## 1 Introduction

Throughout this paper, *G* denotes a simple connected graph with vertex set *V*(*G*) and edge set *E*(*G*). Let *p* pebbles be distributed onto the vertices of a graph *G*. A pebbling move consists of removing two pebbles from one vertex and then placing one pebble at an adjacent vertex. The pebbling number of a connected graph *G*, denoted by *f*(*G*), is the least *n* such that any distribution of *n* pebbles on *G* allows one pebble to be moved to any specified vertex by a sequence of pebbling moves.

If each vertex (except *v*) has at most one pebble, then no pebble can be moved to *v*. Also, if *u* is of distance *d* from *v* and at most 2^{d} −1 pebbles are placed on *u* (and none elsewhere), then no pebble can be moved from *u* to *v*. So it is clear that *f*(*G*) ⩾ max{∣*V*(*G*)∣, 2^{D}}, where ∣*V*(*G*)∣ is the number of vertices of *G*, and *D* is the diameter of *G*. Furthermore, *f*(*K _{n}*) =

*n*and

*f*(

*P*) = 2

_{n}^{n−1}in [1], where

*K*denotes a complete graph with

_{n}*n*vertices and

*P*denotes a path with

_{n}*n*vertices.

Pebbling of graphs was first introduced by Chung [1]. In [2], Pachter et al. obtained the pebbling number of

*Let**G**be a connected graph*. *For**u*, *v* ∈ *V*(*G*), *we denote by**d _{G}*(

*u*,

*v*)

*the distance between*

*u*

*and*

*v*

*in G*.

*The*

*k*-

*th power of*

*G*,

*denoted by*

*G*,

^{k}*is the graph obtained from*

*G*

*by adding the edge uv to*

*G*

*whenever*2 ⩽

*d*(

_{G}*u*,

*v*) ⩽

*k*

*in*

*G*.

*That is*,

*E*(

*G*) = {

^{k}*uv*: 1 ⩽

*d*(

_{G}*u*,

*v*) ⩽

*k*}.

Obviously, *G ^{k}* is the complete graph whenever

*k*is at least the diameter of

*G*. In particular, let

*P*be a path with

_{n}*n*vertices, written 〈

*v*

_{1},

*v*

_{2}, ⋯,

*v*〉, i.e.,

_{n}*P*= 〈

_{n}*v*

_{1},

*v*

_{2}, ⋯,

*v*〉.

_{n}*P*by joining

_{n}*v*

_{i}*v*

_{i+2}(

*i*= 1, 2, ⋯,

*n*− 2).

*We call a graph**G**satisfies the 2*-*pebbling property if two pebbles can be moved to any specified vertex when the total starting number of pebbles is* 2*f*(*G*) − *q* + 1, *where**q**is the number of vertices with at least one pebble*.

[2]

[5] *Let**P _{n}* = 〈

*v*

_{1},

*v*

_{2}, ⋯,

*v*〉.

_{n}*If*

*p*(

*v*

_{1}) + 2

*p*(

*v*

_{2}) + ⋯ + 2

^{i−1}

*p*(

*v*) + ⋯ + 2

_{i}^{n−2}

*p*(

*v*

_{n−1}) ⩾ 2

^{n−1},

*then one pebble can be moved to*

*v*

_{n}*from*

*v*

_{1}.

[3] *Let**P*_{2k+1} = 〈 *v*_{1}, *v*_{2}, …, *v*_{2k+1}〉. *If**has* 2^{k}*pebbles and**v*_{1}*has an even number of pebbles*, *then a pebble can be moved to**v*_{2k+1}.

Define *p*(*H*) to be the number of pebbles on a subgraph *H* of *G* and *p*(*v*) to be the number of pebbles on a vertex *v* of *G*. Moreover, denote by *p͠*(*H*) and *p͠*(*v*) respectively, the number of pebbles on *H* and the number of pebbles on *v* after a specified sequence of pebbling moves, respectively.

## 2 The 2-pebbling property of $\begin{array}{c}{P}_{n}^{2}\end{array}$

Let *P _{n}* = 〈

*v*

_{1},

*v*

_{2}, ⋯,

*v*〉. In this section, we first will show that

_{n}*has the 2*-*pebbling property*.

Suppose that 2^{k+1} − *q* + 1 pebbles are placed arbitrarily at the vertices of *q* is the number of vertices with at least one pebble. Suppose that our target vertex is *v*. We divide into two cases: *p*(*v*) = 1 *p*(*v*) = 0.

- Case 1
*p*(*v*) = 1. Then has 2$\begin{array}{c}{P}_{2k}^{2}\end{array}$ ^{k+1}−*q*pebbles other than one pebble on*v*. Since 2^{k+1}−*q*≥ 2^{k}, by Lemma 1.1, we can move one additional pebble to*v*so that*p͠*(*v*) = 2. - Case 2
*p*(*v*) = 0. - Subcase 2.1If
*q*= 2*k*− 1, then = 2$\begin{array}{c}p\left({P}_{2k}^{2}\right)\end{array}$ ^{k+1}− 2*k*+ 2. For*k*= 2, it is clearly that we can move 2 pebbles to*v*. For*k*≥ 3, we can move one pebble to*v*using at most*k*+ 1 pebbles. This leaves 2^{k+1}− 3*k*+ 1 ≥ 2^{k}pebbles on . By Lemma 1.1, we may move one additional pebble to$\begin{array}{c}{P}_{2k}^{2}\end{array}$ *v*so that*p͠*(*v*) = 2. - Subcase 2.2New we assume
*q*≤ 2*k*− 2. Without loss of generality, we may assume that our target vertex*v*is not*v*_{2k}(otherwise, relabeling). We claim that*p*(*v*_{2k}) ≥ 2^{k}+ 2^{k−1}−*q*+ 1 > 2^{k}. Otherwise,By Lemma 1.1, we can move one pebble to$\begin{array}{c}p\left({P}_{2k-1}^{2}\right)\ge {2}^{k+1}-q+1-({2}^{k}+{2}^{k-1}-q)={2}^{k-1}+1.\end{array}$ *v*. This leavespebbles on$\begin{array}{c}{2}^{k+1}-q+1-({2}^{k-1}+1)={2}^{k}+{2}^{k-1}-q\ge {2}^{k}\end{array}$ . By Lemma 1.1, we can move one additional pebble to$\begin{array}{c}{P}_{2k}^{2}\end{array}$ *v*so that*p͠*(*v*) = 2. Using 2^{k}pebbles on*v*_{2k}, we can move at least one pebble to*v*by Lemma 1.1. This leaves at least 2^{k}−*q*+ 1 pebbles on other than one pebble on$\begin{array}{c}{P}_{2k}^{2}\end{array}$ *v*. Let =$\begin{array}{c}p\left({P}_{2k-1}^{2}\right)\end{array}$ *s*and*s*≤ 2^{k−1}. Thus*p͠*(*v*_{2k}) = 2^{k}−*q*−*s*+ 1 ≥ 2. We can move pebbles to$\begin{array}{c}{2}^{k-1}-\lceil \frac{q+s-1}{2}\rceil \end{array}$ pebbles from$\begin{array}{c}{P}_{2k-1}^{2}\end{array}$ *v*_{2k}so that has$\begin{array}{c}{P}_{2k-1}^{2}\end{array}$ pebbles other than one pebble on$\begin{array}{c}s+{2}^{k-1}-\left(\frac{q+s-1}{2}\right)={2}^{k-1}+\left(\frac{s-q+1}{2}\right)\ge {2}^{k-1}\phantom{\rule{0ex}{0ex}}(s\ge q-1)\end{array}$ *v*and*v*_{2k−1}has an even number of pebbles. By Lemma 1.3, we can move one additional pebble to*v*so that*p͠*(*v*) = 2.□

*has 2*-*pebbling property*.

Suppose that 2^{k+1} −*q* + 3 pebbles are placed arbitrarily at the vertices of *v*. If *p*(*v*) = 1, then ^{k+1} − *q* + 2 pebbles other than one pebble on *v*. Since 2^{k+1} − *q* + 2 ≥ 2^{k} + 1, by Lemma 1.1, we can move one additional pebble to *v* so that *p͠*(*v*) = 2. Hence we assume *p*(*v*) = 0.

If *q* = 2*k*, then ^{k+1} − 2*k* +3. For *k* = 2, it is clearly that we can move 2 pebbles to *v*. For *k* ≥ 3, we can move one pebble to *v* using at most *k* + 1 pebbles. This leaves 2^{k+1} − 3*k* + 2 ≥ 2^{k} + 1 pebbles on *q* ≤ 2*k* − 1. Without loss of generality, we may assume that our target vertex *v* is not *v*_{2k} or *v*_{2k+1} (otherwise, relabeling). We claim that *p*(*v*_{2k}) + *p*(*v*_{2k+1}) ≥ 2^{k} + 2^{k−1} −*q* + 3 > 2^{k} + 1. Otherwise,

By Lemma 1.1, we can move one pebble to *v*. This leaves 2^{k+1} −*q* + 3 − (2^{k−1} + 1) ≥ 2^{k} + 1 pebbles on *v* so that *p͠*(*v*) = 2. Using 2^{k} + 1 pebbles on the set {*v*_{2k}, *v*_{2k+1}}, we can move at least one pebble to *v* by Lemma 1.1. This leaves at least 2^{k} − *q* + 2 pebbles on *v*. Let *s* and *s* ≤ 2^{k−1}. Thus *p͠*(*v*_{2k}) + *p͠*(*v*_{2k+1}) = 2^{k} − *q* − *s* + 3 ≥ 3. We can move *v*_{2k} and *v*_{2k+1} so that

pebbles other than one pebble on *v*. By Lemma 1.1, we can move one additional pebble to *v* so that *p͠*(*v*) = 2.□

## 3 Graham’s conjecture on $\begin{array}{c}{P}_{n}^{2}\end{array}$

Given two graphs *G* and *H*, the Cartesian product of them is denoted by *G* × *H*. Its vertex set

and (*u*, *v*) is adjacent to (*u*′, *v*′) if and only if *u* = *u*′ and *vv*′ ∈ *E*(*H*) or *v* = *v*′ and *uu*′ ∈ *E*(*G*).

We can depict *G* × *H* pictorially by drawing a copy of *H* at every vertex of *G* and joining each vertex in one copy of *H* to the corresponding vertex in an adjacent copy of *H*. We write *u*(*H*) (respectively, *v*(*G*)) for the subgraph of vertices whose projection onto *V*(*G*) is the vertex *u*, (respectively, whose projection onto *V*(*H*) is the vertex *v*). Obviously, *u*(*H*) ≅ *H*, *v*(*G*) ≅ *G*.

**Graham’s conjecture***or any two graphs**G**and**H*, *we have*

[5] *Let**P _{n}*

*be a path with*

*n*

*vertices and*

*G*

*be a graph with the 2*-

*pebbling property*.

*Then*

*f*(

*P*×

_{n}*G*) ≤ 2

^{n−1}

*f*(

*G*).

*Let**P*_{2k} = 〈 *v*_{1}, *v*_{2}, ⋯, *v*_{2k}〉 (*k* ≥ 2) *and**p*(*v _{i}*) =

*j*(1 ≤

_{i}*i*≤ 2

*k*).

*If one of the following cases holds*

*j*_{1}+*j*_{2}+ 1 = 2^{k}*and**j*_{2}> 0*;**j*_{1}+*j*_{2}+ 2 = 2^{k},*j*_{2}> 0*and**p͠*(*v*_{1}) =*j*_{1}+ 1*;**j*_{1}+*j*_{2}+ 2 = 2^{k},*j*_{2}> 0*and**p͠*(*v*_{2}) =*j*_{2}+ 1;*j*_{1}+*j*_{2}+ 3 = 2^{k},*p͠*(*v*_{1}) =*j*_{1}+ 1*and**p͠*(*v*_{2}) =*j*_{2}+ 1,*then we can move one pebble to the vertex**v**of* ,$\begin{array}{c}{P}_{2k}^{2}\end{array}$ *where**v*∈ {*v*_{3},*v*_{4}, ⋯,*v*_{2k}}.

Let *A* = 〈 *v*_{2}, *v*_{3}, *v*_{4}, ⋯, *v*_{2k}〉. Then *A*^{2} ≅

- If
*j*_{1}is odd and*j*_{2}is even, then we can move pebbles to$\begin{array}{c}\frac{{j}_{1}-1}{2}\end{array}$ *v*_{3}from*v*_{1}. Thus we haveBy Lemma 1.3, we are done. If$\begin{array}{c}\stackrel{~}{p}\left({A}^{2}\right)=\frac{{j}_{1}-1}{2}+{j}_{2}=\frac{({j}_{1}+{j}_{2}+1)+({j}_{2}-2)}{2}\ge {2}^{k-1}.\end{array}$ *j*_{1}is even and*j*_{2}is odd, then we can move pebbles to$\begin{array}{c}\frac{{j}_{1}-2}{2}\end{array}$ *v*_{3}and one pebble to*v*_{2}from*v*_{1}so that*p͠*(*v*_{2}) =*j*_{2}+ 1 andBy Lemma 1.3, we are done.$\begin{array}{c}\stackrel{~}{p}\left({A}^{2}\right)=\frac{{j}_{1}-2}{2}+{j}_{2}+1=\frac{({j}_{1}+{j}_{2}+1)+({j}_{2}-1)}{2}\ge {2}^{k-1}.\end{array}$ - Let
*j*_{1}and*j*_{2}be odd. If*p͠*(*v*_{1}) =*j*_{1}+ 1, then we can move one pebble to*v*_{2}and pebbles to$\begin{array}{c}\frac{{j}_{1}-1}{2}\end{array}$ *v*_{3}from*v*_{1}so that*p͠*(*v*_{2}) =*j*_{2}+ 1 andBy Lemma 1.3, we are done. For even numbers$\begin{array}{c}\stackrel{~}{p}\left({A}^{2}\right)={j}_{2}+1+\frac{{j}_{1}-1}{2}=\frac{({j}_{1}+{j}_{2}+2)+({j}_{2}-1)}{2}\ge {2}^{k-1}.\end{array}$ *j*_{1}and*j*_{2}, we can move pebbles to$\begin{array}{c}\frac{{j}_{1}}{2}\end{array}$ *v*_{3}from*v*_{1}so thatBy Lemma 1.3, we are done.$\begin{array}{c}\stackrel{~}{p}\left({A}^{2}\right)={j}_{2}+\frac{{j}_{1}}{2}=\frac{({j}_{1}+{j}_{2}+2)+({j}_{2}-2)}{2}\ge {2}^{k-1}.\end{array}$ - The proof is similar to (2).
- If
*j*_{1}is even and*j*_{2}is odd, then we can move pebbles to$\begin{array}{c}\frac{{j}_{1}}{2}\end{array}$ *v*_{3}from*v*_{1}so thatAnd$\begin{array}{c}\stackrel{~}{p}\left({A}^{2}\right)={j}_{2}+1+\frac{{j}_{1}}{2}=\frac{({j}_{1}+{j}_{2}+3)+({j}_{2}-1)}{2}\ge {2}^{k-1}.\end{array}$ *p͠*(*v*_{2}) =*j*_{2}+ 1, we are done by Lemma 1.3. If*j*_{1}is odd and*j*_{2}is even, then we can move one pebble to*v*_{2}and pebbles to$\begin{array}{c}\frac{{j}_{1}-1}{2}\end{array}$ *v*_{3}from*v*_{1}so that*p͠*(*v*_{2}) =*j*_{2}+ 2 andBy Lemma 1.3, we are done.□$\begin{array}{c}\stackrel{~}{p}\left({A}^{2}\right)={j}_{2}+2+\frac{{j}_{1}-1}{2}=\frac{({j}_{1}+{j}_{2}+3)+{j}_{2}}{2}\ge {2}^{k-1}.\end{array}$

Let *G* be a graph with the 2-pebbling property. Then *f*^{k}*f*(*G*).

For *k* = 1, *P*_{2}. By Lemma 3.1, the conclusion is right. We assume that the inequality *f*^{k}*f*(*G*) is right for *k* > 1. Next, we shall determine the above bound of the pebbling number of *G*.

Suppose that 2^{k+1}*f*(*G*) pebbles have been distributed arbitrarily on the vertices of *G*. Let *p _{i}* =

*p*(

*v*(

_{i}*G*)), where 1 ≤

*i*≤ 2

*k*+ 2. Let

*q*be the number of occupied vertices in

_{i}*v*(

_{i}*G*)(1 ≤

*i*≤ 2

*k*+ 2). Suppose that

*v*is our target vertex in

*G*. Suppose that

*v*= (

*v*,

_{i}*x*) ∈

*v*(

_{i}*G*) (3 ≤

*i*≤ 2

*k*+ 2) for

*x*∈

*V*(

*G*) (relabeling if necessary). For simplicity, let

*A*= 〈

*v*

_{3},

*v*

_{4}, …,

*v*

_{2k+1},

*v*

_{2k+2}〉. By the induction hypothesis,

*f*(

*A*

^{2}×

*G*) ≤ 2

^{k}

*f*(

*G*). We shall divide into the following two cases.

- Case 1
*p*_{1}+*p*_{2}≤ (2^{k+1}− 2)*f*(*G*). Then*A*^{2}×*G*can obtain at least + 2$\begin{array}{c}\frac{1}{2}\left(({2}^{k+1}-2)f\left(G\right)-2f\left(G\right)\right)\end{array}$ *f*(*G*) = 2^{k}*f*(*G*) pebbles by a sequence of pebbling moves and we are done. - Case 2
*p*_{1}+*p*_{2}≥ (2^{k+1}− 2)*f*(*G*) + 1. Thus*p*(*A*^{2}×*G*) < 2*f*(*G*). Let*p*(*A*^{2}×*G*) = (*j*_{0}+*α*_{0})*f*(*G*), where 0 ≤*j*_{0}≤ 1 and 0 ≤*α*_{0}< 1, and let*p*= (_{i}*j*+_{i}*α*)_{i}*f*(*G*), where*j*≥ 0 and 0 ≤_{i}*α*< 1 (_{i}*i*= 1, 2). Now we claim that$\begin{array}{c}\sum _{i=1}^{2}\end{array}$ *q*> (_{i}*j*_{0}+*α*_{0})*f*(*G*). Otherwise, we have$\begin{array}{c}\sum _{i=1}^{2}\end{array}$ *q*≤ (_{i}*j*_{0}+*α*_{0})*f*(*G*). So*p͠*(*A*^{2}×*G*) ≥ + ($\begin{array}{c}\frac{1}{2}\left({2}^{k+1}f\left(G\right)-({j}_{0}+{\alpha}_{0})f\left(G\right)-({j}_{0}+{\alpha}_{0})f\left(G\right)\right)\end{array}$ *j*_{0}+*α*_{0})*f*(*G*) = 2^{k}*f*(*G*) and we are done. Let$\begin{array}{c}\sum _{i=0}^{2}\end{array}$ *α*=_{i}*s*≤ 2. Then$\begin{array}{c}\sum _{i=0}^{2}\end{array}$ *j*= 2_{i}^{k+1}−*s*. Next, we discuss two following cases. Let*B*= 〈(*v*_{2},*x*), (*v*_{3},*x*), ⋯, (*v*_{2k+2},*x*)〉. Then*B*^{2}≅ . We first prove the following claim.$\begin{array}{c}{P}_{2k+1}^{2}\end{array}$

**Claim**: Suppose *j*_{2} = 0. If one of the following cases holds:(1) *j*_{0} = 0, *s* = 1; (2) *j*_{0} = 0, *s* = 2; (3) *j*_{0} = 1, *s* = 1, then we can move one pebble to any vertex of *B*^{2}.

In fact, since *j*_{2} = 0, *p*_{1} = (2^{k+1} − *j*_{0} − *s* + *α*_{1})*f*(*G*). We move (2^{k} − 2)*f*(*G*) pebbles to *v*_{3}(*G*) so that *p͠*_{3} ≥ (2^{k} − 2)*f*(*G*) using (2^{k+1} − 4)*f*(*G*) pebbles on *v*_{1}(*G*), and (1 − *α*_{2})*f*(*G*) to *v*_{2}(*G*) so that *p͠*_{2} = *f*(*G*) using (2−2*α*_{2})*f*(*G*) pebbles on *v*_{1}(*G*). So we can move one pebble to (*v*_{2}, *x*) and (2^{k} − 2) pebbles to (*v*_{3}, *x*) so that *p͠*(*B*^{2}) = 2^{k} − 1. And the remaining (2 − *j*_{0} − *s* + *α*_{1} + 2*α*_{2})*f*(*G*) pebbles on *v*_{1}(*G*), i.e., *p͠*_{1} = (2 − *j*_{0} − *s* + *α*_{1} + 2*α*_{2})*f*(*G*).

*p͠*_{1}= (1 +*α*_{1}+ 2*α*_{2})*f*(*G*). If*α*_{0}≤*α*_{2}, then*p͠*_{1}≥ (1 +*α*_{0}+*α*_{1}+*α*_{2})*f*(*G*) = (1+*s*)*f*(*G*) = 2*f*(*G*). we move two pebbles to (*v*_{1},*x*), and move one pebble to (*v*_{2},*x*) so that*p͠*((*v*_{2},*x*)) = 2. By Lemma 1.3, we are done. If*α*_{0}>*α*_{2}, note that*q*_{1}+*q*_{2}>*α*_{0}*f*(*G*) and*q*_{2}≤*α*_{2}*f*(*G*), then*q*_{1}> (*α*_{0}−*α*_{2})*f*(*G*). Thus*p͠*_{1}+*q*_{1}> (1 +*α*_{0}+*α*_{1}+*α*_{2})*f*(*G*) = 2*f*(*G*). By the 2-pebbling property, we move two pebbles to (*v*_{1},*x*), and move one pebble to (*v*_{2},*x*) so that*p͠*((*v*_{2},*x*)) = 2 and*p͠*(*B*^{2}) = 2^{k}. By Lemma 1.3, we are done.*p͠*_{1}= (*α*_{1}+ 2*α*_{2})*f*(*G*). If*α*_{0}≤*α*_{2}, then*p͠*_{1}≥ (*α*_{0}+*α*_{1}+*α*_{2})*f*(*G*) =*sf*(*G*) = 2*f*(*G*). If*α*_{0}≥*α*_{2}, then*q*_{1}> (*α*_{0}−*α*_{2})*f*(*G*). Thus*p͠*_{1}+*q*_{1}> (*α*_{0}+*α*_{1}+*α*_{2})*f*(*G*) = 2*f*(*G*). By (1), we are done.*p͠*_{1}= (*α*_{1}+ 2*α*_{2})*f*(*G*). Note that*q*_{1}+*q*_{2}> (1 +*α*_{0})*f*(*G*) and*q*_{2}≤*α*_{2}*f*(*G*), so*q*_{1}> (1 +*α*_{0}−*α*_{2})*f*(*G*). Thus*p͠*_{1}+*q*_{1}> (1 +*α*_{0}+*α*_{1}+*α*_{2})*f*(*G*) = (1 +*s*)*f*(*G*) = 2*f*(*G*). By (1), we are done.

Next, we continue to complete the proof of the theorem with the following two cases.

- Case 1
*α*_{0}= 0. Obviously, 0 ≤*s*≤ 1.Suppose that*j*_{0}= 0,*p*_{1}+*p*_{2}= 2^{k+1}*f*(*G*). By Lemma 3.1, we can move 2^{k}pebbles to the vertex (*v*_{2},*x*) of*v*_{2}(*G*) so that*p͠*(*B*^{2}) = 2^{k}. By Lemma 1.3, we are done. Suppose*j*_{0}= 1,*p*_{1}+*p*_{2}= 2^{k+1}*f*(*G*) −*f*(*G*). If*s*= 0, then*p*_{1}+*p*_{2}= (*j*_{1}+*j*_{2})*f*(*G*). This implies that*j*_{1}+*j*_{2}+ 1 = 2^{k+1}. For*j*_{2}= 0, we can move at least (2^{k}− 1)*f*(*G*) pebbles to*A*^{2}×*G*from*v*_{1}(*G*) so thatFor$\begin{array}{c}\stackrel{~}{p}({A}^{2}\times G)={2}^{k}f\left(G\right).\end{array}$ *j*_{2}> 0, note that we can move at least*j*pebbles to the vertex (_{i}*v*,_{i}*x*) of*v*(_{i}*G*) (*i*= 1, 2). By Lemma 3.2, we are done. If*s*= 1, then*p*_{1}+*p*_{2}= (*j*_{1}+*j*_{2}+ 1)*f*(*G*). This implies that*j*_{1}+*j*_{2}+ 2 = 2^{k+1}. For*j*_{2}= 0,*p*_{1}= (2^{k+1}− 2 +*α*_{1})*f*(*G*) pebbles. By Claim, we are done. For*j*_{2}> 0, we claim that at least one of (*α*_{1}*f*(*G*) +*q*_{1}) and (*α*_{2}*f*(*G*) +*q*_{2}) is greater than*f*(*G*). For suppose not, we have ($\begin{array}{c}\sum _{i=1}^{2}\end{array}$ *α*_{i}*f*(*G*) +*q*) ≤ 2_{i}*f*(*G*). ButThis is a contradiction. Suppose$\begin{array}{c}\sum _{i=1}^{2}\left({\alpha}_{i}f\right(G)+{q}_{i})=\sum _{i=1}^{2}{\alpha}_{i}f\left(G\right)+\sum _{i=1}^{2}{q}_{i}>\sum _{i=1}^{2}{\alpha}_{i}f\left(G\right)+f\left(G\right)=(s+1)f\left(G\right)=2f\left(G\right).\end{array}$ *α*_{1}*f*(*G*) +*q*_{1}>*f*(*G*). By the 2-pebbling property we can move at least*j*_{1}+ 1 pebbles to the vertex (*v*_{1},*x*) of*v*_{1}(*G*). Note that we can move at least*j*_{2}pebbles to the vertex (*v*_{2},*x*) of*v*_{2}(*G*). By Lemma 3.2, we are done. Suppose*α*_{2}*f*(*G*) +*q*_{2}>*f*(*G*). Then we can move at least*j*_{2}+ 1 pebbles to the vertex (*v*_{2},*x*) of*v*_{2}(*G*). Note that we can move at least*j*_{1}pebbles to the vertex (*v*_{1},*x*) of*v*_{1}(*G*). By Lemma 3.2, we are done. - Case 20 <
*α*_{0}< 1. Obviously, 1 ≤*s*≤ 2.Suppose that*j*_{0}= 0,*p*_{1}+*p*_{2}= 2^{k+1}*f*(*G*)-*α*_{0}*f*(*G*). If*s*= 1, then*α*_{1}+*α*_{2}= 1 −*α*_{0}. Thus*p*_{1}+*p*_{2}= (*j*_{1}+*j*_{2}+ 1)*f*(*G*) −*α*_{0}*f*(*G*). This implies that*j*_{1}+*j*_{2}+ 1 = 2^{k+1}. For*j*_{2}= 0,*p*_{1}= (2^{k+1}− 1 +*α*_{1})*f*(*G*). By Claim, we are done. For*j*_{2}> 0, note that we can move at least*j*pebbles to the vertex (_{i}*v*,_{i}*x*) of*v*(_{i}*G*) (*i*= 1, 2). By Lemma 3.2, we are done. If*s*= 2, then*α*_{1}+*α*_{2}= 2 −*α*_{0}. Thus*p*_{1}+*p*_{2}= (*j*_{1}+*j*_{2}+ 2)*f*(*G*) −*α*_{0}*f*(*G*). This implies that*j*_{1}+*j*_{2}+ 2 = 2^{k+1}. For*j*_{2}= 0, then*p*_{1}= (2^{k+1}− 2 +*α*_{1})*f*(*G*). By Claim, we are done. For*j*_{2}> 0, now we claim that at least one of (*α*_{1}*f*(*G*) +*q*_{1}) and (*α*_{2}*f*(*G*) +*q*_{2}) is greater than*f*(*G*). For suppose not, we have ($\begin{array}{c}\sum _{i=1}^{2}\end{array}$ *α*_{i}*f*(*G*) +*q*) ≤ 2_{i}*f*(*G*). ButThis is a contradiction. By Case 1, we are done.Suppose that$\begin{array}{c}\sum _{i=1}^{2}\left({\alpha}_{i}f\right(G)+{q}_{i})=\sum _{i=1}^{2}{\alpha}_{i}f\left(G\right)+\sum _{i=1}^{2}{q}_{i}>\sum _{i=1}^{2}{\alpha}_{i}f\left(G\right)+{\alpha}_{0}f\left(G\right)=({\alpha}_{1}+{\alpha}_{2}+{\alpha}_{0})f\left(G\right)=2f\left(G\right).\end{array}$ *j*_{0}= 1,*p*_{1}+*p*_{2}= 2^{k+1}*f*(*G*)−(1 +*α*_{0})*f*(*G*) = (2^{k+1}− 1)*f*(*G*) −*α*_{0}*f*(*G*). If*s*= 1, then*α*_{1}+*α*_{2}= 1 −*α*_{0}. ThusThis implies that$\begin{array}{c}{p}_{1}+{p}_{2}=({j}_{1}+{j}_{2}+1)f\left(G\right)-{\alpha}_{0}f\left(G\right).\end{array}$ *j*_{1}+*j*_{2}+ 2 = 2^{k+1}. For*j*_{2}= 0,*p*_{1}= (2^{k+1}− 2 +*α*_{1})*f*(*G*). By Claim, we are done. For*j*_{2}> 0, we claim that at least one of (*α*_{1}*f*(*G*) +*q*_{1}) and (*α*_{2}*f*(*G*) +*q*_{2}) is greater than*f*(*G*). For suppose not, we have ($\begin{array}{c}\sum _{i=1}^{2}\end{array}$ *α*_{i}*f*(*G*) +*q*) ≤ 2_{i}*f*(*G*). ButThis is a contradiction. By Case 1, we are done.If$\begin{array}{c}\sum _{i=1}^{2}\left({\alpha}_{i}f\right(G)+{q}_{i})=\sum _{i=1}^{2}{\alpha}_{i}f\left(G\right)+\sum _{i=1}^{2}{q}_{i}>\sum _{i=1}^{2}{\alpha}_{i}f\left(G\right)+(1+{\alpha}_{0})f\left(G\right)=(1+{\alpha}_{1}+{\alpha}_{2}+{\alpha}_{0})f\left(G\right)=2f\left(G\right).\end{array}$ *s*= 2, then*α*_{1}+*α*_{2}= 2 −*α*_{0}. Thus*p*_{1}+*p*_{2}= (*j*_{1}+*j*_{2}+ 2)*f*(*G*) −*α*_{0}*f*(*G*). This implies that*j*_{1}+*j*_{2}+ 3 = 2^{k+1}. Note that*α*_{i}*f*(*G*) +*q*< 2_{i}*f*(*G*) for 1 ≤*i*≤ 2. We claim that*α*_{2}*f*(*G*) +*q*_{2}>*f*(*G*) and*α*_{1}*f*(*G*) +*q*_{1}>*f*(*G*). For suppose not, we have ($\begin{array}{c}\sum _{i=1}^{2}\end{array}$ *α*_{i}*f*(*G*) +*q*) ≤ 3_{i}*f*(*G*). ButThis is a contradiction. By the 2-pebbling property we can move at least$\begin{array}{c}\sum _{i=1}^{2}\left({\alpha}_{i}f\right(G)+{q}_{i})=\sum _{i=1}^{2}{\alpha}_{i}f\left(G\right)+\sum _{i=1}^{2}{q}_{i}>\sum _{i=1}^{2}{\alpha}_{i}f\left(G\right)+(1+{\alpha}_{0})f\left(G\right)=(1+{\alpha}_{1}+{\alpha}_{2}+{\alpha}_{0})f\left(G\right)=3f\left(G\right).\end{array}$ *j*_{1}+ 1 pebbles to the vertex (*v*_{1},*x*) of*v*_{1}(*G*) and at least*j*_{2}+ 1 pebbles to the vertex (*v*_{2},*x*) of*v*_{2}(*G*) By Lemma 3.2, we are done.□By Theorems 2.1, 2.2 and 3.3, we have the following

Supported by the Natural Science Foundation of Anhui Higher Education Institutions of China (KJ2016A633, KJ2016B001) and the Quality Engineering Foundation of Huaibei Normal University (2018jpkc04).

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