中文

$C_{10}$与$C_{12}$的Gallai-Ramsey数

组合数学 2018-08-31 v1

摘要

Gallai着色是不含彩虹三角形的完全图边着色,而Gallai kk-着色是使用kk种颜色的Gallai着色。给定整数k1k\ge1与图H1,,HkH_1, \ldots, H_k,Gallai-Ramsey数GR(H1,,Hk)GR(H_1, \ldots, H_k)是最小整数nn,使得完全图KnK_n的每一个Gallai kk-着色都包含某个i{1,,k}i \in \{1, \ldots, k\}下颜色iiHiH_i的单色拷贝。当H=H1==HkH = H_1 = \cdots = H_k时,我们简记为GRk(H)GR_k(H)。我们继续研究偶圈与路的Gallai-Ramsey数。对所有n3n\ge3k1k\ge1,令Gi=P2i+3G_i=P_{2i+3}为所有i{0,1,,n2}i\in\{0,1, \ldots, n-2\}上含2i+32i+3个顶点的路,且Gn1{C2n,P2n+1}G_{n-1}\in\{C_{2n}, P_{2n+1}\}。令对所有j{1,,k}j\in\{1, \ldots, k\}ij{0,1,,n1} i_j\in\{0,1,\ldots, n-1\}i1i2ik i_1\ge i_2\ge\cdots\ge i_k 。Song近期猜想GR(Gi1,,Gik)=3+min{i1,n2}+j=1kijGR(G_{i_1}, \ldots, G_{i_k}) = 3+\min\{i_1, n^*-2\}+\sum_{j=1}^k i_j,其中当Gi1P2n+1G_{i_1}\ne P_{2n+1}n=nn^* =n,当Gi1=P2n+1G_{i_1}= P_{2n+1}n=n+1n^* =n+1。该猜想已对n{3,4}n\in\{3,4\}及所有k1k\ge1被验证成立。本文中,我们证明上述猜想对n{5,6}n \in\{5, 6\}及所有k1k \ge1成立。我们的结果意味着对所有k1k \ge 1,当n{5,6}n\in\{5,6\}GRk(C2n)=GRk(P2n)=(n1)k+n+1GR_k(C_{2n}) = GR_k(P_{2n}) = (n-1)k+n+1,且当1n61\le n \le6 GRk(P2n+1)=(n1)k+n+2GR_k(P_{2n+1})= (n-1)k+n+2

关键词

引用

@article{arxiv.1808.10282,
  title  = {Gallai-Ramsey numbers of $C_{10}$ and $C_{12}$},
  author = {Hui Lei and Yongtang Shi and Zi-Xia Song and Jingmei Zhang},
  journal= {arXiv preprint arXiv:1808.10282},
  year   = {2018}
}

备注

arXiv admin note: text overlap with arXiv:1803.07963