Refined sign-balance on 321-avoiding permutations

European Journal of Combinatorics 26 (2005) 1009–1018

www.elsevier.com/locate/ejc

Refined sign-balance on 321-avoiding permutations

Astrid Reifegerste

Institut für Mathematik, Universität Hannover, Welfengarten 1, D-30167 Hannover, Germany

Available online 23 July 2004

Abstract

The number of even 321-avoiding permutations of lengthn is equal to the number of odd onesif n is even, and exceeds it by then−1

2 th Catalan number otherwise. We present an involution thatproves a refinement of this sign-balance property respecting the length of the longest increasingsubsequence of the permutation. In addition, this yields a combinatorial proof of a recent analogousresult of Adin and Roichman dealing with the last descent. In particular, we answer the questionof how to obtain the sign of a 321-avoiding permutation from the pair of tableaux resulting fromthe Robinson–Schensted–Knuth algorithm. The proof of the simple solution is based on a matchingmethod given by Elizalde and Pak.© 2004 Elsevier Ltd. All rights reserved.

1. Introduction

Let Tn be the set of 321-avoiding permutations in the symmetric groupSn. (Apermutation is called 321-avoidingif it has no decreasing subsequence of length three.)Simion and Schmidt [6] proved the following sign-balance property ofTn: the numberof even permutations inTn is equal to the number of odd permutations ifn is even, andexceeds it by the Catalan numberC1

2 (n−1) otherwise. Very recently, Adin and Roichman

[1] refined this result by taking into account the maximum descent. We give an analogousresult for a further important permutation statistic, the length of the longest increasingsubsequence. In a recent paper, Stanley [8] established the importance of sign-balance.

E-mail address:[email protected].

0195-6698/$ - see front matter © 2004 Elsevier Ltd. All rights reserved.doi:10.1016/j.ejc.2004.06.009

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1010 A. Reifegerste / European Journal of Combinatorics 26 (2005) 1009–1018

For π ∈ Tn, let lis(π) be the length of the longest increasing subsequence inπ . By[2, Theorem 4], the number of permutationsπ ∈ Tn for which lis(π) = k is just the squareof theballot number

b(n, k) = 2k − n + 1

n + 1

(n + 1

k + 1

)

where�n+12 � ≤ k ≤ n. This is an immediate consequence of the Robinson–Schensted–

Knuth correspondence which gives a bijection between permutations and pairs of standardYoung tableaux of the same shape (see, e.g., [7]). It is well-known that the length of thelongest increasing (decreasing) subsequence of a permutation is just the length of the firstrow (column) of its associated tableaux.

Consequently, 321-avoiding permutations correspond to pairs of standard Youngtableaux having at most two rows. Such tableaux can be identified with ballot sequencesin a natural way. (We call a sequenceb1b2 · · · bn with bi = ±1 a ballot sequenceifb1 + b2 + · · · + bj ≥ 0 for all j .) Given a standard Young tableau with at most tworows, setbi = 1 if i appears in the first row andbi = −1 otherwise. By the Ballot theorem(see, e.g., [4]), there areb(n, k) ballot sequences of lengthn having exactlyk componentsequal to 1.

Elizalde and Pak [3] have already made use of the simplicity of the RSK correspondencefor 321-avoiding permutations for their bijection between refined restricted permutations.

Our main result is the following.

Theorem 1.1. For all n ≥ 1, wehave∑

π∈T2n+1

sign(π) · qlis(π) =∑π∈Tn

q2lis(π)+1

∑π∈T2n+2

sign(π) · qlis(π) = (q − 1)∑π∈Tn

q2lis(π)+1.

To prove this theorem, we establish an involution onTn which is defined in terms ofthe corresponding tableau pairs. While it is clear how to see the length of the longestincreasing subsequence from the tableaux, for the sign of a permutation this connection isstill unknown. InSection 2, we solve this problem for 321-avoiding permutations.

The description of the main bijectionΦ is done inSection 3. Modifying Φ slightlyyields a combinatorial proof of Adin–Roichman’s result which is given inSection 4. Weconclude with a simple combinatorial proof of the equidistribution of the last descent andπ−1(n)− 1 overTn asked for by Adin and Roichman.

2. How to obtain the sign from the tableaux

The key problem we are confronted with is to figure out how to see the sign of apermutation by looking at its associated pair of tableaux. For 321-avoiding permutations,there is a simple answer.

Givenπ ∈ Tn, let p andq be the ballot sequencesdefined by the tableauxP and Qwhich weobtain by applying the RSK algorithm toπ . (As usual, wewrite P to denote the

A. Reifegerste / European Journal of Combinatorics 26 (2005) 1009–1018 1011

insertion tableau, andQ for the recording tableau.) Define the statisticsrs to be the sumof the elements of the second row ofP andQ, that is,

srs(π) =∑

pi =−1

i +∑

qi =−1

i .

Proposition 2.1. For anyπ ∈ Tn with lis(π) = k, wehavesign(π) = (−1)srs(π)+n−k.

For theproof we use a method to generate a matching between excedances and anti-excedances described by Elizalde and Pak [3].

For π ∈ Sn, an excedance(anti-excedance) of π is an integeri for which πi > i(πi < i ). Here theelementπi is called anexcedance letter(anti-excedance letter). Itis characteristic for 321-avoiding permutations that both the subword consisting of allexcedance letters and the subword consisting of the remaining letters are increasing. Due tothis condition, wesay that the permutation isbi-increasing. In particular, the fixed pointsof a permutationπ ∈ Tn are just such integersi for which π j < πi for all j < i andπ j > πi for all j > i . Consequently, each longest increasing subsequence ofπ containsall the fixed points ofπ .

Given a permutation π ∈ Tn, let i1 < i2 < · · · < i s be its excedances andj1 < j2 < · · · < jt the anti-excedances. Construct the matching as follows.

(1) Initialize a = b = 1.

(2) While a ≤ s andb ≤ t repeat the procedure:

• If i a > jb, then increaseb by 1.• If πia < π jb, then increasea by 1.• If i a < jb andπia > π jb, then matchi a with jb and increase botha andb by 1.

For more clarity, we representπ ∈ Tn by ann×n array with a dot in each of the squares(i , πi ). We will identify the integeri with the dot(i , πi ), anduse the terms “excedance”and “anti-excedance” correspondingly for the dots as well (seeFig. 1 for an example).

It follows immediately from the descriptions of the RSK algorithm and the matchingthat the elementπi appears in the second row of the tableauP if and only if (i , πi )

is a matched excedance. In particular, thenumber of matched pairs is just the lengthof the second row ofP, see [3, Lemma 5]. Because of the symmetry of the RSKcorrespondence—if(P, Q) is the pair of tableaux associated withπ ∈ Sn, then(Q, P)corresponds to the inverse permutationπ−1—the integerj is contained in the second rowof Q if and only if ( j , π j ) is a matched anti-excedance. For the permutationπ ∈ T12considered inFig. 1, weobtain the tableaux

P = Q =

and hencesrs(π) = 56.The connection between the matched pairs and the inversions of the permutation is not

difficult to see. We call a pair(i , j ) an inversionof π if i < j andπi > π j , andwriteinv(π) to denote the number of inversions ofπ .


Fig. 1. Example of the matching forπ = 4 1 2 5 7 8 3 6 9 12 10 11∈ T12.

Let [i , j ] be a matched pair wherei < j . Denote byc(i , j ) thenumber of integersl forwhich i < l < j andπl > πi or j < l andπ j < πl < πi . Graphically, the dots(l , πl ) arecontained in regions 2 and 3 of the scheme:

Clearly, the first mentioned ones are excedances while the last mentioned are anti-excedances. Forr = 1, . . . ,4, letcr (i , j ) be the number of dots in regionr .

Lemma 2.2. Let π ∈ Tn with lis(π) = k. Furthermore, let[i1, j1], [i2, j2], . . . , [i n−k,

jn−k] be the matched pairs where il < jl for all l. Then we have:

(a) c(i , j ) is even ifand only ifπi + j is even.(b) inv(π) = c(i1, j1)+ c(i2, j2)+ · · · + c(i n−k, jn−k)+ n − k.

Proof. (a) Sinceπ ∈ Tn, we havec1(i , j ) = 0. Moreover, there is no dot northeast of(i , πi ) or southwest of( j , π j ). Hencec2(i , j ) + c3(i , j ) + 2c4(i , j ) = n − πi + n − j .Thus we havec(i , j ) = c2(i , j )+ c3(i , j ) ≡ πi + j mod 2.(b) Sinceπ is bi-increasing, any inversion ofπ has to be a pair(i , j ) with an excedanceiand an anti-excedancej .

First assume thati is a matched excedance, and letj be its match. Furthermore, let(i , j ′) be an inversion. Ifj ≤ j ′ we haveπ j ≤ π j ′ < πi . There are exactly c3(i , j ) + 1such integersj ′. If j ′ < j , the anti-excedancej ′ is matched with an excedancei ′ for whichi ′ < i . (Otherwise,j ′ would be matched withi by the definition of the matching.) Thus(i , πi ) belongs to the dots which are counted byc2(i ′, j ′).


Now let i be an unmatched excedance. If(i , j ) is an inversion, thenj must be matched.(Otherwise, we could matchi with j .) By the construction, the matchi ′ of j satisfiesi ′ < i(and henceπi ′ < πi ). Thus(i , πi ) is contained in the region whose dots are counted byc2(i ′, j ). �

By definition, we havesrs(π) = πi1 + πi2 + · · · + πin−k + j1 + j2 + · · · + jn−k.Therefore, from the lemma we immediately obtain the assertion ofProposition 2.1: a 321-avoiding permutation is even if and only if the sum of the elements of the second row ofthe two tableaux, increased by the length of this row, is even.

3. Proof of Theorem 1.1

In this section, we construct a bijection onTn which proves the main result. Its essentialpart is an involution on the ballot sequences of lengthn having a given numberk of 1s.

We define thesignof a ballot sequenceb to be 1 if the sum of all integersi with bi = −1is even and−1 otherwise, andwrite sign(b) to denote it.

For a ballot sequenceb, let ε(b) be the smallest even integeri for which bi = −bi+1.If there is no such integer setε(b) = 0. By An,k we denote the set of all ballot sequencesb of lengthn havingk 1s and satisfying ε(b) > 0, and byA∗

n,k the set of all sequencesbwith ε(b) = 0.

Consider now the following mapφ on An,k. Let b ∈ An,k satisfyε(b) = j . Then definec = φ(b) to be the sequence which is obtained fromb by exchanging the elementsbj andbj +1, i.e.,cj = −bj , cj +1 = −bj +1, andci = bi otherwise.

Proposition 3.1.

(a) The mapφ is a sign-reversing involution on An,k.(b) For odd n, we have|A∗

n,k| = b(n−12 , k−1

2 ) if k is odd, and|A∗n,k| = 0 otherwise.

(c) For even n, we have|A∗n,k| = b(n

2 − 1, � k−12 �) for all k.

Proof. (a) If b ∈ An,k with ε(b) = j , then the sequencec = φ(b) belongs toAn,k

as well. Note thatb1 + · · · + bj −1 ≥ 1 since j is even, and hencec1 + · · · + cj =b1 + · · · + bj −1 − bj ≥ 0. Obviously,φ is an involution. By construction, the sum of thepositions of the−1s inb andc differs by 1. Hencesign(b) = −sign(c).(b), (c) A sequenceb belongs toA∗

n,k if andonly if b2i = b2i+1 for 1 ≤ i ≤ �n−12 �.

It is easy to see that there is a bijection between the sequences inA∗n,k and all the

ballot sequences of length�n−12 � having a certain number of 1s. In the case of oddn, set

d = b2b4 · · · bn−1. (Notethatb2 = b3 = 1; otherwise, we would haveb1+b2+b3 = −1.)The numberk of 1s in b is always an odd integer sinceb1 = 1. Therefore,d is a ballotsequence withk−1

2 components equal to 1. Ifn is even, then the sequenced is defined asd = b2b4 · · · bn−2. Because we havebn = 1 if k is even andbn = −1 otherwise we mayomit the final element. Hence the number of 1s ind is equal to� k−1

2 �. �

Now we construct the involution that provesTheorem 1.1.Given π ∈ Tn with lis(π) = k, let p and q be the ballot sequences defined by

the tableauxP and Q resulting from the RSK algorithm. Defineσ = Φ(π) to be the


321-avoiding permutation whose associated pair(P′, Q′) of tableaux is given by thepair

(p′,q′) =(φ(p),q) if p ∈ An,k

(p, φ(q)) if p ∈ A∗n,k andq ∈ An,k

(p,q) otherwise

of ballot sequences.Sinceφ preserves the number of 1s in a ballot sequence we havelis(σ ) = lis(π).

By definition, srs(π) is even if and only ifp andq have the same sign. Consequently,Φ reverses the sign of the permutation ifp ∈ An,k or q ∈ An,k. (This follows fromPropositions 2.1and3.1(a).)

Let now π be a fixed point ofΦ, that is,ε(p) = ε(q) = 0. Clearly, the numberof such permutations is just|A∗

n,k|2. In the case of oddn (and hence oddk), we have

sign(π) = (−1)srs(π). By definition of A∗n,k, pi = −1 for any even integeri if and only

if pi+1 = −1. The same relation holds for the sequenceq, too. Therefore,∑

pi =−1 i and∑qi =−1 i , respectively, are the sum ofn−k

2 odd numbers. In particular,p and q are ofthe same sign. Thussrs(π) is even, andsign(π) = 1. By similar reasoning, we obtainsign(π) = (−1)k if n is even.

Summarized,Φ yields the relations

e(2n + 1,2k + 1)− o(2n + 1,2k + 1) = b(n, k)2

e(2n + 1,2k)− o(2n + 1,2k) = 0

e(2n + 2,2k + 1)− o(2n + 2,2k + 1) = −b(n, k)2

e(2n + 2,2k + 2)− o(2n + 2,2k + 2) = b(n, k)2

for all n andk. Heree(n, k) ando(n, k) denote the number of even and odd permutationsin Tn, respectively, whose longest increasing subsequence is exactly of lengthk.

4. A combinatorial proof of Adin–Roichman’s result

In [1], Adin and Roichman proved the sign-balance on the set of 321-avoidingpermutations having a given last descent by a recursion formula for the generating functionand asked for a combinatorial proof. By modifyingΦ slightly, we obtain the desiredinvolution. Moreover, this involution even proves a further refinement of our main resultandTheorem 4.1.

Forπ ∈ Sn, adescentof π is an integeri for whichπi > πi+1. The set of the descentsof π we denote byD(π), its maximum element byldes(π). (Defineldes(id) = 0.)

Since the restrictions of a 321-avoiding permutation to its excedances and non-excedances, respectively, are increasing words, an integeri is a descent ofπ ∈ Tn ifand only if i is an excedance buti + 1 is not. In particular, the maximum descent is themaximum excedance as well.

By [5, Corollary 3.13], there areb(n+d−1,n−1) = n−dn+d

(n+d

d

)permutationsπ ∈ Tn

for which ldes(π) = d.


Theorem 4.1 ([1, Theorem 4.1]). For all n ≥ 1, wehave∑

π∈T2n+1

sign(π) · qldes(π) =∑π∈Tn

q2ldes(π),

∑π∈T2n

sign(π) · qldes(π) = (1 − q)∑π∈Tn

q2ldes(π).

By the description of the RSK algorithm, the last descent is just the maximum elementiin the first row of the tableauQ for which i +1 appears in the second row ofQ. (Moreover,an integeri is a descent ofπ ∈ Tn if andonly if i appears in the first row whilei + 1 is inthe second row ofQ.) Thus the bijectionΦ also preserves the last descent ofπ if ε(p) > 0or ε(q) ≤ ldes(π)− 2. Consequently, the involution we look for will nearly coincide withΦ; weonly have to modify the casesε(p) = 0, ε(q) ≥ ldes(π)− 1 andε(p) = ε(q) = 0.

For a ballot sequenceb, let δ(b) be the greatest integeri for which bi = −bi+1 = 1.(Setδ(11· · ·1) = 0. As mentioned above, we haveδ(q) = ldes(π) if q is the sequencedefined by the recording tableau ofπ .) In addition toAn,k and A∗

n,k, we useBn,k, B∗n,k,

andB×n,k to denote the sets of all ballot sequencesb of lengthn havingk 1s and satisfying

0 < ε(b) < δ(b) − 1,0 < ε(b) = δ(b) − 1, andε(b) ≥ δ(b) > 0, respectively. (The setAn,k is just the union ofBn,k, B∗

n,k, andB×n,k.)

Lemma 4.2. Let π ∈ Tn be a permutation for whichlis(π) = k and ldes(π) = d, and(P, Q) its associatedpair of tableaux. Furthermore, let p and q be the ballot sequencesdefined by P and Q, respectively. Assume that p∈ A∗

n,k and q∈ A∗n,k ∪ B∗

n,k ∪ B×n,k. Then

we have:

(a) If d is even, thenπ is even.(b) If both n and d are odd, thenπ is even ifand only if q∈ A∗

n,k. Moreover, there are asmany sequences q∈ A∗

n,k as sequences q∈ B∗n,k satisfyingδ(q) = d.

(c) If n is even and d is odd, thenπ is even ifand only if q∈ A∗n,k and k is even. Moreover,

for any even k there are as many sequences q∈ A∗n,k as sequences q∈ B∗

n,k withqn = 1 satisfyingδ(q) = d.

Proof. By the definition ofδ, we haveq = q1q2 · · · qd−11(−1)(−1)a1b wherea,b ≥ 0.(Here exponentiation denotes repetition.)

(a) By assumption, we haveε(q) = 0 or ε(q) = δ(q) − 1 or ε(q) ≥ δ(q). Ifδ(q) = d > 0 is even,ε(q) equals necessarilyd. (The integerε(q) is positive sinceqd = −qd+1 and it has to be even by definition.) In particular, all the componentsqi = −1with i < d appear in pairs. First letn be odd. ByProposition 3.1(b), k must be odd as wellsincep ∈ A∗

n,k. As already discussed in the previous section,∑

pi =−1 i is the sum ofn−k2

odd numbers. Sinceq contains the element−1 exactlyn − k times, the exponenta is odd.Thus

∑qi =−1 i is also the sum ofn−k

2 odd numbers. Hencep andq have the same sign,andsrs(π) is even. In the case of evenn, we have to distinguish whetherk is evenor not.By reasoning in a similar way as done for oddn, we can show thatsrs(π) is even if andonly if k is even. Thussign(π) = 1.

(b) Let bothn (and hencek) andd be odd. By definition, we haveq ∈ A∗n,k if qd−1 = 1,

andq ∈ B∗n,k if qd−1 = −1. (Note that the exponenta must be odd in the first case.)


Consider the following mapψ that takesq ∈ A∗n,k to a sequenceq′ ∈ B∗

n,k. Forq ∈ A∗

n,k, let j be the last position of−1 in q. Defineq′ to be the sequence which isobtained fromq by exchanging the elementsqd−1 andqj . Sinced − 1 is even,we haveq1 + · · · + qd−2 ≥ 1, and the exchange is permitted. Obviously,q′ ∈ B∗

n,k andδ(q′) = d.(Note that j must be odd. Thus we may assume thatj > d + 1.) Moreover,ψ reversesthe sign. It is easy to see thatψ is bijective. The inverse map just takesq ∈ B∗

n,k to thesequenceq′ ∈ A∗

n,k which is defined byq′d−1 = qj = 1,q′

j = qd−1 = −1 andq′i = qi

otherwise, wherej is the position of the first 1 to the right ofqd.Similarly to part (a), we can prove that the signs ofp and q coincide if q ∈ A∗

n,k.Applyingψ yields the conversion.

(c) Let n be even now andd odd. For evenk we can prove the assertion by the samearguments as used in part (b). Note that the mapψ is not defined ifq ∈ B∗

n,k ends withan element−1. In particular, we havesign(p) = sign(q) (and hencesign(π) = 1) ifand only ifq ∈ A∗

n,k. If k is odd, thenp andq are of the same sign. On the one side, wehavepn = −1; on the other side either there is an even integeri for which qi = −1 andqi+1 = 1 ord = n − 1. Hencesign(π) = −1. �

Let π ∈ Tn satisfy lis(π) = k and ldes(π) = d. As before, let p andq be the ballotsequences defined by the tableauxP andQ obtained from the RSK correspondence. Defineσ = Ψ (π) to be the 321-avoiding permutation whose associated tableauxP′ and Q′ aregiven by the sequences

(p′,q′) =

(φ(p),q) if p ∈ An,k

(p, φ(q)) if p ∈ A∗n,k andq ∈ Bn,k

(p, ψ(q)) if p ∈ A∗n,k andq ∈ A∗

n,k ∪ B∗∗n,k; n − k is even and

d is odd(p,q) otherwise

whereψ is defined inthe proof of Lemma 4.2(b), andB∗∗n,k denotes the set of all sequences

q ∈ B∗n,k for which qn = 1. By the previous discussion,Ψ is an involution onTn which

preserves bothlis andldes, and reverses the sign if π is not a fixed point.Finally, we enumerate the fixed points ofΨ regarding the last descent.

Proposition 4.3. Let f(n,d) be the number of permutationsπ ∈ Tn for whichΨ (π) = π

and ldes(π) = d.

(a) For odd n, we have f(n,d) = b(n+d−32 , n−3

2 ) if d is even, and f(n,d) = 0 otherwise.(b) For even n, we have f(n,d) = b(�n+d−2

2 �, n−22 ) for all d.

Proof. (a) First let n be odd. In this case,π is a fixedpoint if and only if p ∈ A∗n,k and

ε(q) = d. (On condition thatd is even, this is the only possibility forq ∈ A∗n,k ∪ B∗

n,k ∪B×

n,k.) As shown in the proof ofProposition 3.1(b), the sequencep corresponds in a one-to-

one fashion to a ballot sequence of lengthn−12 with k−1

2 components equal to 1. (Note thatk has to be odd.) On the other side, there is a bijection between ballot sequencesq of lengthn with an oddnumberk of 1s satisfying ε(q) = δ(q) = d and ballot sequencesq′ of lengthn−1

2 having k−12 1s and satisfying δ(q′) = d

2 . Setq′ = q2q4 · · · qd−2qdqd+2 · · · qn−1. Notethat we haveqi = qi+1 for all eveni < d. Furthermore,qi = −1 for d + 1 ≤ i ≤ j


andqi = 1 for j + 1 ≤ i ≤ n, where j must be an even integer becausen − k is even.In particular,qd+2 = −1. Hence we haveδ(q′) = d

2 . Consequently,(p,q) correspondsto a pair of ballot sequences thatis associated with a permutationτ ∈ Tn−1

2for which

ldes(τ ) = d2 (andlis(τ ) = k−1

2 ). Thus f (n,d) = b(n−12 + d

2 − 1, n−12 − 1).

(b) Now consider the case of evenn. If d is even, we haveε(q) = d again. For evenk, the situation is similar to part (a). The sequenceq corresponds to a ballot sequenceq′of length n

2 with k2 elements equal to 1 andδ(q′) = d

2 ; setq′ = q2q4 · · · qdqd+2 · · · qn.If k is odd, we can not assume thatqd+2 = −1. Thus we must add the elementqd+1 toq′ to obtainδ(q′) = d

2 . On theother side, we may omit the final element since we haveqn = 1. Henceq can be identified withq′ = q2q4 · · · qdqd+1qd+2qd+4 · · · qn−2; a ballotsequence of lengthn2 with k−1

2 1s andδ(q′) = d2 . For any fixedpointπ of Ψ , the sequence

p belongs toA∗n,k. By Proposition 3.1(c), p corresponds to a ballot sequence of length

n2 − 1 with � k−1

2 � 1s. Definep′ to be the sequence obtained fromp by this bijectionand adding an element 1 ifk is even and−1 otherwise. Thus(p,q) corresponds in a one-to-one fashion to a pair(p′,q′) of ballot sequences of lengthn2 with � k

2� 1s satisfyingδ(q′) = d

2 . Consequently,f (n,d) counts the number of permutations inTn2

having the last

descentd2 .It is not difficult to see that for any evend there are as many sequencesq with δ(q) = d

as withδ(q) = d + 1 among the ballot sequences arising from the insertion tableau of afixedpoint ofΨ . �

By Lemma 4.2, the fixedpoints ofΨ are odd permutations if and only ifn is even andd is odd. This completes the proof ofTheorem 4.1.

As aby-product we obtain the following refinement ofTheorems 1.1and4.1.

Corollary 4.4. For all n ≥ 1, wehave∑π∈Tn

q2lis(π)+1t2ldes(π)=∑

π∈T∗2n+1

sign(π) · qlis(π)t ldes(π)

∑π∈Tn

q2lis(π)+1t2ldes(π)=∑π∈T∗

2n

sign(π) · qlis(π)t ldes(π)

+ q∑π∈T�2n

sign(π) · qlis(π)t ldes(π).

where T∗n (resp. T�n ) denotes the set of all permutationsπ ∈ Tn whose last descent is even

and whose longest increasing subsequence is of odd (resp. even) length.

5. Final remarks

In their paper, Adin and Roichman investigate a further permutation statistic onTn: theposition of the lettern in π , denoted bylind. Usinggenerating functions, they prove theequidistribution ofldes andlind − 1. We give a simple bijective proof.


Theorem 5.1 ([1, Theorem 3.1]). The statisticsldes andlind−1 are equidistributed overTn. Moreover, for any B⊆ [n − 2] they are equidistributed over the set

{π ∈ Tn : D(π−1) ∩ [n − 2] = B}where[n] means the set{1, . . . ,n}.Proof. Letπ ∈ Tn be a permutation for whichldes(π) = d. (Thend is the last excedanceas well.) Defineσ to be the permutation which is obtained fromπ by deleting the lettern,followed by inserting the elementn between the positionsd andd + 1.

This map is a bijection onTn: all the letters to the right ofn in π are smaller than thepositions they take. (Note that a fixed point has to exceed the elements on its left.) Thusthe erasure ofn yields a bi-increasing permutation of lengthn − 1 whose last descent isat most d. Inserting n afterπd preserves both the order of the excedance letters and non-excedance letters, and we haveσ ∈ Tn. Clearly,lind(σ ) = d + 1. If d is an excedance ofσ thenn is a fixedpoint inπ ; otherwise its original position wasd, and wecan recoverπfrom σ by exchangingσd andσd+1 = n.

Furthermore, we haveD(π−1)∩[n−2] = D(σ−1)∩[n−2]. By definition, the descentset D(π−1) of the inverse ofπ contains all the integersi for which the elementi + 1appears to the left of the elementi in π . For i < n − 1, the relative position ofi andi + 1in σ is the same as that inπ . �

Consequently, we also obtain sign-balance onTn concerning the statisticlind.

Acknowledgement

I would like to thank Herb Wilf for the interest taken in this work.

References

[1] R.M. Adin, Y. Roichman, Equidistribution andsign-balance on 321-avoiding permutations, 2003,math.CO/0304429(preprint).

[2] E. Deutsch, A.J. Hildebrand, H.S. Wilf, Longest increasing subsequences in pattern-restricted permutations,2003,math.CO/0304126(preprint).

[3] S. Elizalde, I. Pak, Bijections for refined restricted permutations, J. Combin. Theory Ser. A 105 (2004)207–219.

[4] W. Feller, An Introduction to Probability Theoryand Its Applications, vol. I, Wiley, New York, 1968.[5] A. Reifegerste, The excedances and descents of bi-increasing permutations, 2002,math.CO/0212247

(preprint).[6] R. Simion, F.W. Schmidt, Restricted permutations, European J. Combin. 6 (1985) 383–406.[7] R.P. Stanley, Enumerative Combinatorics, vol. II, Cambridge University Press, 1999.[8] R.P. Stanley, Some remarks on sign-balanced and maj-balanced posets, 2002,math.CO/0211113(preprint).

http://arxiv.org/pdf/math.CO/0304429




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Refined sign-balance on 321-avoiding permutations