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H D i ftegraph. the of n eoe by denoted , ( scle aitncycle Hamilton a called is = G A and G h S fDprmn of Department of NSF the + ) d onaino hn (No. China of Foundation m ( v 1 h denme,the number, edge the f G e,w sals some establish we per, i = d = [ = ) K A = G 303 China. 230013, i 1 G ( | m,n G + ≤ E G d q 2 G lUniversity, al Let . ( ( stesignless the is ) H a d = G G ( ij eacomplete a be v 2 n G stegraph the is , ) ,i aldto called is ), i lca spec- placian G · · · forder of ] | h degree the ) ≤ · · · ≤ − h number the 1 ∨ l lengths ble and vertices 1 = H D ( sthe is , G G H k be ) G we , d be n n is , . pancyclic if it contains cycles of every length l, 3 l n. Clearly, a bipartite graph ≤ ≤ is not pancyclic. A pancyclic graph is certainly Hamiltonian, but the converse is not true. A cycle of length l is called an l cycle. The problem of deciding whether a − graph is Hamiltonian is one of the most difficult classical problems in graph theory. Indeed, it is NP-complete. Recently, the spectral theory of graphs has been applied to this problem. Firstly, Fiedler and Nikiforov [4] gave tight conditions on spectral radius of a graph and its complement for the existence of Hamiltonian paths and cycles. Next, Bo Zhou [13] gave tight conditions on the signless spectral radius of a graph complement for the existence of Hamiltonian paths and cycles. Yu and Fan [9] established the spectral conditions for a graph to be Hamilton-connected in terms of the spectral radius of the adjacency matrix or signless Laplacian matrix of the graph or its complement. Lu, Liu and Tian [7] gave sufficient conditions for a bipartite graph to be Hamiltonian in terms of the spectral radius of the adjacency matrix of the graph. Since then, many researchers have studied the analogous problems under various spectral conditions; see [2, 3, 5, 6, 8, 10, 14]. But there is no spectral sufficient conditions on pancyclic graphs. In this paper, we first establish a new sufficient conditions for a graph to be pancyclic in terms of the edge number of the graph, then basing on edge number sufficient condition, we give (signless Laplacian) spectral radius sufficient conditions for a graph to be pancyclic.

2 Preliminary

We begin with some definitions. Given a graph G of order n, a vector X Rn is called ∈ to be defined on G, if there is a 1-1 map ϕ from V (G) to the entries of X; simply written Xu = ϕ(u). If X is an eigenvector of A(G) (Q(G)), then X is defined naturally on G, i.e. Xu is the entry of X corresponding to the vertex u. One can find that when λ is an eigenvalue of G corresponding to the eigenvector X if and only if X = 0, 6

λXv = Xu, for each vertex v V (G). (2.1) ∈ u∈XNG(v)

The equation (2.1) is called eigen-equation of G. When q is an signless Laplacian eigenvalue of G corresponding to the eigenvector X if and only if X = 0, one can find 6 that [q dG(v)]Xv = Xu, for each vertex v V (G). (2.2) − ∈ u∈XNG(v) The equation (2.2) is called signless Laplacian eigen-equation of G. Lemma 2.1 [12] Let G be a graph of order n with degree sequence d1 dn, if n ≤···≤ for all positive integers k such that dk k< and dn−k n k, then G is a pancyclic ≤ 2 ≥ − graph or bipartite graph. Lemma 2.2 [11] Let G be a connected graph of order n with m deges. Then

µ(G) 2m n + 1, ≤ − p and the equality holds if and only if G = Kn or G = K1,n−1. Lemma 2.3 [9] Let G be a graph of order n with m deges. Then 2m q(G) + n 2. ≤ n 1 − − If G is connected, the equality holds if and only if G = K1,n−1 or G = Kn. Otherwise, the equality holds if and only if G=Kn−1 + v.

2 3 Main Results

Theorem 3.1 Let G be a connected graph on n( 5) vertices and m deges with ≥ minimum degree δ(G) 2. If ≥ n 2 m − + 4, (3.1) ≥  2  then G is a pancyclic graph unless G is a bipartite graph or G NP1= K2 (Kn−4 + ∈ { ∨ 2K1), K5 6K1, K3 (K2 + 3K1), K3 (K1 + K1,4), K3 (K2 + K1,3), (K2 2K1) ∨ ∨ ∨ ∨ ∨ ∨ 5K1, K4 5K1, K1,2 4K1, K2 (K1 + K1,3), K3 4K1 . ∨ ∨ ∨ ∨ } Proof: Suppose that G is neither a pancyclic graph nor a bipartite graph. By Lemma n 2.1, there exists an positive integer k for dk k< , such that dn−k n k 1. Then ≤ 2 ≤ − − we have k n−k n

2m = di + di + di Xi=1 i=Xk+1 i=nX−k+1 k2 +(n 2k)(n k 1) + k(n 1) ≤ − − − − = n2 n + 3k2 + (1 2n)k − − n 2 = 2 − + 8 (k 2)(2n 3k 7),  2  − − − − thus n 2 (k 2)(2n 3k 7) m − + 4 − − − . (3.2) ≤  2  − 2

n 2 n 2 k− n− k− Since − +4 m − +4 ( 2)(2 3 7) , thus (k 2)(2n 3k 7) 0.  2  ≤ ≤  2  − 2 − − − ≤ Next, we discuss in the follow two cases. Case 1 Assume that (k 2)(2n 3k 7) = 0, i.e., k = 2 or 2n 3k 7 = 0. − − − − − n 2 Then, m = − + 4 and all inequalities in the above arguments should be equal-  2  ities. Case 1.1 If k = 2, then G is a graph with d1 = d2 = 2, d3 = = dn−2 = n 3, · · · − dn−1 = dn = n 1. The two vertices of degree n 1 must be adjacent to every vertex, − − so they induce a K2. The two vertices of degree 2 are not adjacent to other vertices, so they induce a 2K1. For the remaining n 4 vertices of degree n 3. They must be − − adjacent to each other to make sure the requirement of the degree n 3, so they induce − a Kn−4. By the above analysis, we can get the graph G must be K2 (Kn−4 + 2K1). ∨ n Case 1.2 If 2n 3k 7 = 0, then we can get n 13 because k< , and hence − − ≤ 2 n = 11, k = 5 ,or n = 8, k = 3. The corresponding permissible graphic sequence are (5, 5, 5, 5, 5, 5, 10, 10, 10, 10, 10), (3, 3, 3, 4, 4, 7, 7, 7), respectively. For the degree sequence (5, 5, 5, 5, 5, 5, 10, 10, 10, 10, 10). The five vertices of de- gree 10 must be adjacent to every vertex, so they induce a K5. The remaining six vertices now have degree 5, so they induce a 6K1. Then the graph must be K5 6K1. ∨ By the similar discussion, the degree sequence (3, 3, 3, 4, 4, 7, 7, 7) must be correspond to K3 (K2 + 3K1). ∨ Case 2 Assume that (k 2)(2n 3k 7) < 0, i.e., k 3 and 2n 3k 7<0. In this − − − ≥ − − case, we have 6 2k < n 13. ≤ ≤ Case 2.1 If n = 13, then k 6 and 2n 3k 7>0. ≤ − − Case 2.2 If n = 12, then k 5 and 2n 3k 7>0. ≤ − − Case 2.3 If n = 11, then k 5 and 2n 3k 7 0. ≤ − − ≥ Case 2.4 If n = 10, then k 4 and 2n 3k 7>0. ≤ − − Case 2.5 If n = 8, then k 3 and 2n 3k 7 = 0. ≤ − − The above five cases all contradict to 2n 3k 7<0. − −

3 Case 2.6 If n = 9, then k 4. When k = 4, then d4 4, d5 4 and we have 9 ≤ 9 ≤ ≤ 50 di 52 by (3.1) and (3.2). When di = 50, the degree sequence of G is ≤ i=1 ≤ i=1 (3, 4, 4P, 4, 4, 7, 8, 8, 8) or (4, 4, 4, 4, 4, 6, 8, 8, 8) orP (4, 4, 4, 4, 4, 7, 7, 8, 8), it is easy to see that G = K3 (K1 + K1,4) or G = K3 (K2 + K1,3) or G = (K2 2K1) 5K1; 9 ∨ ∨ ∨ ∨ When di = 52, the degree sequence of G is (4, 4, 4, 4, 4, 8, 8, 8, 8), it is easy to see iP=1 that G = K4 5K1. When k = 3, then 2n 3k 7 = 2>0, contradiction with ∨ − − 2n 3k 7<0. − − Case 2.7 If n = 7, then k 3, and because k 3, i.e., k = 3 . Thus d3 3, d4 3 7 ≤ 7 ≥ ≤ ≤ and 28 di 30. When di = 28, the degree sequence of G is (3, 3, 3, 3, 5, 5, 6) ≤ ≤ iP=1 iP=1 or (3,3,3,3,4,6,6) or (2,3,3,3,5,6,6), it is easy to see that G = K1,2 4K1 or G = 7 ∨ K2 (K2 + K1,2) or G = K2 (K1 + K1,3). When di = 30, the degree sequence of ∨ ∨ iP=1 G is (3, 3, 3, 3, 6, 6, 6), it is easy to see that G = K3 4K1. ∨

Table 1: The maximum length cycle (l(G)) of G G l(G) G l(G)

K2 (Kn−4 +2K1) Cn−1 K3 4K1 C6 ∨ ∨ K1,2 4K1 C6 K3 (K2 +3K1) C7 ∨ ∨ K2 (K1 + K1,3) C6 K3 (K2 + K1,3) C8 ∨ ∨ K2 (K2 + K1,2) C7 K4 5K1 C8 ∨ ∨ K3 (K1 + K1,4) C7 K5 6K1 C10 ∨ ∨ (K2 2K1) 5K1 C8 ∨ ∨

In Table 1, G = K2 (K2 + K1,2) contains cycles of every length l, 3 l 7, ∨ ≤ ≤ namely it is pancyclic graph, a contradiction. The other graphs in Table 1 are neither pancyclic nor bipartite. The proof is complete. Theorem 3.2 Let G be a connected graph on n( 5) vertices with minimum degree ≥ δ(G) 2. If ≥ µ(G) n2 6n + 15, ≥ p − then G is a pancyclic graph unless G is a bipartite graph. Proof: Suppose that G with m edges is neither a pancyclic graph nor a bipartite graph. Because Kn is pancyclic and δ(K1,n−1) = 1. By Lemma 2.2,

n2 6n + 15 µ(G) < √2m n + 1, p − ≤ − then n 2 m> − + 4.  2 

By Theorem 3.1, we get G NP1= K2 (Kn−4 + 2K1), K5 6K1, K3 (K2 + ∈ { ∨ ∨ ∨ 3K1), K3 (K1 +K1,4), K3 (K2 +K1,3), (K2 2K1) 5K1, K4 5K1, K1,2 4K1, K2 ∨ ∨ ∨ ∨ ∨ ∨ ∨ (K1 +K1,3), K3 4K1 . According to calculation, when G K2 (Kn−4 +2K1), K5 ∨ } ∈ { ∨ ∨ 6K1, K3 (K2 + 3K1) , ∨ } n 2 m = − + 4,  2 

4 a contradiction. So, G NP2= K3 (K1 + K1,4), K3 (K2 + K1,3), (K2 2K1) ∈ { ∨ ∨ ∨ ∨ 5K1, K4 5K1, K1,2 4K1, K2 (K1 + K1,3), K3 4K1 . ∨ ∨ ∨ ∨ } T For G = (K2 2K1) 5K1, let X = (X1, X2, , X9) be the eigenvector cor- ∨ ∨ · · · responding to µ(G), where Xi(1 i 5) correspond to the vertex of degree 4, ≤ ≤ Xi(6 i 7) correspond to the vertex of degree 7 and Xi(8 i 9) correspond to ≤ ≤ ≤ ≤ the vertex of degree 8. Then by eigen-equation (2.1), we have

X1 = X2 = = X5, X6 = X7, X8 = X9, · · ·  µ(G)X1 = 2X6 + 2X8,   µ(G)X6 = 5X1 + 2X8, µ(G)X8 = 5X1 + 2X6 + X8.   ′ ′ ′ T Transform above into a matrix equation (A (G) µ(G)I)X = 0, where X =(X1, X6, X8) − and 0 2 2 ′ A (G)=  5 0 2  . 5 2 1   Let f(x) := det(xI A′(G)), then f(x)= x3 x2 24x 30, and µ(G) is the largest root − − − − of f(x) = 0. Through calculation, µ(G) = 5.9150 < √92 6 9 + 15, a contradiction. − × Using the same method, we get the spectral radius of the other graphs in NP2, showing in the following Table 2.

Table 2: The spectral radius of G G µ(G) √n2 6n + 15 G µ(G) √n2 6n + 15 − − K1,2 4K1 4.2182 4.6904 K3 (K2 +K1,3) 5.9612 6.4807 ∨ ∨ K2 (K1 +K1,3) 4.3723 4.6904 K3 4K1 4.6056 4.6904 ∨ ∨ K3 (K1 +K1,4) 6.0322 6.4807 K4 5K1 6.2170 6.4807 ∨ ∨

2 From Table 2, all graphs in NP2 satisfy µ(G)<√n 6n + 15, a contradiction. − The proof is complete. Theorem 3.3 Let G be a connected graph on n( 5) vertices with minimum degree ≥ δ(G) 2. If ≥ 10 q(G) + 2n 6, ≥ n 1 − − then G is a pancyclic graph unless G is a bipartite graph or G = K3 4K1. ∨ Proof: Suppose that G is neither a pancyclic graph nor a bipartite graph. Because Kn is pancyclic and δ(K1,n−1) = 1. By Lemma 2.3 10 2m + 2n 6 q(G) < + n 2, n 1 − ≤ n 1 − − − then n 2 m> − + 4.  2 

By Theorem 3.1, we get G NP1= K2 (Kn−4 + 2K1), K5 6K1, K3 (K2 + ∈ { ∨ ∨ ∨ 3K1), K3 (K1 +K1,4), K3 (K2 +K1,3), (K2 2K1) 5K1, K4 5K1, K1,2 4K1, K2 ∨ ∨ ∨ ∨ ∨ ∨ ∨ (K1 +K1,3), K3 4K1 . Because when G K2 (Kn−4 +2K1), K3 (K2 +K1,3), K5 ∨ } ∈ { ∨ ∨ ∨ 6K1 , } n 2 m = − + 4,  2  a contradiction. So, G NP2= K3 (K1 + K1,4), K3 (K2 + K1,3), (K2 2K1) ∈ { ∨ ∨ ∨ ∨ 5K1, K4 5K1, K1,2 4K1, K2 (K1 + K1,3), K3 4K1 . ∨ ∨ ∨ ∨ }

5 T For (K2 2K1) 5K1, let X =(X1, X2, , X9) be the eigenvector corresponding ∨ ∨ · · · to q, where Xi (1 i 5) correspond to the vertex of degree 4, Xi (6 i 7) ≤ ≤ ≤ ≤ correspond to the vertex of degree 7 and Xi (8 i 9) correspond to the vertex of ≤ ≤ degree 8. Then by signless Laplacian eigen-equation (2.2), we have

X1 = X2 = = X5, X6 = X7, X8 = X9, · · ·  (q(G) 4)X1 = 2X6 + 2X8, −  (q(G) 7)X6 = 5X1 + 2X8,  − (q(G) 8)X8 = 5X1 + 2X6 + X8. −   ′ T Transform above into a matrix equation (Q (G) q(G)I)X˜ = 0, where X˜ =(X1, X6, X8) − and 4 2 2 ′ Q (G)=  5 7 2  . 5 2 9   Let g(x) := det(xI Q′(G)), then g(x)= x3 20x2 +103x 116, and q(G) is the largest − − − root of g(x) = 0. Through calculation, q(G) = 12.5052 < 10 + 2 9 6 = 13.2500 9−1 × − . Using the same method, we get the signless Laplacian spectral radius of the other graphs in NP2, showing in the following Table 3,

Table 3: The signless Laplacian spectral radius of G G q(G) 10 +2n 6 G q(G) 10 +2n 6 n−1 − n−1 − K1,2 4K1 8.8965 9.6667 K3 (K2 +K1,3) 12.6769 13.2500 ∨ ∨ K2 (K1 +K1,3) 9.3408 9.6667 K3 4K1 9.7720 9.6667 ∨ ∨ K3 (K1 +K1,4) 12.8381 13.2500 K4 5K1 13.1789 13.2500 ∨ ∨

10 From Table 3, all graphs in NP2 except G = K3 4K1 satisfy q(G) < +2n 6, ∨ n−1 − a contradiction. The proof is complete.

6 4 Appendix

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