Colloquium Mathematicum 139(015, no, 73-98 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 Email: zhihongsun@yahoocom Homepage: http://wwwhytceducn/xsjl/szh School of Science, China University of Mining and Technology, Xuzhou, Jiangsu 1116, PR China Email: wanglinlin 1986@aliyuncom 3 School of Mathematical Sciences, Huaiyin Normal University, Huaian, Jiangsu 3001, PR China Email: 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 3 010 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
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
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
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
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
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
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 1 1 1 [ (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
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 7 3 4 < 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
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
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
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
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
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 1 4 4 8 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
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
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
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
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
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
(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
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 11371163, 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, 47-58 [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, 150-160 [5 Y Guo and L Volkmann, Tree-Ramsey numbers, Australas J Combin 11(1995, 169-175 [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
[8 AF Sidorenko, Asymptotic solution for a new class of forbidden r-graphs, Combinatorica 9(1989, 07-15 [9 ZH Sun, Ramsey numbers for trees, Bull Austral Math Soc 86(01, 164-176 [10 ZH Sun and LLWang, Turán s problem for trees, J Combin Number Theory 3(011, 51-69 [11 M Woźniak, On the Erdős-Sós conjecture, J Graph Theory 1(1996, 9-34 1