Largest Family Without a Pair of Posets on Consecutive Levels of the Boolean Lattice

Suppose k ≥ 2 is an integer. Let Yk be the poset with elements x1,x2,y1,y2,…,yk− 1 such that y1 < y2 < ⋯ < yk− 1 < x1,x2 and let Yk′\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$Y_{k}^{\prime }$\end{document} be the same poset but all relations reversed. We say that a family of subsets of [n] contains a copy of Yk on consecutive levels if it contains k + 1 subsets F1,F2,G1,G2,…,Gk− 1 such that G1 ⊂ G2 ⊂⋯ ⊂ Gk− 1 ⊂ F1,F2 and |F1| = |F2| = |Gk− 1| + 1 = |Gk− 2| + 2 = ⋯ = |G1| + k − 1. If both Yk and Yk′\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$Y^{\prime }_{k}$\end{document} on consecutive levels are forbidden, the size of the largest such family is denoted by Lacn,Yk,Yk′\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$\text {La}_{\mathrm {c}}\left (n, Y_{k}, Y^{\prime }_{k}\right )$\end{document}. In this paper, we will determine the exact value of Lacn,Yk,Yk′\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$\text {La}_{\mathrm {c}}\left (n, Y_{k}, Y^{\prime }_{k}\right )$\end{document}.

A chain of length k is a poset with elements x 1 , x 2 , . . . , x k such that x 1 < x 2 < · · · < x k .
Let (n, k) be the sum of the k largest binomial coefficients of the form n i . Then Sperner's theorem can be extended as follows: Theorem 1.2 (Erdős [3]) Let F be a family of subsets of [n] without a chain of length k. Then |F | ≤ (n, k − 1).
Before stating the next result, we need the following notation. Let 2 ≤ k ≤ n and 0 ≤ r ≤ n be two integers. The following lacunary sum of binomial coefficients was first introduced by Ramus [13] in 1834.
S(n, k, r) = n i=0 i≡r mod k n i .
Another type of generalizations of Sperner Theorem is to determine the largest size of a family of subsets of [n] without a copy of poset Y k and Y k defined below. Let Y k be the poset with elements x 1 , x 2 , y 1 , y 2 , . . . , y k−1 such that y 1 < y 2 < · · · < y k−1 < x 1 , x 2 , and let Y k be the same poset but all relations reversed. Katona and Tarján [9] gave the following result for k = 2. (In their paper, posets Y 2 and Y 2 are denoted by V (the cherry poset) and (the fork poset) respectively.)

Theorem 1.5 (Katona and Tarján [9])
La tn, De Bonis, Katona and Swanepoel [2] studied the case k = 3. We remark that they actually proved a result for the so-called butterfly poset, but their proof implies the following theorem. For the case k ≥ 4, Methuku and Tompkins [12] got the next theorem. Theorem 1.7 (Methuku and Tompkins [12]) For n ≥ k ≥ 4, For other results related to the La function, see [1, 5-7, 11, 15]. Now, we consider the following problem, which is a combination of the two types of problems above. We say that a family of subsets of [n] contains a copy of Y k on consecutive levels if it contains We denote by La c n, Y k , Y k the largest size of a family of subsets of [n] containing neither Y k nor Y k on consecutive levels. We note that this problem is mentioned in [4] in the language of the oriented hypercube. The authors in [4] gave an asymptotic formula of the size of largest family without tree posets in consecutive levels. For exact result, they proved Let m = (n − k)/2 in the rest of this paper. For k ≥ 3, we have the following theorem. [10] showed that S(n, k, m) = min r: 0≤r≤k−1

S(n, k, r).
So our result implies that the trivial construction is the best. Here, a trivial construction consists of all subsets except the ones with size s ≡ m (mod k).
(II) If n = k ≥ 3, then m = 0 and S(n, k, m) = n 0 + n n = 2. In this case, Theorem 1.8 is trivial, since every family F with size 2 n − 1 contains a copy of Y k or Y k on consecutive levels.
The rest of the paper is organized as follows. In the next section, we present some preliminary results. In Section 3, we will prove our main theorem (Theorem 2.1), which implies Theorem 1.8.

Preliminary Results
A cyclic permutation σ of [n] is a cyclic ordering a 1 , a 2 , . . . , a n , a 1 , where a i ∈ [n] for i = 1, 2, . . . , n. Let F be a family of subsets of [n] containing neither Y k nor Y k on consecutive levels. We say a set F ∈ σ if F is an interval along the cyclic permutation σ . Now, we double count the sum S of φ F over all cyclic permutations σ and F ∈ F such that F ∈ σ , where For any F ∈ (F \ {∅, [n]}), we have |F |!(n − |F |)! cyclic permutations σ satisfying F ∈ σ . If ∅ ∈ F or [n] ∈ F, all cyclic permutations satisfy the above condition, and the number of cyclic permutations is (n − 1)!. So On the other hand, for any cyclic permutation σ , let I(n) σ be the family of intervals along σ . The i th level of I(n) σ is the collection of its elements of size i. (See Fig. 1 for an example of I(6) σ , where σ = 1, 2, 3, 4, 5, 6, 1.) For 0 ≤ i ≤ n, let x i be the number of subsets in the i th level of F ∩ I(n) σ . Then  Supposing that we know Theorem 2.1, then Since |F | is an integer, we have as desired. So by (II) of Remark 1.9, it is sufficient to prove Theorem 2.1. The following two lemmas are needed to give constraints of x 0 , x 1 , . . . , x n .
Before starting the proofs of the two lemmas above, we introduce some notations. Let σ = a 1 , a 2 , . . . , a n , a 1 , where a j ∈ [n] for 1 ≤ j ≤ n. In the rest of this paper, we view I(n) σ (a subfamily of I(n) σ ) as a graph. The vertex set of this graph is I(n) σ (the subfamily of I(n) σ ), and two vertices are connected by an edge if they are from consecutive levels and they have inclusion relation (see Fig. 2). We denote by I s t the vertex {a t+1 , a t+2 , . . . , a t+s } (with addition taken modulo n) of I(n) σ . Clearly, Fig. 2).

Fig. 2 Vertices I s t of I(n) σ
In order to prove the two kinds of constraints of k consecutive x i 's above, we consider some typical structures in k consecutive levels of I(n) σ . If the k levels are the levels from 0 th to (k − 1) st , then every Y k on these levels must contain ∅. In this case, we consider a special kind of Y k , which we denote by Fig. 3 for an example of Y k (1).) If the k levels are the levels from (i + 1) st to (i + k) th (the middle part of Fig. 2), then we introduce a new kind of structure on k consecutive levels. A family of k + 2 subsets is called X k if it is Fig. 3 for two types of examples of t = 2 and t = k respectively.) In each X k , we call the 3 elements in the (i + k) th level and the (i + k − 1) st level a cherry, and the 3 elements in the (i + 1) st level and the (i + 2) nd level a fork. Proof of Lemma 2.2 First, we show x 0 + x 1 + · · · + x k−1 ≤ (k − 1)n. Since x 0 ≤ 1 and x i ≤ n for 1 ≤ i ≤ k − 1, we have x 0 + x 1 + · · · + x k−1 ≤ (k − 1)n + 1. If x 0 + x 1 + · · · + x k−1 = (k − 1)n + 1, every subset of the levels from 0 th to (k − 1) st is in Fig. 3 Examples of structures on k consecutive levels F, then one can easily find a copy of Y k on consecutive levels. (See Y k (1) for an example in Fig. 3.) Similarly, we have x n−k+1 + x n−k+2 + · · · + x n ≤ (k − 1)n. Now, we prove that On the other hand, we will prove below the claim that in each X k , the sum of the weight function over its edges is at least |{I ∈ F ∩ X k | |I | = i + 2}| + |{I ∈ F ∩ X k | |I | = i + k − 1}|. Then if we sum over all 2n X k 's, T is at least 2(x i+2 + x i+k−1 ). Now, we divide the cherries of all 2n X k 's into 4 types (see Fig. 4), according to whether its 3 elements are in F or not. We call a cherry Type 1 (2 or 3) if the middle element and both (one or none) of its neighbors are in F. The rest of the cases are Type 4, namely that the middle element is not in F. Note that in Type 4, we do not distinguish the cases if two neighbors of the middle element are in F or not. Similarly, we divide all forks into Types 5, 6, 7 and 8 (see Fig. 4).
To prove the claim, we distinguish several cases by the types defined above. Note that in each X k , in order to avoid copies of Y k and Y k in F, either at least 2 of the 4 vertices in the (i + 1) st level and the (i + k) th level are not in F, or at least one of the k − 2 vertices in the levels from (i + 2) nd to (i + k − 1) st is not in F. If the cherry and the fork in some X k are Types 1 and 5 respectively, then we must have some vertex I not in F in the chain connecting the cherry and the fork in this X k . Then along this chain in both directions from I , take the first vertices in F respectively. Thus, we can find at least two edges with weight 1. Note that such vertices exist since in Types 1 and 5, the vertices in the (i + 2) nd level and the (i + k − 1) st level are in F. The same argument works for the pairs Types 1 and 6, and Types 2 and 5. If they are Types 1 and 7, then we already have two desired edges in the fork. If they are Types 1 and 8, then we find the first vertex in F along the chain from the fork to the cherry, this vertex and its neighbor below in the chain will give the edge we want. The same argument works for Types 4 and 5. In the rest cases, it is easy to find the desired edges in the cherry and the fork. (If k = 3, Types 1 or 2 or 3 and 8 (5 or 6 or 7 and 4) cannot appear in any X k , since i + 1 = i + k − 2. If k = 3 or 4, Types 1 and 5 (1 and 6, or 2 and 5) cannot appear in any X k , since they will form a Y k or a Y k in F.) Therefore, we have That is, as required.
Proof of Lemma 2.3 By symmetry, it is enough to prove the first part of the lemma. That is, we need to prove We distinguish three cases to prove it.
Note that x i ≤ n for i = 1, 2, . . . , k − 2, then by the assumption x 0 = 1, we have if n is odd; and if n is even.
Case 2 x k−1 = n/2 when n is even.
Suppose that the result is not true. That is, Note that x k−1 = n/2 in this case and x i ≤ n for i = 1, 2, . . . , k − 2. Then by the assumption x 0 = 1, we have x 1 = x 2 = · · · = x k−2 = n. Therefore, by the assumption x 1 + x 2 + · · · + x k = (k − 1)n, we have x k = n/2. Since x k−1 = n/2 and x k = n/2, we can find 2 vertices such that they are in the k th level and the (k − 1) st level respectively, they are in F, and they have inclusion relation. Then a copy of Y k on the levels from 1 st to k th can be found, a contradiction.
Case 3 x k−1 ≥ n/2 + 1. Fig. 5). To avoid a copy of Y k (j ) in F, at least one vertex of Y k (j ) in the levels from 1 st to (k − 2) nd is not in F. So by (I) of Remark 2.4, we have

The number of pairs of vertices
This completes the proof of this case and the lemma.
Before starting the proof of Theorem 2.1, we need the following notations and lemmas for helping us to use the constraints above. Recall the definition of S(n, k, r) and m = (n − k)/2 . Now let z be an integer such that z ≡ r (mod k). Then we denote

Lemma 2.5
Let t ≤ n be an integer. We have the following equalities.
Proof (I) follows from the definition of S(n, k, t | t). (II) follows from (I) if 0 ≤ t ≤ k − 1, and the definition of S(n, k, t | t) if k ≤ t ≤ n. (IV) follows from (II) and (III). Now, we distinguish two cases to prove (III).
If t > n − k, then we have t + k > n and n − k − t < 0. Hence, S(n, k, t|t) = S(n, k, t) and S(n, k, n − k − t|n − k − t) = 0 by (I), and this completes the proof of this case. If since r < k.