Turán s problem and Ramsey numbers for trees. Zhi-Hong Sun 1, Lin-Lin Wang 2 and Yi-Li Wu 3

Transcription

1 Colloquium Mathematicum 139(015, no, Turán s problem and Ramsey numbers for trees Zhi-Hong Sun 1, Lin-Lin Wang and Yi-Li Wu 3 1 School of Mathematical Sciences, Huaiyin Normal University, Huaian, Jiangsu 3001, PR China zhihongsun@yahoocom Homepage: School of Science, China University of Mining and Technology, Xuzhou, Jiangsu 1116, PR China wanglinlin 1986@aliyuncom 3 School of Mathematical Sciences, Huaiyin Normal University, Huaian, Jiangsu 3001, PR China yiliwu@16com Abstract Let T 1 n (V, E 1 and T n (V, E be the trees on n vertices with V {v 0, v 1,, v n 1, E 1 {v 0 v 1,, v 0 v n 3, v n 4 v n, v n 3 v n 1, and E {v 0 v 1,, v 0 v n 3, v n 3 v n, v n 3 v n 1 In this paper, for p n 5 we obtain explicit formulas for ex(p; T 1 n and ex(p; T n, where ex(p; L denotes the maximal number of edges in a graph of order p not containing L as a subgraph Let r(g 1, G be the Ramsey number of the two graphs G 1 and G In this paper we also obtain some explicit formulas for r(t m, T i n, where i {1, and T m is a tree on m vertices with (T m m Mathematics Subject Classification: 05C55, 05C35, 05C05 Key words and phrases: Ramsey number, tree, Turán s problem 1 Introduction In this paper, all graphs are simple graphs For a graph G (V (G, E(G let e(g E(G be the number of edges in G and let (G be the maximal degree of G For a forbidden graph L, let ex(p; L denote the maximal number of edges in a graph of order p not containing any copies of L The corresponding Turán problem is to evaluate ex(p; L For a graph G of order p, if G does not contain any copies of L and e(g ex(p; L, we say that G is an extremal graph In this paper we also use Ex(p; L to denote the set of extremal graphs of order p not containing L as a subgraph Let N be the set of positive integers Let p, n N with p n For a given tree T n on n vertices, it is difficult to determine the value of ex(p; T n The famous Erdős-Sós conjecture asserts that ex(p; T n (n p For the progress on the Erdős-Sós conjecture, see for example [8, 11 Write p k(n 1 r, where k N and r {0, 1,, n Let P n be the path on n vertices In [4 Faudree and Schelp showed that ( ( n 1 r (11 ex(p; P n k 1 (n p r(n 1 r

2 Let K 1,n 1 denote the unique tree on n vertices with (K 1,n 1 n 1, and let T n denote the unique tree on n vertices with (T n n For n 4 let T n (V, E be the tree on n vertices with V {v 0, v 1,, v n 1 and E {v 0 v 1,, v 0 v n 3, v n 3 v n, v n v n 1 In [10 we determine ex(p; K 1,n 1, ex(p; T n and ex(p; T n For i 1, let T i n (V, E i be the tree on n vertices with V {v 0, v 1,, v n 1, E 1 {v 0 v 1,, v 0 v n 3, v n 4 v n, v n 3 v n 1, E {v 0 v 1,, v 0 v n 3, v n 3 v n, v n 3 v n 1 In this paper, for p n 5 we obtain explicit formulas for ex(p; Tn 1 and ex(p; Tn (see Theorems 1 and 31 For a graph G, as usual G denotes the complement of G Let G 1 and G be two graphs The Ramsey number r(g 1, G is the smallest positive integer p such that, for every graph G with p vertices, either G contains a copy of G 1 or else G contains a copy of G Let n N, n 6 and let T n be a tree on n vertices As mentioned in [7, recently Zhao proved the following conjecture of Burr and Erdős [: r(t n, T n n Let m, n N In 1973 Burr and Roberts [3 showed that for m, n 3, { m n 3 if mn, (1 r(k 1,m 1, K 1,n 1 m n if mn In 1995, Guo and Volkmann [5 proved that for n > m 4, { m n 3 if m(n 1, (13 r(k 1,m 1, T n m n 4 if m(n 1 Recently the first author evaluated the Ramsey number r(t m, Tn for T m {P m, K 1,m 1, T m, Tm In particular, he proved that (see [9 for n > m 7, { m n 3 if m 1 n 3, (14 r(k 1,m 1, Tn m n 4 if m 1 n 3 Suppose m, n N and i, j {1, In this paper, using the formula for ex(p; T i n and the method in [9 we evaluate r(t m, T i n for T m {K 1,m 1, T m, T m, T j m In particular, we have the following typical results: r(t i n, T j n n 6 (1 ( 1 n /, r(p n, T j n n 7 for n 17, r(t i n, T n r(t i n, T n n 5 for n 8, r(k 1,m 1, T i n m n 4 for n > m 7 and mn, r(t i m, T j n m n 5 for m 7, n (m 3 3 and m 1 n 4, and for n > m 16, m n 4 if m 1 n 4, r(t m, Tn i m n 6 if n m 1 1 (mod, m n 5 otherwise

3 In addition to the notation introduced above, throughout the paper we also use the following symbols: [x is the greatest integer not exceeding x, d(v is the degree of the vertex v in a graph, Γ(v is the set of vertices adjacent to the vertex v, d(u, v is the distance between the two vertices u and v in a graph, K n is the complete graph on n vertices, G[V 0 is the subgraph of G induced by vertices in the set V 0 (we write G[v 1,, v m instead of G[{v 1,, v m, G V 0 is the subgraph of G obtained by deleting vertices in V 0 and all edges incident to them, and finally e(v 1 V 1 is the number of edges with one endpoint in V 1 and another endpoint in V 1 Evaluation of ex(p; T 1 n Lemma 1 Let p, n N with p n 1 1 Then ex(p; K 1,n 1 [ (n p This is a known result See for example [10, Theorem 1 Lemma Let p, n N, p n 7 and G Ex(p; Tn 1 Suppose that G is connected Then (G n 4 and e(g [ (n 4p Proof Since a graph not containing K 1,n 3 as a subgraph implies that the graph does not contain Tn 1 as a subgraph, by Lemma 1 we have [ (n 4p (1 e(g ex(p; Tn 1 ex(p; K 1,n 3 If (G n 5, using Euler s theorem we see that e(g 1 (n 5p v V (G d(v This together with (1 yields (n 4p 1 [ (n 4p e(g (n 5p This is impossible Hence (G n 4 Now we show that (G n 4 Suppose q n and q k(n 1 r with k N and r {0, 1,, n Then clearly kk n 1 K r does not contain any copies of Tn 1 and so ex(q; Tn 1 e(kk n 1 K r For q n we see that e(kk n 1 K r e(k n 1 K 1 (n 1(n > n 1 For q n 1 we see that (n 6q (n 6(n 1 > ( n 1 (n q r(n 1 r (n q ( n 1 > q 1 Hence and so e(kk n 1 K r k(n 1(n ( ex(q; T 1 n e(kk n 1 K r > q 1 for q n r(r 1 Suppose v 0 V (G, d(v 0 (G m and Γ(v 0 {v 1,, v m If m p 1, as G does not contain T 1 n as a subgraph, we see that G[v 1,, v m does not contain K as a subgraph and hence e(g[v 1,, v m m 1 Therefore (3 e(g d(v 0 e(g[v 1,, v m m m 1 p 3 By (, we have e(g ex(p; T 1 n > p 1 and we get a contradiction Hence m < p 1 Suppose that u 1,, u t are all vertices in G such that d(u 1, v 0 d(u t, v 0 Then t 1 Assume u 1 v 1 E(G with no loss of generality If m p, then V (G {v 0, v 1,, v m, u 1 and v i v j E(G for i < j m If v 1 v i E(G for some i {, 3,, m, then u 1 v j E(G for all j 1, i Hence ex(p; T 1 n e(g max{m, m 3 m p 4, which contradicts ( By the above, m < p We first assume m n As G does not contain any copies of T 1 n, we see that {v,, v m is an independent set, u i v j E(G for any i {, 3,, t and j {, 3,, m, and u i v 1 E(G for any i 1,,, t Set V 1 {v 0, v, v 3,, v m Then 3

4 e(g[v 1 m 1 If u 1 is adjacent to at least two vertices in {v, v 3,, v m, then v 1 v j / E(G for any j, 3,, m If v 1 is adjacent to at least two vertices in {v, v 3,, v m, then u 1 v j / E(G for any j, 3,, m Hence there are at most m edges with one endpoint in V 1 and another endpoint in G V 1 Therefore, (4 e(g e(g[v 1 m e(g V 1 m 1 e(g V 1 For m {n, n 1 let G 1 K m Then clearly e(g 1 m(m 1 > m 1 For m k(n 1 r n with k N and 0 r n let G 1 kk n 1 K r Then G 1 does not contain any copies of T 1 n and e(g 1 > m 1 by ( Thus, by (4 we have e(g m 1 e(g V 1 < e(g 1 (G V 1 for m n This contradicts the fact G Ex(p; T 1 n Suppose m n 3 and d(v 1 n 3 Then v 1 v s E(G for some s {, 3,, n 3 We claim that V (G {v 0, v 1,, v m, u 1,, u t Otherwise, there exists w V (G such that d(v 0, w 3 As d(v 1 n 3, we see that the subgraph induced by {v 1, v s, w Γ(v 1 contains a copy of T 1 n This contradicts the assumption G Ex(p; T 1 n Hence the claim is true and so V (G p n t Since p n we have t For i 1,,, t and j, 3,, n 3 we have u i v j E(G, u i v 1 E(G and so t1 d(v 1 n 3 Therefore t n 4 and hence e(g e(g[v 0, v, v 3,, v n 3 d(v 1 e(g[u 1,, u t ( ( ( ( n 3 t n t n 3 Clearly K n 1 K t 1 does not contain Tn 1 and ( ( ( n 1 t 1 n e(k n 1 K t 1 ( t n 1 t > e(g This contradicts the assumption G Ex(n t; T 1 n Now suppose m n 3 and d(v 1 n 4 If t 1, setting V {v 0, v 1,, v n 3, u 1 we see that e(g e(g[v 0, v, v 3,, v n 3 d(v 1 d(u 1 1 e(g V ( n 3 n 4 n 4 e(g V n 3n 4 e(g V < e(k n 1 (G V This contradicts the assumption G Ex(p; Tn 1 Hence t For i 1,,, t and j, 3,, n 3 we see that u i v j E(G and u i v 1 E(G Let V 3 {v 0, v 1,, v n 3 Then e(g d(v 1 e(g[v 0, v, v 3,, v n 3 e(g V 3 ( n 3 n 4 e(g V 3 n 5n 4 e(g V 3 < e(k n (G V 3 Since G is an extremal graph, we get a contradiction 4

5 Summarizing all the above we obtain (G n 4 and so e(g (n 4p v V (G d(v This together with (1 yields e(g [ (n 4p, which completes the proof Lemma 3 Let n, n 1, n N with n 1 < n 1 and n < n 1 Then ( ( n1 n { ( ( ( n 1 n n 1 n1 n n 1 < min, Proof It is clear that ( ( n1 n (n 1 n (n 1 n 1 n 1 n and (n ( ( ( 1 n1 n n 1 n1 n (n 1(n (n 1 n n 1(n 1 n n (n 1 n 1 (n 1 n > 0 ( n1 n < (n 1 n (n 1 n 1 n 1 n Thus the lemma is proved Lemma 4 Suppose that p N, p 6, and G is a connected graph of order p that does not contain any copies of T6 1 Then e(g p 3 Proof Clearly (T6 1 3 Suppose v 0 V (G, d(v 0 (G m and Γ(v 0 {v 1,, v m If (G m 3, using Euler s theorem we see that e(g 3p p 3 From now on we assume (G m 4 If d(v for all v V (G {v 0, then e(g 1 v V (G d(v 1 ( m (p 1 3(p 1 < p 3 So the result is true Now we assume d(v 3 for some v V (G {v 0 We may choose a vertex u 0 V (G so that u 0 v 0, d(u 0 3 and d(u 0, v 0 is as small as possible We first assume d(u 0, v 0 1 and u 0 v 1 with no loss of generality That is, d(v 1 3 Suppose Γ(v 1 {v 0, v 1,, v m Since d(v 1 3 and G does not contain any copies of T6 1, we see that V (G {v 0,, v m, m p 1 5 and G[v 1,, v m does not contain any copies of K Thus e(g d(v 0 m 1 m 1 (m 1 3 p 3 Now assume Γ(v 1 {v 0, v 1,, v m {w 1,, w t Since d(v 0 m 5, d(v 1 3 and G does not contain any copies of T6 1, we see that V (G {v 0, v 1,, v m, w 1,, w t and {v,, v m is an independent set For t, we have e(g[w 1,, w t 1 and v i w j / E(G for any i {, 3,, m and j {1,,, t Thus e(g d(v 0 d(v 1 11 m < (m1t 3 p 3 Now assume t 1 Then v 1 v i E(G for some i {, 3,, m and v j w 1 / E(G for j {, 3,, m {i Hence e(g d(v 0 d(v 1 11 m < (m 3 p 3 Next we assume d(u 0, v 0 Then {v 1,, v m is an independent set If Γ(u 0 {v 1,, v m, then V (G {v 0,, v m, u 0 and so e(g d(v 0 d(u 0 mm < (m 3 p 3 If Γ(u 0 {v,, v m {v 1, w 1,, w t, we see that V (G {v 0, v 1,, v m, u 0, w 1,, w t and so e(g d(v 0 d(u 0 e(g[w 1,, w t m m1 < (m t 3 p 3 Finally we assume d(u 0, v 0 3 Suppose that v 0 v 1 u 1 u u k u 0 is the shortest path in G between v 0 and u 0, and Γ(u 0 {w 1,, w t, u k Since G is connected and G does not contain any copies of T6 1, it is easily seen that V (G {v 0, v 1,, v m, u 1,, u k, u 0, w 1,, w t, 5

6 d(v d(v m 1, d(v 1 d(u 1 d(u k and e(g[w 1,, w t 1 Clearly G is a tree or a graph obtained by adding an edge to a tree Hence e(g p < p 3 Summarizing all the above proves the lemma Theorem 1 Suppose p, n N, p n 1 4 and p k(n 1 r, where k N and r {0, 1,, n Then {[ (n p ex(p; Tn 1 max [ (n 1 r, (n p r(n 1 r (n p (n 1 r if n 16 and 3 r n 6 or if 13 n 15 and 4 r n 7, (n p r(n 1 r otherwise Proof Clearly ex(n 1; Tn 1 e(k n 1 (n (n 1 Thus the result is true for p n 1 From now on we assume p n Since T5 1 P 5, by (11 we obtain the result in the case n 5 Now we assume n 6 Suppose G Ex(p; Tn 1 and G 1,, G t are all components of G with V (G i p i and p 1 p p t Then clearly G i Ex(p i ; Tn 1 for i 1,,, t We first consider the case n 6 If p i 5, then clearly G i Kpi and e(g i ( p i If p i 6 and p i 5k i r i with k i N and 0 r i 4, from Lemma 4 we have e(g i p i 3 p i r i(5 r i e(k i K 5 K ri Since k i K 5 K ri does not contain any copies of T6 1 and G i Ex(p i ; T6 1, we see that e(g i e(k i K 5 K ri and so e(g i e(k i K 5 K ri Therefore, there is a graph G Ex(p; T6 1 such that G a 1 K 1 a K a 3 K 3 a 4 K 4 a 5 K 5, where a 1,, a 5 are nonnegative integers If a 1 a a 3 a 4 1, then ex(p; T6 1 e(g e(a 5 K 5 K r k ( ( 5 r If a1 a a 3 a 4 > 1, then a 1 3a 3a 3 a 4 > 3 r(5 r and so e(a 1 K 1 a K a 3 K 3 a 4 K 4 a 3a 3 6a 4 < (a 1 a 3a 3 4a 4 r(5 r ( 5 (k a 5 ( r Thus, ex(p; T6 1 e(g e(a 1 K 1 a K a 3 K 3 a 4 K 4 e(a 5 K 5 < k ( ( 5 r Since kk 5 K r does not contain any copies of T6 1, we get a contradiction Thus ex(p; T 6 1 e(kk 5 K r k ( ( 5 r p r(5 r This proves the result for n 6 From now on we assume n 7 If t 1, then G is connected Thus, by Lemma we have [ (n 4p (5 e(g for t 1 Now we assume t We claim that p i n 1 for i Otherwise, p 1 p < n 1 and so G 1 G Kp1 K p If p 1 p < n, by Lemma 3 we have e(g 1 G e(k p1 K p ( p 1 ( p ( < p1 p e(kp1 p Since K p1 p does not contain Tn 1 and G 1 G Ex(p 1 p ; Tn 1 we get a contradiction Hence p 1 p n Using Lemma 3 again we see that ( ( p1 p e(g 1 G e(k p1 K p ( ( n 1 p1 p n 1 < e(k n 1 K p1 p n1 6

7 Since p 1 p < n 1, we have p 1 p n 1 < n 1 Therefore K n 1 K p1 p n1 does not contain Tn 1 As G 1 G is an extremal graph without Tn, 1 we also get a contradiction Thus, the claim is true Next we claim that p i n 1 for all i 1,,, t 1 If p t 1 n, by Lemma we have [ (n 4pt 1 e(g t 1 G t e(g t 1 e(g t [ (n 4pt [ (n 4(pt 1 p t Let H Ex(p t 1 p t n 1; K 1,n 3 As p t 1 p t n 1 p t 1 n 1, we have e(h [ (n 4(p t 1p t n1 by Lemma 1 Clearly K n 1 H does not contain any copies of Tn 1 and ( [ n 1 (n 4(pt 1 p t n 1 e(k n 1 H e(k n 1 e(h [ (n 4(pt 1 p t n 1 > e(g t 1 G t Since G t 1 G t Ex(p t 1 p t ; T 1 n, we get a contradiction Hence p 1 p p t 1 n 1 Combining this with the previous assertion that p t p n 1 we obtain (6 p 1 n 1, p p t 1 n 1 and p t n 1 As G is an extremal graph, we must have (7 G 1 Kp1, G Kn 1,, G t 1 Kn 1 If p t n 1, then G t Kn 1 By (7, G K p1 (t 1K n 1 kkn 1 K r Thus, ( ( n 1 r (n p r(n 1 r (8 e(g k for t and p t n 1 Now we assume p t n By Lemma, e(g t [ (n 4pt Since p 1 n 1, we have G 1 Kp1 and so e(g 1 e(k p1 ( p 1 Let H1 Ex(p 1 p t ; K 1,n 3 Then H 1 does not contain Tn 1 as a subgraph By Lemma 1, for p 1 n 4 we have [ [ [ (n 4(p1 p t (n 4pt (n 4p1 e(h 1 [ (n 4pt (n 4(p [ (n 4pt > p 1(p 1 1 e(g 1 G t This contradicts G 1 G t Ex(p 1 p t ; Tn 1 Hence n 3 p 1 n 1 For p 1 {n 3, n and p t n, we have p 1 (p 1 (n 3 n 4 and so ( [ p1 (n 4pt e(g 1 G t e(g 1 e(g t p 1(p 1 1 (n 4p t p 1(p 1 (n 3 (n 4(p 1 p t 7

8 n 4 (n 4(p ( 1 p t n 1 ( n 1 [ (n 4(p1 p t n 1 < (n 4(p 1 p t n 1 Let H Ex(p 1 p t n 1; K 1,n 3 Then K n 1 H does not contain any copies of Tn 1 Since p 1 p t n1 p 1 1 n, applying Lemma 1 we have e(h [ (n 4(p 1p t n1 Thus, we have e(k n 1 H ( n 1 [ (n 4(p 1 p t n1 > e(g 1 G t This contradicts G 1 G t Ex(p 1 p t ; Tn 1 By the above, for t and p t n we have p 1 p p t 1 n 1 If p t n, setting H 3 Ex(p t (n 1; K 1,n 3 and then applying Lemmas 1 and we find that [ ( [ (n 4pt n 1 (n 4(pt (n 1 e(g t < e(k n 1 H 3 This contradicts the fact G t Ex(p t ; Tn 1 Hence n p t < n and so r 1 Note that p k(n 1 r (k 1(n 1 n 1 r and n n 1 r < n Hence t k, p t n 1 r and therefore ( [ n 1 (n 4(n 1 r e(g e((k 1K n 1 e(g t (k 1 (9 [ (n p (n 1 r for t and p t n Since G Ex(p; Tn, 1 by comparing (5, (8 and (9 we get { [ [ (n 4p (n p r(n 1 r (n p e(g max,, (n 1 r Observe that p k(n 1 r n 1 r We see that [ (n 4p [ (n p p [ (n p (n 1 r and therefore { [ (n p r(n 1 r (n p ex(p; Tn 1 e(g max, (n 1 r (10 { (n p r(n 1 r [ r(n 3 r (n 1 max 0, For 7 n 1 we have r(n 3 r (n 1 (n 3 4 (n 1 (n < 0 For r {0, 1,, n 5, n 4, n 3, n we see that r(n 3 r (n 1 < 0 Suppose n 13 and 3 r n 6 For 4 r n 7 we have r n 3 n 11 and so r(n 3 r (n 1 n 14n 17 4 n 14n 17 4 ( r n 3 ( n 11 n 6 0 For r {3, n 6 we have r(n 3 r (n 1 3(n 6 (n 1 n 16 Now combining the above with (10 we deduce the result 8

9 Corollary 1 Suppose p, n N, p n 5 and n 1 p Then (n p (n 1 8 ex(p; Tn 1 (n (p 1 Proof Suppose p k(n 1 r with k N and r {0, 1,, n Then r 1 Clearly (n 1 4 r(n 1 r ( n 1 ( n 1 r ( n 1 ( n 1 1 n and n 1 r > n Thus, from Theorem 1 we deduce that ex(p; T 1 n (n p (n and ex(p; T 1 n (n p r(n 1 r (n p (n 1 /4 This proves the corollary 3 Evaluation of ex(p; T n Lemma 31 Let p, n N, p n 7 and G Ex(p; T n Suppose that G is connected Then (G n 3 Moreover, for p < n we have (G n 4 Proof Since a graph does not contain K 1,n 3 implies that the graph does not contain T n, by Lemma 1 we have [ (n 4p (31 e(g ex(p; Tn ex(p; K 1,n 3 Suppose that v 0 V (G, d(v 0 (G m and Γ(v 0 {v 1,, v m If V (G {v 0, v 1,, v m, then m p 1 n 1 Since G does not contain Tn, we see that G[v 1,, v m does not contain K 1, and hence e(g[v 1,, v m m Therefore e(g d(v 0 e(g[v 1,, v m m m 3(p 1 (n 4p 3 < [ (n 4p This contradicts (31 Thus p > m 1 Suppose that u 1,, u t are all vertices such that d(u 1, v 0 d(u t, v 0 Then t 1 We may assume without loss of generality that v 1,, v s are all vertices in Γ(v 0 adjacent to some vertex in the set {u 1,, u t Then 1 s m Let V 1 {v 0, v 1,, v m, V 1 V (G V 1 and let e(v 1 V 1 be the number of edges with one endpoint in V 1 and another endpoint in V 1 Since G does not contain T n, for m n 3 each v i (1 i s has one and only one adjacent vertex in the set {u 1,, u t Thus, for m n 3 we must have e(v 1 V 1 s t If m n 1, since G does not contain Tn as a subgraph, we see that d(v i for i 1,, m and so e(g[v 1 d(v 0 e(g[v s1,, v m m m s Hence e(g e(g[v 1 e(v 1 V 1 e(g V 1 3m s s e(g V 1 m e(g V 1 Suppose m 1 k(n 1 r with k N and 0 r n Set G 1 kk n 1 K r Since m 1 n, by ( we have e(g 1 > (m 1 1 > m Thus, e(g 1 (G V 1 e(g 1 e(g V 1 > m e(g V 1 e(g As G 1 does not contain any copies of T n and G is an extremal graph, we get a contradiction Hence (G m n Suppose m n As G does not contain T n as a subgraph, we see that d(v 1 9

10 d(v s and so e(g[v 1 n ( n s Since 1 s m n n 8, we have e(g e(g[v 1 e(v 1 V 1 e(g V 1 ( n s n s e(g V 1 (n (n 1 s(n 7 s e(g V 1 ( n 1 < e(g V 1 e(k n 1 (G V 1 This is impossible since G is an extremal graph By the above, (G n 3 We first assume (G n 3 We claim that d(v i n 4 for i 1,,, s If i {1,,, s and d(v i n 3, let u j be the unique adjacent vertex of v i in {u 1,, u t and let V {v 0, v 1,, v n 3, u j Then there is at most one vertex adjacent to u j in G V Hence e(g V 1 1 e(g V Since each v r (1 r s is adjacent to one and only one vertex in {u 1,, u t and (G[V 1 n 3, we see that e(g[v 1 1 n 3 d G[V1 (v r r0 s(n 4 (n s(n 3 Note that s (G n 3 From the above we deduce that (n (n 3 s e(g e(g[v 1 e(v 1 V 1 e(g V 1 e(g[v 1 s e(g V 1 (n (n 3 s e(g[v 1 s 1 e(g V s 1 e(g V (n (n 3 s (n (n 3 n 1 e(g V e(g V (n 1(n < e(g V e(k n 1 (G V Since K n 1 (G V does not contain T n and G is an extremal graph, we get a contradiction Hence the claim is true Thus, for (G n 3 we have d G[V1 (v i n 5 for i 1,,, s and so (3 e(g[v 1 1 n 3 d G[V1 (v i i0 s(n 5 (n s(n 3 (n (n 3 Now we assume p < n and p n 1 r Then 1 r < n 1 By the above, (G n 3 Assume (G n 3 Then V (G V 1 p (n r 1 < n, (G V 1 n 3 and so e(g V 1 min{ ( r1, (r1(n 3 Since e(g[v 1 (n (n 3 s by (3, we deduce that e(g e(g[v 1 e(v 1 V 1 e(g V 1 (n (n 3 { r(r 1 (r 1(n 3 s s min, ( (n (n 3 r 1 if r n 3 (n (n 3 (n 3(n 1 if r n ( ( n 1 r < e(k n 1 K r 10 s

11 This is impossible since G is an extremal graph Thus, (G n 4 for p < n Now the proof is complete Lemma 3 Let p, n N, p n 7 and G Ex(p; Tn Suppose that G is connected Then p < n Proof By Lemma 31, we have (G n 3 and so e(g (n 3p Assume that p k(n 1 r with k N and r {0, 1,, n Let G 1 Ex(n 1 r; K 1,n 3 Then e(g 1 [ (n 4(n 1r by Lemma 1 Hence, if (k (n 1 r, then ( n 1 [ (n 4(n 1 r e((k 1K n 1 G 1 (k 1 (p r (n 1(n [ (n 4(n 1 r [ (n 3p p r (n 1 [ (n 3p (k (n 1 r [ (n 3p > e(g This is impossible since (k 1K n 1 G 1 does not contain Tn as a subgraph and G Ex(p; Tn Thus (k (n 1 r 1 If k 3, then r n and p 3(n 1n 4n 5 and so [ (n 3p (n 3(4n 5 e(g 4n 17n 15 ( ( < 4n 14n 1 n 1 n 3 e(3k n 1 K n Since 3K n 1 K n does not contain Tn and G Ex(p; Tn, we get a contradiction Thus k For p (n 1 r with r {0, 1,, n 4, n 3, n we see that r(n r < n and so e(k n 1 K r (n 1(n r(r 1 > (n 3(n r e(g This contradicts the assumption G Ex(p; Tn Now suppose p (n 1r with 3 r n 5 If (G n 4, then e(g (n 4p From previous argument we have ( n 1 [ (n 4(n 1 r [ (n 3p r e(k n 1 G 1 [ (n 4p (n 4p n 1 > e(g Since K n 1 G 1 does not contain Tn as a subgraph and G Ex(p; Tn, we get a contradiction Hence (G n 3 Suppose v 0 V (G, d(v 0 n 3, Γ(v 0 {v 1,, v n 3, V 1 {v 0, v 1,, v n 3 and V 1 V (G V 1 Suppose also that there are exactly s vertices in Γ(v 0 adjacent to some vertex in V 1 Then 1 s n 3 By (3, e(g[v 1 (n (n 3 s As G does not contain any copies of Tn, we see that e(v 1 V 1 s Since V (G V 1 V 1 p (n n r and G V 1 does not contain any copies of Tn we see that e(g V 1 ex(n r; Tn We claim that { (n 4(n r ex(n r; Tn max, 11 (n 1(n r(r 1

12 for 3 r n 5 Let G Ex(n r; Tn If G is connected, using Lemma 31 we have (G n 4 and so e(g (n 4(nr Now suppose that G is not connected If n 1, n {1,,, n, from Lemma 3 we have e(k n1 K n < e(k n1 n for n 1 n < n and e(k n1 K n < e(k n 1 K n1 n (n 1 for n 1 n n Thus, G G 1 G, where G 1 and G are components of G with V (G 1 p 1 < n 1 and V (G p n 1 For p n we have p 1 r n 3 and so e(g 1 p 1 (p 1 1 (n 4p 1 For p n we also have (G n 4 and so e(g (n 4p by Lemma 31 Hence for p n we find that e(g e(g 1 e(g (n 4p 1 (n 4p (n 4(nr Now assume p n 1 Then p 1 r 1 and e(g e(k n 1 K r1 Hence the claim is true and so { (n 4(n r e(g V 1 ex(n r; Tn max, Thus, (n 1(n r(r 1 (n 1(n r(r 1 e(g e(g[v 1 e(v 1 V 1 e(g V 1 (n (n 3 { (n 4(n r (n 1(n r(r 1 s s max, ( n 1 { (n 4(n 1 r n (n 1(n r(r 1 max, (n r ( n 1 {[ (n 4(n 1 r (n 1(n r(r 1 < max, { max e(k n 1 G 1, e(k n 1 K r This is impossible since G is an extremal graph By the above we must have k 1 and so p k(n 1 r < n as asserted Lemma 33 Let p, n N, p n 7 and G Ex(p; Tn Suppose that G is connected Then (G n 4 and e(g [ (n 4p Proof By (31, e(g [ (n 4p If (G n 5, using Euler s theorem we see that e(g 1 (n 5p v V (G d(v Hence (n 4p 1 [ (n 4p e(g (n 5p This is impossible Thus (G n 4 By Lemmas 31 and 3, (G n 4 Therefore (G n 4 and so e(g 1 (n 4p v V (G d(v Recall that e(g [ (n 4p Then e(g [ (n 4p as asserted Lemma 34 Let p and k be nonnegative integers, p 5k r and r {0, 1,, 3, 4 Suppose that G is a graph of order p without T6 r(5 r Then e(g p Proof Clearly (T6 3 We prove the lemma by induction on p For p 5 we have e(g p(p 1 p r(5 r Now suppose that p 6 and the lemma is true for all graphs of order p 0 < p without T6 If (G 3, then e(g 1 v V (G d(v 3p p 3 p r(5 r Suppose (G m 4, v 0 V (G, d(v 0 m, Γ(v 0 {v 1,, v m, V 1 {v 0, v 1,, v m and V 1 V (G V 1 If G[V 1 is a component of G, then e(g[v 1 e(k 5 10 for m 4, 1

13 and e(g[v 1 m m 3m for m 5 since d(v i for i 1,,, m By the inductive hypothesis, e(g[v 1 (p m 1 r 1(5 r 1, where r 1 {0, 1,, 3, 4 is given by p m 1 r 1 (mod 5 Thus, for m 4 we have e(g e(g[v 1 e(g[v 1 10 (p 5 r(5 r p r(5 r, and for m 5 we have e(g e(g[v 1 e(g[v 1 3m (p m 1 r 1(5 r 1 p m r(5 r p 3 p From now on we assume that G[V 1 is not a component of G and m (G 4 Hence there is a vertex u 1 such that d(u 1, v 0 and u 1 v 1 E(G with no loss of generality Then v 1 v i / E(G for i, 3,, m For m 4 we see that e(g[v 1 e(v 1 V For m 5 we see that d(v i for i 1,,, m and so e(g[v 1 e(v 1 V 1 m i1 d(v i m Hence, for m 4 we have e(g e(g[v 1 e(v 1 V 1 e(g[v 1 m e(g[v 1 By the inductive hypothesis, e(g[v 1 (p m 1 r 1(5 r 1, where r 1 {0, 1,, 3, 4 is given by p m 1 r 1 (mod 5 Thus, e(g m (p m 1 r 1(5 r 1 p r 1(5 r 1 For r 1 1 we have e(g p < p r(5 r For r 1 0 and r 0, 1, 4 we have e(g p p r(5 r Therefore, we only need to consider the case p m 1, 3 (mod 5 Now assume p m 1, 3 (mod 5 and Γ(u 1 {v 1,, v m {w 1,, w t As m 4 we have m 6 Set V {v 0, v 1,, v m, u 1 and V V (G V Since d(v i for i 1,,, m, we see that e(g e(g[v e(v V e(g[v m i1 d(v i t e(g[v m t e(g[v Note that p m 4 (mod 5 and e(g[v 4(5 4 (p m by the inductive hypothesis We then have e(g m t (p m p t 6 For t 3 we get e(g p t 6 p 3 p r(5 r For t 4 set V 3 {v 0, v 1,, v m, u 1, w 1,, w t and V 3 V (G V 3 Since d(v i for i 1,,, m and d(w j for j 1,,, t, using the inductive hypothesis we see that e(g e(g[v 3 e(v 3 V 3 e(g[v 3 m d(v i i1 t j1 d(w j e(g[v 3 m t e(g[v 3 m t (p m t p 4 r(5 r < p By the above, the lemma has been proved by induction Theorem 31 Let p, n N, p n 1 4 and p k(n 1 r, where k N and r {0, 1,, n Then {[ (n p ex(p; Tn max [ (n 1 r, (n p r(n 1 r (n p (n 1 r if n 16 and 3 r n 6 or if 13 n 15 and 4 r n 7, (n p r(n 1 r otherwise Proof Clearly ex(n 1; Tn e(k n 1 (n (n 1 Thus the result is true for p n 1 Now we assume p n Since T5 T 5, taking n 5 in [10, Theorem 31 we obtain the result 13

14 in the case n 5 For n 6 we see that ex(p; T6 e(kk 5 K r 10k r(r 1 p r(5 r This together with Lemma 34 gives the result in this case Applying Lemmas 33, 3 and replacing Tn 1 with Tn in the proof of Theorem 1 we deduce the result for n 7 Corollary 31 Suppose p, n N, p n 5 and n 1 p Then (n p (n 1 8 ex(p; Tn (n (p 1 4 The Ramsey number r(t i n, T n Lemma 41 ([9, Lemma 1 Let G 1 and G be two graphs Suppose p N, p max{ V (G 1, V (G and ex(p; G 1 ex(p; G < ( p Then r(g1, G p Proof Let G be a graph of order p If e(g ex(p; G 1 and e(g ex(p; G, then ex(p; G 1 ex(p; G e(g e(g ( p This contradicts the assumption Hence, either e(g > ex(p; G 1 or e(g > ex(p; G Therefore, G contains a copy of G 1 or G contains a copy of G This shows that r(g 1, G V (G p So the lemma is proved Lemma 4 ([9, Lemma 3 Let G 1 and G be two graphs with (G 1 d 1 and (G d Then (i r(g 1, G d 1 d (1 ( 1 (d 1 1(d 1 / (ii Suppose that G 1 is a connected graph of order m and d 1 < d m Then r(g 1, G d 1 d 1 d (iii If G 1 is a connected graph of order m, d 1 m 1 and d > m, then r(g 1, G d 1 d Theorem 41 Let n N and i, j {1, (i If n is odd with n 17, then r(tn, i Tn j n 7 (ii If n is even with n 1, then r(tn, i Tn j n 6 Proof Suppose n 1 Since (Tn i (Tn j n 3, from Lemma 4 we know that r(tn, i Tn j n 7 for odd n, and r(tn, i Tn j n 6 for even n If n is odd with n 17, using Theorems 1 and 31 (with k 1 and r n 6 we see that ex(n 7; Tn i (n (n 7 1 (n 4(n 7 (n 7 < 1 ( n 7 and so ex(n 7; Tn i ex(n 7; Tn j < ( n 7 Thus, by Lemma 41 we have r(t i n, Tn j n 7 Hence (i is true From Theorems 1 and 31 (with k 1 and r n 5 we see that for n 1, ex(n 6; Tn i (n (n 6 4(n 5 n 7n 16 < n 13 n 1 1 ( n 6 Thus, by Lemma 41 we have r(t i n, Tn j and so ex(n 6; T i n ex(n 6; T j n < ( n 6 n 6 Hence r(t i n, T j n n 6 for even n, proving (ii Lemma 43 Let n N, n 5 and i {1, Let G n be a connected graph of order n such that ex(n 5; G n < n 5n 4 Then r(t i n, G n n 5 Thus, Proof By Theorems 1 and 31, ex(n 5; T i n (n (n 5 3(n 4 n 6n 11 ( n 5 ex(n 5; G n ex(n 5; Tn i < n 5n 4 n 6n 11 14

15 Appealing to Lemma 41 we obtain r(tn, i G n n 5 Lemma 44 ([10, Theorem 31 Let p, n N with p n 5 Let r {0, 1,, n be given by p r (mod n 1 Then [(n (p 1 r 1 if n 7 and r n 4, ex(p; T n (n p r(n 1 r otherwise Theorem 4 Let n N, n 8 and i {1, Then r(t i n, T n r(t i n, T n n 5 Proof Let T n {T n, T n As K n 3 does not contain any copies of T i n and K n 3 K n 3,n 3 does not contain any copies of T n, we see that r(t i n, T n 1 (n 3 n 5 Taking p n 5 and r n 4 in Lemma 44 we find that [ (n (n 6 (n 4 1 ex(n 5; T n By [10, Theorem 41, n 11 n 15 < n 5n 4 ex(n 5; T n (n (n 5 3(n 4 n 6n 11 < n 5n 4 Thus, applying Lemma 43 we obtain r(tn, i T n n 5 Hence r(tn, i T n n 5 as asserted Remark 41 Let n N, n 5 and i {1, From [5, Theorem 31(ii we know that r(k 1,n 1, Tn i n 3 Theorem 43 Let n N and i {1, Then r(p n, Tn i n 7 for n 17, r(p n 1, Tn i n 7 for n 13, r(p n, Tn i n 7 for n 11 and r(p n 3, Tn i n 7 for n 8 Proof Suppose n 8 and s {0, 1,, 3 From Lemma 4(ii we have r(p n s, Tn i (n 3 1 n 7 By (11, (n (n 7 5(n 6 (n 4(n 7 16 n (n 3(n 7 3(n 5 (n 4(n 7 8 n ex(n 7; P n s (n 4(n 7 (n 4 (n 5(n 7 (n 5 (n 4(n 7 1 3n if s 0, if s 1, if s, if s 3 By Theorems 1 and 31, (n 4(n 7 [ if n 16, ex(n 7; Tn i (n (n 7 5(n 6 (n 4(n 7 16 n if n < 16 For n 17, 13, 11 or 8 according as s 0, 1, or 3, from the above we find ex(n 7; P n s ex(n 7; Tn i < ( n 7 and so r(pn s, Tn i n 7 by Lemma 41 This completes the proof 15

16 5 The Ramsey number r(t i m, T n for m < n Proposition 51 (Burr[1 Let m, n N with m 3 and m 1 n Let T m be a tree on m vertices Then r(t m, K 1,n 1 m n Proposition 5 (Guo and Volkmann [5, Theorem 31 Let m, n N, m 3 and n k(m 1 b with k N and b {0, 1,, m \ { Let T m K 1,m 1 be a tree on m vertices Then r(t m, K 1,n 1 m n 3 Moreover, if k m b, then r(t m, K 1,n 1 m n 3 Lemma 51 ([6, Theorem 83, pp11-1 Let a, b, n N If a is coprime to b and n (a 1(b 1, then there are two nonnegative integers x and y such that n ax by Theorem 51 Let m, n N, n > m 5, m 1 n and i {1, Then r(tm, i K 1,n 1 m n 3 or m n 4 Moreover, if n (m 3 1 or m n 4 (m 1x (m y for some nonnegative integers x and y, then r(t m, K 1,n 1 m n 3 for any tree T m K 1,m 1 of order m Proof Let T m K 1,m 1 be a tree on m vertices From Proposition 5 we know that r(t m, K 1,n 1 m n 3 By Lemma 4(iii, r(tm, i K 1,n 1 m 3 n 1 Thus, r(tm, i K 1,n 1 mn 3 or mn 4 If n (m 3 1, then mn 4 (m (m 3 and so mn 4 (m 1x(m y for some nonnegative integers x and y by Lemma 51 If m n 4 (m 1x (m y for x, y {0, 1,,, setting G xk m 1 yk m we see that G does not contain any copies of T m and G does not contain any copies of K 1,n 1 Thus r(t m, K 1,n 1 1 V (G mn 3 Now putting all the above together we obtain the theorem Theorem 5 Let m, n N, n > m 6, m 1 n 3 and i {1, Then r(tm, i T n m n 3 Proof By Theorems 1 and 31, ex(mn 3; Tm i (m (mn 3 (m < (m (mn 3 Thus applying [9, Theorem 51 we obtain the conclusion Theorem 53 Suppose i {1,, m, n N, n > m 7 and m 1 (n 3 Then m n 5 r(tm, i T n m n 4 and m n 6 r(tm, i Tn m n 4 Moreover, if n k(m 1 b q(m a, k, q N, a {0, 1,, m 3, b {0, 1,, m and one of the following conditions holds: (1 b {1,, 4, ( b 0 and k 3, (3 n (m 3, (4 n m 1 b(m, (5 a 3 and n (a 4(m 1 4, then r(tm, i Tn r(tm, i T n m n 4 Proof By Lemma 4 we have r(tm, i T n m 3 n and r(tm, i Tn m 3 n 3 Since m 1 n 3, we have m 1 m n 4 From Corollaries 1 and 31 we find ex(mn 4; Tm i (m (mn 5 Hence, by [9, Lemma 5 we have r(tm, i T n mn 4, and by [9, Lemma 4 we have r(tm, i Tn m n 4 Now applying [9, Theorems 44 and 54 we deduce the remaining assertion 16

17 6 The Ramsey number r(g m, T j n for m < n Theorem 61 Let m, n N, m 5, n 8, n > m and j {1, Then r(k 1,m 1, Tn j m n 4 or m n 5 Moreover, if mn, then r(k 1,m 1, Tn j m n 4 Proof From Lemma 4 we deduce that r(k 1,m 1, Tn j m 1 n 3 (1 ( 1 (m (n 4 / mn 4 (1 ( 1 mn / So, it suffices to prove that r(k 1,m 1, Tn j m n 4 By Lemma 1, ex(m n 4; K 1,m 1 [ (m (mn 4 By Theorems 1 and 31, we have [ (n 4(m n 4 ex(m n 4; Tn j (n (m n 4 (m 3(n m or Since [ (m (mn 4 [ (n 4(mn 4 (mn 6(mn 4 < ( mn 4 and (m (m n 4 (n (m n 4 (m 3(n m (m n 4(m n 5 (m 4(n m m 4 ( m n 4 <, we see that ex(m n 4; K 1,m 1 ex(m n 4; Tn j < ( mn 4 and so r(k1,m 1, Tn j m n 4 by Lemma 41 This completes the proof Theorem 6 Let m, n N, m 4, n 7, m 1 n 4 and j {1, (i If G m is a connected graph of order m with ex(m n 4; G m (m (mn 5, then r(g m, Tn j m n 4 (ii r(t m, Tn j r(tm, 1 Tn j r(tm, Tn j m n 4 for m 5, r(tm, Tn j m n 4 for m 6, and r(p m, Tn j m n 4 Proof Set t (n 4/(m 1 Suppose that G m is a connected graph of order m with ex(mn 4; G m (m (mn 5 Then clearly ((t 1K m 1 t(m 1 n 4 Thus, (t1k m 1 does not contain any copies of G m and (t 1K m 1 does not contain any copies of Tn j Hence r(g m, Tn j 1 (t 1(m 1 m n 4 By Theorems 1 and 31, [ (n 4(m n 4 ex(m n 4; Tn j or If ex(m n 4; T j n [ (n 4(mn 4, then (n (m n 4 (m 3(n m ex(m n 4; G m ex(m n 4; Tn j (m (m n 5 (n 4(m n 4 ( m n 4 < If ex(m n 4; T j n (n (mn 4 (m 3(n m, then ex(m n 4; G m ex(m n 4; Tn j (m (m n 5 (n (m n 4 (m 3(n m ( ( m n 4 (m 4(n m 1 m n 4 < 17

18 Therefore, by Lemma 41 we always have r(g m, T j n m n 4 and hence r(g m, T j n m n 4 This proves (i Now consider (ii Note that m n 4 1 (mod m 1 By (11, we have ex(m n 4; P m (m (mn 5 By Lemma 44, ex(m n 4; T m (m (mn 5 for m 5 By [10, Theorem 4, ex(m n 4; T m (m (mn 5 for m 6 By Theorems 1 and 31, ex(m n 4; T i m (m (mn 5 for i {1, and m 5 Thus from (i and the above we deduce (ii The proof is complete Lemma 61 Let j {1,, m, n N, m 7 and m 1 n 4 Assume n m1 1 or n max {m, 19 m (i If G m is a connected graph of order m with ex(m n 5; G m (m (mn 6, then r(g m, T j n m n 5 (ii For T m {P m, T m, T m, T 1 m, T m we have r(t m, T j n m n 5 Proof Since m n 5 n 1 m 4, by Theorems 1 and 31 we have [ (n 4(m n 5 ex(m n 5; Tn j (n (m n 5 (m 4(n 1 (m 4 or If n m 1, then (m 4(n 3 (m 4 (n 5 If n m, then 3 m 4 n 6 and so (m 4(n 3 (m 4 ( n 3 (m 4 n 3 ( n 3 (n 6 n 3 3(n 6 Thus, (n 4(m n 5 m (n (m n 5 (m 4(n 1 (m 4 (m 4(n 3 (m 4 n m (n 5 n m m 10 > 0 if n m 1 1, 3(n 6 n m n 10 m 8 > 0 if n max {m, 19 m Therefore, from the above we deduce that (61 ex(m n 5; T j n < (n 4(m n 5 m Hence, if G m is a connected graph of order m with ex(m n 5; G m (m (mn 6, then ex(m n 5; G m ex(m n 5; Tn j (m (m n 6 (n 4(m n 5 m < ( m n 5 Applying Lemma 41 we obtain (i Now we consider (ii Since m 1 (m n 5, by Corollaries 1 and 31 we have ex(mn 5; T i m (m (mn 6 for i {1, By (11, ex(mn 5; P m (m (mn 6 By Lemma 44, ex(m n 5; T m (m (mn 6 By [10, Theorems 41-45, ex(m n 5; T m (m (mn 6 Thus, from the above and (i we deduce (ii This proves the lemma Theorem 63 Let m N and j {1, 18

19 (i We have r(t m, T j m1 { m 4 if m and m 9, m 5 if m and m 16 (ii If n N, m 7, n max {m, 19 m and m 1 n 4, then r(t m, T j n mn 5 Proof We first assume m and m 9 By Lemma 4(i, we have r(t m, T j m1 m m m 4 By Lemma 44, ex(m 4; T m (m (m 4 (m 3 m 5m7 By Theorems 1 and 31, ex(m 4; T j m1 (m 1(m 4 4(m 4 m 5m 10 Thus, ex(m 4; T m ex(m 4; T j m1 ( m 4 m 5m 7 m 5m 10 m 10m 17 < m 9m 10 Hence, by Lemma 41 we obtain r(t m, T j m1 m 4 and so r(t m, T j m1 m 4 Now we assume m and m 16 By Lemma 4(i, r(t m, T j m1 m m 1 m 5 By Lemma 44, ex(m 5; T m [ (m (m 6 (m 3 m 11m14 By Theorems 1 and 31, ex(m 5; T j m1 [ (m 1(m 5 (m 5 m 11m14 Thus, ( ex(m 5; T m ex(m 5; T j m 5 m1 m 11m 14 < m 11m 15 Hence, by Lemma 41 we obtain r(t m, T j m1 m 5 and so r(t m, T j m1 m 5 This proves (i Now we consider (ii Suppose n N, m 7 and n max {m, 19 m By Lemma 61(ii, r(t m, Tn j m n 5 By Lemma 4, we have r(t m, Tn j m n 3 Thus, r(t m, Tn j m n 5 This proves (ii The proof is complete Theorem 64 Let j {1,, m, n N, m 7 and m 1 n 4 Suppose n m1 1 or n max {m, 19 m Assume that G m {P m, Tm, Tm, 1 Tm or G m is a connected graph of order m such that ex(m n 5; G m (m (mn 6 If n (m 3 3 or m n 6 (m 1x (m y for some nonnegative integers x and y, then r(g m, Tn j m n 5 Proof If n (m 3 3, then m n 6 (m (m 3 and so m n 6 (m 1x (m y for some x, y {0, 1,, by Lemma 51 Now suppose m n 6 (m 1x(m y, where x, y {0, 1,, Set G xk m 1 yk m Then (G n 4 Thus, G does not contain any copies of G m and G does not contain any copies of Tn j Hence r(g m, Tn j 1 V (G m n 5 On the other hand, by Lemma 61 we have r(g m, Tn j m n 5 Thus r(g m, Tn j m n 5 This proves the theorem Corollary 61 Let m, n N, m 7, m 1 n b, b {, 3, 5, n max {m, 19 m and j {1, Assume that G m {P m, Tm, Tm, 1 Tm or G m is a connected graph of order m with ex(m n 5; G m (m (mn 6 Then r(g m, Tn j m n 5 Proof Set k (n b/(m 1 Then k N For b we have k Since (k (m 1 3(m if b, m n 6 (k 1(m 1 (m if b 3, (k 1(m 1 if b 5, the result follows from Theorem 64 19

20 Theorem 65 Let m N, m 1 and i, j {1, Then r(tm, i T j m1 r(t m, T j m1 m 5 Proof Let T m {T i m, T m By Theorems 1, 31 and [10, Theorem 41, (m (m 5 3(m 4 ex(m 5; T m, ex(m 5; T j (m 1(m 5 5(m 5 [ (m 3(m 5 m1 or Since (m (m 5 3(m 4 (m 3(m 5 (m 5(m 67 m < ( m 5 and (m (m 5 3(m 4 (m 1(m 5 5(m 5 ( m 5 m 1m 6 < m 11m 15, we see that ex(m 5; T m ex(m 5; T j m1 < ( m 5 Hence, applying Lemma 41 we deduce that r(t m, T j m1 m 5 Since (T m m 3 and (T j m1 m, by Lemma 4(i we have r(t m, T j m1 m 3 m m 5 Hence r(t m, T j m1 m 5 This proves the theorem Acknowledgements The first author is supported by the National Natural Science Foundation of China (grant No , and the second author is supported by the Fundamental Research Funds for the Central Universities (grant No 014QNA58 References [1 SA Burr, Generalized Ramsey theory for graphs a survey, in: Graphs and Combinatorics, RA Bari and F Harary (eds, Lecture Notes in Math 406, Springer, Berlin, 1974, 5-75 [ SA Burr and P Erdős, Extremal Ramsey theory for graphs, Util Math 9(1976, [3 SA Burr and JA Roberts, On Ramsey numbers for stars, Util Math 4(1973, 17-0 [4 RJ Faudree and RH Schelp, Path Ramsey numbers in multicolorings, J Combin Theory Ser B 19(1975, [5 Y Guo and L Volkmann, Tree-Ramsey numbers, Australas J Combin 11(1995, [6 LK Hua, Introduction to Number Theory, Springer, Berlin, 198 (translated from the Chinese by P Shiu [7 SP Radziszowski, Small Ramsey numbers, Revision #14, Electron J Combin 014, Dynamic Survey DS1, 94pp 0

21 [8 AF Sidorenko, Asymptotic solution for a new class of forbidden r-graphs, Combinatorica 9(1989, [9 ZH Sun, Ramsey numbers for trees, Bull Austral Math Soc 86(01, [10 ZH Sun and LLWang, Turán s problem for trees, J Combin Number Theory 3(011, [11 M Woźniak, On the Erdős-Sós conjecture, J Graph Theory 1(1996,