MOMENTS OF ORTHOGONAL POLYNOMIALS AND COMBINATORICS
SYLVIE CORTEEL, JANG SOO KIM AND DENNIS STANTON
Abstract. This paper is a survey on combinatorics of moments of orthogonal polynomials and linearization coe cients. This area was started by the seminal work of Flajolet followed by Viennot at the beginning of the 1980’s. Over the last 30 years, several tools were conceived to extract the combinatorics and compute these moments. A survey of these techniques is presented, with applications to polynomials in the Askey scheme.
1. Introduction
In this article we will consider polynomials Pn(x) in one variable x with coe cients in a commutative ring K indexed by a non-negative integer n. In general K will be C or R. Sometimes these polynomials will depend on formal variables a, b, c, d and q. Given a sequence µn n 0 of elements of K, there exists a unique linear functional on the { } n L space of polynomials, : K[x] K such that µn = (x ). The sequence µn n 0 is called the moment sequence of thisL linear! functional. L { }
Definition 1.1. Asequence Pn(x) n 0 is called an orthogonal polynomial sequence (OPS) with respect to a linear functional{ }if L (1) Pn(x) is a polynomial of degree n for each n 0, (2) (P (x)P (x)) = K for all n, m 0, where K = 0. L n m n n,m n 6 In this case we also call Pn(x) an orthogonal polynomial.
Note that if is a linear function for an OPS, then any linear functional 0 which is obtained by multiplyingL a nonzero constant to is also a linear functional for the OPS.L Conversely, if L L and 0 are linear functionals for an OPS, then = c 0 for a nonzero constant c.Thus,thereis a uniqueL linear functional for an OPS such thatL ·L(1) = 1. In this case we say that and its corresponding moments areLnormalized. Several booksL were written on the subject [2, 5,L 31, 27]. Here we focus on the moments of these polynomials and their relation to combinatorics started by Flajolet [12] and Viennot [33]. The Hankel determinant of the moment sequence µn n 0 is { } µ0 µ1 µn µ µ ··· µ 1 2 ··· n+1 n = . . . . . . . .. . µ µ µ n n+1 ··· 2n Theorem 1.2. [31] A linear functional has an OPS if and only if =0for all n 0. n Moreover, an OPS is uniquely determinedL up to a constant multiple. Indeed,6 the polynomials
The first author is funded by the IDEX Universit´eSorbonne Paris Cit´eProject “ALEA Sorbonne” and by the city of Paris: project Emergences “Combinatoire `aParis”. The second 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 third author is partially supported by NSF grant DMS-1148634. 1 2 SYLVIE CORTEEL, JANG SOO KIM AND DENNIS STANTON
Pn(x) in monic form, are given explicitly by :
µ0 µ1 µn µ µ ··· µ 1 2 ··· n+1 1 . . . . Pn(x)= . . .. . . n 1 µn 1 µn µ2n 1 ··· 1 x xn ··· One of the most important theorems in the classical theory of orthogonal polynomials is the three-term recurrence relation.
Theorem 1.3. Any monic OPS Pn(x) n 0 satisfies a three-term recurrence relation { } Pn+1(x)=(x bn)Pn(x) nPn 1(x), for n 1, for some sequences b ,b ,... and , ,..., where =0. 0 1 1 2 n 6 For non-monic orthogonal polynomials we can always rescale them to get monic ones. The three-term recurrence relation and the moments can be easily obtained as follows.
Proposition 1.4. Suppose that Pn(x) is an orthogonal polynomial with a three-term recurrence relation
Pn+1(x)=(Anx Bn)Pn(x) CnPn 1(x). Then the rescaled orthogonal polynomial Pn(x)=anPn(tx) satisfies
Pn+1(x)=(Anx Bn)Pn(x) CnPn 1(x), e where aen+1 e ean+1e e e an+1 An = tAn, Bn = Bn, Cn = Cn. an an an 1 Moreover, if µn n e0 and µn n 0 aree moment sequences ofe Pn(x) and Pn(x), respectively, then there exists{ a} constant {c such} that for all n 0, e e µ = c tnµ . n · n The orthogonality of orthogonal polynomials with real coe cients is defined by a measure e w(x), that is a functional : R[x] R defined by the Stieltjes integral L ! 1 (P (x)) = P (x)dw(x). L Z 1 The most general classical orthogonal polynomials are called the Askey-Wilson polynomials [3]. All of the other classical polynomials can be obtained as a specialization or a limit of these polynomials. The Askey-Wilson polynomials depend upon five parameters: a, b, c, d, and q and are defined by basic hypergeometric series. We use the usual notation k 1 m (a; q) = (1 aqi), (a ,...,a ; q) = (a ; q) . k 1 m k i k i=0 i=1 Y Y Definition 1.5. The Askey-Wilson polynomials are defined by n n n 1 (ab, ac, ad; q)n (q ,abcdq , az, a/z; q)k k (1) pn(x; a, b, c, d q)= n q | a (q, ab, ac, ad; q)k kX=0 1 z+z with x = 2 . MOMENTS OF ORTHOGONAL POLYNOMIALS AND COMBINATORICS 3
These functions are polynomials in x of degree n due to the relation k 1 (az, a/z; q) = (1 2axqj + a2q2j). k j=0 Y A representing measure for the Askey-Wilson polynomials may be given for 0
2. Path interpretation of the polynomials and the moments
For general orthogonal polynomials which satisfy Theorem 1.3, the moments µn are always polynomials in the three-term recurrence coe cients with non-negative integral coe cients. For example, 2 3 µ0 =1,µ1 = b0,µ2 = b0 + 1,µ3 = b0 +2b0 1 + b1 1. The orthogonal polynomials pn(x) may also be given as polynomials in x and the three-term recurrence coe cients, but not necessarily positive. For example, p (x)=1,p(x)=x b ,p(x)=x2 (b + b )x +(b b ). 0 1 0 2 0 1 0 1 1 The purpose of this section is to give combinatorial interpretations for these two phenomena using paths. Paths live in the quarter plane ⇧= N N.Apath ! is a sequence ! =(s0,...,sn)where s =(x ,y ) ⇧. The s ’s are the vertices⇥ of the path. The starting vertex is s ,theending i i i 2 i 0 vertex is sn and (si,si+1) is an elementary step of the path. The step (si,si+1) is called North if x = x and y +1=y ; • i i+1 i i+1 North-North if xi = xi+1 and yi +2=yi+1; • North-East if x +1=x and y +1=y ; • i i+1 i i+1 South-East if xi +1=xi+1 and yi 1=yi+1; and • East if x +1=x and y = y . • i i+1 i i+1 4 SYLVIE CORTEEL, JANG SOO KIM AND DENNIS STANTON
The paths are weighted. There is a partial application wt : ⇧ ⇧ K which associates to ⇥ ! each step (si,si+1)aweight wt(si,si+1) K.Theweight of the path ! is the product of the weight of the steps : 2 n 1 wt(!)= wt(si,si+1). i=0 Y Definition 2.1. [33] A Favard path is a path ⌘ =(s0,...,sn) starting at s0 =(0, 0) with three types of elementary steps: North, North-North and North-East. The weight of the elementary step (s ,s )isx if the step is North-East and b (resp. ) if the step is North (resp. i i+1 k k+1 North-North) and the y-coordinate of si is k.Thelength of the path is the y-coordinate of sn.
Let Favn be the set of Favard paths of length n.
Lemma 2.2. [33] Let Pn(x) be a sequence of polynomials which satisfies the recurrence of Theorem 1.3 and with boundary{ } conditions P (x)=1and P (x)=x b . Then 0 1 0 Pn(x)= wt(!). ! Fav 2X n A Motzkin path of length n is a path consisting of North-East steps, East steps, and South- East steps which lies in the first quadrant. Let Motn,k,` be the set of Motzkin paths from (0,k) to (n, `), and let Motn = Motn,0,0. Suppose that two sequences b = bn n 0 and = n n 1 are given. For a Motzkin path { } { } P ,wedefinewtb, (P ) to be the product of the weight of every step in P , where the weight of each step is defined as follows. (1) A North-East step has weight 1. (2) An East step has weight bk,wherek is the y-coordinate of starting point. (3) A South-East step has weight k,wherek is the y-coordinate of starting point.
For the remainder of this section we assume that Pn(x) are orthogonal polynomials given by P 1(x) = 0, P0(x) = 1, and for n 1, Pn+1(x)=(x bn)Pn(x) nPn 1(x), with the normalized linear functional and the nth moment µ = (xn). L n L Theorem 2.3. [33] We have
µn = wtb, (!). ! Mot 2X n We will sketch a proof of a generalization of Theorem 2.3 that also gives a combinatorial proof of the orthogonality. Theorem 2.4. [33] For all n, k, `, (2) (xnP (x)P (x)) = wt(!). L k ` 1 ··· ` ! Mot 2 Xn,k,` Proof. (Sketch [33].) The proof of this theorem uses the combinatorial interpretation of the polynomials as Favard paths. Let En,k,` be the set of triplets made of A Favard path f of length k • A Favard path g of length ` • A Motzkin path ! of length n + p(f)+p(g)wherep(f) is the number of North-East • steps of f. MOMENTS OF ORTHOGONAL POLYNOMIALS AND COMBINATORICS 5
Let Fn,k,` be the subset of En,k,` where p(f)=k, p(g)=` and ! starts with k North-East steps and ends with ` South-East steps. Proving the theorem is equivalent to defining a sign-reversing involution ✓ on En,k,` whose fixed points are exactly the elements in the set Fn,k,`. Let j be the minimal index such that (sj,sj+1) is an East step or a South-East step in !.Ifj
n 1 µnz = = C0(z); z2 n 0 1 X 1 b0z 2 2z 1 b z 1 . ..
` (xnP (x))zn = z` C (z); L ` 1 ··· ` i n 0 i=0 X Y min(k,`) ` k (xnP (x)P (x))zn = C (z) zC (z) zC (z). L ` k 1 ··· j j i i i i n 0 j=0 i=j+1 i=j+1 X X Y Y The last equation in Corollary 2.5 can be proved by observing the fact that we can uniquely decompose a Motzkin path ! Mot as 2 n,k,` ! = !k jD !1D!0U!10 U!`0 j, ··· ··· where j is a nonnegative integer, each !i and !i0 is a Motzkin path in which the starting and ending vertices have the smallest y-coordinates, D is a South-East step, and U is a North-East step. BIJECTIONS ON TWO VARIATIONS OF NONCROSSING PARTITIONS
JANG SOO KIM
Abstract. We find bijections on 2-distant noncrossing partitions, 12312-avoiding partitions, 3-Motzkin paths, UH-free Schr¨oder paths and Schr¨oder paths with- out peaks at even height. We also give a direct bijection between 2-distant noncrossing partitions and 12312-avoiding partitions.
E-mail address: [email protected]
Date:June18,2015. 2000 Mathematics Subject Classification. 05A18, 05A15. Key words and phrases. noncrossing partition, Motzkin path, Schr¨oder path. 6 SYLVIE CORTEEL, JANG SOO KIM AND DENNIS STANTON
1 2 3 4 5 6 7 8 9
Figure 1. The standard representation of ( 1, 4, 8 , 2, 5, 9 , 3 , 6, 7 ). Figure 1. The diagram representing the set partition{ }1,{4, 7 , }2,{5,}9 {, 3 }, 6, 7 . 2 1 2 {{ } { } { } { }} 2 2 3. Combinatorics 2 0 1 2 1 2 In this0 section we show that the moments of Hermite, Charlier, and Laguerre polynomials are equal to the numbers of perfect matchings, set partitions, and permutations respectively. A set partition of a set X is a collectionf of mutually disjoint nonempty subsets of X 0 whose union is X. We denote by ⇧n the set of set partitions of [n]= 1, 2,...,n . Let ⇡ = B ,B ,...,B ⇧ . Each element B of ⇡ is called a block.Thesize{ of a block} B is { 1 2 k}2 n 0 2 2 i 2 2 2 0 i the cardinality Bi . An edge of ⇡ is a pair (i, j) of integers such that i 3.1. Hermite polynomials. The Hermite polynomials Hn(x) are defined by H 1(x)=0,H0(x)=1, and Hn+1(x)=xHn(x) nHn 1(x), for n 0. A Dyck path of length n is a Motzkin path in Motn without East steps. A Hermite history of length n is a Dyck path of length n in which every South-East step between the lines y = i and y = i 1 is labeled by an integer in 0, 1,...,i 1 . { } By Theorem 2.3, the nth moment µn of the Hermite polynomials is µn = wtb, (P ), P Mot 2X n where b = bn n 0 and = n n 1 are given by bn = 0 for all n 0 and n = n for n 1. Let P Mot{ }.Thenwt (P{ )=0if} P has an East step. Otherwise, P is a Dyck path and 2 n b, BIJECTIONS ON TWO VARIATIONS OF NONCROSSING PARTITIONS JANG SOO KIM Abstract. We find bijections on 2-distant noncrossing partitions, 12312-avoiding partitions, 3-Motzkin paths, UH-free Schr¨oder paths and Schr¨oder paths with- out peaks at even height. We also give a direct bijection between 2-distant noncrossing partitions and 12312-avoiding partitions. E-mail address: [email protected] Date:June18,2015. 2000 Mathematics Subject Classification. 05A18, 05A15. Key words and phrases. noncrossing partition, Motzkin path, Schr¨oder path. MOMENTS OF1 ORTHOGONAL2 3 4 POLYNOMIALS5 6 7 AND8 COMBINATORICS9 7 1 2 3 4 5 6 7 8 9 10 11 12 2 1 0 1 0 0 FigureFigure 2. 1.AnThe example standard of the representation bijection H ofsending ( 1, 4, 8 perfect, 2, 5 matchings, 9 , 3 , 6 to, 7 Her-). mite histories. { } { } { } { } 2 1 2 2 2 wtb, (P ) is equal to the number of Hermite histories whose underlying2 Dyck0 path1 is P .Thus µn is equal to the number2 of Hermite histories of length n. 1 2 We now0 give a bijection : (n) from the set of perfect matchings of [n] H Mn !HH Mn to the set H (n) of Hermite histories of length n, which appears in [11, Section 2]. It su ces f0 to considerH the case when n is even since both sets are empty otherwise. Let ⇡ be a 2M2n perfect matching of [2n]. Then the corresponding Hermite history H (⇡) is defined as follows. 0 2 2 1 2 1 0 For i =1, 2,...,2n,ifi is an opener of ⇡,thentheith step of H (⇡) is a North-East step. If i is a closer0 of0⇡,thenthe1 ith step of H (⇡) is a South-East step labeled by the2 number of crossings (a, b), (c, d) of ⇡ with a The rescaled discrete q-Hermite polynomials hn(x; q) are defined by h 1(x; q) = 0, h0(x; q)= 1, and for n 0, n 1 hn+1(x; q)=xhn(x; q)e q [n]qhn 1(x; q). e e Again the odd moments are 0. The analogous version to Theorem 3.2 is the next result. e e e Theorem 3.3. [30, p. 310, (5.4)] The 2nth moment of the rescaled discrete q-Hermite polyno- mials hn(x; q) is µ = [1] [3] [2n 1] = qcrossing(⇡)+2 nesting(⇡), e 2n q q ··· q ⇡ 2XM2n where nesting(⇡) is the number of pairs of two edges (a, b) and (c, d) of ⇡ such that a 3.2. Charlier polynomials. The Charlier polynomials Cn(x) are defined by Cn+1(x)=(x n 1)Cn(x) nCn 1(x), for n 1 with initial conditions C 1(x) = 0 and C0(x) = 1. Theorem 3.4. The nth moment of the Charlier polynomials is equal to the number of set partitions of [n]. Theorem 3.4 may be derived using the representing measure for the Charlier linear functional 1 1 1 (p(x)) = p(n) . LCharlier e n! n=0 X We will instead prove a generalization of the above theorem using the q-Charlier polynomials Cn(x, a; q) given by Cn+1(x, a; q)=(x a [n]q)Cn(x, a; q) a[n]qCn 1(x, a; q). A Charlier history of length n is a Motzkin path of length n in which every South-East step between the lines y = i and y = i 1 is labeled by an integer in 0, 1,...,i 1 and every East on the line y = i is either unlabeled or labeled by an integer in {0, 1,...,i 1}. { } By Theorem 2.3, the nth moment µn of the Charlier polynomials is µn = wtb, (P ), P Mot 2X n where b = bn n 0 and = n n 1 are given by bn = n + a for all n 0 and n = an for n 1. Let {P }Mot .Then{ } 2 n unlabeled(⌘) ⌘ wtb, (P )= a q| |, ⌘ X where the sum is over all Charlier histories ⌘ with underlying Motzkin path P , unlabeled(⌘)is the number of unlabeled steps of ⌘ and ⌘ is the sum of labels in ⌘. | | We now give a bijection C :⇧n C (n) from the set ⇧n of set partitions of [n]tothe set (n) of Charlier histories of length!Hn. This map can be considered as a special case of HC 2 JANG SOO KIM 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 10 11 12 2 1 0 1 MOMENTS OF ORTHOGONAL POLYNOMIALS0 AND COMBINATORICS0 9 1 2 3 4 5 6 7 8 9 10 11 12 13 14 2 1 0 1 0 0 0 Figure 3. An example of the bijection C sending set partitions to Charlier histories. 1 2 3 4 5 6 7 8 9 Foata and Zeilberger’s map [13], which will be described in the next subsection. Let ⇡ ⇧ be 2 n a set partition of [n]. Then the corresponding Charlier history C (⇡)=⌘ is defined as follows. For i =1, 2,...,n,ifi is an opener of ⇡,thenthe2 +2ith+1 step of ⌘ is a North-East step. If i is a closer of ⇡,thentheith step of ⌘ is a South-East step labeled(0, 1) by the number of crossings (a, b), (c, d) of ⇡ with a 1 x (p(x)) = p(x)e dx. LLaguerre Z0 We will consider the following generalization of the above theorem. The q-Laguerre polynomials Ln(x; q) are defined by 2 Ln+1(x; q)=(x [n]q y[n + 1]q)Ln(x; a) y[n]qLn 1(x; q). A Laguerre history of length n is a Motzkin path of length n in which every North-East or South-East step between the lines y = i and y = i 1 is labeled by an integer in 0, 1,...,i 1 and every East step on the line y = i is either unlabeled or labeled by an integer{ in i, i +} 1,..., 1, 0, 1,...,i 1 . { } 2 JANG SOO KIM 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 10 11 12 2 1 0 1 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 2 1 0 1 0 0 0 10 SYLVIE CORTEEL, JANG SOO KIM AND DENNIS STANTON 1 2 3 4 5 6 7 8 9 10 11 12 13 14 +2 0 0 0 2 1 1 0 1 1 1 0 0 Figure 4. An example of the bijection L sending permutations to Laguerre histories. The matching North-East steps and South-East steps are connected by dotted lines. 1 2 3 4 5 6 7 8 9 By Theorem 2.3, the nth moment µn of2 the Laguerre+2 +1 polynomials is (0, 1) µn = wtb, (P ), P Mot 2X n (0, 0) where b = bn n 0 and = n n 1 are given by bn = y[n + 1]q +[n]q for all n 0 and = y[n]2 for{ n} 1. Let P {Mot} . Then it is easy to see that n q 2 n Figure 1. The standard representation ofw(⌘ () 1⌘, 4, 8 , 2, 5, 9 , 3 , 6, 7 ). wtb, (P )= y {q| |, } { } { } { } ⌘ X where the sum is over all Laguerre histories ⌘ with underlying Motzkin path P , w(⌘)isthe total number of North-East steps, unlabeled East steps, and East steps labeled by a nonnegative integer, and ⌘ is the sum of labels in ⌘. | | We now give a bijection L : Sn L(n) from the set Sn of permutations of [n]totheset (n) of Laguerre histories of length!Hn, which is due to Foata and Zeilberger [13]. We need HL the following definitions to describe L(⇡). Let ⇡ Sn be a permutation of [n]. An upper edge of ⇡ is a pair (i, j) such that i<⇡(i)=j. A lower2 edge of ⇡ is a pair (i, j) such that i = ⇡(j) For i =1, 2,...,n,ifi is an opener of ⇡,thentheith step of ⌘ is a North-East step whose label will be determined later. If i is a closer of ⇡,thentheith step of ⌘ is a South-East step. In this case suppose that the jth step and ith step are the matching North-East step and South-East step. Then the label of the jth step (resp. ith step) is the number of upper crossings (resp. lower crossings) (a, b), (c, d) of ⇡ with a Theorem 3.7. The q-Laguerre polynomials Ln(x; q) has the nth moment wex(⇡) crossing(⇡) µn = y q . ⇡ S X2 n Corteel et al. [8] showed that the moment has several interpretations in terms of permuta- tions, permutation tableaux, matrix anszatz, etc. Simion and Stanton [30] consider octabasic Laguerre polynomials (whose recurrence relation has eight independent q’s) and express their moments using various statistics of permutations. 4. The odd-even trick In this section we give a relation between moments for two di↵erent sets of orthogonal polynomials. We call this occurrence the odd-even trick. It occurs for the three cases considered in Section 3, and o↵ers alternative combinatorial interpretations for the moments. This idea appears in [5, p. 40]. Definition 4.1. For an orthogonal polynomial satisfying the Theorem 2.4, let µn = µn( bk k 0, k k 1) { } { } denote the nth moment. From Theorem 2.4, µn is a positive polynomial function of the three-term recurrence coe - cients. The odd-even trick is the observation that if bk = 0 for all k, the polynomials pn(x) are alternately even and odd. Moreover if 2 p2n(x)=en(x),p2n+1(x)=xon(x ), the even and polynomials en(x) and on(x) are themselves orthogonal polynomials. The moment sequences for en(x) and on(x) are related to the moment sequence of pn(x). Proposition 4.2. Given a sequence k with 0 =0we have µ2n(0, k k 1)=µn( 2k + 2k+1 k 0, 2k 1 2k k 1), { } { } { } µ2n+2(0, k k 1)= 1µn( 2k+1 + 2k+2 k 0, 2k 2k+1 k 1). { } { } { } Proof. There are several ways to prove this. We give a sketch of the combinatorial proof due to Viennot [34]. The first statement is equivalent to a bijection between 12 SYLVIE CORTEEL, JANG SOO KIM AND DENNIS STANTON Dyck paths of length 2n where South-East steps starting at y-coordinate k are weighted • by k, Motzkin paths of length n where the East steps (resp. South-East steps) starting at • y-coordinate k are weighted by 2k + 2k+1 (resp. 2k 2k 1). Starting with a Dyck path of length 2n,(s ,...,s ). For i from 0 to n 1, suppose that 0 2n s2i =(2i, 2k) If (s2i,s2i+1) and (s2i+1,s2i+2) are North-East steps, change them to one North-East • step; If (s ,s ) and (s ,s ) are South-East steps, change them to one South-East • 2i 2i+1 2i+1 2i+2 step weighted by 2k 2k 1; If (s ,s ) is a South-East step and (s ,s ) is a North-East step, change them • 2i 2i+1 2i+1 2i+2 to one East step weighted by 2k; If (s ,s ) is a North-East step and (s ,s ) is a South-East step, change them • 2i 2i+1 2i+1 2i+2 to one East step weighted by 2k+1. This gives a Motzkin path of length n with the correct weights. The second statement is equivalent to a bijection between Dyck paths of length 2n + 2 where South-East steps starting at y-coordinate k are • weighted by k Motzkin paths of length n where the East steps (resp. South-East step) starting at • y-coordinate k are weighted by 2k+1 + 2k+2 (resp. 2k+1 2k). One needs to delete the first and last step of the Dyck path and apply the same bijection and adjust the weights accordingly. ⇤ Example 4.3. The Hermite polynomials have bk = 0, k = k,so (2n 1)!! = µ2n(0, k k 1)=µn( 4k +1 k 0, 2k(2k 1) k 1). { } { } { } 2 Example 4.4. The Laguerre polynomials have bk =2k + 1, k = k , so 2 n!=µn( 2k +1 k 0, k k 1)=µ2n(0, (k + 1)/2 k 1). { } { } {b c} Example 4.5. The q-Charlier polynomials have bk = a +[k]q, k = a[k]q, so in this case µn( b k 0, k k 1)=µ2n(0, ⇤k k 1), { } { } { } where [k/2]q, if k is even, ⇤k = (a, if k is odd. Example 4.6. The renormalized Askey-Wilson polynomials Qn(y) (see Proposition 5.1) 1 n y = a + a 2x, Q (y)=( 1) P (x) n n satisfy Theorem 1.3 with bk = Ak 1 + Ck, k = AkCk. In this case µn( bk k 0, k k 1)=µ2n(0, ⇤k k 1), { } { } { } where Ck/2, if k is even, ⇤k = (A(k 1)/2, if k is odd. The next Theorem follows from Example 4.6. MOMENTS OF ORTHOGONAL POLYNOMIALS AND COMBINATORICS 13 Theorem 4.7. Let ✓2n = µ2n(0, ⇤k k 1). { } The Askey-Wilson moments satisfy n n n s n s 2 µn(a, b, c, d q)= (a +1/a) ( 1) ✓2n 2s. | s s=0 X ✓ ◆ Using the Proposition 4.2 we see Corollary 4.8. Given a sequence a for k 0 with a =0we have k 0 µn( a2k+1 + a2k k 0, a2ka2k 1 k 1)=a1µn 1( a2k+2 + a2k+1 k 0, a2k+1a2k k 1). { } { } { } { } Example 4.9. The Laguerre polynomials have b =2k + 1, = k2, so a = k/2 k k k d e 2 (n + 1)! = µn+1( 2k +1 k 0, k k 1)=µn( 2k +2 k 0, k(k + 1) k 1). { } { } { } { } 5. Askey-Wilson moments In this section we consider the Askey-Wilson polynomials and explicit forms for their mo- ments. Recall that these polynomials in x have five parameters: a, b, c, d, and q.InSub- section 5.1 we give explicit formulas for the moments. In Subsection 5.2 we use a non-linear change of variables on the parameters to obtain four new parameters ↵, , , and . Using these parameters we give a combinatorial interpretation for the moments as weights of special tableaux. Recall from (1.5) that the Askey-Wilson polynomial pn(x; a, b, c, d q) is not monic and has n | the leading term (abcd; q)n(2x) . It is true, but not obvious, that the Askey-Wilson polynomial pn(x; a, b, c, d q) is symmetric in all four parameters a, b, c, and d, not just b, c, and d. Let | 1 Pn(x)= pn(x; a, b, c, d q) (abcd; q)n | be the normalized Askey-Wilson (AW) polynomial whose leading term is (2x)n. The moment sequences for Pn(x) and pn(x; a, b, c, d q) are the same. For historical reasons, and to agree with the literature, we keep the 2x. | Proposition 5.1. The normalized Askey-Wilson polynomials Pn(x) satisfy the three-term re- currence relation : Pn+1(x)=(2x bn)Pn(x) nPn 1(x) for n 0 with P 1(x)=0and P0(x)=1, 1 bn =(a + a An Cn), n = An 1Cn, where n n n n 1 (1 abq )(1 acq )(1 adq )(1 abcdq ) An = 2n 1 2n , a(1 abcdq )(1 abcdq ) n n 1 n 1 n 1 a(1 q )(1 bcq )(1 bdq )(1 cdq ) C = . n (1 abcdq2n 2)(1 abcdq2n 1) In this section, we will start by giving enumeration formulas for the moments of these poly- nomials. We will then present the combinatorics coming from staircase tableaux. We will end the section with some special cases. 14 SYLVIE CORTEEL, JANG SOO KIM AND DENNIS STANTON 5.1. Enumeration formulas. The moments of the AW polynomials, denoted µn(a, b, c, d q), are not polynomials in a, b, c, d and q. But it was recently proven [25, Proposition 2.1]| that 2n(abcd; q) µ (a, b, c, d q) are polynomials in a, b, c, d, q with integer coe cients. Several n n | explicit formulas for µn(a, b, c, d q) are known. The following is the simplest known expression as a double sum. | Theorem 5.2. [9, Theorem 1.13] We have 2 1 n (ab, ac, ad; q) m q j a 2j(aqj + q j/a)n µ (a, b, c, d q)= m qm . n | 2n (abcd; q) (q, q1 2j/a2; q) (q, q2j+1a2; q) m=0 m j=0 j m j X X Theorem 5.2 was proved using techniques built in [17, 18]. The formula has two clear defects. n It does not demonstrate the polynomiality of 2 (abcd; q)nµn(a, b, c, d q)ina, and it not obviously symmetric in the four parameters a, b, c, and d. | A fivefold sum formula was found where the polynomiality is clear. Corollary 5.3. [25, Theorem 5.6] We have n n n n u v w x 2 µn(a, b, c, d q)= n k n k a b c d | 2 2 1 kX=0 ✓✓ ◆ ✓ ◆◆ u+v+wX+x+2t=k (ac; q)v(bd; q)w t t+1 u + v + w + t v + w + x + t u + x + t ( 1) q( 2 ) , ⇥ (abcd; q) u v, w, x + t x v+w q q q where the second sum is over all integers u, v, w, x 0 and k t k/2 satisfying u + v + w + x +2t = k. A combinatorial approach to the proof of Corollary 5.3 is given in Section 7. When d = 0, the n moments 2 µn(a, b, c, 0 q) are polynomials and there is an explicit polynomial formula which also establishes symmetry.| Corollary 5.4. [25, Theorem 2.3] The Askey-Wilson moments for d =0are n n n n 2 µn(a, b, c, 0 q)= n k n k | 2 2 1 kX=0 ✓✓ ◆ ✓ ◆◆ u v w t t+1 u + v + t v + w + t w + u + t a b c ( 1) q( 2 ) , ⇥ v q w q u q u+v+Xw+2t=k where the second sum is over all integers u, v, w 0 and k t k/2 satisfying u+v+w+2t = k. There is a result showing symmetry using very-well poised basic hypergeometric series. Theorem 5.5. [25, Theorem 2.10] For an arbitrary A, n (aA, bA, cA, dA; q) n+1 n n 2nµ (a, b, c, d q)= m ( q)m W (m) n | (A2,abcd; q) 8 7 s s 1 m=0 m s=0 X X ✓✓ ◆ ✓ ◆◆ n 2s m n+2s+2p m + p n 2s p m( n+2s+p)+(m) A q 2 , ⇥ m m p=0 q q X where 2 m m 8W7(m)=8W7(A /q; A/a, A/b, A/c, A/d, q ; q, abcdq ), MOMENTS OF ORTHOGONAL POLYNOMIALS AND COMBINATORICS 15 and [14, Chap. 2.1] 5 1 (1 a q2k) (a ; q) (a ; q) W (a ; a ,...,a ; q, z)= 0 0 k i k zk. 8 7 0 1 5 (1 a ) (q; q) (qa /a ; q) 0 k i=1 0 i k kX=0 Y n The expression for 2 (abcd; q)nµn(a, b, c, d q) in Theorem 5.5 is clearly symmetric in a, b, c, and d. From TheoremMOMENTS 5.5 OFit may ORTHOGONAL be shown that POLYNOMIALS| it is a polynomial AND COMBINATORICS in these four parameters. 9 We do not give the details here. Instead we o↵er another representation which shows polynomiality but breakswhere symmetry 1/py + py + bn An 1Cn Bn = , ⇤n = (aA, bA, cA, dA; q) m1 mq (1 q)2 (3) m W (m)= (cd)j(A/c, A/d; q ) (ab; q) are given(A2; q by) the Askey-Wilson8 7 recurrencej with the substitutj ionsj a a/py, b m j=0 q ! ! bpy, c c/py, d dpy. X ! ! j j (Aaq , Abq ,cd; q)m j. We also know an enumeration formula : ⇥ The details appear in [25]. Theorem 6.4. [?] The results in Corollaries 5.3 and 5.4, and Theorem 5.5, all involve a product of di↵erences Z (y; ↵, , , ; q)= of binomial coe cients and q-binomialn coe cients. The reason for this unusual behavior is the formThis of Theorem formula 5.2. is a The polynomial power in in they; ↵ numerator, , , ; q with leads 4 tonn!coe binomial cients. coe Thiscients, is notwhile the denominatoreasy to terms see from lead the to previous di↵erences formula. of q-binomial Thereare coe severalcients. easy This pr dioof↵erence that this can is be true switched to thefor binomialy = q =1. coe cients, giving the di↵erences that are displayed. The proofsWhen ofy = theseq = results 1 this generating are analytic polynomial and use properties of the staircase of the tableau Askey-Wilson of size n functionalis AW . There is a combinatorial proof ofn the1 c = 0 case of Corollary 5.4 using weighted Motzkin L paths(3) in [25]. Josuat-Verg`esZn(1, ↵, , also, , 1) did = this( case↵ + (which+ + is + thei(↵ Al-Salam-Chihara+ )( + )) polynomials), see [20, Theorem 6.1.1]. The di↵erencesi=0 of binomial coe cients occur naturally in these com- Y binatorialWe approaches, give here a see simple (14). bijective A combinatorial proof based proof on of [? a]: general result remains open. 5.2. CombinatoricsDefinition 6.5. of the moments.An inversionWe define table here of size a combinatorialn is a table objectT such that that was for defined • to study the stationaryi 1, ..., ndistribution, T [i] is a ofnon-negative the asymmetric integer exclusion less than processi. with open boundaries 2 { } [32]. A colored inversion table of size n is a table T such that for i 1, ..., n , • 2 { } T [i]=(i 1)x with x ↵, , , or T [i]=jx,y with 0 j Figure 5.1. A staircase tableau of size 7 SeeProposition the left entry 6.6. of FigureThere 5 exists for an a weightexample preserving of a staircase bijection tableau. between staircase tableaux (with q =1)ofsizen and colored inversion tables of size n. Proof. Start with a staircase tableau of size n and number the columns from 1 to n from left to right and the rows from 1 to n from top to bottom. The bijection is obtained doing the following steps. For each column i, look at the topmost Greek letter in column i and count • the number of cells j directly to its left that does not contain any Greek letter. 16 SYLVIE CORTEEL, JANG SOO KIM AND DENNIS STANTON Definition 5.7. [10] The weight wt( ) of a staircase tableau is a monomial in ↵, , , , and q, which we obtain as follows. SomeT blank boxes of are assignedT a q, based on the label of the closest labeled box to its right in the same row andT the label of the closest labeled box below it in the same column, such that: every blank box which sees a to its right gets assigned a q; • every blank box which sees an ↵ or to its right, and a or below it, gets assigned • a q. After this assignment, the weight of ,wt( ) is then defined as the product of all labels in all boxes. T T The right entry of Figure 5 shows that this staircase tableau has weight ↵3 2 3 3q9.The type of the tableau is the number of ↵’s plus the number of ’s on the diagonal. We denote it by t( ). T Let T t( ) (4) Z (y, ↵, , , , q)= wt( )y T , n T XT where the sum is taken on the staircase tableaux of size n. We now link this generating polynomial to the moments of the Askey-Wilson polynomials. Let 1 q ↵ + + (1 q ↵ + )2 +4↵ a = , p 2↵ 1 q ↵ + (1 q ↵ + )2 +4↵ c = , p 2↵ 1 q + + (1 q + )2 +4 b = , p 2 1 q + (1 q + )2 +4 d = . p 2 Proposition 5.8 (see [9]; Theorem 1.11). The generating polynomial Zn(y; ↵, , , ; q) is (abcd; q) pyn(↵ )n µ n ⇥ n where µn are the moments of the orthogonal polynomials defined by Gn+1(x)=(x Bn)Gn(x) ⇤nGn 1(x), where 1/py + py + bn An 1Cn B = , ⇤ = n 1 q n (1 q)2 are given by the Askey-Wilson recurrence with the substitutions a a/py, b bpy, c c/py, ! ! ! d dpy. ! Remark 5.9. This proposition uses a result of [10] that relates the partition function of the ASEP and the generating polynomial of the staircase tableaux. The proof is quite complicated. It is still an open problem to find a simple combinatorial proof. n Remark 5.10. The Askey-Wilson moments 2 (abcd; q)nµn(a, b, c, d q) do not have positive coe cients as polynomials in a, b, c, d, and q. The change of variables| in the parameters miraculously does make these moments positive. MOMENTS OF ORTHOGONAL POLYNOMIALS AND COMBINATORICS 17 We also know an enumeration formula : Theorem 5.11. [9, Theorem 1.14] n n ↵ (ab, ac/y, ad; q)k k Zn(y; ↵, , , ; q)=(abcd; q)n q 1 q (abcd; q)k ✓ ◆ kX=0 k k j j k n (k j)2 2 j k (1 + y + q a + q y/a) q (a /y) . ⇥ (q, q2j 2k+1y/a2; q) (q, a2q1 2j+2k/y; q) j=0 k j j X This formula is a polynomial in y, ↵, , , , q with positive coe cients whose sum of coe - cients is 4nn!. This is not easy to see from Theorem 5.11. The special case y = q = 1 does make this clear. Proposition 5.12. The generating polynomial of the staircase tableau of size n satisfies n 1 (5) Zn(1,↵, , , ,1) = (↵ + + + + i(↵ + )( + )). i=0 Y We give here a simple bijective proof based on [7]. A colored inversion table of size n is a table T such that for i 1,...,n , T [i]=(i 1)x with x ↵, , , or T [i]=jx,y with 0 j If this letter, say x, is topmost and leftmost, record T [i]=ix. • Otherwise let y be the first Greek letter to the left of x and let z be the first Greek • letter under y.ThenT [i]=jx,z. We present the inverse of the algorithm. Start from a colored inversion table T of size n. Mark all the cells of the tableau as free. For i from n down to 1, if Ti =(i 1)x, put an x in in the ith column, as high as possible. Mark all the cells • to its left as reserved. Otherwise, Ti is equal to some jx,y. Put an x in the ith column, as high as possible. • Mark j boxes to its immediate left as reserved. Suppose that x is inserted in row k. Insert a y in column i j 1 as high as possible but in a row with a label larger than • k (such a cell always exists as one can check that the diagonal box of column i j 1 is still free). Mark all the cells to its left as reserved. ⇤ For example, we obtain the inversion table T =(0 , 1 , 2↵, 1↵, , 2↵, , 2 , , 1 , ) from the tableau on the left of Figure 5, . Note that the proof of Proposition 5.13 gives a proof of (5). 18 SYLVIE CORTEEL, JANG SOO KIM AND DENNIS STANTON 5.3. Special cases. We give alternative proofs of special cases for Zn(y; ↵, , , ; q)provenin [9, Table 1]. These were proven using the staircase tableaux and the matrix ansatz. We use here the moments and the explicit three-term recurrence relation. The first result is new and does not have (yet) a combinatorial proof. Proposition 5.14. If = /y, then n 1 j Zn(y; ↵, , , ; q)= (↵y + q ). j=0 Y Proof. First let’s check that the choice = /y forces ⇤1 =0, so that in Proposition 5.8 n µn = B0 . In general we have bd = / . So Proposition 5.1 and Proposition 5.8 imply that ⇤1 has the numerator factor (1 bdy )=(1+y / )=0. Since abcd = /↵ ,wehave Z (y; ↵, , , ; q)=( /↵ ; q) pyn(↵ )n Bn. n n ⇥ 0 Take the version of Proposition 5.1 which uses b instead of a. In this case A(0) = C(0) = 0, and 1/py + py + bpy +1/bpy py B = = . 0 1 q ⇤ Proposition 5.12 may be shown using the recurrence relation instead of the bijection of Proposition 5.13. We need the fact, which follows from Proposition 1.4, that the rescaled Laguerre polynomials which have 2 bn = A(2n + ✓ + 1), n = A n(n + ✓), have moments of n n µn = A (✓ + i). i=1 Y We do not give these details. The next result can be proven bijectively by adapting the bijection between staircase tableaux and inversion tables given in the preceding subsection, see [7]. Proposition 5.15. If ↵ = =0, then n 1 j Zn(y; ↵, , , ; q)= ( + [j]q + q ). j=0 Y Proof. This analogously comes from the q-Laguerre polynomials studied by de M´edicis and Viennot [11] which have n n 2 2n 1 n 1 b =Aq ([n] + B + q + + q ) , = A q [n] B + q + + q , n q ··· n q ··· n n j 1 µ =A (B + q + + q ). n ··· j=1 Y ⇤ MOMENTS OF ORTHOGONAL POLYNOMIALS AND COMBINATORICS 19 6. Modified moments The Askey-Wilson polynomials have the polynomial representation in Definition 1.5 using the Askey-Wilson basis of polynomials, n 1 (x; a)= (1 2axqk + a2q2k)=(az, a/z; q) ,z= ei✓,x= cos ✓. n n kY=0 However it is not clear how Theorem 1.2 and Theorem 5.2 lead to the explicit formula in Definition 1.5. One may ask if an analogue of Theorem 1.2 exists for the Askey-Wilson basis n n(x; a) instead of x . We shall see that for the Askey-Wilson polynomials it does exist. It was given by Wilson in his 1978 thesis and is implicit in [35]. Suppose that polynomial sequences (x) 1 and (x) 1 are given with { k }k=0 { k }k=0 deg( k(x)) = deg( k(x)) = k. For the linear functional define a matrix M by L M = ( ). i,j L i j Proposition 6.1. The orthogonal polynomial p (x) for is n L M0,0 M0,1 M0,n M M ··· M 1,0 1,1 ··· 1,n 1 . . . . pn(x)= . . .. . , Bn Mn 1,0 Mn 1,1 Mn 1,n ··· (x) (x) (x) 0 1 ··· n where M0,0 M0,1 M0,n 1 ··· M1,0 M1,1 M1,n 1 ··· Bn = . . . . . . . .. . Mn 1,0 Mn 1,1 Mn 1,n 1 ··· The determinant Bn in Proposition 6.1 is non-zero and pn(x) is a polynomial of exact degree n with leading term n(x). To prove this, let Dn be the matrix of the determinant n in Theorem 1.2. Let Y and Z be the n n non-singular lower triangular matrices defined by i s j ⇥ t T i(x)= s=0 Yisx , j(x)= t=0 Zjtx . Then Mi,j is the ij-entry of the matrix YDnZ , which is invertible. P P k Theorem 1.2 is the special case k(x)= k(x)=x of Proposition 6.1. Whenever we have two bases of polynomials for which Mi,j may be explicitly computed, we have a determinantal formula for pn(x). The explicit representation for the polynomials is reduced to evaluating the determinant det(Mi,j) 0 i n 1 , 0 j n,j =k 6 n 1 k (6) pn(x)= ( 1) k(x)det(Mij) 0 i n 1 Bn 0 j n,j =k kX=0 6 20 SYLVIE CORTEEL, JANG SOO KIM AND DENNIS STANTON The Askey-Wilson bases provide such an example for the Askey-Wilson polynomials. We use a normalized representing measure for 0 (ab, ac; q)j(bd, cd; q)i(ad; q)i+j AW ( i j)=Mi,j = . L (abcd; q)i+j Proof. Finding ( ) amounts to shifting a to aqj and d to dqi in the Askey-Wilson LAW i j integral. ⇤ To find the coe cient of k(x; a) for the Askey-Wilson polynomials in Definition 1.5, the following determinant evaluation of Wilson ([35, p. 1155]) is used in (6). Proposition 6.3. If 0 k n, we have i (Aq ; q)j det i = nn⇡k/⇡n, (Dq ; q)j 0 i n 1,0 j n,j=k ✓ ◆ 6 where n 1 (q, D/A; q)r r r(r 1) nn = r A q (Dq ; q)n 1 r=0 Y n n 1 (q ,Dq ; q)k k ⇡k = ( q) (q, A; q)k Proposition 6.3 with A = ad and D = abcd and (6) give the explicit formula in Definition 1.5 for the Askey-Wilson polynomials. Another example occurs with the Hahn polynomials Qn(x; ↵, ,N), 0 n N, whose measure is purely discrete (the hypergeometric distribution), and is located at the integers x =0, 1, ,N, ··· ↵+x +N x x N x w(x; ↵, ,N)= ↵+ +1+N . N In this case the choices of (x)=( x) =( x)( x + 1) ( x + i 1), i i ··· (x)=( N + x) =( N + x)( N + x + 1) ( N + x + j 1) j j ··· yield (↵ + 1) ( + 1) N! M = i j ( 1)i+j, for i + j N. i,j (↵ + + 2) (N i j)! i+j Here another determinant evaluation leads to the explicit hypergeometric representation [27] for Qn(x; ↵, ,N). MOMENTS OF ORTHOGONAL POLYNOMIALS AND COMBINATORICS 21 7. Linearization coefficients Suppose that Pn(x) n 0 is an orthogonal polynomial sequence with respect to a linear { } functional : K[x] K.Since Pn(x) n 0 is a basis of the ring K[x], we can express the L ! { } product Pn(x)Pm(x) of two polynomials in the basis as follows: n+m k (7) Pn(x)Pm(x)= cn,mPk(x). kX=0 If we multiply both sides of (7) by P`(x) and apply the linear functional ,thenbythe orthogonality we obtain L (P (x)P (x)P (x)) c` = L n m ` . n,m (P (x)2) L ` 2 By Theorem 2.4, (P`(x) ) is obtained immediately once we know the three-term recur- L ` rence relation. Thus computing the coe cients cn,m is equivalent to computing the quantity (P (x)P (x)P (x)). Although it is more common to call c` the linearization coe cient, for L n m ` n,m brevity, we will instead call the quantity (P (x)P (x) P (x)) the linearization coe cient L n1 n2 ··· nk of the orthogonal polynomials Pn(x). Using the three-term recurrence relation, Lemma 2.2 allows us to consider orthogonal polyno- mials Pn(x) as generating functions of Favard paths. By Theorem 2.3, there is a combinatorial meaning to the moments µ = (xn) of P (x). Therefore it is possible to understand the n L n linearization coe cients of Pn(x) combinatorially. When Pn(x) are q-Hermite, q-Charlier, or q- Laguerre polynomials, there is a nice combinatorial expression for the linearization coe cients. In this section we will consider the linearization coe cients of the Hermite polynomials. Then we will see a connection between the linearization coe cients of the q-Hermite polynomials and the moments of Askey-Wilson polynomials. Recall that the Hermite polynomials Hn(x) are defined by H 1(x) = 0, H0(x) = 1, and for n 0, (8) Hn+1(x)=xHn(x) nHn 1(x), n and the moment µn = (x ) is equal to the number of perfect matchings of [n]. Using the three-term recurrence relationL (8), one can easily see that (9) H (x)= ( 1)edge(⇡)xfix(⇡), n ⇡ Matching(n) 2 X where fix(⇡) is the number of singletons in ⇡, which are also called fixed points. Recall that n is the set of perfect matchings of [n]. Suppose that n = n1 + n2 + + nk and for 1 i Mk,let ··· (10) Si = ai 1 +1,ai 1 +2,...,ai 1 + ni , { } where ai 1 = n1 + n2 + + ni 1 for i 2 and a0 = 0. Then [n] is a disjoint union of Si’s. An ··· edge (i, j) of ⇡ n is called homogeneous if i, j Sr for some r [k], and inhomogeneous otherwise. If every2M edge is inhomogeneous in ⇡ 2, then we call ⇡2an inhomogeneous perfect 2Mn matching. We denote by (n1,n2,...,nk) the set of inhomogeneous perfect matchings of [n]. The following theoremM is due to Azor, Gillis, and Victor [4] and Godsil [15]. Theorem 7.1. The linearization coe cient of the Hermite polynomials is equal to the number of inhomogeneous perfect matchings: (H (x) H (x)) = (n ,n ,...,n ) . L n1 ··· nk |M 1 2 k | 22 SYLVIE CORTEEL, JANG SOO KIM AND DENNIS STANTON Proof. By (9), H (x) H (x) is equal to n1 ··· nk edge(⇡ )+ +edge(⇡ ) fix(⇡ )+ +fix(⇡ ) ( 1) 1 ··· k x 1 ··· k . ··· ⇡ Matching(n ) ⇡ Matching(n ) 12 X 1 k2 X k Since (xm) is the number of perfect matchings of [m], we have L edge(⇡ )+ +edge(⇡ ) (H (x) H (x)) = ( 1) 1 ··· k , L n1 ··· nk (⇡ ,...,⇡ ,⇡ ) X 1 Xk 0 2 where X is the set of all (k + 1)-tuples (⇡1,...,⇡k,⇡0) such that ⇡ Matching(n ),...,⇡ Matching(n ), 1 2 1 k 2 k and ⇡0 is a perfect matching of the union of the sets of fixed points in ⇡1,⇡2,...,⇡k. As before we say that an edge (a, b)ishomogeneous if (a, b) S for some 1 i k, and inhomogeneous 2 i otherwise, where Si is given in (10). Note that all edges in ⇡1,...,⇡k are homogeneous and ⇡0 may have both homogeneous and inhomogeneous edges. We will construct a sign-reversing involution ⇢ on X. For (⇡ ,...,⇡ ,⇡ ) X,wedefine 1 k 0 2 ⇢((⇡1,...,⇡k,⇡0)) = (⇡10 ,...,⇡k0 ,⇡0) as follow. If ⇡1,...,⇡k are all empty and ⇡0 has only inhomogeneous edges, then ⇡0 = ⇡ for all 0 i k. Otherwise, there is a homogeneous edge i i (a, b) in one of ⇡1,⇡2,...,⇡k or ⇡0. Take the homogeneous edge (a, b) such that b is minimal. If (a, b) ⇡ for some 1 i k,thenlet⇡0 = ⇡ (a, b) and ⇡0 = ⇡ (a, b) , and ⇡0 = ⇡ 2 i i i \{ } 0 0 [{ } j j for j =0,i.If(a, b) ⇡ ,thenleti be the index for which a, b S and let ⇡0 = ⇡ (a, b) 6 2 0 2 i i i [{ } and ⇡0 = ⇡0 (a, b) , and ⇡j0 = ⇡j for j =0,i. It is not hard\{ to check} that ⇢ is a sign-reversing6 involution on X whose fixed points are the (k+1)-tuples (⇡1,...,⇡k,⇡0) X such that ⇡1,...,⇡k are all empty and ⇡0 is an inhomogeneous perfect matchings of S 2 S . This completes the proof. 1 [···[ k ⇤ From now on we put x = cos ✓. The generating function for the q-Hermite polynomials H (x q) is given by n | 1 H (cos ✓ q) 1 H(cos ✓, z):= n | zn = . (q; q) (zei✓; q) (ze i✓; q) n=0 n X 1 1 The orthogonality for the q-Hermite polynomials is ⇡ H (cos ✓ q)H (cos ✓ q)v(cos ✓ q)d✓ =0,n= m, n | m | | 6 Z0 where (q; q) 2i✓ 2i✓ v(cos ✓ q)= 1 (e ; q) (e ; q) . | 2⇡ 1 1 We now look at the Askey-Wilson polynomials defined in (1.5). The total mass (q; q) ⇡ (abcd; q) (11) I0(a, b, c, d)= 1 w(cos ✓, a, b, c, d; q)d✓ = 1 2⇡ 0 (ab, ac, ad, bc, bd, cd; q) Z 1 MOMENTS OF ORTHOGONAL POLYNOMIALS AND COMBINATORICS 23 of the measure is called the Askey-Wilson integral. Observe that the Askey-Wilson integral is the generating function for linearization coe cients of the q-Hermite polynomials ⇡ I (a, b, c, d)= H(cos ✓, a)H(cos ✓, b)H(cos ✓, c)H(cos ✓, d)v(cos ✓ q)d✓ 0 | Z0 1 = (H (x q)H (x q)H (x q)H (x q)) L n1 | n2 | n3 | n4 | n ,n ,n ,n =0 1 2X3 4 an1 bn2 cn3 dn4 , ⇥ (q; q)n1 (q; q)n2 (q; q)n3 (q; q)n4 where is the linear function for H (x q)definedby L n | ⇡ (p(x)) = p(cos ✓)v(cos ✓ q)d✓, L | Z0 Using this observation Ismail, Stanton, and Viennot [19] evaluated the Askey-Wilson integral combinatorially. More precisely, they considered the rescaled q-Hermite polynomials, which are more suitable to work with combinatorially, and showed the following generalization of Theorem 7.1. Theorem 7.2. [19, Theorem 3.2] Let be the normalized linear functional for H (x q). Then L n | (H (x q) H (x q)) = qcrossing(⇡), L n1 | ··· nek | e ⇡ (n ,...,n ) 2M X1 k where (n ,...,n ) ise thee set of inhomogeneouse perfect matchings of [n ] [n ]. M 1 k 1 ]···] k This idea can be extended to compute the moments of the Askey-Wilson polynomials as follows. Let ⇡ (q; q) n (12) I (a, b, c, d)= 1 (cos ✓) w(cos ✓, a, b, c, d; q)d✓. n 2⇡ Z0 Then the normalized moment µn(a, b, c, d; q) of the Askey-Wilson polynomials is (13) µn(a, b, c, d; q)=In(a, b, c, d)/I0(a, b, c, d). By the same observation as above we have ⇡ I (a, b, c, d)= (cos ✓)nH(cos ✓, a)H(cos ✓, b)H(cos ✓, c)H(cos ✓, d)v(cos ✓ q)d✓ n | Z0 1 = (xnH (x q)H (x q)H (x q)H (x q)) L n1 | n2 | n3 | n4 | n ,n ,n ,n =0 1 2X3 4 an1 bn2 cn3 dn4 . ⇥ (q; q)n1 (q; q)n2 (q; q)n3 (q; q)n4 Using the above formula Kim and Stanton [25] found the formula for µn(a, b, c, d; q) in Corol- lary 5.3. In what follows we briefly explain the idea of their proof. First, in [25] they consider the rescaled q-Hermite polynomials and show the following theo- rem. Theorem 7.3. [25, Theorem 5.1] Let be the normalized linear functional for H (x q). Then L n | (xnH (x q) H (x q)) = qcrossing(⇡), L n1 | ··· nek | e ⇡ (n;n ,...,n ) 2M X1 k e e e 24 SYLVIE CORTEEL, JANG SOO KIM AND DENNIS STANTON where (n; n ,...,n ) is the set of perfect matchings ⇡ of S S S such that if ⇡ M 1 k 1 [···[ k [ k+1 contains a homogeneous edge (a, b) Si, then i = k +1. Here, Si is given in (10) for 1 i k, and 2 S = n + + n +1,n + + n +2,...,n + + n + n . k+1 { 1 ··· k 1 ··· k 1 ··· k } In order to evaluate the sum in the left hand side of the above theorem, they [25] decompose ⇡ (n; n1,n2,n3,n4) as a pair (⇡0,⇡1) of a matching ⇡0 of [n]withm fixed points and an2 inhomogeneousM perfect matching ⇡ (m, n ,n ,n ,n ) for some integer m.Inthis 1 2M 1 2 3 4 decomposition we have crossing(⇡) = crossing⇤(⇡0) + crossing(⇡1), where crossing⇤(⇡0)isthe number of pairs (e1,e2) such that e1 = a, b and e2 = c, d are edges of ⇡0 with a (n m)/2 crossing⇤(⇡) (14) (1 q) q ⇡ (n,m) 2MX⇤ k+m n n (k m)/2+1 (k m)/2 ( 2 ) 2 = n k n k ( 1) q k m , 1 k 0 ✓✓ 2 ◆ ✓ 2 ◆◆ 2 q X where ⇤(n, m) is the set of matchings of [n]withm fixed points. TheM appearance of the di↵erence of binomial coe cients in (14) can be explained as follows. A Dyck prefix is a path obtained from a Dyck path by taking the first m steps for some m.By a similar argument using Hermite histories, the left hand side of (14) is the generating function for Dyck prefixes whose South-East steps are labeled by 1 or qi,wherei is the y-coordinate of the starting point. Then we can use Penaud’s idea [29] which decomposes such a labeled Dyck prefix into a Dyck prefix and a certain labeled Dyck path. The number of Dyck prefixes is given by a di↵erence of binomial coe cients. There are analogous combinatorial interpretations for the linearization coe cients of q- Charlier polynomials and q-Laguerre polynomials. Theorem 7.4. [1, p. 127] Let C be the normalized linear functional for the q-Charlier poly- a L nomials Cn(x; q). Then a a block(⇡) crossing(⇡) C (C (x; q) C (x; q)) = a q , L n1 ··· nk ⇡ ⇧(n ,...,n ) 2 X1 k where ⇧(n1,...,nk) is the set of partitions of S1 Sk which do not have homogeneous edges. Here, S is given in (10) for 1 i k. [···[ i Kim, Stanton and Zeng [24] found a combinatorial proof of Theorem 7.4. Kasraoui, Stanton, and Zeng [23] showed the following theorem using recurrence relations. Theorem 7.5. [23, Theorem 5] Let be the normalized linear functional for the q-Laguerre LL polynomials Ln(x; q). Then (L (x; q) L (x; q)) = ywex(⇡)qcrossing(⇡), LL n1 ··· nk ⇡ D(n ,...,n ) 2 X1 k where D(n1,...,nk) is the set of permutations ⇡ of S1 Sk such that if ⇡(a)=b then a and b are in di↵erent S ’s. Here, S is given in (10) for[1···[i k. i i MOMENTS OF ORTHOGONAL POLYNOMIALS AND COMBINATORICS 25 Note that in Theorem 7.5, if n = = n = 1, then D(n ,...,n ) becomes the set of 1 ··· k 1 k derangements of [k], i.e., permutations with no fixed points. Permutations in D(n1,...,nk) are called multi-derangements. There is a nice generating function expression for the number of multi-derangements, see [16, p. 563] and references therein. 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LIAFA, CNRS et UniversiteParisDiderot,Case7014,75205ParisCedex13France´ E-mail address: [email protected] Department of Mathematics, Sungkyunkwan University (SKKU), 2066 Seobu-ro, Jangan-gu, Su- won, Gyeonggi-do 440-746, South Korea E-mail address: [email protected] School of Mathematics, University of Minnesota, Minneapolis, Minnesota 55455 USA E-mail address: [email protected]