= ki+1 ki+2 . . . kn , define the roll-attaching operation on Bn and B as follows: attach the rolling B

of B to B = β k2 . . . ki . The hexagonal chain obtained by the rollattaching operation is written as β k2 . . . ki−1 k i k¯ i+1 . . . k¯ n , where k i ∈ {α , β , γ }. If k i = α , then si ≡ a, ri ≡ b; if k i = β , then si ≡ b, ri ≡ c; if k i = γ , then si ≡ c, ri ≡ d as shown in Fig. 2 [16]. 3. Auxiliary results To derive our main results, we show first some consequences on the permanental polynomials. Next, we give some auxiliary results on the coefficients sum of the permanental polynomial of a hexagonal chain. The graph G considered here are simple graphs.  Proposition 3.1 ([13]). Let G be a bipartite graph with π (G, x ) = nk=0 bk xn−k . If n is even, then b0 = 1, b2k ≥ 0 and b2k−1 = 0 for k ≥ 1. Thus π (G, 1) > 0. Proposition 3.2 ([9]). Let Pn be a path on n vertices. Then the permanental polynomial of Pn is given as





2  n − k n−2k π (Pn , x ) = x . k n

k=0

Theorem 3.1 ([1]). Let G be the union of two disjoint graphs G1 and G2 . Then

π (G, x ) = π (G1 , x )π (G2 , x ). Theorem 3.2 ([1]). Let e = uv be any edge of G. Then

π (G, x ) = π (G − uv, x ) + π (G − u − v, x ) + 2



(−1 )|V (C )| π (G − V (C ), x ),

C∈Ce (G )

where Ce (G ) is the set of cycles in G containing e, V(C) is the set of vertices in C and |V(C)| is the number of vertices in V(C). It follows from Theorem 3.2 that Corollary 3.1 ([1]). If an edge e = uv is not in any cycle of G, then

π (G, x ) = π (G − uv, x ) + π (G − u − v, x ). Let G1 be a graph with vertices u1 and v1 . Let G2 be a graph with vertices u2 and v2 . Then the bridge graph G1  G2 through edges e1 and e2 is the one obtained from G1 and G2 by connecting u1 and u2 with an edge e1 and connecting v1 and v2 with an edge e2 . See Fig. 3.

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Fig. 3. The bridge graph G = G1  G2 .

Theorem 3.3 ([9]). For the bridge graph G = G1  G2 through edges e1 = u1 u2 and e2 = v1 v2 , the following result holds.

π (G, x ) = π (G1 , x )π (G2 , x ) + π (G1 − u1 , x )π (G2 − u2 , x ) + π ( G1 − v1 , x )π ( G2 − v2 , x ) + π ( G1 − u1 − v1 , x )π ( G2 − u2 − v2 , x )  +2 (−1 )|V (C )| π (G − V (C ), x ). C∈Ce1 (G )

In the following, we give some useful lemmas. Lemma 3.1. Let H be a hexagonal chain with n hexagons and e = uv an edge of H. Then

π (H, 1 ) − π (H − u, 1 ) − π (H − u − v, 1 ) > 0. Proof. When n = 1, from [9] we have

π (H, x ) = x6 + 6x4 + 9x2 + 4. By Proposition 3.2, we have

π (H − u, x ) = π (P5 , x ) = x5 + 4x3 + 3x and

π (H − u − v, x ) = π (P4 , x ) = x4 + 3x2 + 1. Thus for the case n = 1, it holds π (H, 1 ) − π (H − u, 1 ) − π (H − u − v, 1 ) = 20 − 8 − 5 = 7 > 0. When n ≥ 2, it follows from Theorem 3.2 that



π (H, x ) = π (H − uv, x ) + π (H − u − v, x ) + 2

(−1 )|V (C )| π (H − V (C ), x ),

C∈Ce (H )

i.e.



π (H, x ) − π (H − uv, x ) − π (H − u − v, x ) = 2

(−1 )|V (C )| π (H − V (C ), x ).

C∈Ce (H )

Taking x = 1, we get



π (H, 1 ) − π (H − uv, 1 ) − π (H − u − v, 1 ) = 2

(−1 )|V (C )| π (H − V (C ), 1 ).

C∈Ce (H )

Since H is a hexagonal chain, the cycles C containing e = uv must be of even orders. Thus (−1 )|V (C )| = 1 holds. However, H − V (C ) has two components H1 and H2 , and both of them are bipartite graphs with even orders. For all the cycles  C ∈ Ce (H ), at least one component of H − V (C ) is not empty. Then by Proposition 3.1, we have C∈Ce (H ) (−1 )|V (C )| π (H − V (C ), 1 ) > 0. Thus

π (H, 1 ) − π (H − uv, 1 ) − π (H − u − v, 1 ) > 0.

(3)

Next we show that π (H − uv, 1 ) > π (H − u, 1 ). Case 1: u is a vertex of degree 2 in the hexagonal chain H. Let u be another neighbor except v of u in H. Then H − uv has no cycles containing uu . Thus by Corollary 3.1, we have

π (H − uv, x ) = π (H − uv − uu , x ) + π (H − uv − u − u , x ) = xπ (H − u, x ) + π (H − u − u , x ). Setting x = 1, we get π (H − uv, 1 ) = π (H − u, 1 ) + π (H − u − u , 1 ). Since H − u − u is a nonempty bipartite graph with an even order, from Proposition 3.1 we have π (H − u − u , 1 ) > 0. Thus π (H − uv, 1 ) − π (H − u, 1 ) > 0, i.e. π (H − uv, 1 ) > π (H − u, 1 ).

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W. Li et al. / Applied Mathematics and Computation 291 (2016) 30–38

Fig. 4. A zigzag chain Zn .

Case 2: u is a vertex of degree 3 in the hexagonal chain H. Let u , u be another two neighbors of u except v in H. From Theorem 3.2, we have

π (H − uv, x ) = π (H − uv − uu , x ) + π (H − uv − u − u , x )  +2 (−1 )|V (C )| π (H − uv − V (C ), x ) C∈Cuu (H−uv )

= [π (H − uv − uu − uu

, x ) + π (H − uv − uu − u − u

, x )]  + π ( H − uv − u − u , x ) + 2 (−1 )|V (C )| π (H − uv − V (C ), x ) C∈Cuu (H−uv )

= xπ (H − u, x ) + π (H − u − u , x ) + π (H − u − u , x )

+2



(−1 )|V (C )| π (H − uv − V (C ), x ).

C∈Cuu (H−uv )

Let x = 1. Then

π (H − uv, 1 ) = π (H − u, 1 ) + π (H − u − u

, 1 ) + π (H − u − u , 1 )  +2 (−1 )|V (C )| π (H − uv − V (C ), 1 ). C∈Cuu (H−uv )

Since uu is an edge of a hexagon in a hexagonal chain, the cycles containing uu must be of even orders. Thus

)| = 1. On the other hand, according to Proposition 3.1, we know that π (H − u − u

, 1 ) > 0, π (H − u − u , 1 ) > (−1 )|V (C |V (C )| π (H − uv − V (C ), 1 ) > 0. It follows that π (H − uv, 1 ) − π (H − u, 1 ) > 0, i.e. π (H − uv, 1 ) > 0 and C∈Cuu (H−uv ) (−1 ) π (H − u, 1 ). Comparing with Eq. (3), we have that π (H, 1 ) − π (H − u, 1 ) − π (H − u − v, 1 ) > 0. 

Let Zn = C1C2 . . . Cn be a zigzag chain and v1 u1 the common edge of C1 and C2 . For 2 ≤ i ≤ n, let vi , ui , ci , di be the vertices of V (Ci ) − V (Ci−1 ) such that ui−1 vi , vi ui , ui ci , ci di , di vi−1 are the edges of Ci , as depicted in Fig. 4. The following results hold for a zigzag chain Zn . Lemma 3.2. Let Zn be a zigzag chain. Then (a) for n = 1, π (Z1 − u1 , 1 ) = π (Z1 − v1 , 1 ); (b) for n ≥ 2, π (Zi − ui , 1 ) < π (Zi − vi , 1 ) with 2 ≤ i ≤ n. Proof. (a) For n = 1, it follows from Proposition 3.2 that

π (Z1 − u1 , x ) = π (P5 , x ) = x5 + 4x3 + 3x and

π (Z1 − v1 , x ) = π (P5 , x ) = x5 + 4x3 + 3x. Thus π (Z1 − u1 , 1 ) = π (Z1 − v1 , 1 ) = 8. (b) For n ≥ 2 and 2 ≤ i ≤ n, we can see that Zi − ui is the bridge graph of Zi−1 and K1 ∪ P2 . From Theorem 3.3, we have

W. Li et al. / Applied Mathematics and Computation 291 (2016) 30–38

35

Fig. 5. (a) B1n ; (b) B2n ; (c) B3n ; (d) B4n , where A is a hexagonal chain with i − 1 hexagons, B is a hexagonal chain with n − i hexagons and B is the rolling of B.

π (Zi − ui , x ) = (x3 + x )π (Zi−1 , x ) + (x2 + 1 )π (Zi−1 − ui−1 , x ) + x2 π (Zi−1 − vi−1 , x ) + xπ (Zi−1 − ui−1 − vi−1 , x ). Further by Corollary 3.1, we obtain

π (Zi − vi , x ) = (x3 + 2x )π (Zi−1 , x ) + (x2 + 1 )π (Zi−1 − vi−1 , x ). Thus we have

π (Zi − ui , x ) − π (Zi − vi , x ) = (x2 + 1 )π (Zi−1 − ui−1 , x ) + xπ (Zi−1 − ui−1 − vi−1 , x ) − xπ (Zi−1 , x ) − π (Zi−1 − vi−1 , x ) = [π (Zi−1 − ui−1 , x ) − π (Zi−1 − vi−1 , x )] + [x2 π (Zi−1 − ui−1 , x ) + xπ (Zi−1 − ui−1 − vi−1 , x ) − xπ (Zi−1 , x )]. Taking x = 1, we derive

π (Zi − ui , 1 ) − π (Zi − vi , 1 ) = [π (Zi−1 − ui−1 , 1 ) − π (Zi−1 − vi−1 , 1 )] + [π (Zi−1 − ui−1 , 1 ) + π (Zi−1 − ui−1 − vi−1 , 1 ) − π (Zi−1 , 1 )]. Thus we get

[π (Zi − ui , 1 ) − π (Zi − vi , 1 )] − [π (Zi−1 − ui−1 , 1 ) − π (Zi−1 − vi−1 , 1 )] = π (Zi−1 − ui−1 , 1 ) + π (Zi−1 − ui−1 − vi−1 , 1 ) − π (Zi−1 , 1 ). It follows from Lemma 3.1 that π (Zi−1 − ui−1 , 1 ) + π (Zi−1 − ui−1 − vi−1 , 1 ) − π (Zi−1 , 1 ) < 0. Thus [π (Zi − ui , 1 ) − π (Zi − vi , 1 )] − [π (Zi−1 − ui−1 , 1 ) − π (Zi−1 − vi−1 , 1 )] < 0. Following this, it holds that π (Zi − ui , 1 ) − π (Zi − vi , 1 ) < π (Zi−1 − ui−1 , 1 ) − π (Zi−1 − vi−1 , 1 ) < · · · < π (Z1 − u1 , 1 ) − π (Z1 − v1 , 1 ) = 0. Therefore, we get π (Zi − ui , 1 ) − π (Zi − vi , 1 ) < 0 i.e. π ( Zi − ui , 1 ) < π ( Z i − vi , 1 ).  The next lemma shows the relation between the coefficients sum of the permanental polynomial of a hexagonal chain and the one of its roll-attaching chain. Lemma 3.3. Let B1n = β k2 · · · ki−1 β ki+1 · · · kn as shown in Fig. 5(a). Let B3n = β k2 . . . ki−1 α k¯ i+1 . . . k¯ n as shown in Fig. 5(c). Then π (B1n , 1 ) < π (B3n , 1 ). Proof. We show that π (B1n , 1 ) − π (B3n , 1 ) < 0. From Theorems 3.2, 3.3 and Corollary 3.1, we have



π (B1n , x ) = π (B1n − pa, x ) + π (B1n − p − a, x ) + 2

C 1 ∈C

=

(−1 )|V (C )| π (B1n − V (C 1 ), x ) 1

( )

1 pa Bn

π (A, x )[x2 π (B, x ) + xπ (B − r, x ) + xπ (B − s, x ) + π (B − r − s, x )] + π (A − p, x )[xπ (B, x ) + π (B − s, x )] + π (A − q, x )[xπ (B, x ) + π (B − r, x )]  1 + π (A − p − q, x )π (B, x ) + 2 (−1 )|V (C )| π (B1n − V (C 1 ), x ) C 1 ∈C pa (B1n )

and

π (B3n , x ) = π (B3n − ps, x ) + π (B3n − p − s, x ) + 2

 C 3 ∈C ps (B3n )

(−1 )|V (C )| π (B3n − V (C 3 ), x ) 3

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W. Li et al. / Applied Mathematics and Computation 291 (2016) 30–38

=

π (A, x )[(x2 + 1 )π (B, x ) + xπ (B − r, x )] + π (A − p, x )[(x2 + 1 )π (B − s, x ) + xπ (B − r − s, x )] + π (A − q, x )[xπ (B, x ) + π (B − r, x )] + π (A − p − q, x )  3 × [xπ (B − s, x ) + π (B − r − s, x )] + 2 (−1 )|V (C )| π (B3n − V (C 3 ), x ). C 3 ∈C ps (B3n )

From Fig. 5(a) and (c), we know that the edges pa and ps are both in the ith hexagon of the hexagonal chain. Since B and B are isomorphism, it holds that



(−1 )|V (C )| π (B1n − V (C 1 ), x ) = 1

C 1 ∈C pa (B1n )



(−1 )|V (C )| π (B3n − V (C 3 ), x ). 3

C 3 ∈C ps (B3n )

Thus

π (B1n , x ) − π (B3n , x ) = [π (A, x ) − xπ (A − p, x ) − π (A − p − q, x )] [xπ (B − s, x ) + π (B − r − s, x ) − π (B, x )]. Taking x = 1, we have

π (B1n , 1 ) − π (B3n , 1 ) = [π (A, 1 ) − π (A − p, 1 ) − π (A − p − q, 1 )] [π (B − s, 1 ) + π (B − r − s, 1 ) − π (B, 1 )]. From Lemma 3.1 we get that π (B1n , 1 ) − π (B3n , 1 ) < 0. Thus π (B1n , 1 ) < π (B3n , 1 ).



Lemma 3.4. Let B2n = β k2 . . . ki−1 γ ki+1 . . . kn as shown in Fig. 5(b). Let B3n = β k2 . . . ki−1 α k¯ i+1 . . . k¯ n β k2 . . . ki−1 α ki+1 . . . kn as shown in Fig. 5(c) and (d). It holds that (a) if π (A − p, 1 ) < π (A − q, 1 ) and π (B − r, 1 ) ≥ π (B − s, 1 ), then π (B2n , 1 ) < π (B3n , 1 ); (b) if π (A − p, 1 ) < π (A − q, 1 ) and π (B − r, 1 ) ≤ π (B − s, 1 ), then π (B2n , 1 ) < π (B4n , 1 ). Proof. (a) From Theorem 3.2 and Corollary 3.1, we have

π (B2n , x ) = π (B2n − qs, x ) + π (B2n − q − s, x ) + 2



(−1 )|V (C )| π (B2n − V (C 2 ), x ) 2

C 2 ∈Cqs (B2n )

=

π (A, x )[(x + 1 )π (B, x ) + xπ (B − r, x )] + π (A − p, x )[xπ (B, x ) + π (B − r, x )] + π (A − q, x )[(x2 + 1 )π (B − s, x ) + xπ (B − s − r, x )] + π (A − p − q, x )  2 × [xπ (B − s, x ) + π (B − r − s, x )] + 2 (−1 )|V (C )| π (B2n − V (C 2 ), x ). 2

C 2 ∈Cqs (B2n )

Combining with the proof of the above Lemma, we derive

π (B2n , x ) − π (B3n , x ) = [π (A − p, x ) − π (A − q, x )][xπ (B, x ) + π (B − r, x ) − (x2 + 1 )π (B − s, x ) − xπ (B − r − s, x )]. Take x = 1. Then

π (B2n , 1 ) − π (B3n , 1 ) = [π (A − p, 1 ) − π (A − q, 1 )] × [π (B, 1 ) + π (B − r, 1 ) − 2π (B − s, 1 ) − π (B − r − s, 1 )]. By the given condition, π (A − p, 1 ) < π (A − q, 1 ) and π (B − r, 1 ) ≥ π (B − s, 1 ). It follows from Lemma 3.1 that

π (B, 1 ) + π (B − r, 1 ) − 2π (B − s, 1 ) − π (B − r − s, 1 ) = π (B − r, 1 ) − π (B − s, 1 ) + π (B, 1 ) − π (B − s, 1 ) − π (B − r − s, 1 ) > 0. Thus π (B2n , 1 ) − π (B3n , 1 ) < 0, i.e. π (B2n , 1 ) < π (B3n , 1 ). (b) From Theorem 3.2 and Corollary 3.1, we can see

π (B4n , x ) = π (B4n − pr, x ) + π (B4n − p − r, x ) + 2



C 4 ∈C

=

(−1 )|V (C )| π (B4n − V (C 4 ), x ) 4

( )

4 pr Bn

π (A, x )[(x + 1 )π (B, x ) + xπ (B − s, x )] + π (A − p, x )[(x2 + 1 )π (B − r, x ) + xπ (B − r − s, x )] + π (A − q, x )[xπ (B, x ) + π (B − s, x )] + π (A − p − q, x )  4 × [xπ (B − r, x ) + π (B − r − s, x )] + 2 (−1 )|V (C )| π (B4n − V (C 4 ), x ). 2

C 4 ∈C pr (B4n )

Similarly, we obtain

π (B2n , x ) − π (B4n , x ) = [π (B − r, x ) − π (B − s, x )] × [xπ (A, x ) − x2 π (A − p, x ) − xπ (A − p − q, x )] + [π (A − p, x ) − π (A − q, x )] × [xπ (B, x ) − x2 π (B − s, x ) − xπ (B − r − s, x )].

and

B4n =

W. Li et al. / Applied Mathematics and Computation 291 (2016) 30–38

37

Set x = 1. We get

π (B2n , 1 ) − π (B4n , 1 ) = [π (B − r, 1 ) − π (B − s, 1 )][π (A, 1 ) − π (A − p, 1 ) − π (A − p − q, 1 )] + [π (A − p, 1 ) − π (A − q, 1 )][π (B, 1 ) − π (B − s, 1 ) − π (B − r − s, 1 )]. Since π (B − r, 1 ) ≤ π (B − s, 1 ), combining Lemma 3.1 we have

[π (B − r, 1 ) − π (B − s, 1 )][π (A, 1 ) − π (A − p, 1 ) − π (A − p − q, 1 )] ≤ 0. Similarly, since π (A − p, 1 ) < π (A − q, 1 ), by Lemma 3.1 we obtain

[π (A − p, 1 ) − π (A − q, 1 )][π (B, 1 ) − π (B − s, 1 ) − π (B − r − s, 1 )] < 0. Thus we have π (B2n , x ) − π (B4n , x ) < 0, and so π (B2n , x ) < π (B4n , x ).



4. Main results In this section, we show the hexagonal chains that attain the minimum and maximum values of coefficients sums of their permanental polynomials. Theorem 4.1. For the hexagonal chains Bn ∈ Bn , the linear chain Ln attains the minimum value of coefficients sum of the permanental polynomial. Proof. We know that B1 = {L1 = Z1 = H1 }, B2 = {L2 = Z2 = H2 } and B3 = {L3 , Z3 = H3 }. Thus it suffices to consider the case n ≥ 3. Let Bn ∈ Bn be a hexagonal chain with minimum value of coefficients sum of the permanental polynomial. We show Bn = Ln . Suppose to the contrary that Bn = Ln . Let ki be the first element of k2 , k3 , , kn such that ki = β , i.e. Bn = ββ . . . β ki . . . kn , where ki = α or γ . Without loss of generality, let ki = α (otherwise, replace Bn by Bn ). Then Bn = ββ . . . βα ki+1 . . . kn . Take B n = ββ . . . ββ k¯ i+1 . . . k¯ n . By Lemma 3.3, we know that π (B n , 1 ) < π (Bn , 1 ), which contradicts the hypothesis that Bn attains the minimum value of coefficients sum of the permanental polynomial. Thus when n ≥ 3, Ln attains the minimum value of π (Bn , 1). When n = 1, for any B1 ∈ B1 , B1 = L1 . When n = 2, for any B2 ∈ B2 , B2 = L2 . Thus, for all hexagonal chains Bn ∈ Bn , Ln has the minimum coefficients sum of the permanental polynomial.  Theorem 4.2. For the hexagonal chains Bn ∈ Bn , the zigzag chain Zn attains the maximum value of coefficients sum of the permanental polynomial. Proof. Since B1 = {L1 = Z1 = H1 }, B2 = {L2 = Z2 = H2 } and B3 = {L3 , Z3 = H3 }, we only need to consider the case n ≥ 3. Let Bn ∈ Bn be the hexagonal chain that has the maximum value of coefficients sum of permanental polynomial. We prove Bn = Zn . Suppose to the contrary that Bn = Zn . Let ki be the first element of k2 , k3 , . . . , kn such that ki = β or ki = ki−1 ∈ {α , γ }. Case 1: ki = β . If i = 2, then Bn = ββ k3 . . . kn . Let B n = βα k¯ 3 . . . k¯ n . Then by Lemma 3.3, we have π (B n , 1 ) > π (Bn , 1 ). This contradicts that π (Bn , 1) is maximum. Assume that i ≥ 3. Without loss of generality, let Bn = β k2 k3 . . . ki−2 γ β ki+1 . . . kn (otherwise, replace Bn by Bn ). Let B n = β k2 k3 . . . ki−2 γ α k¯ i+1 . . . k¯ n . From Lemma 3.3, we know that π (B n , 1 ) > π (Bn , 1 ), which contradicts that π (Bn , 1) is maximum. Case 2: ki−1 = ki ∈ {α , γ }. In this case, i ≥ 3. Without loss of generality, let ki−1 = ki = γ . Then Bn = β k2 . . . ki−2 γ γ ki+1 . . . kn (otherwise, replace Bn by Bn ). Denote Bn by Aγ B, where A = C1C2 . . . Ci−1 = β k2 . . . ki−3 αγ = Zi−1 and B = ki+1 . . . kn . According to the labeling of vertices in Fig. 4, let p = ui−1 and q = vi−1 , where i − 1 ≥ 2. By Lemma 3.2 (b), we get that π (A − p, 1 ) < π (A − q, 1 ), which satisfies the first hypothesis of Lemma 3.4 (a) and (b). If π (B − r, 1 ) ≥ π (B − s, 1 ), then let B n = β k2 . . . ki−2 γ α k¯ i+1 . . . k¯ n . By Lemma 3.4 (a), we obtain that π (B n , 1 ) > π (Bn , 1 ), which contradicts that π (Bn , 1) is maximum. If π (B − r, 1 ) ≤ π (B − s, 1 ), then let B n = β k2 . . . ki−2 γ α ki+1 . . . kn . By Lemma 3.4 (b), we know that π (B n , 1 ) > π (Bn , 1 ), contradicting to the choice of Bn . Thus when n ≥ 3, Zn attains the maximum value of π (Bn , 1). When n = 1, for any B1 ∈ B1 , B1 = Z1 . When n = 2, for any B2 ∈ B2 , B2 = Z2 . In conclusion, for all hexagonal chains Bn ∈ Bn , Zn has the maximum coefficients sum of the permanental polynomial.  Acknowledgments We are grateful to the anonymous referees’ helpful suggestions for improving our paper. The first author is supported by NSFC (Grant No. 11501448), the Natural Science Basic Research Plan in Shaanxi Province of China (Grant No. 2015JQ1025) and the Fundamental Research Funds for the Central Universities (Grant No. 3102015ZY071). The third author is supported by NSFC (Grant No. 11371180).

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