Suppose k ≥ 2 is an integer. Let Y_k be the poset with elements x₁,x₂,y₁,y₂,…,y_k− 1 such that y₁ < y₂ < ⋯ < y_k− 1 < x₁,x₂ and let \(Y_{k}^{\prime }\) 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 F₁,F₂,G₁,G₂,…,G_k− 1 such that G₁ ⊂ G₂ ⊂⋯ ⊂ G_k− 1 ⊂ F₁,F₂ and |F₁| = |F₂| = |G_k− 1| + 1 = |G_k− 2| + 2 = ⋯ = |G₁| + k − 1. If both Y_k and \(Y^{\prime }_{k}\) on consecutive levels are forbidden, the size of the largest such family is denoted by \(\text {La}_{\mathrm {c}}\left (n, Y_{k}, Y^{\prime }_{k}\right )\). In this paper, we will determine the exact value of \(\text {La}_{\mathrm {c}}\left (n, Y_{k}, Y^{\prime }_{k}\right )\).

假设ķ ≥2是一个整数。设Y _k为元素x ₁，x ₂，y ₁，y ₂，…，y _{k − 1的}摆放，使得y ₁ < y ₂ <⋯< y _{k − 1} < x ₁，x ₂，令\（ Y_ {k} ^ {\ prime} \）是相同的位姿，但所有关系都相反。我们说[ n ]的子集族包含Y _k的副本在连续的水平，如果它包含ķ + 1个子集˚F ₁，˚F ₂，G ^ ₁，G ^ ₂，...，g ^ _{ķ - 1}，使得G ^ ₁ ⊂ ģ ₂ ⊂⋯⊂ ģ _{ķ - 1} ⊂ ˚F ₁，˚F ₂和| F ₁ | = | F ₂ | = | G _{k − 1} | +1 = | G _{k − 2} | + 2 =⋯= | G ₁ | + k− 1.如果同时禁止连续级别上的Y _k和\（Y ^ {\ prime} _ {k} \），则此类族的最大大小由\（\ text {La} _ {\ mathrm { c}} \ left（n，Y_ {k}，Y ^ {\ prime} _ {k} \ right）\）。在本文中，我们将确定\（\ text {La} _ {\ mathrm {c}} \ left（n，Y_ {k}，Y ^ {\ prime} _ {k} \ right）\的确切值\ ）。