#### Abstract

It has been conjectured by Erdős, Faudree, Rousseau, and Schelp that for all satisfying that (except . In this paper, we prove this for the case and .

#### 1. Introduction

All graphs considered in this paper are undirected and simple. and stand for cycle, path, complete, and star graphs on vertices, respectively. The graph is obtained by adding an additional vertex to the path and connecting this new vertex to each vertex of . The number of edges in a graph is denoted by . Further, the minimum degree of a graph is denoted by . An independent set of vertices of a graph is a subset of the vertex set in which no two vertices are adjacent. The independence number of a graph , is the size of the largest independent set. The neighborhood of the vertex is the set of all vertices of that are adjacent to , denoted by . denote to . For vertex-disjoint subgraphs and of we let . Let be a subgraph of the graph and , is defined as . Suppose that and is nonempty, the subgraph of whose vertex set is and whose edge set is the set of those edges of that have both ends in is called the subgraph of induced by , denoted by .

The cycle-complete graph Ramsey number is the smallest integer such that for every graph of order , contains or . The graph shows that . In one of the earliest contributions to graphical Ramsey theory, Bondy and Erdős [1] proved the following result: for all . The above restriction was improved by Nikiforov [2] when he proved the equality for . Erdős et al. [3] gave the following conjecture.

*Conjecture 1. *, for all except .

The conjecture was confirmed by Faudree and Schepl [4] and Rosta [5] for in early work on Ramsey theory. Yang et al. [6] and Bollobás et al. [7] proved the conjecture for and , respectively. The conjecture was proved by Schiermeyer [8] for . Jaradat and Baniabedalruhman [9, 10] proved the conjecture for and . Later on, Chena et. al. [11] proved the conjecture for . Recently, Jaradat and Al-Zaleq [12] and Y. Zhang and K. Zhang [13], independently, proved the conjecture in the case . In a related work, Radziszowski and Tse [14] showed that and . In [15] Jayawardene and Rousseau proved that . Also, Schiermeyer [16] proved that . For more results regarding the Ramsey numbers, see the dynamic survey [17] by Radziszowski.

Until now, the conjecture is still open. Researchers are interested in determining all the values of the Ramsey number . In this paper our main purpose is to determine the values of which confirm the conjecture in the case and . The follwoing known theorem will be used in the sequel.

Theorem 1.1. *Let be a graph of order without a path of length . Then
**
Further, equality holds if and only if its components are complete graphs of order .*

#### 2. Main Result

In this paper we confirm the Erdős, Faudree, Rousseau, and Schelp conjecture in the case and . In fact, we prove that . It is known, by taking , that . In this section we prove that this bound is exact in the case and . Our proof depends on a sequence of 8 lemmas.

Lemma 2.1. *Let be a graph of order that contains neither nor an 8-elemant independent set. Then .*

*Proof. * Suppose that contains a vertex of degree less than 8, say . Then . Since , as a result has independent set consists of 7 vertices. This set with the vertex is an 8-elemant independent set of vertices of . That is a contradiction.

Throughout all Lemmas 2.2 to 2.8, we let be a graph with minimum degree that contains neither nor an 8-elemant independent set.

Lemma 2.2. *If contains , then .*

*Proof. * Let be the vertex set of , Let and for each . Since for all . Since there is a path of order 8 joining any two vertices of , as a result for all (otherwise, if for some , then the concatenation of the -path of order 8 with , is a cycle of order 9, a contradiction). Similarly, since there is a path of order 7 joining any two vertices of , as a result for all and for all and we have that (otherwise, if there are such that , and , then the concatenation of the -path of order 7 with , is a cycle of order 9, a contradiction). Also, since there is a path of order 6 joining any two vertices of , as a result, , (otherwise, if there are such that , then the concatenation of the -path of order 6 with , is a cycle of order 9 where , and , a contradiction). Therefore . Thus, .

Lemma 2.3. *If contains , then contains .*

*Proof. *Let be the vertex set of where the induced subgraph of is isomorphic to . Without loss of generality we may assume that . Let and for each . Then, as in Lemma 2.2, we have the following:(1) for all except possibly for and .(2) for all .(3) for all .(4) for all .

Since , as a result at least five of the induced subgraphs , are complete. Since , it implies that these complete graphs contain . Hence, contains .

Lemma 2.4. *If contains , then contains or .*

*Proof. *Let be the vertex set of . Let and for each . Since , for all . Now we consider the following two cases.*Case 1. *. for some , say . Then it is clear that contains . In fact, the induced subgraph contains .*Case 2. *. for all . Note that between any two vertices of there are paths of order 5,6 and 7. Thus, as in Lemma 2.2, for all , we have the following.(1). (2). (3).

Since , we have that the induced subgraphs , are complete. Since , it implies that these complete graphs contain . Hence, contains .

Lemma 2.5. *If contains , then contains .*

*Proof. * Let be the vertex set of where and . Let and for each . Now we have the following two cases.*Case 1. *. Since , for all . Now we have the following.(1)for all except possibly for since otherwise a cycle of order 9 is produced, a contradiction.(2) for all .(3) for all .(4) for all .

((2), (3), and (4) follows easily from being that contains paths of order 7, 6, and 5 between any two vertices and , ). Since , as a result at least three of the induced subgraphs , are complete graphs. Now we have the following two assertions.

and so for each . The following is the proof of assertion for .

Since , . Let and is adjacent to . Then we have the following.(i)If , then is a , this is a contradiction.(ii)If , then is a , this is a contradiction.(iii)If , then is a , this is a contradiction.(iv)If , then is a , this is a contradiction.(v)If , then is a , this is a contradiction.(vi)If , then is a , this is a contradiction.(vii)If , then is a , this is a contradiction.

Since , , and so . Hence, . By a similar argument as above and using the symmetry of , one can easily show that .

If there is such that , then and so for any with . The following is the proof of assertion (ii).

Assume that . By (1) for all except possibly . Thus, for , is adjacent to and to at most and . Now we show that . Assume . By (1) for all except possibly . Thus, for any , is adjacent to and to at most , and . Now, we have the following.(A)If adjacent to both and , then is a .(B)If adjacent to both and , then is a .(C)If adjacent to both and , then is a .

Thus, is adjacent to at most one of , and , and so . We now show that . As above assume . By (1), for all except possibly . Thus, for , is adjacent to , and . Hence, is a , which implies that is adjacent to at most one of and and so .

Now, by using the same argument as above and taking into account that is symmetry, we can easily see that if , then both of and are greater than or equal 6. So we need to consider the case when . As above, for all except possibly and 7. Thus, for any , is adjacent to and to at most and . Now, assume that . By using (A), (B) and (C) as above and using the same arguments to get the same contradiction. Similarly, by symmetry we get that .

Therefore, from *(i)* and *(ii),* at least four of the induced subgraphs , contain 7 vertices and so at least two of them contain . Thus, contains .*Case 2. *, say . For simplicity, in the rest of this case we consider where and let . Then (otherwise, contains ) and . Hence . Now we have the following assertions (see the Appendix).(1) for all and .(2) for all and .(3) for all and .(4) for all and .

Since , as a result at least five of the induced subgraphs , are complete graphs. Since and contains no , and so for each . Therefore at least three of the induced subgraphs , contain . Thus, contains .

Lemma 2.6. *If contains , then contains or .*

*Proof. *Let be the vertex set of where and . Let and for each . Since for all . Now we have the following cases.*Case 1. * for all . Then we have the following.(1) for all except possibly for .(2) for all .(3) for all .

Since , as a result at least one of the induced subgraphs , is complete. Since , it implies that this complete graph contains .*Case 2. * for some , say . In the rest of this case we have the following subcases:*Subcase 2.1. *. For simplicity, in the rest of this subcase we consider where and let . Since , then , for all . Now we have the following assertions.(1) for all with .(2) for all with .(3) for all with .(4) for all with .

Since , as a result at least one of the induced subgraphs is complete.Since and for each (because otherwise contains ), it implies that this complete graph contains .*Subcase 2.2. *. Then, by the symmetry, we have a subcase similar to Subcase 2.1.

Lemma 2.7. *If contains , then contains or .*

*Proof. * Let be the vertex set of . Let and for each . Since , for all . Now we split our work into the following two cases.*Case 1. *There are such that , then contains .*Case 2. * for all . Then we consider the following subcases.*Subcase 2.1. * for all . Since between any two vertices of there are paths of order 5 and 6, as a result and for each . Therefore, since , at least five of the induced subgraphs , are complete graphs. Since , these complete graphs contain . Thus, contains .*Subcase 2.2. * for some , say and and is a path. For simplicity, in the rest of this subcase we consider where . Since , then , for all . Now we have the following.(1) for all .(2) for all .(3) for all .(4) for all .

Therefore, since , at least five of the induced subgraphs , are complete graphs. Since , it implies that these complete graphs contain . Thus, contains .

Lemma 2.8. *If be a graph of order , then contains or .*

*Proof. * Suppose that contains neither nor . Then we have the following claims.*Claim 1. * for any .*Proof. * Suppose that is a vertex with . Let where is a component for each . has minimum number of independent vertices if it has a maximum number of edges. Thus, by Theorem 1.1 must be a complete graph for each . But contains no . Thus, must be a complete graph of order at most 5. Also contains no , thus must be a complete graph of order at most 4. Hence, the minimum number of independent vertices of occurs only if contains either a 7 tetrahedrons and an isolated vertex or 6 tetrahedron, a triangle and a or 6 tetrahedrons and 2 triangles. In any of these cases . This is a contradiction. The proof of the claim is complete.*Claim 2. *. *Proof. *Since and contains no and since ,. But has no 8-element independent set, so . Thus, . The proof of the claim is complete.

Now, for any 7 independent vertices , and , set . Analogously, we set , . Let , , and .*Claim 3. *. *Proof. * Suppose that . Then . And so . But , so contains a vertex, say , which is not adjacent to any of , and . Thus, is an 8-element independent set. Therefore, . That is a contradiction. The proof of the claim is complete.

Now, by Lemma 2.1, and so by Claim 1, we have that . Thus, if **, **then . By a similar argument as in Claim 1, we have that and . Note that for any is greater than or equal to 8. And so . Now we have the following cases.

*Case 1. *, then by a similar argument as in Claim 1, we have that and . Then, has an independent set which consists of 7 vertices. This set with the vertex is an 8-element independent set of vertices of . That is a contradiction.

*Case 2. *, then . By a similar argument as in Claim 1, we have that and . Now we have the following two subcases.*Subcase 2.1. *. Then we have a subcase similar to Case 1.*Subcase 2.2. *. The best case of such subgraph is the graph that shown in Figure 1. Now we have the following two subcases.*Subcase 2.2.1*. There is a vertex of , say , that is not adjacent to at least one vertex of each , say . Then is an 8-elemant independent set of vertices of . And so . That is a contradiction.*Subcase 2.2.2.* For each vertex of there is such that this vertex is adjacent to all vertices of . Then contains or . That is a contradiction.

*Case 3. *, Then by using the same argument as in Case 2, we have the same contradiction.

Theorem 2.9. *. **Proof. *Suppose that there exists a graph of order 57 that contains neither nor an 8-elements independent set. Then by Lemma 2.1, and by Lemma 2.8, contains or . Thus, by Lemmas 2.7, 2.6, 2.5, 2.4, 2.3, and 2.2, we have that . That is a contradiction. The proof is complete.