#### Abstract

A pebbling move on a graph consists of taking two pebbles off one vertex and placing one pebble on an adjacent vertex. The pebbling number of a connected graph , denoted by , is the least such that any distribution of pebbles on allows one pebble to be moved to any specified but arbitrary vertex by a sequence of pebbling moves. This paper determines the pebbling numbers and the 2-pebbling property of the middle graph of fan graphs.

#### 1. Introduction

Pebbling on graphs was first introduced by Chung . Consider a connected graph with a fixed number of pebbles distributed on its vertices. A pebbling move consists of the removal of two pebbles from a vertex and the placement of one of those pebbles on an adjacent vertex. The pebbling number of a vertex in a graph is the smallest number with the property that from every placement of pebbles on , it is possible to move a pebble to by a sequence of pebbling moves. The pebbling number of a graph , denoted by , is the maximum of over all the vertices of .

In a graph , if each vertex (except ) has at most one pebble, then no pebble can be moved to . Also, if is of distance from and at most pebbles are placed on (and none elsewhere), then no pebble can be moved from to . So it is clear that , where is the number of vertices of and is the diameter of .

Throughout this paper, let be a simple connected graph with vertex set and edge set . For a distribution of pebbles on , denote by and the number of pebbles on a subgraph of and the number of pebbles on a vertex of , respectively. In addition, denote by and the number of pebbles on and the number of pebbles on after a specified sequence of pebbling moves, respectively. For , refers to taking pebbles off and placing pebbles on . Denote by the path with vertices in order.

We now introduce some definitions and give some lemmas, which will be used in subsequent proofs.

Definition 1. A fan graph, denoted by , is a path plus an extra vertex connected to all vertices of the path , where .

Definition 2. The middle graph of a graph is the graph obtained from by inserting a new vertex into every edge of and by joining by edges those pairs of these new vertices which lie on adjacent edges of .

Now one creates the middle graph of . Edges of are the inserted new vertices in the sequence, and edges of are the inserted new vertices , respectively. By joining by edges those pairs of these inserted vertices which lie on adjacent edges of , this obtains the middle graph of (see Figure 1).

Definition 3. A transmitting subgraph is a path such that there are at least two pebbles on , and after a sequence of pebbling moves, one can transmit a pebble from to .

Lemma 4 (see ). Let . If then is a transmitting subgraph.

Definition 5. The -pebbling number, , of a connected graph, , is the smallest positive integer such that from every placement of pebbles, pebbles can be moved to a specified target vertex by a sequence of pebbling moves.

Lemma 6 (see ). If is the complete graph with    vertices, then .

Lemma 7 (see ). Consider .

Chung found the pebbling numbers of the -cube , the complete graph , and the path (see ). The pebbling number of was determined in . In [6, 7], Ye et al. gave the number of squares of cycles. Feng and Kim proved that and (see ). Liu et al. determined the pebbling numbers of middle graphs of , , and (see ). In , Ye et al. proved that and , where denotes the middle graph of . Motivated by these works, we will determine the value of the pebbling number and the 2-property of middle graphs of .

#### 2. Pebbling Numbers of

In this section, we study the pebbling number of . Let , and let . Obviously, the subgraph induced by is a complete graph with vertices. For , . Hence we have the following theorem.

Theorem 8 (see ). Consider .

Lemma 9. Let . If pebbles are placed on , then one pebble can be moved to any specified vertex of by a sequence of pebbling moves.

Proof. Let be our target vertex, and let , where . We may assume that (after relabeling if necessary). Let . If , then by Lemma 6, and we can move one pebble to . If , then . We delete to obtain the subgraph with pebbles, thus we can move one pebble to . If , then we delete to obtain the subgraph with at least pebbles and we are done.

Theorem 10. Consider .

Proof. We place 7 pebbles on and one pebble on each vertex of the set , other vertices have no pebble, then no pebble can be moved to . So . We now place 11 pebbles on . We assume that is our target vertex and . Recall and .(1)Consider . By Theorem 8 and Lemma 9, we can move one pebble to .(2)Consider (or ). Let , let , and let . If , then . Thus we can move at least pebbles from to so that for . By Lemma 6, and we can move one pebble to . If , then . By Lemma 7, we can move one pebble to . If , then at least one of and can obtain one pebble from every placement of 3 pebbles on by a sequence of pebbling moves. If , then . So is a transmitting subgraph. If , then . By Lemma 6, and . So is a transmitting subgraph. If , then . So is a transmitting subgraph.(3)Consider . If or , then we are done with (2). If , then or . So or is a transmitting subgraph.(4)Consider (or ). If or , then we are done with (2). If , then . Obviously, we are done if or . Next suppose that and . Thus . So is a transmitting subgraph.

Theorem 11. Consider .

Proof. We place 7 pebbles on and one pebble on each vertex of except , and . In this configuration of pebbles, we cannot move one pebble to . So . Next, let us use induction on to show that . For , our theorem is true by Theorem 10. Suppose that if . Now pebbles are placed arbitrarily on the vertices of . Suppose that is our target vertex and .
(1) Consider . By induction and Theorem 8, we can move one pebble to .
(2) Consider (or ). Obviously, . Otherwise, . can obtain one pebble. Let   .
If , then we delete to obtain the subgraph with at least pebbles. By induction, we can move one pebble to . If , then . Thus we delete to obtain the subgraph with pebbles. By induction, we are done.
Next, suppose that . By Lemma 6, . If , then is a transmitting subgraph. If , then , and we are done. If there exists some with   , then , and we are done. Thus we assume that , , and for .
Now, we consider (). Clearly, if , then we are done. Suppose that there exists some with (). It is clear that if one of the three cases ((i) (), (ii) , and (iii) ()) happens, then we can move one pebble to . Thus we assume that (), , , and . If there are sets such that for , then for . Let , let , and let . If for all and for all , then and . Recall that for all and . Then and . Thus is a transmitting subgraph. So there is at least some in such that or at least some in such that . If there are two and in such that and or two and in such that and or some in such that and some in such that , then . By Lemma 6, . Hence is a transmitting subgraph.
Finally, there are two remaining cases, (i) there is only some in such that , and (ii) there is only some in such that . So . If , then is a transmitting subgraph. If , then, in , and . For (i), we have for . By the transmitting subgraph , and we are done. Suppose that (ii) holds. If , then we can move one pebble from to so that , and we are done. If , then and we are done.
(3) Consider (or ). Obviously, and   . Otherwise, one pebble can be moved to . The proof is similar to (2).
(4) Consider (or ) and . Let , and let . If , then delete to obtain the subgraph with at least pebbles. By induction, we can move one pebble to . If , then we can move one pebble from to , after deleting to obtain the subgraph with pebbles. Hence we assume that . According to symmetry, . Therefore we are done.

#### 3. The 2-Pebbling Property of

For a distribution of pebbles on , let be the number of vertices with at least one pebble. We say a graph satisfies the 2-pebbling property if two pebbles can be moved to any specified vertex when the total starting number of pebbles is . Next we will discuss the 2-pebbling property of . For the convenience of statement, let , and let , where , , , and . In this section let , where and are the number of vertices with at least one pebble in and , respectively.

Lemma 12. Suppose that and . If    and   , then one can move pebbles to .

Proof. Let (or ). Since and , so . Without loss of generality, there exists a positive integer    such that the corresponding vertex with and for . Thus . Using the remaining pebbles on vertices , we can move at least pebbles to so that . By Lemma 6, . So we can move one additional pebble from to so that .

Lemma 13. Suppose that and . If    and   , then one can move pebbles to .

Proof. Let   (or ). Since , we see that there is only some vertex    with . Without loss of generality, there exists a positive integer    such that the corresponding vertex with and for . If    or , then we can move one pebble to by the transmitting subgraph or . Now using the remaining at least pebbles on the set , we can move pebbles from the to so that . By Lemma 6, and we can move one additional pebble from to so that .
Suppose that    and . If , then . Thus there must exist one vertex    with and for . Using the transmitting subgraph or , we can move one pebble to . Now, using the remaining pebbles on the set , we can move pebbles from the set to so that . By Lemma 6, and we are done. Next, suppose that .
(1) Consider . We divide into three subcases.
(1.1) Consider . We delete vertices to obtain the subgraph with two sets and , and and . Thus we can move pebbles from to so that . By Lemma 6, and we can move two pebbles from to .
(1.2) Consider . Suppose that or (). Let . Obviously, . If , then We delete to obtain the subgraph with two sets and . So and . We can move pebbles from to so that . By Lemma 6, and we are done. If and , then We delete to obtain the subgraph with two sets and . So and . We can move pebbles from to so that . By Lemma 6, and we are done. If and , then . We delete to obtain the subgraph with two sets and . So and . We can move pebbles from to so that . By Lemma 6, .
(1.3) Consider for . Thus . We delete to obtain the subgraph with two sets and . So and . pebbles can be moved from to so that . By Lemma 6, . So we can move one additional pebble from to .
(2) Consider ; that is, . We divide into three subcases.
(2.1) Consider . We delete to obtain the subgraph with two sets and . One has and . We can move pebbles from to so that . By Lemma 6, and we can move two pebbles from to .
(2.2) Consider . We have We delete vertices to obtain the subgraph with two sets and . So and (except one moved pebble on ). We can move pebbles from to so that (except one moved pebble on ). By Lemma 6, we can move 3 additional pebbles to so that .
(2.3) for . Thus . Deleting to obtain the subgraph with two sets and . One has and (except one moved pebble on ). We can move pebbles from to so that (except one moved pebble on ). By Lemma 6, we can move 3 additional pebbles to so that .

Theorem 14. has the 2-pebbling property.

Proof. Suppose that is our target vertex. If , then the number of pebbles on except one pebble on is   . By Theorem 11, we can move one additional pebble to so that . Next we assume that .
(1) Consider (). If there exists some vertex with   , then . Using the remaining pebbles, we can move one additional pebble to so that . If for , then . Thus we can move at least pebbles from to so that . By Lemma 6, we can move two pebbles to .
(2) Consider (). Let (or ). If , then we can put one pebble on . After that, the remaining    pebbles on suffice to put one additional pebble on by Theorem 11. Next we assume .
(2.1) Suppose that . If there is some vertex with   , then . The remaining    pebbles on will suffice to put one additional pebble on so that . Next we assume that for . Obviously, and . If , then . Thus we can move at least pebbles from to so that . By Lemma 6, we can move 3 additional pebbles to so that and we are done. If , then, by Lemma 12, we are done.
(2.2) Suppose that . If we can find some vertex with or find two vertices with and with , then using 4 pebbles on or two pebbles on and two pebbles on we can move one pebble to . Then the remaining    pebbles on will suffice to put one additional pebble to so that .
Consider only some vertex with . If , then , , and . When and , we can move at least pebbles from to so that except for one pebble on . By Lemma 6, we can put 3 additional pebbles on so that . When , we are done with Lemma 12. If , then , , and . When and , we can move at least pebbles from to so that except for one pebble on . By Lemma 6, we can put 3 additional pebbles on so that . When and , we are done with Lemmas 12 and 13.
Consider for . Obviously, and . When and , we can move at least pebbles from to so that . By Lemma 6, and we are done. When and , we are done with Lemmas 12 and 13.

#### Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

#### Acknowledgments

This work is supported by National Natural Science Foundation of China (no. 10971248) and Anhui Provincial Natural Science Foundation (nos. 1408085MA08 and KJ2013Z279).