The Structural Properties of ( 2, 6 ) -Fullerenes

: A ( 2,6 ) -fullerene F is a 2-connected cubic planar graph whose faces are only 2-length and 6-length. Furthermore, it consists of exactly three 2-length faces by Euler’s formula. The ( 2,6 ) -fullerene comes from Došli´c’s ( k ,6 ) -fullerene, a 2-connected 3-regular plane graph with only k -length faces and hexagons. Došli´c showed that the ( k ,6 ) -fullerenes only exist for k = 2, 3, 4, or 5, and some of the structural properties of ( k ,6 ) -fullerene for k = 3, 4, or 5 were studied. The structural properties, such as connectivity, extendability, resonance, and anti-Kekulé number, are very useful for studying the number of perfect matchings in a graph, and thus for the study of the stability of the molecular graphs. In this paper, we study the properties of ( 2,6 ) -fullerene. We discover that the edge-connectivity of ( 2,6 ) -fullerenes is 2. Every ( 2,6 ) -fullerene is 1-extendable, but not 2-extendable ( F is called n - extendable ( | V ( F ) | ≥ 2 n + 2) if any matching of n edges is contained in a perfect matching of F ). F is said to be k - resonant ( k ≥ 1) if the deleting of any i (0 ≤ i ≤ k ) disjoint even faces of F results in a graph with at least one perfect matching. We have that every ( 2,6 ) -fullerene is 1-resonant. An edge set, S , of F is called an anti-Kekulé set if F − S is connected and has no perfect matchings, where F − S denotes the subgraph obtained by deleting all edges in S from F . The anti-Kekulé number of F , denoted by ak ( F ) , is the cardinality of a smallest anti-Kekulé set of F . We have that every ( 2,6 ) -fullerene F with | V ( F ) | > 6 has anti-Kekulé number 4. Further we mainly prove that there exists a ( 2,6 ) -fullerene F having f F hexagonal faces, where f F is related to the two parameters n and m .

A (2,6)-fullerene F is a cubic planar graph such that every face is either 2-length or 6-length.A graph with two vertices and n parallel edges joining them is denoted by n × K 2 .The smallest (2,6)-fullerene is 3 × K 2 .A plane graph is a graph that can be embedded in the plane such that its edges intersect only at their ends.Any such embedding divides the plane into connected regions called f aces.Two different faces, f 1 , f 2 , are adjacent if their boundaries have an edge in common.A face is said to be incident with the vertices and edges in its boundary, and vice versa.An edge is said to be incident with the ends of the edge, and vice versa.Two vertices that are incident with a common edge are adjacent, and two distinct adjacent vertices are neighbors.If S is a set of vertices in a graph, F, the set of all neighbors of the vertices in S is denoted by N(S), and |N(S)| denotes the number of neighbors of S.
Let F be a (2, 6)-fullerene graph with vertex-set V(F) and edge-set E(F).We denote the number of vertices and edges in F by |V(F)| and |E(F)|.For H ⊆ H, we let F − H be the subgraph of F obtained from F by removing the elements in H.A matching of F is a set of disjoint edges, M, of F. A perfect matching of F is a matching M that covers all vertices of F. A perfect matching of a graph coincides with a Kekulé structure of some molecular graph in organic chemistry.A set, H , of disjoint even faces of a graph, F, is a resonant pattern if F has a perfect matching M such that the boundary of each face in H is an M-alternating cycle.F is said to be k-resonant any matching of n edges is contained in a perfect matching of F. F is bicritical if F contains an edge and F − u − v contains a perfect matching, for every pair of distinct vertices u, v ∈ V(F).In this paper, we show that every (2, 6)-fullerene is 1-extendable, 1-resonant but not 2-extendable, bicritical.
The anti-Kekulé set of a (2, 6)-fullerene F with perfect matchings is an edge set, S ⊆ E(F), such that F − S is connected and has no perfect matchings.The anti-Kekulé number of F, denoted by ak(F), is the cardinality of a smallest anti-Kekulé set of F. It is NP-complete to find the smallest anti-Kekulé set of a graph.Moreover, it has been shown that the anti-Kekulé set of a graph significantly affects the whole molecule structure by the valence bond theory.We know the (5,6), (4,6), and (3,6)-fullerenes have the anti-Kekulé numbers 4, 4, and 3, respectively.In this paper, We show that every (2,6)-fullerene F has the anti-Kekulé number 4, with |V(F)| > 6 .

Main Results
An edge-cut of F is a subset of edges E ⊆ E(F) such that F − E is disconnected.An k-edge-cut is an edge-cut with k edges.The edge-connectivity of F, denoted by κ (F), is equal to the minimum cardinality of edge-cuts.F is k-edge-connected if F cannot be separated into at least two components by removing fewer than k edges.Proof.Since every edge of F is incident with a 2-length face or a 6-length face, there is no cut edge in F. Therefore, F is 2-edge-connected.For one 2-length face, C, in F, we denote it by C = xyx.Then either F ∼ = 3 × K 2 or the two edges incident with x and y, respectively, other than xy form an 2-edge-cut of F. Therefore, κ (F) = 2, where |V(F)| > 2.
Proof.Let F be a (2, 6)-fullerene.If |V(F)| = 2, then F ∼ = 3 × K 2 , and the conclusion holds as F has no 2-edge-cut.So, next we suppose |V(F)| > 2. By Lemma 1, F has an 2-edge-cut.Let E = {e 1 , e 2 } be an 2-edge-cut whose deletion separates F into two components, F and F .Then E is a matching of F, as F is 3-regular and has edge-connectivity 2. Suppose every edge, e i , has one endpoint, say x i , on F , and the other endpoint, say y i , on F , i = 1, 2. Suppose the outer face of F is exactly the outer face of F, thus F lies in some inner face of F .Then, there are two hexagons, denoted by f 1 and f 2 , such that both f 1 and f 2 are incident with x 1 , x 2 , y 1 , and y 2 .If one of F and F contains a cut edge, without losing generality, assume that F contains a cut edge, e = uv, then F − e has two connected components, say F 1 and F 2 .Then, both e 1 and e 2 cannot be incident to the same component F i (i = 1, 2), otherwise there exists a cut edge, e, in F, which is a contradiction.Then, That is, all of F 1 , F 2 , and F are 2-length faces and we obtain a (2, 6)-fullerene with six vertices, and thus the conclusion holds.If both F and F contain cut edges, then there is a face with length more than 6, which is a contradiction.If neither F nor F has a cut edge, then F and F are 2-edgeconnected, and in each of them there is only one face that is not 2-length or 6-length, and we denote these two boundaries of the exceptional faces by C and C , respectively.Let v , e , and f be the number of vertices, edges, and faces in F , respectively.Let l be the length of C , and f 2 and f 6 be the number of 2-length faces and 6-length faces in F , respectively.By Euler's formula and the structure of F , it follows that Since F has no face with length more than 6, each of the two faces, f 1 and f 2 , has at most two additional vertices on C .Hence, 2 ≤ l ≤ 6.By (2), we can obtain 1 ≤ f 2 ≤ 2. If f 2 = 1, we have l = 2, which means that F is a 2-length face, and thus the conclusion holds.If f 2 = 2, then l = 6 and there are no additional vertices on C , which implies that F is a 2-length face, and thus the conclusion holds.Therefore, every 2-edge-cut of a (2, 6)-fullerene isolates a 2-length face.
The resonance of faces of a plane bipartite graph is closely related to a 1-extendable property.It was revealed that every face (including the infinite one) of a plane bipartite graph G is resonant if and only if G is 1-extendable [20].Combing with Lemma 3, we know that every (2, 6)-fullerene is 1-resonant.
Proof.Let F be a (2, 6)-fullerene graph.Let f be a 2-length face of F with the boundary v 1 v 2 v 1 .By the definition of extendability, we know that |V(F)| ≥ 4.Then, there exist two vertices, u 1 and u 2 , of F, which are different from v 1 and v 2 such that u 1 v 1 ∈ E(F) and u 2 v 2 ∈ E(F).Since the four vertices, u 1 , u 2 , v 1 , and v 2 , must be contained in the same hexagon of F, there is a vertex, is a matching and cannot be contained in a perfect matching of F. Thus, no (2, 6)-fullerene is 2-extendable.
Similarly, we can show no (2, 6)-fullerene is bicritical.Proof.Let F be a (2, 6)-fullerene graph, and f be a 2-length face of F with the boundary Theorem 3 (Tutte's Theorem [21]).A graph G has a perfect matching if and Theorem 4 (Hall's Theorem [21]).Let F be a bipartite graph with bipartition W and B. Then F has a perfect matching if and only if |W| = |B| and for any U ⊆ W, |N(U)| ≥ |U| holds.
Proof.Let F be a (2, 6)-fullerene.For any vertex, u, in 1b the graph F 6 ).We can easily see that both 3 × K 2 and F 6 cannot exist the anti-Kekulé set.On the other hand, if we let n and f 6 be the number of vertices and the hexagons of F, respectively, then, by Euler's formula and the formula of degree sum, we can obtain n = 2 f 6 + 2. Thus, if f 6 = 0, then n = 2 and F ∼ = 3 × K 2 .If f 6 = 1, then n = 4, which is impossible as every hexagonal face must contain six vertices.If f 6 = 2, then n = 6 and F ∼ = F 6 (see Figure 1b the graph F 6 ).Therefore, when |V(F)| ≤ 6, there is no anti-Kekulé set in F.  Next, we discuss the anti-Kekulé number of F with |V(F)| > 6.Then, there is a vertex, u, in F and |N(u)| = 3.Let x, y, and z be the three neighbors of u.Let e 1 and e 2 be two edges incident with x other than ux, and let e 3 and e 4 be two edges incident with y other than uy.Since every face of F is 2-length or 6-length and F is 2-edge-connected, the four edges e 1 , e 2 , e 3 , and e 4 are pairwise different.We claim that {e 1 , e 2 , e 3 , e 4 } is an anti-Kekulé set.It is obvious that F − {e 1 , e 2 , e 3 , e 4 } has no perfect matchings as the two vertices, x, y, cannot be contained in the same perfect matching.If F − {e 1 , e 2 , e 3 , e 4 } is no connected, then we obtain a cut edge, uz, in F, contradicting Lemma 1.Then, we find an anti-Kekulé set of size 4, and so ak(F) ≤ 4.
In the following, we show ak(F) ≥ 3. Let A be an anti-Kekulé set of size ak(F).Then F = F − A is connected and has no perfect matchings.According to Theorem 3, there exists S ⊆ V(F ) such that c 0 (F − S) > |S|.If we choose such an S with the maximum size, then F − S has no even components.On the contrary, suppose that F − S has an even component, H. Let F i be the odd components of F − S, where 1 ≤ i ≤ |S| + 2. For F i ⊆ F, let d(F i ) denote the number of the set of edges with one end in F i and the other end in F − F i .Denote the number of edges between S and the odd components by N. Since F is cubic, S sends out at most 3|S| to N. In addition, the above inequality gives ak(F) ≥ 3.
We find that the anti-Kekulé number of F is either 3 or 4. Suppose, on the contrary, that ak(F) = 3.Then there exists an anti-Kekulé set, A = {e 1 , e 2 , e 3 }, of cardinality three in F, such that F − A is connected and has no perfect matchings.Assume W and B are the bipartition of F. By Hall's theorem, there exists U ⊆ W such that where N F−A (U) means N(U) in F − A. Moreover, for e i ∈ A, since A is the smallest anti-Kekulé set, F − A + e i has a perfect matching.Immediately, by Theorem 4, for the above subset U, for i = 1, 2, and 3, where In addition, the neighbors of U will be increased by at most one if we add an edge, e i , to Combining inequalities (4)-( 6), we have |U| = |N F−A (U)| + 1, and e i is incident with the vertices of U and B − N F−A+e i (U) in F − A + e i .Thus, the edges going out from U ⊆ V(F) either go into A or go into the edges going out from N F−A (U).Then, the number of edges between U and that is, there is no edge between N F−A (U) and W − U in F − A. As a result, A is an edge-cut, which is a contradiction to the definition of an anti-Kekulé set.
In [23], Grünbaum and Motzkin showed that (5, 6)-fullerene and (4, 6)-fullerene having n hexagonal faces exist for every non-negative integer, n, satisfying n = 1, and gave a similar result for (3, 6)-fullerene.Therefore, we consider whether (2, 6)-fullerene having n hexagonal faces also exists for any n.We tried to give a positive answer to this question, but we found that the conclusion seems quite elusive.Therefore, in this part, we mainly prove that there exists a (2, 6)-fullerene F having f F hexagonal faces, where f F is related to the two parameters n and m.
Let F be a (2, 6)-fullerene.A f ragment, H, of F is a subgraph of F consisting of a cycle together with its interior and every inner face of H is also a face of F. We define ∂(H) as the boundary of the exterior face of H.A face, f , of F is a neighboring f ace of H if f is not a face of H and f has at least one edge in common with H.A path of length k (the number of edges) is called a k-path.Denote by f H the number of hexagons of H. Proposition 1.In all the (2, 6)-fullerenes, there exists a fragment, say G n , such that f G n = n 2 + n, n ∈ Z.
Proof.Let G 0 be a 2-length face and f 11 and f 12 be two neighboring faces of G 0 (see Figure 2a).Then f G 0 = 0. Suppose that f 11 and f 12 are hexagons.Set G 1 = G 0 { f 11 , f 12 }, suppose both f 11 and f 12 are inner faces of G 1 , and let f 21 , f 22 , f 23 , and f 24 be four neighboring faces of G 1 along the clockwise direction, such that f 21 is incident with the two consecutive 2-degree vertices on ∂(G 1 ) (see Figure 2b).Then f G 1 = 2. Suppose that f 21 , f 22 , f 23 , and f 24 are hexagons, pairwise different, and intersecting if and only if f 2i and f 2,i+1 are intersecting at only one edge for i = 1, 2, 3, 4, Suppose f 21 , f 22 , f 23 , and f 24 are the inner faces of G 2 , and let f 31 , f 32 , f 33 , f 34 , f 35 , and f 36 be six neighboring faces of G 2 along the clockwise direction, such that f 31 is incident with the two consecutive 2-degree vertices on ∂(G 2 ) (see Figure 2c).Then Suppose that the proposition holds for any integer less than n, where n > 2. According to the induction hypothesis, f G n−1 = n 2 − n and f n1 , f n2 , . . ., f n,2n are 2n neighboring faces of G n−1 along the clockwise direction, such that f n1 is incident with the two consecutive 2-degree vertices on ∂(G n−1 ).Suppose that f n1 , . . ., f n,2n are hexagons, pairwise different, and intersecting if and only if f ni and f n,i+1 are intersecting at only one edge for i = 1, 2, . . ., 2n, f n,2n+1 = f n1 .Set G n = G n−1 { f n1 , . . ., f n,2n }.Suppose f n1 , . . ., f n,2n are all inner faces of G n (see Figure 3).Then Proposition 2. In all the (2, 6)-fullerenes, there exists a fragment, say C n , such that f C n = n, n ∈ Z.
Proof.Let C 0 be 3 × K 2 , then f C 0 = 0. Let d 1 and d 2 be two 2-length faces.Its boundary, . Let P i be a path that connects two vertices, v 1i and v 2i (i = 1, 2), and V(P 1 ) V(P 2 ) = ∅.If both P 1 and P 2 are 2-paths, then, as F is 2-connected, there is a hexagon, say f 1 , such that f 1 contains the paths P 1 and P 2 and the edges v 11 v 12 and v 21 v 22 .Set C 1 = d 1 d 2 f 1 , without loss of generality, suppose f 1 is the inner face of C 1 (see Figure 4a).Thus, f C 1 = 1.If both P 1 and P 2 are 4-paths, then all whose internal vertices are denoted by x 1 , x 2 , x 3 and y 1 , y 2 , y 3 , respectively, such that P 1 = v 11 x 1 x 2 x 3 v 21 and P 2 = v 12 y 1 y 2 y 3 v 22 .Let x 2 y 2 ∈ E(F), then there are 2 hexagons, denoted by f 1 and also suppose f 1 and f 2 are two inner faces of C 2 (see Figure 4b), then f C 2 = 2. Suppose P 1 and P 2 are 2n-paths, n ∈ N + .Let P 1 = v 11 x 1 . . .x 2n−1 v 21 and P 2 = v 12 y 1 . . .y 2n−1 v 22 .Suppose that x i y i ∈ E(F) (i = 2, 4, . . ., 2n − 2), then there are n hexagons between P 1 and P 2 , denoted by f 1 , f 2 . . .f n .Set C n = d 1 d 2 n i=1 f i , such that f 1 , f 2 . . .f n are the inner faces of C n (see Figure 4c).Therefore, C n is a fragment and f C n = n, n ∈ N + .Thus, there exists a fragment, C n , such that f C n = n, n ∈ Z. Proposition 3. In all the (2, 6)-fullerenes, there exists a fragment, say L m n , such that f Proof.Let G n and G n be two fragments, as indicated in Figure 3.By Proposition 1, we know that G n and G n both have n 2 + n hexagons.Suppose n is a positive integer.Since there are 2n + 2 2-degree vertices on ∂(G n ), we can record them clockwise as u 1 , u 2 , . . ., u 2n+2 , such that u 1 and u 2n+2 are adjacent.Similarly, 2n + 2 2-degree vertices on ∂(G n ) are denoted by v 1 , v 2 , . . ., v 2n+2 along the anticlockwise direction of G n , such that v 1 and v 2n+2 are adjacent.For G 1 and G 1 .Let e 1 = u 1 v 1 , e 2 = u 2 v 2 , then e 1 and e 2 are contained in the hexagon, say f 1 .Set K 1 = G 1 G 1 f 1 (see Figure 5a), then f K 1 = 5.For G n and G n .Let e i = u i v i , i = 1, 2 . . .n + 1, then e i and e i+1 are contained in the hexagon, say f i , i = 1, 2 . . .n.
Set K n = G n G n n i=1 f i , suppose all of f 1 . . .f n are the inner faces of K n (see Figure 5b, the embedding of K n ), then Next, we construct the fragment L m n from K n as follows.We replace each edge, e i = u i v i , by a path, P i , such that P i = u i x i1 x i2 . . .x i,2m v i , i = 1, 2 . . .n + 1, m ∈ Z. Suppose that x i2 x i+1,1 , x i4 x i+1,3 . . .x i,2m x i+1,2m−1 be the edges of F, i = 1, 2 . . .n. Therefore, there are m + 1 hexagons between P i and P i+1 , denoted by f i1 , f i2 . . .
Proof.Let H 0 ∼ = F 6 be the (2, 6)-fullerene with six vertices.Without loss of generality, suppose the exterior face of H 0 is a 2-length face with the boundary u 1 v 1 u 1 , and the remaining two 2-length faces are connected by an edge, u 2 v 2 (see Figure 6a the embedding of H 0 and the labelling of u 1 , v 1 , u 2 , v 2 ).Next, we construct the fragment H n from H 0 as follows: we replace the two parallel edges, u 1 v 1 , and one edge, u 2 v 2 , by two paths, P 1 and P 3 , and one path, P 2 , such that P i = u i x i1 x i2 . . .x i,2n v i , i = 1, 2, and P 3 = u 1 x 31 x 32 . . .x 3,2n v 1 , n ∈ N + .Suppose that x i2 x i+1,1 , x i4 x i+1,3 . . .x i,2n x i+1,2n−1 be the edges of F, i = 1, 2. We construct n + 1 hexagons between P i and P i+1 , denoted by f i1 , f i2 . . .
are the inner faces of H n , n ∈ N + (see Figure 6b).Therefore, H n is a fragment and f H n = 2(n + 1) = 2n + 2, n ∈ N + .Thus, there exists a fragment, H n , such that f By Propositions 1-4, we can find a (2, 6)-fullerene F having f F hexagonal faces that relates to the parameters n and m.Theorem 7.There exists a (2, 6)-fullerene F such that f F = n 2 + 2n, n ∈ Z.
Proof.Let G n be a fragment of F, as shown in Figure 3.Its boundary, ∂(G n ), is labelled u 1 , u 2 , . . ., u 4n+2 along the clockwise direction, where u 1 and u 2 are two consecutive 2degree vertices.Let C n be a fragment of F, as shown in Figure 4c.Its boundary, ∂(C n ), is labelled v 1 , v 2 , . . ., v 4n+2 along the clockwise direction, where v 1 and v 2 are two consecutive 3-degree vertices.Next, we assume each of the graphs G n and C n drawn on a hemisphere, with the boundary as equator.
Proof.Let G m+n+1 be a fragment of F, as shown in Figure 3.Its boundary, ∂(G m+n+1 ), is labelled u 1 , u 2 , . . ., u 4m+4n+6 along the clockwise direction, where u 1 and u 2 are two consecutive 2-degree vertices.Let L m n be a fragment of F, as shown in Figure 5c.Its boundary, ∂(L m n ), is labelled v 1 , v 2 , . . ., v 4m+4n+6 along the clockwise direction, where v 1 and v 2 are two consecutive 3-degree vertices.Next, we assume each of the graphs G m+n+1 and L m n drawn on a hemisphere, with the boundary as equator.
Proof.Let G n be a fragment of F, as shown in Figure 3.Its boundary, ∂(G n ), is labelled u 1 , u 2 , . . ., u 4n+2 along the clockwise direction, where u 1 and u 2 are two consecutive 2degree vertices.Let H n be a fragment of F, as shown in Figure 6b.Its boundary, ∂(H n ), is labelled v 1 , v 2 , . . ., v 4n+2 along the clockwise direction, where v 1 and v 2 are two consecutive 3-degree vertices.Next, we assume each of the graphs G n and H n drawn on a hemisphere, with the boundary as equator.If ∂(G n ) = ∂(H n ), then set F = G n H n .By Propositions 1 and 4, then f F = f G n + f H n = n 2 + 3n + 2, n ∈ Z.

Conclusions
In this paper, we characterized the structural properties of (2, 6)-fullerene F. By the conclusion of Došlić on (k, 6)-fullerenes, we know that every (2, 6)-fullerene is 1-extendable, and then, by means of the relationship between the extendable and resonance, we know that every (2, 6)-fullerene is 1-resonant.But, by the definition of n-extendable, we find that every (2, 6)-fullerene is not 2-extendable, and then not k-extendable (k ≥ 3).Moreover, we find that no (2, 6)-fullerene is bicritical.On the other hand, for the anti-Kekulé number of (2, 6)-fullerene F, we firstly obtain the upper bound 4 of the anti-Kekulé number of F by the definition of an anti-Kekulé set.Secondly, with the help of Tutte's Theorem, and combing with the structure of (2, 6)-fullerene, we obtain the lower bound 3 of the anti-Kekulé number of F. Finally, by analyzing that the anti-Kekulé number of F cannot be 3, we find that the anti-Kekulé number of F is 4.
Grünbaum and Motzkin showed that (5, 6)-fullerene and (4, 6)-fullerene having n hexagonal faces exist for every non-negative integer, n, satisfying n = 1, and showed a similar result for (3, 6)-fullerene.Therefore, at the end of the paper, we consider whether (2, 6)-fullerene having n hexagonal faces also exists for any n.We try to give a positive answer to this question, but we find that the conclusion seems quite elusive.So, we mainly prove that there exists a (2, 6)-fullerene F having f F hexagonal faces, where f F is related to the two parameters n and m.There are, however, still several important open questions.
which is a contradiction to the choice of S. Since |V(F )| is even, then c 0 (F − S) ≥ |S| + 2 by parity.For any edge e ∈ A, adding e to F − S will connect at most two odd components, then c o (F + e − S) ≥ c 0 (F − S) − 2. Since A is the smallest anti-Kekulé set of F, then F + e has a perfect matching for any edge e ∈ A. Hence, by Theorem 3, for the above subset S, c o (F + e − S) ≤ |S|.Therefore, |S| ≥ c o (F + e − S) ≥ c 0 (F − S) − 2 ≥ |S|.We obtain c 0 (F − S) = |S| + 2, and the edge, e, connects exactly two components of F − S.

Figure 3 .
Figure 3.The fragment G n .