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Generalized Dumont-Foata polynomials and alternative tableaux

Dumont and Foata introduced in 1976 a three-variable symmetric refinement of Genocchi numbers, which satisfies a simple recurrence relation. A six-variable generalization with many similar properties was later considered by Dumont. They generalize a lot of known integer sequences, and their ordinary generating function can be expanded as a Jacobi continued fraction. We give here a new combinatorial interpretation of the six-variable polynomials in terms of the alternative tableaux introduced by Viennot. A powerful tool to enumerate alternative tableaux is the so-called"matrix Ansatz", and using this we show that our combinatorial interpretation naturally leads to a new proof of the continued fraction expansion.

They show that the polynomial F n is symmetric in x, y, and z, with non-negative coefficients, and such that F n (1, 1, 1) = G 2n+2 . Another nice property is that the generating function ∞ n=1 F n t n can be expanded as a J-fraction (more precision will be given further in this introduction). Gessel and Zeng [7] showed that F n+1 is the nth moment of some orthogonal polynomials known as continuous dual Hahn polynomials, which are an important sequence in the Askey-Wilson hierarchy [8].
Dumont [4] conjectured that we have the following J-fraction for Γ n t n : where the parameters b n and λ n are defined by: b n = (x + n)(ȳ + n) + (y + n)(z + n) + (z + n)(x + n) − n(n + 1), λ n = n(x + y + n − 1)(ȳ + z + n − 1)(z + x + n − 1). (5) This was proved independently by Randrianarivony [10] and Zeng [14] (and of course this implies the J-fraction expansion for F n t n ). More precisely, Randrianarivony's method consists in the study of a Stieltjes tableau and Zeng's method consists in calculations of Hankel determinants.
The main goal of this article is to give a new combinatorial interpretation of Γ n in terms of alternative tableaux [9,13] and six statistics on them, and obtain as a consequence a new proof of the continued fraction expansion (4). Alternative tableaux were introduced by Viennot [13] in the context of a model of statistical physics called Partially Asymmetric Simple Exclusion Process (PASEP), and previous work of Corteel and Williams [2]. The "matrix Ansatz" first appeared in [3], as a way to obtain the stationary distribution of the PASEP. In the combinatorial context, it is a method to enumerate these alternative tableaux in terms of operators satisfying certain relations. We will also describe an analog of the matrix Ansatz to enumerate escaliers, which are the combinatorial objects used by Dumont to define Γ n in [4]. This article is organized as follows. In Section 2 we give definitions and known facts about alternative tableaux and the matrix Ansatz. In Section 3 we prove the new combinatorial interpretation of Γ n in terms of alternative tableaux using the recurrence (3). Section 4 contains our new proof of the continued fraction expansion (4). In Section 5, we describe the analog of the matrix Ansatz to enumerate escaliers.

Alternative tableaux
Throughout this article we use the French convention for Young diagrams, and Young diagrams may contain rows or columns of size 0. Any Young diagram is characterized by its upper-right boundary, which is a sequence of unit steps going left or going down. We will encode this sequence by a word in the two letters D and E, so that D corresponds to the step → and E corresponds to the step ↓. For example, DDEDE is the Young diagram with two rows of respective lengths 2 and 3.
Definition 2.1. Let λ be a Young diagram. An alternative tableau of shape λ is a filling of λ such that each cell is either empty, contains an arrow ← or an arrow ↓, and each arrow has a clear view to the boundary. More precisely, all cells below a ↓ in the same column (or to the left of a ← in the same row) are empty. A column (respectively, row) of an alternative tableau is free if it contains no ↓ (respectively, no ←). We denote by fr(T ) (respectively, fc(T )) the number of free rows (respectively, free columns) of T . See Figure 1 for examples. We use here the notation with arrows as introduced by Nadeau [9]. Alternative tableaux of a given shape can be enumerated via a method called matrix Ansatz. This method appeared in the context of a model of statistical physics (the partially asymmetric simple exclusion process), where it is used to derive the stationary probabilities of any state of the process. Proposition 2.2 (Corteel-Williams [2]). Let W | be a row vector, |V a column vector, and D and E matrices such that: Let w be a word in the two letters D and E, then we have where the sum is over alternative tableaux T of shape w.
The result of Corteel and Williams was actually stated in terms of permutation tableaux, which are slightly different objects. We refer to Viennot [13] and Nadeau [9] for the corresponding statement in terms of alternative tableaux, and for the bijection between permutation tableaux and alternative tableaux. We have chosen to use alternative tableaux in this work because of their symmetry. Indeed, there is an elementary involution on alternative tableaux which is conjugation. To conjugate a tableau, take the image of the whole picture with respect to the South-West to North-East axis symmetry (in particular, the ← and ↓ are exchanged). See [9] for details.
Note that relations (6) ensure that W |w|V is well-defined and can be computed explicitly. Indeed, we can use DE − ED = D + E to obtain some coefficients c i,j such that w = i,j c i,j E i D j , and from the other relations we can obtain W |w|V . We refer to [2,1] for more details.
Although not necessary to compute W |w|V for a given word w, it is useful to have explicit matrices satisfying the PASEP matrix Ansatz. It can be checked [3] that the following N × N-matrices: y +xx + 1 y +x + 1x + 2 They are essentially a particular case of matrices defined by Derrida & al [3] in the context of the PASEP. As for the vectors, we can take W | = (1, 0, 0, . . . ) and |V = (1, 0, 0, . . . ) * , and all relations in (6) are satisfied.

The new combinatorial interpretation of Γ n
It is known [13] that G 2n+2 is the number of alternative tableaux whose shape is the staircase with n rows and columns, i.e. the Young diagram corresponding to the word (DE) n . In [1], we have given three statistic in staircase alternative tableaux to give a combinatorial interpretation of F n (x, y, z). These are: the number of free rows, the number of free columns, and the number of corners containing ← or ↓. Here, we give six statistics for the more general case of Γ n . Another difference is that in [1], the combinatorial interpretation was derived from the J-fraction for F n t n , but here we use the recurrence relation (3) to prove the result.
Definition 3.1. A column (respectively, row) of an alternative tableau is empty if it contains no ↓ nor ←. Let T be an alternative tableau. We denote by: • emr(T ), the number of empty rows in T , • fnc(T ), the number of free non-empty columns in T , • dco(T ), the number of corners containing a ↓ in T , • fnr(T ), the number of free non-empty rows in T , • emc(T ), the number of empty columns in T , • lco(T ), the number of corners containing a ← in T .
Moreover let T n be the set of alternative tableaux whose shape is the staircase Young diagram with n rows and n columns.
For example, the rightmost tableau in Figure 1 is in T 5 , and the six statistics that we have just defined are respectively 2, 2, 2, 0, 1, 0. The main new result of this article is the following: Proof. Both sides are equal to 1 when n = 1, so it suffices to show that the righthand side satisfies the recurrence relation (3). We distinguish six kinds of tableaux in the set T n−1 , according to the content of their leftmost column and upper left corner. Assuming that the theorem is true for n − 1, we will show that these six kinds of tableaux have generating functions which add up to the right-hand side of (3). This is summarized in the following table.
The leftmost column is: The upper left corner contains: For example we will show that the tableaux of the fourth kind (case 4), i.e. those having an empty leftmost column, have generating function xȳΓ n−1 . The three cells containing a × in this table do not correspond to any tableaux.
Recursive construction of staircase alternative tableaux.
• Case 1. When the upper-left corner contains a ↓, there is no other arrow in the leftmost column. This corresponds to the first picture in Figure 2. In this first kind of tableaux, the topmost row is free non-empty, and the upper left corner contains a ↓, so this gives a factorxz. After removing the leftmost column, there can remain any tableau in T n−2 , hence the factor Γ n−1 . So the first kind of tableaux gives indeed the termxzΓ n−1 .
• Case 2. There is a factor yz since we assume that there is a ← in the upper left corner, and that the leftmost column is free and non-empty. These tableaux can be obtained the following way: consider any tableau T in T n−2 , add a column to its left with a ← in the topmost cell of the added column. We color some of the other cells in the added column in gray as in the second picture in Figure 2, such that a cell is colored if there is no ← to its right. Then, decide whether each gray cell is empty or contains a ←. All tableaux of the second kind can be obtained this way, and the gray cells are in correspondence with the free rows of T . At the level of generating functions, this amounts to substitute x with x + 1 andx withx + 1. Indeed, an empty (respectively, free non-empty) row remains so if we add nothing in the gray cell, but becomes non-free if we add a ←.
• Case 3. This corresponds to the second picture in Figure 2, but with the assumption that there is a ↓ in one of the gray cells. Let us consider the set S of tableaux of the second kind such that there is at least a ← in some gray cell. This set has generating function yz(Γ + n−1 −Γ n−1 ), indeed we have already yzΓ + n−1 for the all tableaux of the second kind and the term −yzΓ n−1 removes the cases where all gray cells are empty. Then, there is a bijection between this set S and the tableaux of the third kind. Indeed, let T ∈ S, consider the bottommost ← in the leftmost column of T , and replace this ← with a ↓. This way, we obtain exactly the tableaux of the third kind. Replacing the ← with ↓ gives a factor xy −1 at the level of generating functions. Thus we obtainxz(Γ + n−1 − Γ n−1 ) for the third kind of tableaux.
• Case 4. This is similar to case 1, and corresponds to the third picture in Figure 2. Here, removing the first column gives a factor xȳ since we assume that the leftmost column is empty, and hence the upper row is empty. There can remain any tableau in T n−2 , hence the factor Γ n−1 . Thus we obtain xȳΓ n−1 for the fourth kind of tableaux.
• Case 5. This corresponds to the fourth picture in Figure 2, with the assumption that the gray cells contain no ↓ and at least a ←. Here the gray cells are obtained exactly as in case 2 above. Proceeding similar to case 3 above, we obtain xy(Γ + n−1 − Γ n−1 ) for the fifth kind of tableaux. • Case 6. There is a bijection between the fifth kind and the sixth kind of tableaux, similar to the bijection used in case 3. From a tableau of the fifth kind, consider the bottommost ← in the leftmost column, and replace it with a ↓. Replacing the ← with ↓ gives a factorxy −1 at the level of generating functions. Thus we obtain xx(Γ + n−1 − Γ n−1 ) for the sixth kind of tableaux. Adding the six terms in the above table, we get the right-hand side of (3). This shows that the right-hand side of (9) satisfies the same recurrence as Γ n , and completes the proof.
As previously mentioned, there is a simple bijection between alternative tableaux and permutation tableaux and it is possible to derive a combinatorial interpretation of Γ n in terms of permutation tableaux, but the result is much more natural with the alternative tableaux. In particular, the conjugation of alternative tableaux gives an easy way to prove a symmetry property of Γ n , which has been first given by Randrianarivony [10] and Zeng [14]. • if σ has signature 1, then Γ n (u, v, w,ū,v,w) = Γ n (x, y, z,x,ȳ,z), • if σ has signature -1, then Γ n (u, v, w,ū,v,w) = Γ n (x,ȳ,z, x, y, z). In particular, F n (x, y, z) = Γ n (x, y, z, x, y, z) is symmetric in x, y, and z.
From the combinatorial interpretation in (9) and using the conjugation of alternative tableaux, we have All symmetries given in the can be obtained by combining (10) and (11).
It is rather curious that one symmetry is obvious on the combinatorial interpretation, and another one in the recurrence relation. In the model in terms of escaliers (see Section 5), only one symmetry is obvious, and it is both in the combinatorial interpretation and in the recurrence relation. Note that any symmetry of a generating function necessarily appear in the coefficients of its expansion as a J-fraction, and indeed it is straightforward to check that the coefficients b n and λ n defined in (5) have the same symmetries as Γ n .

The continued fraction expansion
In this section, we show that Γ n can be calculated via the matrix Ansatz, and we derive as a consequence a new proof of the continued fraction expansion for Γ n t n . We consider the matrix where D, E are defined in (8), and I is the identity matrix. It turns out that we can exploit Proposition 2.2 to obtain the following: Proposition 4.1. For any n ≥ 0, we have To prove this, we need a few helpful definitions and lemmas.
Definition 4.2. Let T * n be the set of pairs (T, X) where T ∈ T n and X is a subset of the empty rows and columns of T . Such a pair (T, X) is called an extended tableau, and will be represented the following way: from a picture of T , each row or column in X is distinguished by a dashed line going through it. See Figure 3 for some examples. Given U = (T, X) ∈ T * n , we define • hr(U) as the number of dashed rows, • hc(U) as the number of dashed columns. For any statistic "stat" on alternative tableaux, and U = (T, X) ∈ T * n we define stat(U) = stat(T ). For any extended tableau U, we define the weight w(U) as the latest equality following from emr(T ) + fnr(T ) = fr(T ) and emc(T ) + fnc(T ) = fc(T ) for any alternative tableau T .  Proof. In the sum U ∈T * n w(U), we have distinguished two kinds of empty rows (dashed or non-dashed) with respective weights x −x andx instead of one kind of empty row with weight x. Similarly we have distinguished two kinds of empty columns (dashed or non-dashed) with respective weightsȳ − y and y instead of one kind of empty column with weightȳ. By an elementary argument, it is clear that these distinctions do not change the generating function. Here the corners are numbered from the upper left one to the lower right one. For example, the first extended tableau in Figure 3 has profile (1, 5, 1, 4, 6), and the second one has profile (5, 2, 3, 1, 4). Proof. Let w be the word obtained from i 1 . . . i n through the substitution 1 → ED, 2 → D, 3 → D, 4 → E, 5 → E, 6 → ǫ (ǫ being the empty word). The main point is that there is a bijection φ between elements in T * n of profile (i 1 , . . . , i n ) and alternative tableaux of shape w. Indeed, to build an extended tableau, once the contents of the corners are specified it remains only to choose an alternative tableau of a smaller shape. More precisely, the bijection can be done the following way: • for each empty corner of the extended tableau, remove the corresponding cell in the Young diagram, • shrink each dashed row or column, • for each corner of the extended tableau containing a ← (respectively, ↓), shrink the row (respectively, column) containing it.  See Figure 4 for an example, where we give the image of the two extended tableaux in Figure 3.
The weight of an extended tableau U of profile (i 1 , . . . , i n ) is the product where: • a is the number of non-dashed free columns in U, An important property of the bijection φ is that the free rows (respectively, columns) of φ(U) are in correspondence with non-dashed free rows (respectively, columns) of U. It follows that where the first sum is over extended tableau of profile (i 1 , . . . , i n ) and the second one is over alternative tableau of shape w. Now, examine the product M i 1 . . . M in . The factors D and E in this product readily gives the word w, and the other factors readily gives z bzd (x −x) e (ȳ − y) f , so: Using Proposition 2.2, the result follows from (18)   Using Equation (15) in Lemma 4.3 and Theorem 3.2, this is equal to Γ n+1 .
From the definitions of D and E in (8), the matrix M defined in (12) can be calculated explicitly and we obtain the following statement.
where b i and λ i are defined in (5).

Proof. We have
We have also Hence, M i,i+1 M i+1,i = λ i+1 . It is straightforward to check that other coefficients in M are 0, and this completes the proof.
As a direct consequence of Propositions 4.1 and 4.6, let us give a new proof of the continued fraction expansion given in (4). First, note that W |M n |V is the top-left coefficient (M n ) 0,0 of the matrix M n . This coefficient can be obtained by expanding the product M n and we obtain Since the matrix M is tridiagonal, we can restrict the sum to the case where two successive indices differ by at most 1, i.e. |i j − i j+1 | ≤ 1 for any j ∈ {0, . . . , n − 1} where i 0 = i n = 0. These indices thus define the successive heights of a Motzkin path. Then (23) shows that Γ n+1 can be seen as the generating function of Motzkin paths of n steps, with a weight b i for a level step at height i, and a weight λ i for a step ր between height i − 1 and i. By a standard argument [6] this implies the continued fraction given in (4).

The matrix Ansatz for escaliers
In the previous section, we have applied the link between alternative tableaux and matrices D and E satisfying DE − ED = D + E to obtain the continued fraction. In this section, we consider escaliers, which are the combinatorial objects used by Dumont [4] to define Γ n . We will show that these objects can be enumerated by a similar method, but with matrices B and A satisfying BA − AB = A + I.
We will denote a Young diagram by a word in B and A in the same way that we did with D and E (B and A respectively correspond to steps → and ↓ in the North-East boundary of the Young diagram).
Definition 5.1. A surjective pretableau is a partial filling a Young diagram with ×, such that there is at least one × in each row and at most one × in each column. A surjective tableau is a surjective pretableau such that there is exactly one × in each column. An escalier (of size n) is a surjective tableau of shape (BBA) n . See Figure 5 for some examples. The definition of Γ n by Dumont [4] is given in terms of escaliers of size n and six statistics on them. There is an obvious bijection between escaliers of size n and surjective pretableau of shape (BBA) n−1 (remove the bottom row of the escalier), so that Dumont's definition is equivalent to the following (as mentioned in the introduction, this is also known to be equivalent with Definition 1.1).

Definition 5.2.
A co-corner of the Young diagram (BBA) n is a cell which is the left neighbor of a corner (for example the upper-left cell is a co-corner). Let S n be the set of surjective pretableau of shape (BBA) n . A column is empty if it contains no ×. A × is doubled if there is another × in the same row. We denote by: • mi(T ), the number of empty columns of odd index, For example, the values of the six statistics on the first surjective pretableau in Figure 5 are 1, 1, 1, 2, 1, and 0. As for the second one, the values are 0, 1, 2, 2, 0, and 1. The fact that these objects also follow the recurrence (3) is seen by distinguishing several kinds of elements in S n−1 according to the content of the bottom row [10,14].
The analog of the matrix Ansatz for escaliers is given in the following proposition. The proof is similar to the case of alternative tableaux, and various examples of this kind of results were given in [1]. Proposition 5.3. Let W | be a row vector, |V a column vector, and A and B matrices such that: Let w be a word in the two letters B and A, then the number of surjective tableaux of shape w is W |w|V .
Proof. This is done by a recurrence on the number of cells in the Young diagram.
If there is no cell, then w = A i B j for some i and j, so W |w|V equals 0 if i > 0 or j > 0 and 1 otherwise. Since there is at least a × in each row and column of a surjective tableau, there is no such tableau of shape A i B j if i > 0 or j > 0, but we do have the "empty" surjective tableau in the "empty Young diagram" when i = j = 0. Next, consider a word w which is not in the form A i B j . It means we can factorize it in w = w 1 BAw 2 , and the factor BA corresponds to a corner of the Young diagram. We can distinguish three kinds of surjective tableaux of shape w, depending on the content of this corner.
• If the corner is empty, we can remove it and obtain any surjective tableaux of shape w 1 ABw 2 . By the recurrence assumption, their number is W |w 1 ABw 2 |V . • If the corner contains a doubled ×, we can delete the corner and its column, and obtain any surjective tableau of shape w 1 Aw 2 . Their number is W |w 1 Aw 2 |V . • If the corner contains a non-doubled ×, we can remove the corner and its row and column, and obtain any surjective tableau of shape w 1 w 2 . Their number is W |w 1 w 2 |V .
It follows that the number of surjective tableaux of shape w is W |w 1 ABw 2 |V + W |w 1 Aw 2 |V + W |w 1 w 2 |V = W |w 1 BAw 2 |V = W |w|V , and this completes the recurrence.
It is easily checked that the following N × N-matrices: We keep the definition of W | and |V as in the previous sections, since this also ensures that we have W |A = 0 and B|V = 0. To see how to use this result in the case of Γ n , let us begin with the particular case y = z =ȳ =z = 1. We know that the number of surjective tableaux of shape (BBA) n is W |(BBA) n |V . If we want to count surjective pretableaux, we have to authorize empty columns, and this is done by replacing B with B +I. Indeed, in the expansion of the product W |((B + I)(B + I)A) n |V , the choice of B or I in some factor corresponds to the choice of leaving a column empty or not. So the number of surjective pretableaux of shape (BBA) n is W |((B +I)(B +I)A) n |V . If we want to follow the empty columns of odd (respectively, even) index by the parameter x (respectivelyx), it suffices to mark the terms I, and we obtain Γ n+1 (x, 1, 1,x, 1, 1) = W |((B + xI)(B +xI)A) n |V .
As for the general case, in the same way that we have obtained Proposition 4.1 from the combinatorial interpretation in terms of alternative tableaux, we can obtain the following from the combinatorial interpretation in terms of escaliers. As in the previous section, we need some helpful definitions and lemmas.
Definition 5.5. The profile of T ∈ S n is the sequence (i 1 , . . . , i n ), where • i k = 1 if the kth co-corner and kth corner are empty, • i k = 2 if the kth co-corner and kth corner both contain a ×, • i k = 3 if the kth co-corner contains a non-doubled ×, • i k = 4 if the kth co-corner contains a doubled × and the kth corner is empty, • i k = 5 if the kth corner contains a non-doubled ×, • i k = 6 if the kth corner contains a doubled × and the kth co-corner is empty.
Proof. We follow the same scheme as in Proposition 4.5, and use Proposition 5.3. Here, the surjective pretableaux of a given profile are in bijection with surjective tableaux of a particular shape. Rather than giving a formal detailed proof, we sketch how to understand the matrices N 1 through N 6 , having in mind the proof of Proposition 5.3 and the way surjective tableaux are build recursively.
• The kth co-corner and kth corner correspond to the kth factor BBA in the word (BBA) n . If these are empty (i.e. i k = 1), we can remove the two cells and replace the factor BBA with ABB. With the terms xI andxI as seen before, we see that this case correspond to the matrix N 1 = A(B + xI)(B +xI).
• If i k = 2, i.e. the kth co-corner and kth corner both contain a ×, we can remove the two columns containing these ×, and remove the factor BB in BBA. But we need to distinguish two cases, depending on wether there is a third × in the same row or not, and if there is not, we also remove the row. This gives a factor A + I, and there is a weight yz because of the doubled × in the corner and co-corner. Hence this case correspond to the matrix N 2 = yz(A + I).
• If i k = 3, i.e. the kth co-corner contains a non-doubled ×, we can remove its row and column, so the kth factor BBA becomes a B +xI. There is a weight z for the non-doubled × in the co-corner. Hence this case gives the matrix N 3 = z(B +xI).
• If i k = 4, the difference with the previous case is that the × in the kth co-corner is doubled, so we do not remove its row. So there remains a factor A, and there is a weightz instead of z. Hence this case gives the matrix N 4 =zA(B +xI).
• If i k = 5, this is similar to the case when i k = 3. But the weight isȳ instead of z for the non-doubled × in the corner, and there remains a column of odd index so this gives a factor B + xI instead of B +xI. Hence this case gives the matrix N 5 =ȳ(B + xI).
• If i k = 6, this is similar to the case when i k = 4. But the weight is y instead of z for the doubled × in the corner, and there remains a column of odd index so this gives a factor B + xI instead of B +xI. Hence this case gives the matrix N 6 = yA(B + xI).
When we form the product N i 1 . . . N in , it is clear that the matrix N i k will impose the conditions on the kth co-corner and corner, and hence W |N i 1 . . . N in |V is the generating function for elements in S n of profile (i 1 , . . . , i n ). Now, we can prove Proposition 5.4.

Proof.
We have N = 6 i=1 N i , hence using Lemma 5.6: W |N n |V = This completes the proof.
From the definitions of B and A in (26), the matrix N defined in (27) can be calculated explicitly and we obtain the following statement.
Proposition 5.7. The matrix N = (N i,j ) i,j∈N is tridiagonal, and such that for any i ≥ 0 we have N i,i = b i and N i,i+1 N i+1,i = λ i+1 .
As in the case of alternative tableaux in the previous section, the previous two propositions means that the continued fraction expansion given in (4) can be derived from the combinatorial interpretation in terms of escaliers, and the matrix Ansatz for escaliers given in Proposition 5.3.
Remark 5.8. Observe that N = M since the non-diagonal coefficients are not the same, but the two matrices are equal after a permutation of the variables (x, y, z,x,ȳ,z). However, there is a priori no simple way to link the matrices M i 's with the N i 's, so we tend to think that despite the similarities in the method, there are two really different ways to obtain the continued fraction from a combinatorial model using the matrix Ansatz approach.
It is natural to ask if there is a bijection between T n and S n preserving the six statistics for Γ n . From the fact that the recurrence relation is checked on both sets, in theory it might be possible to describe recursively such a bijection. It would be quite interesting to give a better answer to this question by providing a direct bijection between our alternative tableaux and Dumont's escaliers.

Conclusion
Our main result is the new combinatorial interpretation of Γ n with the alternative tableaux. We obtain two new proofs of Dumont's conjecture, with the same method applied to the two different combinatorial interpretations of Γ n . What it is interesting about these proofs is that they fit in the general framework settled in [1], linking J-fractions, operators satisfying certain commutation relations, and combinatorial objects.