The Combinatorics of Associated Laguerre Polynomials

The explicit double sum for the associated Laguerre polynomials is derived combinatorially. The moments are described using certain statistics on permutations and permutation tableaux. Another derivation of the double sum is provided using only the moment generating function.


Introduction
The classical orthogonal polynomials of Hermite, Laguerre, and Jacobi have explicit representations as terminating hypergeometric series. In the Askey scheme of orthogonal polynomials, each polynomial is given by either a hypergeometric or basic hypergeometric series. These polynomials have been studied extensively combinatorially.
The associated versions of the classical polynomials are expressed as terminating hypergeometric double sums. For example, the coefficient of x k in the associated Laguerre polynomial of degree n is given by a terminating 3 F 2 series. The analogous expansion for the associated Askey-Wilson polynomials uses a 10 φ 9 series, see [5].
The combinatorics of the classical orthogonal polynomials extends easily to include the association parameter. However it is not clear how these double sums occur in the explicit representations. The purpose of this paper is to derive these representations in the Laguerre case, Theorem 2.2, by giving combinatorial interpretations for the polynomials. We also find combinatorial expressions for the moments, see Theorem 3.3 and Theorem 3.4. We use the standard methods of Motzkin paths and permutation statistics to carry this out. Along the way we show that certain statistics on permutation tableaux describe the moments, Theorem 4.1.
As a bonus, in Section 5 we verify an explicit double sum expression for associated Laguerre polynomials from knowledge of the moment generating function. An explicit measure or a verification of the 3-term recurrence is not necessary. This method could be used in other settings to find explicit representations for polynomials when we know neither the measure nor the moments.
For the purposes of this paper it is convenient to replace the parameters α and c by X = c and Y = α + c + 1.
Date: January 19, 2015. The first author was partially supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education (NRF-2013R1A1A2061006). The second author was supported by NSF grant DMS-1148634. 1 Definition 2.1. The associated Laguerre polynomials L n (x; X, Y ) are given by (1) with b n = 2n + X + Y, λ n = (n + X)(n + Y − 1).
The double sum expression for L n (x; X, Y ) we shall prove is the following. We use the notation (a) k = a(a + 1)(a + 2) . . . (a + k − 1). Theorem 2.2. For any non-negative integer n, We shall prove Theorem 2.2 using the even-odd polynomials in [1, p. 40]. A bonus of this approach is that a closely related set of associated Laguerre polynomials, referred to as Model II [4, p. 158] may be simultaneously studied.  n (x; X, Y ) are given by (1) Theorem 2.4. For any non-negative integer n, We now review [1, p. 40], [10, p. 311], the basic setup for the even-odd polynomials.
Suppose that a set of monic orthogonal polynomials P n (x) satisfies (1) with b n = 0 and λ n = Λ n arbitrary. Then P n (x) are even-odd polynomials, that is, for some monic polynomials e n (x) and o n (x).
We see that the choices of give the associated Laguerre polynomials (2) o n (x) = L n (x; X, Y ), e n (x) = L (2) n (x; X, Y ). We shall give combinatorial interpretations for o n (x) and e n (x) which will prove Theorem 2.2. Let Note that we have the following recursive relations.
In order to give combinatorial interpretations for O n,k and E n,k , we need several definitions.
Definition 2.6. A k-marked permutation is a permutation π in which the integers in an increasing subsequence of length k are marked.
We now give combinatorial interpretations for the coefficients E n,k and O n,k . Proposition 2.11. Let E n,k and O n,k be determined by the following recursions with initial conditions E n,k = O n,k = 0 if n < k, k < 0, or n < 0; E n,k = O n,k = 1 if n = k ≥ 0: Proof. We will show that satisfy the recurrence relations. Let π ∈ S E n,k . If n is marked in π, then we can consider π as an element of S O n−1,k−1 . Otherwise, π is obtained from an element σ in S E n−1,k by inserting an unmarked integer n in n different ways. Then we always have RLmin (π) = RLmin (σ) and LRmin (π) = LRmin (σ) unless we insert n at the end in which case we have RLmin (π) = RLmin (σ) + 1 and LRmin (π) = LRmin (σ). This implies that A n,k = B n−1,k−1 + (n − 1 + X)A n−1,k . Now let π ∈ S O n,k . If the last element of π is the marked n + 1, then by removing n + 1 we can consider π as an element of S E n,k without changing RLmin (π) and LRmin (π). Otherwise, let j be the last element of π. Since marked integers form an increasing sequence and n + 1 is marked, j must be unmarked. By removing j and relabeling in order-preserving way, we can consider π as an element σ in S O n−1,k . Then we have RLmin (π) = RLmin (σ) and LRmin (π) = LRmin (σ) unless j = 1 in which case we have RLmin (π) = RLmin (σ) and LRmin (π) = LRmin (σ) + 1.
The next Proposition constitutes proofs of Theorems 2.2 and 2.4.
Suppose that m + 1 = a 1 < min(w 0 ). In order to construct such π = w 0 a * 1 w 1 a * 2 w 2 · · · a * k w k we choose a 2 , a 3 , . . . , a k in n−m−1 k−1 ways, select a permutation σ of [m], shuffle σ with a 2 , a 3 , . . . , a k , insert a 1 = m + 1 at the beginning, and insert the remaining n − k − m integers. Clearly, LRmin (π) = LRmin(σ). In this construction, suppose that b 1 < b 2 < · · · < b n−k−m are the remaining integers to be inserted. We insert b 1 first, and then b 2 , and so on. Then during the ith insertion process RLmin (π) is increased by 1 if and only if b i is inserted just before a j+1 for the unique integer j with a j < b i < a j+1 . Thus inserting b i contributes a factor of (X + k + m + i − 1). Summarizing these we obtain that the sum of X RLmin (π) Y LRmin (π) for such π's is equal to Here we used the well known fact σ∈Sm Y RLmin(σ) = (Y ) m , see [11]. Similarly, one can show that the sum of X RLmin (π) Y LRmin (π) for π = w 0 a * 1 w 1 a * 2 w 2 . . . a * k w k with m + 1 = min(w 0 ) < a 1 is equal to Thus we have which is an expression for E n,k as sum of two terminating 3 F 2 (1)'s. However a 3 F 2 (1) contiguous relation, which is equivalent to the binomial relation gives the stated formula for E n,k as a single 3 F 2 (1). By the same argument, we obtain one can obtain the single sum formula for O n,k .
Remark 2.13. The 3 F 2 in Theorem 2.2 which is the coefficient of x k is not equal to the 3 F 2 given in [4, (5.6.18)]. These two 3 F 2 's are related by a 3 F 2 (1) transformation.

Moments of associated Laguerre polynomials
In this section we interpret the moments for the measures of the associated Laguerre polynomials. This is a routine application of Viennot's [12] theory, details are not provided. We need the following definition in order to give a combinatorial description for the moments.
Definition 3.1. Let π = π 1 π 2 · · · π n be a permutation. If π i is both a left-to-right maximum and a right-to-left minimum, then π i is called a pivot of π. We denote by pivot(π) the number of pivots of π. Note that if π i is a pivot, then π i = i, since π i is greater than all of π 1 , π 2 , . . . , π i−1 and smaller than all of π i+1 , π i+2 , . . . , π n .
Suppose that the orthogonal polynomials p n (x) satisfy (1). Then the n th moment, µ n , of a measure for p n (x) is a polynomial in the sequences {b k } k≥0 and {λ k } k≥1 . We denote this moment by The first result uses a special choice for b 0 .
Proof. This can be proved using one of the bijections between Laguerre histories and permutations, for instance, see [9].
When Z = 1/Y in Lemma 3.2, we obtain moments for the Model II associated Laguerre polynomials.
It is more natural to write the n th moment of the associated Laguerre polynomial as a sum over S n+1 . The next lemma, which is a restatement of the even-odd polynomial construction in Section 2, accomplishes this.
Theorem 3.6. The moments for the associated Laguerre polynomials are Proof. Let a 2k−1 = k − 1 + X and a 2k = k − 1 + Y for k ≥ 1 and a 0 = 0. Then by Lemma 3.5, we have By applying Theorem 3.3 to the above equation, we obtain the desired identity.
As a corollary of the proof above, we obtain the following generating function identity.

Permutation tableaux
In this section we reinterpret the weights on permutations of the previous section to be weights on another set of objects also counted by n!, called permutation tableaux. Permutation tableaux were introduced by Postnikov [8] in the study of totally nonnegative Grassmanian and studied extensively.
A permutation tableau is a filling of a Ferrers diagram using 0's and 1's such that each column contains at least one 1 and there is no 0 which has a 1 above in the same column and a 1 to the left of it in the same row. We allow a Ferrers diagram to have empty rows.
The length of a permutation tableau is the number of rows plus the number of columns. Let PT n denote the set of permutation tableau of length n. If a permutation tableau has length n, we label the southeast border of a Ferrers diagram with 1, 2, . . . , n starting from northeast to southwest. The rows and columns are also labeled by the same labels as in the border steps contained in the rows and columns. An example of a permutation tableau is shown on the left of Figure 1.
In a permutation tableau, a topmost 1 is a 1 which is the topmost 1 in the column containing it. A restricted 0 is a 0 which has a 1 above it in the same column. A rightmost restricted 0 is the rightmost restricted 0 in the row containing it. A row is called restricted if it has a restricted 0, and unrestricted otherwise. We denote by urr(T ) the number of unrestricted rows in T . Similarly, a 0 in T ∈ PT n is called c-restricted if it has a 1 to the left of it in the same row. A column is called restricted if it has a c-restricted 0, and unrestricted otherwise. We denote by urc(T ) the number  Figure 1, then the unrestricted rows are rows 1,6,8,9,16, and the unrestricted columns are columns 3,5,7,10,13,15. Corteel and Nadeau [3] found a bijection Φ : PT n → S n such that if Φ(T ) = π, then urr(T ) = RLmin(π). Corteel and Kim [2] found a generating function for permutation tableaux according to the number of unrestricted columns.
The alternative representation of a permutation tableau is the filling of the same Ferrers diagram obtained by replacing each topmost 1 by an up arrow, each rightmost restricted 0 by a left arrow, and leaving the other entries empty. See Figure 1. The diagram obtained in this way is called an alternative tableau. Equivalently, an alternative tableau can be defined as a filling of a Ferrers diagram with uparrows and leftarrows such that each column contains exactly one uparrow and there is no arrow pointing to another arrow. Alternative tableaux were first considered by Viennot [13] and studied further by Nadeau [7].
In this section we prove the following theorem.
Theorem 4.1. The moments for the associated Laguerre polynomials are For the remainder of this section we shall prove the above theorem. It is easy to see that in the alternative representation of T ∈ PT n , a row is unrestricted if and only if it has no left arrow. There is also a simple characterization for unrestricted columns in the alternative representation. Proof. Suppose that a column, say column c, is restricted. Then column c has a c-restricted 0. Suppose that this c-restricted 0 has a 1 in column c to the left of it in the same row. Then by the condition for permutation tableaux, column c cannot have a 1 above the c-restricted 0. Thus the topmost 1 in column c is strictly to the northwest of the topmost 1 in column c. Since topmost 1s correspond to uparrows in the alternative representation, the uparrow in column c has an arrow strictly to the northwest of it. The converse can be proved similarly.
We will use the following property of the map Φ. If Φ(T ) = π, then i is the label of a column if and only i is followed by a smaller integer in π. Proof. From the construction of the bijection Φ, the labels of unrestricted rows of T are the right-to-left minima of π. Since π and π * have the same right-to-left minima, we obtain the first item.
We now prove the second item. Suppose that column c is an unrestricted column of T . Let row r be the row containing the uparrow of column c. By Lemma 4.2, row r must be unrestricted. Let r 1 < r 2 < · · · < r k be the labels of unrestricted rows of T , and let r = r i . By the construction of the map Φ, before inserting integers from column c, the integers that have been inserted are located to the right of r i−1 . Then after inserting integers from column c, we have c to the left of r i with no integers between them. Since the remaining integers to be inserted are less than c, every integer greater than c is located either to the right of r i or between r i−1 and c in π = Φ(T ). Thus c is a left-to-right maximum of π * . Since c is to the left of r i and c < r i , c is not a right-to-left minimum of π * . Now let c be a left-to-right maximum of π * that is not a right-to-left minimum. Let r 1 < r 2 < · · · < r k be the right-to-left minima of π (equivalently π * ). Suppose that c is located between r i−1 and r i . It is easy to see that c is followed by a smaller number in π. Thus c is the label of a column in T . If c is a restricted column, then by Lemma 4.2, there is an arrow strictly to the northwest of the uparrow in column c. Then the column containing such an arrow has label c > c. When we insert integers from column c in the construction of Φ(T ), c is located to the left of r i−1 . Then c cannot be a left-to-right maximum of π * , which is a contradiction. Thus column c is unrestricted.
The above lemma implies the following.

Proof via the moment generating function
In this section we prove the following equivalent version of Theorem 2.2 using only the recurrence (7) for the moments.
The recurrence (7) is equivalent to knowledge of the moment generating function, which is a Jacobi continued fraction.
First we note that the moment generating function is a quotient of hypergeometric series. This appears in [6, Theorem 6.5, p. 212], or may easily be verified using contiguous relations for 2 F 0 (x) to find the relevant Jacobi continued fraction.
Proposition 5.1. The generating function, as a formal power series in x, for the associated Laguerre moments is .
Then Proposition 5.1 is equivalent to Let L be the linear functional which uses the moments θ n (X, Y ), i.e., L(x n ) = θ n (X, Y ). We prove (6) by showing that the linear functional L satisfies (8) L(x s L n (x; X, Y )) = 0, for 0 ≤ s ≤ n − 1.
We can rewrite L n (x; X, Y ) as First we establish the s = 0 case of (8). Applying the linear functional L and using (7), we obtain, for n ≥ 1, We'll finish proving (8) for 0 < s < n by obtaining, at the last step, an n th difference of a polynomial of degree s, which is zero.