Enumeration on row-increasing tableaux of shape $2 \times n$

Recently O. Pechenik studied the cyclic sieving of increasing tableaux of shape $2\times n$, and obtained a polynomial on the major index of these tableaux, which is a $q$-analogue of refined small Schr\"{o}der numbers. We define row-increasing tableaux and study the major index and amajor index of row-increasing tableaux of shape $2 \times n$. The resulting polynomials are both $q$-analogues of refined large Schr\"{o}der numbers. For both results we give bijective proofs.


Introduction
Let n be a positive integer and λ a partition of n. A semistandard (Young) tableau (SSYT) of shape λ is an array T of positive integers of shape λ that is strictly increasing in every row and weakly increasing in every column. If an SSYT is strictly increasing in both rows and columns, it is called a standard Young tableau (SYT). (Note that in many other literatures such as [10], an SSYT is defined to be strictly increasing in every column and weakly increasing in every row instead, but we prefer to define it in this way for later convenience.) We denote by SYT(λ) the set of standard Young tableaux of shape λ. Throughout this paper we will identify a partition λ with its Young diagram; hence the notations SYT(m × n) and SYT(n m ) are equivalent.
A descent of an SSYT T is an integer i such that i + 1 appears in a lower row of T than i, and define the descent set D(T ) to be the set of all descents of T . The major index of T is defined by maj(T ) = i∈D(T ) i. Similarly, we define an ascent of T to be an integer i such that i + 1 appears in a higher row of T than i, and define the ascent set A(T ) to be the set of all ascents of T . The amajor index of T is defined by amaj(T ) = i∈A(T ) i. For example, for the tableau T shown in Figure 1 The most well-known result on enumerating the major indices of Young tableaux is the following q-hook length formula: [10, p.376] For any partition λ = i λ i of n, we have . (1.1) Applying the above result to SYTs of shape 2 × n, we get the following result. (1. 2) The above result is a q-analogue of the well-known result that SYTs of shape 2 × n is counted by the n-th Catalan number C n = 1 n+1 2n n . The famous RSK algorithm [10] is a bijection between permutations of length n and pairs of SYT of order n of the same shape. Under this bijection, the descent set of a permutation is transferred to the descent set of the corresponding "recording tableau. Therefore many problems involving the statistic descent or major index of pattern-avoiding permutations can be translated to the study of descent and major index of standard tableaux [1,3].
In this paper we study the major (amajor) index polynomial of increasing and row-increasing tableaux. An increasing tableau is an SSYT such that both rows and columns are strictly increasing, and the set of entries is an initial segment of positive integers (if an integer i appears, positive integers less than i all appear). We denote by Inc k (λ) the set of increasing tableaux of shape λ whose entries are {1, 2, . . . , n − k}, i.e., Inc k (λ) denotes the set of increasing tableaux of shape λ, with exactly k numbers appear twice. Figure 1 shows an increasing tableau T ∈ Inc 3 (2 × 6). In [4] O. Pechenik studied increasing tableaux in Inc k (2×n) and obtain the following result.
Motivated by the above results, we define and study row-increasing tableaux. Here a rowincreasing tableau is an SSYT with strictly increasing rows and weakly increasing columns, and the set of entries is a consecutive segment of positive integers. We denote by RInc m k (λ) the set of row-increasing tableaux of shape λ with set of entries {m + 1, m + 2, . . . , m + n − k}. We will also denote RInc 0 k (λ) as RInc k (λ). It is obvious that Inc k (λ) ⊆ RInc k (λ). It is not hard to see (will be explained in Section 4) that RInc k (2 × n) is in bijection with Schröder n-paths with k F steps, and these two sets are both counted by Here r(n, k) is considered as a refinement of the large Schröder number and is the sequence A006318 in OEIS [7].
Our main results are the following formulas involving major index and amajor index of SSYTs in RInc k (2 × n).

Corollary 1.5 For any positive integer n, we have
The organization of the paper is as follows. In Section 2 we give a bijection between rowincreasing tableaux in RInc k (2 × n) \ Inc k (2 × n) and increasing tableaux in Inc k−1 (2 × n). While this bijection does not preserve the descent set, by considering the change of the descent sets, we get a recurrence formula of R q (n, k) in terms of S q (n, k), and hence prove Theorem 1.3 by applying Theorem 1.2. In Section 3 we give a bijection Φ :

Thus we proved Theorem 1.4 by showing that
Finally in Section 4 we review some related work on enumerating major index of Schröder paths.
2 Counting major index for RInc k (2 × n) Theorem 2.6 For any positive integer n, k, we have r(n, k) = s(n, k) + s(n, k − 1). (2.1) Proof. Note that RInc k (2 × n) can be split into two disjoint sets: those that are increasing tableaux (Inc k (2×n)) and those that contains at least one column of identical entries (RInc k (2× n) \ Inc k (2 × n)). By definition the former one is counted by s(n, k). And we will prove (2.1) by providing a bijection between RInc k (2 × n) \ Inc k (2 × n) and Inc k−1 (2 × n).
For any SSYT T , we use T i,j to denote the entry in row i and column j of T . Now we define find the minimal integer j such that T 1,j = T 2,j , i.e., the j-th column is the leftmost column of T with two identical entries. Now we first delete the entry T 2,j , then move all the entries on the right of T 2,j one box to the left and set the last entry as 2n − k + 1, and define the resulting tableau to be f (T ). Note that for any i, The map f is reversible. Given S ∈ Inc k−1 (2 × n). Find the rightmost column j ′ such that S 1,j ′ +1 = S 2,j ′ + 1. (If such a column does not exist, then set j ′ = 0.) Now we first delete the entry S 2,n , then move all the entries S 2,j ′ +1 , S 2,j ′ +2 , . . . , S 2,n−1 one box to the right, and set S 2,j ′ +1 = S 1,j ′ +1 . We denote the resulting tableau as T . It is obvious that Theorem 2.7 For any positive integer n, k with k < n, we have , there are two cases.
1. If T 1,n = T 2,n . In this case we have T 1,n = T 2,n = 2n − k and 2n − k / ∈ D(T ). We will show that the sum of q maj(T ) over all these tableaux is 1) The n-th column is the only column of T with identical entries. In this case the last column of T consist of two identical entries 2n − k and 2n − k / ∈ D(T ). And the sum of q maj(T ) over these tableaux is S q (n − 1, k − 1).
2) There is at least one column with identical entries in T besides the n-th column. Now let T ′ be the tableau obtained by deleting the last column from T . Clearly we have T ′ ∈ RInc k−1 (2 × (n − 1)) \ Inc k−1 (2 × (n − 1)). There are two cases for the last column of T ′ .
, and all the other descents of T ′ are also descents of T . Thus Moreover when we apply f to Thus the sum of q maj(T ) over these tableaux of case a) and b) is S q (n − 1, k − 2).
Combining the Case 1) and Case 2) we have Hence (2.2) is proved.
Proof of Theorem 1.3: From Theorem 1.2 we have Applying the above equation to Theorem 2.7 we have 3 Counting amajor index for RInc k (2 × n) In this section we will prove Theorem 1.4 by showing that Our main idea is to establish a bijection Φ : RInc k (2 × n) → RInc k (2 × n) which satisfies maj(Φ(T )) = amaj(T ) + n − k.
Before establishing the map we need some definitions. A row-increasing tableau T is prime if for each integer j satisfies T 1,j+1 = T 2,j + 1, T 2,j+1 also appears in row 1 in T . We use pRInc m k (λ) to denote the set of all prime row-increasing tableaux of shape λ with set of entries {m + 1, m + 2, . . . , m + n − k}.
For each T ∈ pRInc m k (2 × n), we define two k-element sets A and B as the following: A = {a 1 , a 2 , . . . , a k } ≤ is the set of numbers that appear twice in T . B = {b 1 , b 2 , . . . , b k }, here b i is the the number appears immediately left of a i in the second row of T in cyclic order (if a 1 = T 2,1 , then b k = T 2,n (= m + 2n − k)). Let g(T ) be the tableau of shape 2 × n obtained by first deleting all elements in A from the first row and then inserting all elements in B into the first row and list them in increasing order, and keep the entries in row 2 unchanged. (See Figure 3 and Figure 4 for examples.) Lemma 3.8 The map g is an injection from pRInc m k (2 × n) to RInc m k (2 × n) which satisfies the following: 1) If T 2,1 appears only once in T , then g(T ) 1,i+1 ≤ g(T ) 2,i for each i, 1 ≤ i ≤ n − 1; 2) T 2,1 appears twice in T if and only if g(T ) 1,n = g(T ) 2,n .
Proof. It is obvious that the map g is invertible. Next we will prove that for any T ∈ pRInc m k (2 × n), g(T ) ∈ RInc m k (2 × n) and satisfies corresponding conditions according to the following cases.
1. When T 2,1 appears only once in T .
In this case for each i, 1 ≤ i ≤ n, b i is immediately to the left of a i in the second row of T , hence b i < a i . Note that it is impossible that T 1,i+1 > T 2,i + 1 (in which case there will be no place for the number T 2,i + 1), we now first prove that g(T ) 1,i+1 ≤ g(T ) 2,i for each i, 1 ≤ i ≤ n − 1 according to the following two cases.

according to the definition of prime row-increasing tableaux
we know that in this case T 2,i+1 appears twice in T . Suppose there are exactly x numbers among T 2,2 , T 2,2 , . . . , T 2,i that appear twice in T (x could be 0), then under the map g, exactly x numbers are deleted from the left of T 1,i+1 in the first row and then x + 1 numbers (including T 2,i ) are inserted to the left of T 1,i+1 , therefore we have g(T ) 1,i+1 = T 2,i = g(T ) 2,i for each i, 1 ≤ i ≤ n − 1.
In this case we have a 1 = T 2,1 and b 1 = T 2,n . LetT be the tableau of shape 2× n obtained from T by first deleting {a 2 , . . . , a k } from the first row and then inserting {b 2 , . . . , b k } into the first row and list them in increasing order, and keep the entries in row 2 unchanged. (See Figure 4 for an example.) for each i, j ≤ i ≤ n − 1 and g(T ) 1,n = T 2,n = g(T ) 2,n . Similar to the argument in case 1, we can prove thatT 1,i+1 ≤T 2,i = T 2,i for each i, 1 ≤ i ≤ n − 1. Hence we have that g(T ) 1,i ≤ g(T ) 2,i for each i, 1 ≤ i ≤ n − 1, i.e., g(T ) is weakly increasing in each column. And it is obvious from the definition of g that g(T ) 1,n = g(T ) 2,n = T 2,n if and only if T 2,1 appears twice in T .
Proof. Given T ∈ pRInc m k (2 × n), let T 0 be the skew shape tableau obtained by deleting the numbers in A from row 1 of T and push all the remaining numbers to the right, and keep the second row unchanged. (See Example 3.10 for an example.) We will prove Equation (3.1) by verifying the following facts. On the other hand, for each i, 1 ≤ i ≤ n, i ∈ D(g(T )) \ D(T 0 ) if only if that in g(T ), i + 1 appears in row 2 and i appears in both rows. In this case i + 1 is immediately to the right of i in g(T ), and hence i + 1 appears in both rows of T and i appears in row 2 of T , therefore we have i ∈ A(T ) \ A(T 0 ).
Combining the above two facts we get Equation (3.1).
Example 3.10 Figure 5 shows an example of T , T 0 and g(T ) with n = 6, m = 4, and k = 2.  Proof. Given T ∈ RInc k (2×n), there is a unique way to decompose T into prime row-increasing tableaux: suppose i 1 , i 2 , . . . , i l−1 are all the positive integers such that T 2,i j + 1 = T 1,i j +1 and T 2,i j +1 appear only once in T , we break T between column i j and i j +1 and get a decomposition T 1 T 2 · · · T l of T into prime row-increasing tableaux. Now set Φ(T ) = g(T 1 )g(T 2 ) · · · g(T l ) (see Example 3.12).
From Lemma 3.8 we know that Φ(T ) ∈ RInc k (2 × n). Next we show that Equation 3.2 holds. Suppose T j ∈ pRInc m j k j (2 × n j ) for integers m j , k j , n j with m j , k j ≥ 0, and n j > 0. Then we have n 1 + n 2 + · · · + n l = n, k 1 + k 2 + · · · + k l = k.
From the definition of prime row-increasing tableaux we know it is impossible that the two numbers of the first column of T j is identical when j > 1. Moreover, since T ∈ RInc k (2 × n) we have m 1 = 0. Hence from Proposition 3.9 we know that for each j, 1 ≤ j ≤ l, maj(g(T j )) = amaj(T j ) + m j + n j − k j , always holds. Therefore we have It remains to show that Φ is a bijection. Since g is an injection from pRInc m k (2 × n) to RInc m k (2 × n), it is sufficient to show that given S = Φ(T ) ∈ RInc k (2 × n), we can decompose S at the right places to get S 1 S 2 · · · S l such that S i = g(T i ) for each i, 1 ≤ i ≤ l. And the decomposition is as follows, we first find the largest number j such that S 1,j = S 2,j , i.e., column j is the rightmost column with identical entries. If such a column does not exist, we set j = 0. Now for each i, j ≤ i ≤ n − 1, we break S between column i and i + 1 if S 1,i+1 > S 2,i (Note that when 0 < j < n, S 1,j+1 = S 2,j + 1 always holds) and get a decomposition S 1 S 2 · · · S t of S. From Lemma 3.8 it is clear that such a decomposition guarantees t = l and S i = g(T i ) for each i, 1 ≤ i ≤ l. Therefore we proved that Φ is a bijection.
Example 3.12 Let n = 13, k = 6, and l = 3. Figure 6 shows an example of the map Φ.

Major index of Schröder n-paths
Let P be a Schröder n-path that goes from the origin (0, 0) to (n, n) with k F steps, we can associate with P a word w = w(P ) = w 1 w 2 · · · w 2n−k over the alphabet {0, 1, 2} with exactly We say that w has a descent in position i, 1 ≤ i ≤ n − 1, if w i > w i+1 . The descent set of w is the set of all positions of the descents of w, D(w) = {i : 1 ≤ i ≤ n − 1, w i > w i+1 }. The major index of w is defined as maj(w) = i∈D(w) i. And define maj(P ) = maj(w(P )).
In [2], Bonin, Shapiro and Simion study the major index for Schröder paths and gave the following result: here the sum is over all Schröder n-paths with exactly k F steps. Note that Equation (4.1) differs from Equation (1.6) and (1.7) only by a factor of a power of q. Readers might wonder if there is some simple explanation on these relations. In fact there is an obvious bijection θ between SSYTs in RInc k (2 × n) and Schröder words of length 2n − k that contain exactly k 1s. Given T ∈ RInc k (2 × n), we read the numbers from 1 to 2n − k in increasing order, if i appears only in row 1 (2), we set w i = 0(2), otherwise we set w i = 1, and define θ(T ) = w 1 w 2 · · · w 2n−k . A naive thinking is that if i is an ascent of T , then i is a descent of θ(T ), but this is not always true. When i and i + 1 both appear in row 1 and row 2 of T , then we have i ∈ A(T ) but i / ∈ D(θ(T )). It would be interesting if one can find a simple combinatorial explanation on the relations between descent (ascent) sets of a row-increasing tableaux in RInc k (2 × n) and the descent sets of the correspongding Schröder n-paths with k F steps.