The Combinatorics of Associated Laguerre Polynomials

0.35 MB·English

by  Jang Soo Kim

Symmetry, Integrability and Geometry: Methods and Applications SIGMA 11 (2015), 039, 12 pages
The Combinatorics
of Associated Laguerre Polynomials(cid:63)
Jang Soo KIM † and Dennis STANTON ‡
† Department of Mathematics, Sungkyunkwan University, Suwon 440-746, South Korea
E-mail: [email protected]
‡ School of Mathematics, University of Minnesota, Minneapolis, MN 55455, USA
5 E-mail: [email protected]
1
0 Received January 30, 2015, in final form May 06, 2015; Published online May 11, 2015
2 http://dx.doi.org/10.3842/SIGMA.2015.039

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

] Key words: associated Laguerre polynomial; moment of orthogonal polynomials, permuta-
O
tion tableau
C
2010 Mathematics Subject Classification: 05E35; 05A15
.
h
t
a
m
1 Introduction
[

2
The classical orthogonal polynomials of Hermite, Laguerre, and Jacobi have explicit represen-
v
0 tations as terminating hypergeometric series. In the Askey scheme of orthogonal polynomials,
8
each polynomial is given by either a hypergeometric or basic hypergeometric series. These
8
polynomials have been studied extensively combinatorially.
3
0 The associated versions of the classical polynomials are defined by replacing n by n+c in
.
1 the three term recurrence relation, where c is an arbitrary parameter. The associated classical
0
polynomials may be expressed as terminating hypergeometric double sums. For example, the
5
1 coefficient of xk in the associated Laguerre polynomial of degree n is given by a terminating
: F series. The analogous expansion for the associated Askey–Wilson polynomials uses a φ
v 3 2 10 9
i series, see [8].
X
The combinatorics of the classical orthogonal polynomials extends easily to include the as-
r
a sociation parameter c. 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, Theorems 2.2 and 2.4, by giving combinatorial interpretations for the polynomials. We
also find combinatorial expressions for the moments, see Theorems 3.3 and 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 La-
guerre polynomials from knowledge of the moment generating function. An explicit measure or
a verification of the 3-term recurrence is not necessary. This method may be useful in other
settings to find explicit representations for polynomials when we know neither the measure nor
the moments.
(cid:63)This paper is a contribution to the Special Issue on Exact Solvability and Symmetry Avatars in honour of
Luc Vinet. The full collection is available at http://www.emis.de/journals/SIGMA/ESSA2014.html
2 J.S. Kim and D. Stanton
2 The associated Laguerre polynomials
The monic Laguerre polynomials p (x) = ( 1)nn!Lα(x), whose weight function is
n n
−
xαe−xdx, x 0,
≥
may be defined by the three term recurrence relation
p (x) = (x b )p (x) λ p (x), p (x) = 1, p (x) = 0, (2.1)
n+1 n n n n−1 0 −1
− −
where
b = 2n+α+1, λ = n(n+α).
n n
The associated Laguerre polynomials satisfy the same recurrence with n replaced by n+c,
b = 2n+α+2c+1, λ = (n+c)(n+α+c).
n n
For the purposes of this paper it is convenient to replace the parameters α and c by X = c and
Y = α+c+1.
Definition 2.1. The associated Laguerre polynomials L (x;X,Y) are given by (2.1) with
n
b = 2n+X +Y, λ = (n+X)(n+Y 1).
n n
−
The double sum expression for L (x;X,Y) we shall prove is the following. We use the
n
notation
(n) = n(n+1)(n+2) (n+k 1).
k
··· −
Theorem 2.2. For any non-negative integer n,
n (cid:18) (cid:19) (cid:18) (cid:19)
(cid:88) n k n,k+1,Y 1
Ln(x;X,Y) = (X +k+1)n−k k 3F2 −n,X +k+−1 ;1 (−1)n−kxk.
k=0 −
We shall prove Theorem 2.2 and its relative Theorem 2.4 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 [7, p. 158] may be simultaneously studied.
(2)
Definition 2.3. The Model II associated Laguerre polynomials L (x;X,Y) are given by (2.1)
n
with
b = X, b = 2n+X +Y 1, n > 0, λ = (n 1+X)(n 1+Y).
0 n n
− − −
Theorem 2.4. For any non-negative integer n,
n (cid:18) (cid:19) (cid:18) (cid:19)
(cid:88) n k n,k,Y 1
L(n2)(x;X,Y) = (X +k)n−k k 3F2 −n,X +−k ;1 (−1)n−kxk.
k=0 −
We now review [1, p. 40], [13, p. 311], the basic setup for the even-odd polynomials.
Suppose that a set of monic orthogonal polynomials P (x) satisfies (2.1) with b = 0 and
n n
λ = Λ arbitrary. Then P (x) are even-odd polynomials, that is,
n n n
P (x) = e (cid:0)x2(cid:1), P (x) = xo (cid:0)x2(cid:1),
2n n 2n+1 n
for some monic polynomials e (x) and o (x).
n n
The Combinatorics of Associated Laguerre Polynomials 3
Proposition 2.5. The polynomials e and o are both orthogonal polynomials satisfying
n n
e (x) = (x b (e))e (x) λ (e)e (x),
n+1 n n n n−1
− −
o (x) = (x b (o))o (x) λ (o)o (x),
n+1 n n n n−1
− −
where
b (e) = Λ +Λ , λ (e) = Λ Λ and
n 2n+1 2n n 2n−1 2n
b (o) = Λ +Λ , λ (o) = Λ Λ .
n 2n+2 2n+1 n 2n+1 2n
Here we define Λ = 0, i.e., b (e) = Λ .
0 0 1
We see that the choices of Λ = n 1+Y, Λ = n+X, give the associated Laguerre
2n 2n+1
−
polynomials
o (x) = L (x;X,Y), e (x) = L(2)(x;X,Y).
n n n n
We shall give combinatorial interpretations for o (x) and e (x) which will prove Theorem 2.2
n n
and Theorem 2.4. Let
n n
(cid:88) (cid:88)
e (x) = E xk( 1)n−k, o (x) = O xk( 1)n−k.
n n,k n n,k
− −
k=0 k=0
Note that we have the following recursive relations
O = E +Λ O , E = O +Λ E .
n,k n,k 2n n−1,k n,k n−1,k−1 2n−1 n−1,k
In order to give combinatorial interpretations for O and E , we need several definitions.
n,k n,k
Definition 2.6. A k-marked permutation is a permutation π in which the integers in an in-
creasing subsequence of length k are marked.
Example 2.7. π = ˆ3421ˆ65 is a 2-marked permutation of length 6, where 3 and 6 are marked.
Definition 2.8. Let π = π π π be a word with distinct integers. Then π is called a right-
1 2 n i
···
to-left minimum of π if π < π for all j = i+1,i+2,...,n. We denote by RLMIN(π) the set of
i j
right-to-left minima of π. Similarly, we define RLMAX(π), LRMIN(π), and LRMAX(π), which
are the sets of right-to-left maxima, left-to-right minima, and left-to-right maxima, respectively.
In other words,
RLMIN(π) = π : π < π for all j = i+1,i+2,...,n ,
i i j
{ }
RLMAX(π) = π : π > π for all j = i+1,i+2,...,n ,
i i j
{ }
LRMIN(π) = π : π < π for all j = 1,2,...,i 1 ,
i i j
{ − }
LRMAX(π) = π : π > π for all j = 1,2,...,i 1 .
i i j
{ − }
LetRLmin(π),RLmax(π),LRmin(π),andLRmax(π)bethecardinalitiesofthesetsRLMIN(π),
RLMAX(π), LRMIN(π), and LRMAX(π), respectively.
Definition 2.9. Let π = w aˆw aˆw aˆw be a k-marked permutation, where a < a <
0 1 1 2 2 k k 1 2
···
< a are the marked integers. Let
k
···
LRMIN(cid:48)(π) = LRMIN(π) LRMIN(w ) a ,a ,...,a ,
0 1 2 k
\ \{ }
k
(cid:91)
RLMIN(cid:48)(π) = RLMIN(w ) [a ,a ],
i i i+1
∩
i=0
where a = 0, a = , and [a ,a ] = a ,a +1,...,a . We also denote LRmin(cid:48)(π) =
0 k+1 i i+1 i i i+1
∞ { }
LRMIN(cid:48)(π) and RLmin(cid:48)(π) = RLMIN(cid:48)(π) .
| | | |
4 J.S. Kim and D. Stanton
Example 2.10. Let π = 68ˆ2415ˆ739 with k = 2, a = 2, a = 7, w = 68, w = 415, and
1 2 0 1
w = 39. Then
2
RLMIN(π) = 1,3,9 , RLMAX(π) = 9 ,
{ } { }
LRMIN(π) = 6,2,1 , LRMAX(π) = 6,8,9 ,
{ } { }
and
LRMIN(cid:48)(π) = 6,2,1 6 2,7 = 1 ,
{ }\{ }\{ } { }
RLMIN(cid:48)(π) = (cid:0) 6,8 [0,2](cid:1) (cid:0) 1,5 [2,7](cid:1) (cid:0) 3,9 [7, ](cid:1)
{ }∩ ∪ { }∩ ∪ { }∩ ∞
= ∅ 5 9 = 5,9 .
∪{ }∪{ } { }
Definition 2.11. Let SE be the set of k-marked permutations of [n]. Let SO be the set of
n,k n,k
(k+1)-marked permutations of [n+1] with n+1 marked.
We now give combinatorial interpretations for the coefficients E and O .
n,k n,k
Proposition 2.12. Let E and O be determined by the following recursions with initial
n,k n,k
conditions E = O = 0 if n < k, k < 0, or n < 0; E = O = 1 if n = k 0:
n,k n,k n,k n,k
≥
E = O +(n 1+X)E , O = E +(n 1+Y)O .
n,k n−1,k−1 n−1,k n,k n,k n−1,k
− −
Then
E = (cid:88) XRLmin(cid:48)(π)YLRmin(cid:48)(π), O = (cid:88) XRLmin(cid:48)(π)YLRmin(cid:48)(π).
n,k n,k
π∈SE π∈SO
n,k n,k
Proof. We will show that
A = (cid:88) XRLmin(cid:48)(π)YLRmin(cid:48)(π), B = (cid:88) XRLmin(cid:48)(π)YLRmin(cid:48)(π)
n,k n,k
π∈SE π∈SO
n,k n,k
satisfy the recurrence relations.
Let π SE . If n is marked in π, then we can consider π as an element of SO .
∈ n,k n−1,k−1
Otherwise, π is obtained from an element σ in SE by inserting an unmarked integer n in n
n−1,k
different ways. Then we always have RLmin(cid:48)(π) = RLmin(cid:48)(σ) and LRmin(cid:48)(π) = LRmin(cid:48)(σ)
unlessweinsertnattheendinwhichcasewehaveRLmin(cid:48)(π) = RLmin(cid:48)(σ)+1andLRmin(cid:48)(π) =
LRmin(cid:48)(σ). This implies that A = B +(n 1+X)A .
n,k n−1,k−1 n−1,k
−
Now let π SO . If the last element of π is the marked n+1, then by removing n+1 we can
∈ n,k
consider π as an element of SE without changing RLmin(cid:48)(π) and LRmin(cid:48)(π). Otherwise, let j
n,k
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 SO . Then we have RLmin(cid:48)(π) = RLmin(cid:48)(σ) and LRmin(cid:48)(π) = LRmin(cid:48)(σ)
n−1,k
unless j = 1 in which case we have RLmin(cid:48)(π) = RLmin(cid:48)(σ) and LRmin(cid:48)(π) = LRmin(cid:48)(σ)+1.
This implies that B = A +(n 1+Y)B . (cid:4)
n,k n,k n−1,k
−
The next proposition constitutes proofs of Theorems 2.2 and 2.4.
Proposition 2.13. For n,k 0, we have
≥
(cid:18) (cid:19) (cid:18) (cid:19)
n k n,k+1,Y 1
On,k = (X +k+1)n−k k 3F2 −n,X +k+−1 ;1 ,
−
and
(cid:18) (cid:19) (cid:18) (cid:19)
n k n,k,Y 1
En,k = (X +k)n−k k 3F2 −n,X +−k ;1 .
−
The Combinatorics of Associated Laguerre Polynomials 5
Proof. By Proposition 2.12 we have
E = (cid:88) XRLmin(cid:48)(π)YLRmin(cid:48)(π).
n,k
π∈SE
n,k
Let π = w aˆw aˆw aˆw SE . Then min(w ,a ,a ,...,a ) = m+1 for an integer
0 1 1 2 2··· k k ∈ n,k 0 1 2 k
0 m n k. Since a < a < < a , we have two cases m + 1 = a < min(w ) and
1 2 k 1 0
≤ ≤ − ···
m+1 = min(w ) < a .
0 1
Suppose that m+1 = a < min(w ). In order to construct such π = w aˆw aˆw aˆw we
1 0 0 1 1 2 2 k k
choosea ,a ,...,a in(cid:0)n−m−1(cid:1)ways,selectapermutationσ of[m],shuffleσ witha ·,·a· ,...,a ,
2 3 k k−1 2 3 k
insert a = m + 1 at the beginning, and insert the remaining n k m integers. Clearly,
1
− −
LRmin(cid:48)(π) = LRmin(σ). In this construction, suppose that b < b < < b are the
1 2 n−k−m
···
remainingintegerstobeinserted. Weinsertb first, andthenb , andsoon. Thenduringtheith
1 2
insertion process RLmin(cid:48)(π) is increased by 1 if and only if b is inserted just before a for the
i j+1
uniqueintegerj witha < b < a . Thusinsertingb contributesafactorof(X+k+m+i 1).
j i j+1 i
Summarizing these we obtain that the sum of XRLmin(cid:48)(π)YLRmin(cid:48)(π) for such π’s is equal to−
(cid:18) (cid:19)(cid:18) (cid:19)
n m 1 m+k 1
− − − (Y) (X +k+m) .
m n−k−m
k 1 m
−
Here we used the well known fact (cid:80) YRLmin(σ) = (Y) , see [14, p. 125, Exercise 127].
m
σ∈Sm
Similarly, one can show that the sum of XRLmin(cid:48)(π)YLRmin(cid:48)(π) for π = w aˆw aˆw aˆw
0 1 1 2 2 k k
···
with m+1 = min(w ) < a is equal to
0 1
(cid:18) (cid:19)(cid:18) (cid:19)
n m 1 m+k 1
− − − (Y) X(X +k+m+1) .
m n−k−m−1
k m
Thus we have
n−k (cid:20)(cid:18) (cid:19)(cid:18) (cid:19)
(cid:88) n m 1 m+k 1
E = − − − (Y) (X +k+m)
n,k m n−k−m
k 1 m
m=0 −
(cid:18) (cid:19)(cid:18) (cid:19) (cid:21)
n m 1 m+k 1
+ − − − (Y) X(X +k+m+1) ,
m n−k−m−1
k m
whichisanexpressionforE assumoftwoterminating F (1)’s. Howevera F (1)contiguous
n,k 3 2 3 2
relation, which is equivalent to the binomial relation
(cid:18) (cid:19)(cid:18) (cid:19) (cid:18) (cid:19)(cid:18) (cid:19)
n m 1 m+k 1 n m 1 m+k 1
− − − (Y) + − − − (Y)
m m
k 1 m k m
−
(cid:18) (cid:19)(cid:18) (cid:19) (cid:18) (cid:19)(cid:18) (cid:19)
n m m+k 2 n m m+k 1
− − (k+m 1)(Y) = − − (Y 1) ,
m−1 m
− k m 1 − k m −
−
gives the stated formula for E as a single F (1).
n,k 3 2
By the same argument, we obtain
n−k (cid:20)(cid:18) (cid:19)(cid:18) (cid:19)
(cid:88) n m 1 m+k
O = − − (Y) (X +k+m+1)
n,k m n−k−m
k 1 m
m=0 −
(cid:18) (cid:19)(cid:18) (cid:19) (cid:21)
n m 1 m+k 1
+ − − − (Y) X(X +k+m+1) .
m n−k−m−1
k m
6 J.S. Kim and D. Stanton
Using
(cid:18) (cid:19)(cid:18) (cid:19) (cid:18) (cid:19)(cid:18) (cid:19)
n m 1 m+k n m 1 m+k
− − (Y) + − − (Y)
m m
k 1 m k m
−
(cid:18) (cid:19)(cid:18) (cid:19) (cid:18) (cid:19)(cid:18) (cid:19)
n m m+k 1 n m m+k
− − (k+m)(Y) = − (Y 1) ,
m−1 m
− k m 1 k m −
−
one can obtain the single sum formula for O . (cid:4)
n,k
Remark 1. The F in Theorem 2.2 which is the coefficient of xk is not equal to the F given
3 2 3 2
in [7, equation (5.6.18)]. These two F ’s are related by a F (1) transformation.
3 2 3 2
3 Moments of associated Laguerre polynomials
InthissectionweinterpretthemomentsforthemeasuresoftheassociatedLaguerrepolynomials.
This is a routine application of Viennot’s [15] theory, details are not provided. We need the
followingdefinitionofpivotsinordertogiveacombinatorialdescriptionforthemoments. These
are also called strong fixed points in [14, p. 125, Exercise 128(b)].
Definition 3.1. Let π = π π π be a permutation. If π is both a left-to-right maximum
1 2 n i
···
and a right-to-left minimum, then π is called a pivot of π. We denote by pivot(π) the number of
i
pivots of π. Note that if π is a pivot, then π = i, since π is greater than all of π ,π ,...,π
i i i 1 2 i−1
and smaller than all of π ,π ,...,π .
i+1 i+2 n
Suppose that the orthogonal polynomials p (x) satisfy (2.1). Then the nth moment, µ , of
n n
a measure for p (x) is a polynomial in the sequences b and λ . We denote this
n k k≥0 k k≥1
{ } { }
moment by
(cid:0) (cid:1)
µ b , λ or µ (b ,λ ).
n k k≥0 k k≥1 n k k
{ } { }
The first result uses a special choice for b .
0
Lemma 3.2. For an integer n 0, we have
≥
(cid:88)
µ (b ,(k 1+X)(k 1+Y)) = XRLmin(π)YLRmax(π)Zpivot(π),
n k
− −
π∈Sn
where b = 2k 1+X +Y for k > 0, and b = XYZ.
k 0
−
Proof. This can be proved using one of the bijections between Laguerre histories and permu-
tations, for instance, see [12]. (cid:4)
When Z = 1/Y in Lemma 3.2, we obtain moments for the Model II associated Laguerre
polynomials.
(2)
Theorem 3.3. The moments of the Model II associated Laguerre polynomials L (x;X,Y) are
n
(cid:88)
µ (2k 1+X +Y1−δk,0,(k 1+X)(k 1+Y)) = XRLmin(π)YLRmax(π)−pivot(π).
n
− − −
π∈Sn
Using the substitution (X,Y,Z) (X +1,Y, X+Y ) in Lemma 3.2, we obtain
→ (X+1)Y
Theorem 3.4. The moments of the Model I associated Laguerre polynomials L (x;X,Y) are
n
µ (2k+X +Y,(k+X)(k 1+Y))
n
−
(cid:88)
= (X +1)RLmin(π)−pivot(π)(X +Y)pivot(π)YLRmax(π)−pivot(π).
π∈Sn
The Combinatorics of Associated Laguerre Polynomials 7
If X = Y = 1 in Theorem 3.4, we have
µ (2k+2,k(k+1)) = (n+1)!.
n
It is therefore more natural to write the nth moment of the Model I associated Laguerre polyno-
mialasasumoverS insteadofS . Thenextlemma, whichisarestatementoftheeven-odd
n+1 n
polynomial construction in Section 2, accomplishes this.
Lemma 3.5. Given a sequence a with a = 0, we have
k k≥0 0
{ }
µ (0,a ) = µ (a +a ,a a ),
2n k n 2k 2k+1 2k−1 2k
µ (0,a ) = a µ (a +a ,a a ).
2n+2 k 1 n 2k+1 2k+2 2k 2k+1
Theorem 3.6. The moments of the Model I associated Laguerre polynomials L (x;X,Y) are
n
(cid:88)
µ (2k+X +Y,(k+X)(k 1+Y)) = XRLmin(π)−1YLRmax(π)−pivot(π).
n
−
π∈Sn+1
Proof. Let a = k 1+X and a = k 1+Y for k 1 and a = 0. Then by Lemma 3.5,
2k−1 2k 0
− − ≥
we have
µ (0,a ) = µ (cid:0)2k 1+X +Y1−δk,0,(k 1+X)(k 1+Y)(cid:1),
2n k n
− − −
µ (0,a ) = Xµ (2k+X +Y,(k+X)(k 1+Y)).
2n+2 k n
−
Thus, we have
Xµ (2k+X +Y,(k+X)(k 1+Y))
n
−
= µ (cid:0)2k 1+X +Y1−δk,0,(k 1+X)(k 1+Y)(cid:1).
n+1
− − −
By applying Theorem 3.3 to the above equation, we obtain the desired identity. (cid:4)
As a corollary of the proof above, we obtain the following generating function identity.
Corollary 3.7. For any non-negative integer n, we have
(cid:88)
(X +1)RLmin(π)−pivot(π)(X +Y)pivot(π)YLRmax(π)−pivot(π)
π∈Sn
(cid:88)
= XRLmin(π)−1YLRmax(π)−pivot(π).
π∈Sn+1
We may analogously push the n! terms for the Model II moments to S .
n−1
(2)
Corollary 3.8. The moments of the Model II associated Laguerre polynomials L (x;X,Y) are
n
µ (b ,(k 1+X)(k 1+Y))
n k
− −
(cid:88)
= (X +1) (X +2)RLmin(π)−pivot(π)(X +Y +1)pivot(π)YLRmax(π)−pivot(π),
π∈Sn−1
where b = 2k 1+X +Y for k > 0, and b = X.
k 0
−
8 J.S. Kim and D. Stanton
15 14 13 10 7 5 4 3 15 14 13 10 7 5 4 3
1 0 0 0 0 0 1 0 1 1
↑ ↑
2 0 0 0 0 0 0 1 1 2
← ↑
6 0 0 0 0 1 6
↑
8 0 0 0 0 8
9 1 0 1 1 9
↑ ↑ ↑
11 0 0 0 11
←
12 0 1 1 12
← ↑
16 16
Figure 1. A permutation tableau on the left and its alternative representation on the right.
4 Permutation tableaux
In this section we reinterpret the weights on permutations of the previous section to be weights
onanothersetofobjectsalsocountedbyn!, calledpermutationtableaux. Permutationtableaux
were introduced by Postnikov [11] in the study of totally nonnegative Grassmanian and studied
extensively.
Apermutation tableau isafillingofaFerrersdiagramusing0’sand1’ssuchthateachcolumn
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 denotethesetofpermutationtableauoflengthn. Ifapermutationtableauhaslengthn,we
n
label the southeast border ofa Ferrersdiagram with 1,2,...,nstarting fromnortheast to south-
west. 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 Fig. 1.
In a permutation tableau, a topmost 1 is a 1 which is the topmost 1 in the column containing
it. Arestricted 0isa0whichhasa1aboveitinthesamecolumn. Arightmost restricted 0isthe
rightmost restricted 0 in the row containing it. A row is called restricted if it has a restricted 0,
andunrestricted otherwise. Wedenotebyurr(T)thenumberofunrestrictedrowsinT. Similarly,
a 0 in T PT is called c-restricted if it has a 1 to the left of it in the same row. A column
n
∈
is called restricted if it has a c-restricted 0, and unrestricted otherwise. We denote by urc(T)
the number of unrestricted columns in T. For example, if T is the permutation tableau in
Fig. 1, then the unrestricted rows are rows 1, 6, 8, 9, 16, and the unrestricted columns are
columns 3, 5, 7, 10, 13, 15.
CorteelandNadeau[4]foundabijectionΦ: PT S suchthatifΦ(T) = π, thenurr(T) =
n n
→
RLmin(π). Corteel and Kim [3] 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 Fig. 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 [16] and studied further by Nadeau [10].
In this section we prove the following theorem.
The Combinatorics of Associated Laguerre Polynomials 9
Theorem 4.1. The moments for the associated Laguerre polynomials are
(cid:88)
µ (2k+X +Y,(k+X)(k 1+Y)) = Xurr(T)−1Yurc(T).
n
−
T∈PTn+1
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 , a row is unrestricted if
n
∈
and only if it has no left arrow. There is also a simple characterization for unrestricted columns
in the alternative representation.
Lemma 4.2. In the alternative representation of T PT , a column is unrestricted if and only
n
∈
if there is no arrow that is strictly to the northwest of the uparrow in the column.
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(cid:48) 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(cid:48) 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. (cid:4)
Forapermutationπ = π π , wedefineπrev = π π π . Wealsodefineπ∗ asfollows.
1 n n n−1 1
··· ···
If π = w r w r w r , where r ,r ,...,r are the right-to-left minima of π, then π∗ =
1 1 2 2 k k 1 2 k
···
wrevr wrevr wrevr . Note that π∗ and π have the same right-to-left minima and (π∗)∗ = π.
1 1 2 2··· k k
For example, if
π = 4,2,5,3,1,7,6,8,14,12,15,11,13,10,9,16,
where the right-to-left minima are overlined, then
π∗ = 3,5,2,4,1,7,6,8,10,13,11,15,12,14,9,16,
where the right-to-left minima are overlined and the left-to-right maxima are underlined.
We now describe the bijection Φ: PT S of Corteel and Nadeau [4] using the alternative
n n
→
representation. Let T PT . Suppose that r < r < < r are the labels of the unrestricted
n 1 2 k
∈ ···
rows of T. We first set π to be r r r . For each column of T starting from left to right we
1 2 k
···
will insert integers to π as follows. If column i contains the uparrow in row j and the leftarrows
in rows i < < i , then insert i ,...,i ,i in this order before j in π. If column i does not
1 r 1 r
···
have any leftarrows, we insert only i before j. Observe that we have j < i < i < < i < i.
1 2 r
···
Thus inserting integers in this way does not change the right-to-left minima. We define Φ(T)
to be the resulting permutation π.
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.
Lemma 4.3. Let Φ(T) = π. Then we have the following.
An integer r is the label of an unrestricted row of T if and only if it is a right-to-left
•
minimum of π∗.
An integer c is the label of an unrestricted column of T if and only if it is a left-to-right
•
maximum but not a pivot of 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.
10 J.S. Kim and D. Stanton
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 < r < < r be the labels of unrestricted rows of T, and let r = r . By
1 2 k i
···
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 . Then after inserting integers from column c, we
i−1
havectotheleftofr withnointegersbetweenthem. Sincetheremainingintegerstobeinserted
i
are less than c, every integer greater than c is located either to the right of r or between r
i 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 and
i
c < r , c is not a right-to-left minimum of π∗.
i
Now let c be a left-to-right maximum of π∗ that is not a right-to-left minimum. Let r <
1
r < < r be the right-to-left minima of π (equivalently π∗). Suppose that c is located
2 k
···
between r and r . It is easy to see that c is followed by a smaller number in π. Thus c is the
i−1 i
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(cid:48) > c. When we insert integers from column c(cid:48) in the construction of Φ(T), c(cid:48) is located
to the left of r . Then c cannot be a left-to-right maximum of π∗, which is a contradiction.
i−1
Thus column c is unrestricted. (cid:4)
The above lemma implies the following.
Proposition 4.4. We have
(cid:88) (cid:88)
Xurr(T)Yurc(T) = XRLmin(π)YLRmax(π)−pivot(π).
T∈PTn π∈Sn
Theorem 4.1 follows from Theorem 3.6 and Proposition 4.4.
5 Proof via the moment generating function
In this section we prove the following equivalent version of Theorem 2.2
Ln(x;X,Y) = (Y)n(X +n!1)n(−1)n (cid:88)n (Y)(−(Xn)+k 1) 3F2(cid:18)Yk+−kn,,XX,+Yk−+11;1(cid:19)xk. (5.1)
k k
k=0
We use only specialized knowledge of the moments. What we need is a recurrence relation
for the moments. For ease of notation let’s put
θ (X,Y) = µ (2k+X +Y,(k+X)(k 1+Y)).
n n
−
Proposition 5.1. The moments of the associated Laguerre polynomials satisfy the recurrence
relation
J
(cid:88) (Y 1)J−k(X)J−k (Y)J(X +1)J
θ (X,Y) − = , J 0. (5.2)
k
(J k)! J! ≥
k=0 −
Proof. The recurrence (5.2) is equivalent to the explicit form of the moment generating func-
tion, which is Proposition 5.2, a quotient of hypergeometric series. This appears in [9, Theo-
rem 6.5, p. 212], or may easily be verified using contiguous relations for F (x) to find the
2 0
relevant Jacobi continued fraction. (cid:4)
Proposition 5.2. The generating function, as a formal power series in x, for the associated
Laguerre moments is
∞
(cid:88)µ (2k+X +Y,(k+X)(k 1+Y))xn = 2F0(Y,X +1;x).
n
− F (Y 1,X;x)
2 0
n=0 −