Largest family without a pair of posets on consecutive levels of the Boolean lattice

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 $\mathrm{La}_{\mathrm{c}}(n, Y_k, Y'_k)$.

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 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 [11] in 1834.
S(n, k, r) = n i=0 i≡r mod k n i .
The first author [6] published the next two theorems which are analogs of the two theorems above. Theorem 1.3 (Katona [6]). Let F be a family of subsets of [n] such that no two of the subsets F i , F j satisfy F i ⊂ F j and |F j | − |F i | < k. Then |F| ≤ max{S(n, k, r) | r ∈ {0, . . . , k − 1}}.
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 [7] gave the following result for k = 2. (In their paper, posets Y 2 and Y ′ 2 are denoted by V (the cheery poset) and Λ (the fork poset) respectively.) Theorem 1.5 (Katona and Tarján [7]).
De Bonis, Katona and Swanepoel [2] studied the case k = 3. We remark that they actually proved a result for a poset so called butterfly, but their proof implies the following theorem.
For the case k ≥ 4, Methuku and Tompkins [10] got the next theorem.
Theorem 1.7 (Methuku and Tompkins [10]). For n ≥ k ≥ 4, For other results related to the La function, see [1,5,9,13]. 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] without both Y k and 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.
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).
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 without Y k and 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 Supposing that we know Theorem 2.1, then Since |F| is an integer, we have |F| ≤ n · 2 n − S(n, k, m) + n − 1 n = 2 n − S(n, k, m), as desired. So 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. We denote by I s t the interval {a t+1 , a t+2 , . . . , a t+s } (with addition taken modulo n). Clearly, |I s t | = s, I s−1   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 consecutive levels must contain ∅. In this case, we consider a special kind of Y k , which we denote by Figure 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 Figure 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 Figure 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. (II) For k ≥ 4, the edges of the 2n X k 's cover the edges within the (i + 1) st level and the (i + 2) nd level twice, the edges within the levels from (i + 2) nd to (i + k − 1) st once, and the edges within the (i + k − 1) st level and the (i + k) th level twice. Now, we are ready to start the proofs of Lemmas 2.2 and 2.3.
On the one hand, recall that every vertex in the s th level of I(n) σ have two neighbors in the (s + 1) st level and two neighbors in the (s − 1) st level, and note that only the case when one of the two vertices in the edge is not in F counts nonzero. So On the other hand, we will show T ≥ 2(x i+2 + x i+k−1 ). When k = 3, since every vertex of the (i + 2) nd level in F has at least 2 neighbors not in F, it follows that T ≥ 2 · 2 · x i+2 .
If k ≥ 4, we consider the 2n X k 's, and change the weight function of an edge to 1 if it originally counts 2. Then by (II) of Remark 2.4, T is equal to the weighted sum over all edges in these 2n X k 's. We will prove below the claim that in each X k , the sum of the changed weight function over its edges is at least |{I ∈ F ∩ X k | |I| = i + 2 or |I| = i + k − 1}|.
Then if we summarize it 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 Figure 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 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 Figure 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 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, ✷ 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. Case 1. x k−1 ≤ ⌊n/2⌋ when n is odd or x k−1 ≤ (n/2) − 1 when n is even.
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.
In this case, we count the number of pairs of vertices of the (k − 1) st level in F which are neighbors (having a common neighbor vertex in the (k − 2) nd level). Now, we view the n vertices in the (k − 1) st level around a cycle (see Figure 5), and call a set of vertices a part if these vertices are in F and consecutive in the cycle. Note that the number of the vertices in the (k − 1) st level which are not in F is n − x k−1 , and these vertices cut the cycle into at most n − x k−1 parts. In every part P , the number of the pairs we want is equal to |P | − 1. So in total, we have at least Figure 3). Then in each of such 2x k−1 − n candidates (Y k (j)'s), at least one vertex in the levels from 1 st to (k − 2) nd is not in F. So by (I) of Remark 2.4, we have 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 S(n, k, r | z) = n i=0 i≡r mod k i≤z n i ; 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.
We assign the weight w i (see the definition and properties of w i in Section 2) to the constraint x i + x i+1 + · · · + x i+k−1 ≤ (k − 1)n for every 0 ≤ i ≤ n − k + 1. By (I) of Lemma 2.6, w i > 0 for 0 ≤ i ≤ n − k + 1, so we have Then by (I), (II) and (III) of Lemma 2.6, the LHS of (1) is equal to (S(n, k, m) − S(n, k, j)) x j .
On the other hand, by (IV) of Lemma 2.6, the RHS of (1) is equal to (k − 1)n · S(n, k, m).