论文标题
在连续的布尔晶格上，最大的家庭没有一对Posets
Largest family without a pair of posets on consecutive levels of the Boolean lattice
论文作者
论文摘要
假设$ k \ ge 2 $是整数。令$ y_k $为带有元素的poset $ x_1，x_2，y_1，y_2，\ ldots，y_ {k-1} $，以便$ y_1 <y_1 <y_2 <\ cdots <y_ {k-1} <x_1，x_1，x_1，x_2 $ and let $ y_k'$是相同的poset，但所有关系reversed。我们说，如果包含$ k+1 $ 1 $ f_1，f_2，g_1，g_1，g_2，\ ldots，g_ {k-1} $，则连续级别的$ [n] $子集中包含$ y_k $的副本$ | f_1 | = | f_2 | = | g_ {k-1} | +1 = | g_ {k-2} |+2 = \ cdots = | g_ {1} |+k-1 $。如果连续级别上都禁止$ y_k $和$ y'_k $，则最大的家庭的大小由$ \ m athrm {la} _ {\ mathrm {c}}（n，y_k，y_k，y'_k）$表示。在本文中，我们将确定$ \ mathrm {la} _ {\ mathrm {c}}}}}}}（n，y_k，y'_k）$的确切值。
Suppose $k \ge 2$ is an integer. Let $Y_k$ be the poset with elements $x_1, x_2, y_1, y_2, \ldots, y_{k-1}$ such that $y_1 < y_2 < \cdots < y_{k-1} < x_1, x_2$ and let $Y_k'$ be the same poset but all relations reversed. We say that a family of subsets of $[n]$ contains a copy of $Y_k$ on consecutive levels if it contains $k+1$ subsets $F_1, F_2, G_1, G_2, \ldots, G_{k-1}$ such that $G_1\subset G_2 \subset \cdots \subset G_{k-1} \subset F_1, F_2$ and $|F_1| = |F_2| = |G_{k-1}|+1 =|G_{k-2}|+ 2= \cdots = |G_{1}|+k-1$. If both $Y_k$ and $Y'_k$ on consecutive levels are forbidden, the size of the largest such family is denoted by $\mathrm{La}_{\mathrm{c}}(n, Y_k, Y'_k)$. In this paper, we will determine the exact value of $\mathrm{La}_{\mathrm{c}}(n, Y_k, Y'_k)$.
