Descent polynomials for permutations with bounded drop size

Transcription

1 FPSAC 2010, San Francsco, USA DMTCS proc. AN, 2010, Descent polynomals for permutatons wth bounded drop sze Fan Chung 1, Anders Claesson 2, Mark Dukes 3 and Ronald Graham 1 1 Unversty of Calforna at San Dego, La Jolla CA 92093, USA. 2 The Mathematcs Insttute, School of Computer Scence, Reykjavík Unversty, 103 Reykjavík, Iceland. 3 Scence Insttute, Unversty of Iceland, 107 Reykjavík, Iceland. Abstract. Motvated by jugglng sequences and bubble sort, we examne permutatons on the set {1, 2,..., n} wth d descents and maxmum drop sze k. We gve explct formulas for enumeratng such permutatons for gven ntegers k and d. We also derve the related generatng functons and prove unmodalty and symmetry of the coeffcents. Résumé. Motvés par les sutes de jonglere et le tr à bulles, nous étudons les permutatons de l ensemble {1, 2,..., n} ayant d descentes et une talle de défcence maxmale k. Nous donnons des formules explctes pour l énumératon de telles permutatons pour des enters k et d fxés, ans que les fonctons génératrces connexes. Nous montrons auss que les coeffcents possèdent des proprétés d unmodalté et de symétre. Keywords: Permutatons, descent polynomal, drop sze, Euleran dstrbuton. 1 Introducton There have been extensve studes of varous statstcs on S n, the set of all permutatons of {1, 2,..., n}. For a permutaton π n S n, we say that π has a drop at f π < and that the drop sze s π. We say that π has a descent at f π > π +1. One of the earlest results [8] n permutaton statstcs states that the number of permutatons n S n wth k drops equals the number of permutatons wth k descents. A concept closely related to drops s that of excedances, whch s just a drop of the nverse permutaton. In ths paper we focus on drops nstead of excedances because of ther connecton wth our motvatng applcatons concernng bubble sort and jugglng sequences. Other statstcs on a permutaton π nclude such thngs as the number of nversons (pars (, j such that < j and π > π j and the major ndex of π (the sum of for whch a descent occurs. The enumeraton of and generatng functons for these statstcs can be traced back to the work of Rodrgues n 1839 [9] but was manly nfluenced by McMahon s treatse n 1915 [8]. There s an extensve lterature studyng the dstrbuton of the above statstcs and ther q-analogs, see for example Foata and Han [4], or the papers of Shareshan and Wachs [10, 11] for more recent developments. Ths jont work orgnated from ts connecton wth a paper [2] on sequences that can be translated nto jugglng patterns. The set of jugglng sequences of perod n contanng a specfc state, called the ground AC and MD were supported by grant no from the Icelandc Research Fund c 2010 Dscrete Mathematcs and Theoretcal Computer Scence (DMTCS, Nancy, France

2 116 Fan Chung, Anders Claesson, Mark Dukes and Ronald Graham state, corresponds to the set B n,k of permutatons n S n wth drops of sze at most k. As t turns out, B n,k can also be assocated wth the set of permutatons that can be sorted by k operatons of bubble sort. These connectons wll be further descrbed n the next secton. We note that the maxdrop statstc has not been treated n the lterature as extensvely as many other statstcs n permutatons. As far as we know, ths s the frst tme that the dstrbuton of descents wth respect to maxdrop has been determned. Frst we gve some defntons concernng the statstcs and polynomals that we examne. Gven a permutaton π n S n, let Des(π = {1 < n : π > π +1 } be the descent set of π and let des(π = Des(π be the number of descents. We use maxdrop(π to denote the value of the maxmum drop (or maxdrop of π, maxdrop(π = max{ π( : 1 n }. Let B n,k = {π S n : maxdrop(π k}. It s known, and also easy to show, that B n,k = k!(k+1 n k ; e.g., see [2, Thm. 1] or [7, p. 108]. Let b n,k (r = {π B n,k : des(π = r}, and defne the (k-maxdrop-restrcted descent polynomal B n,k (x = r 0 b n,k (rx r = π B n,k x des(π. Examnng the case of k = 2, we observed the coeffcents b n,2 (r of B n,2 (x appear to be gven by every thrd coeffcent of the smple polynomal (1 + x 2 (1 + x + x 2 n 1. Lookng at the next two cases, k = 3 and k = 4, yelded more mysterous polynomals: b n,3 (r appeared to be every fourth coeffcent of and b n,4 (r every ffth coeffcent of (1 + x 2 + 2x 3 + x 4 + x 6 (1 + x + x 2 + x 3 n 2 (1 + x 2 + 2x 3 + 4x 4 + 4x 5 + 4x 7 + 4x 8 + 2x 9 + x 10 + x 12 (1 + x + x 2 + x 3 + x 4 n 3. After a ferce battle wth these polynomals, we were able to show that b n,k (r s the coeffcent of u r(k+1 n the polynomal P k (u ( 1 + u + + u k n k (1 where P k (u = k A k j (u k+1 (u k+1 1 j j=0 k =j ( u, (2 j and A k denotes the kth Euleran polynomal (defned n the next secton. Further to ths, we gve an expresson for the generatng functon B k (z, y = n 0 B n,k(yz n, namely ( k t ( k A t (y (y 1 1 A t (y z t t=1 =1 B k (z, y = k+1 (. (3 k z (y 1 1 =1

3 Descent polynomals for permutatons wth bounded drop 117 We also gve some alternatve formulatons for P k whch lead to some denttes nvolvng Euleran numbers as well as provng the symmetry and unmodalty of the polynomals B n,k (x. Many questons reman. For example, s there a more natural bjectve proof for the formulas that we have derved for B n,k and B k? Why do permutatons that are k-bubble sortable defne the aforementoned jugglng sequences? 2 Descent polynomals, bubble sort and jugglng sequences We frst state some standard notaton. The polynomal A n (x = π S n x des(π s called the nth Euleran polynomal. For nstance, A 0 (x = A 1 (x = 1 and A 2 (x = 1 + x. Note that B n,k (x = A n (x for k n 1, snce maxdrop(π n 1 for all π S n. The coeffcent of x k n A n (x s denoted n k and s called an Euleran number. It s well known that [5] 1 w e (w 1z w = k,n 0 The Euleran numbers are also known to be gven explctly as [3, 5] n n k = =0 We defne the operator bubble whch acts recursvely on permutatons va bubble(lnr = bubble(lrn. n w k zn k n!. (4 ( n+1 (k+1 n ( 1. In other words, to apply bubble to a permutaton π n S n, we splt π nto (possbly empty blocks L and R to the left and rght, respectvely, of the largest element of π (whch ntally s n, nterchange n and R, and then recursvely apply ths procedure to L. We wll use the conventon that bubble( = ; here denotes the empty permutaton. Ths operator corresponds to one pass of the classcal bubble sort operaton. Several nterestng results on the analyss of bubble sort can be found n Knuth [7, pp ]. We defne the bubble sort complexty of π as bsc(π = mn{k : bubble k (π = d}, the number of tmes bubble must be appled to π to gve the dentty permutaton. The followng lemma s easy to prove usng nducton. Lemma 1 ( For all permutatons π we have maxdrop(π = bsc(π. ( The bubble sort operator maps B n,k to B n,k 1. The class of permutatons B n,k appears n a recent paper [2] on enumeratng jugglng patterns that are usually called steswaps by (mathematcally nclned jugglers. Suppose a juggler throws a ball at tme so that the ball wll be n the ar for a tme t before landng at tme t +. Instead of an nfnte sequence, we wll consder perodc patterns, denoted by T = (t 1, t 2,..., t n. A jugglng sequence s just one n whch two balls never land at the same tme. It s not hard to show [1] that a necessary and suffcent condton for a sequence to be a jugglng sequence s that all the values t + (mod n are dstnct. In

4 118 Fan Chung, Anders Claesson, Mark Dukes and Ronald Graham partcular, t follows that that the average of t s just the numbers of balls beng juggled. Here s an example: If T = (3, 5, 0, 2, 0 then at tme 1 a ball s thrown that wll land at tme = 4. At tme 2 a ball s thrown that wll land at tme = 7. At tme 3 a ball s thrown that wll land at tme = 3. Alternatvely one can say that no ball s thrown at tme 3. Ths s represented n the followng dagram Repeatng ths for all ntervals of length 5 gves For a gven jugglng sequence, t s often possble to further decompose nto shorter jugglng sequences, called prmtve jugglng sequences, whch themselves cannot be further decomposed. These prmtve jugglng sequences act as basc buldng blocks for jugglng sequences [2]. However, n the other drecton, t s not always possble to combne prmtve jugglng sequences nto a longer jugglng sequence. Nevertheless, f prmtve jugglng sequences share a common state (whch one can thnk of as a landng schedule, then we can combne them to form a longer and more complcated jugglng sequences. In [2] prmtve jugglng sequences assocated wth a specfed state are enumerated. Here we menton the related fact concernng B n,k : There s a bjecton mappng permutatons n B n,k to prmtve jugglng sequences of perod n wth k balls that all share a certan state, called the ground state. The bjecton maps π to φ(π = (t 1,..., t n wth t = k + π. As a consequence of the above fact and Lemma 1, we can use bubble sort to transform a jugglng sequence usng k balls to a jugglng sequence usng k 1 balls. To make ths more precse, let T = (t 1,..., t n be a jugglng sequence that corresponds to π B n,k, and suppose that T = (s 1,..., s n s the jugglng sequence that corresponds to bubble(π. Assume that the ball B thrown at tme j s the one that lands latest out of all the n throws. In other words, t j + j s the largest element n {t + } n =1. Now, wrte T = Lt jr where L = (t 1,..., t j 1 and R = (t j+1,..., t n. Then we have T = f k (T = f k (LRs, where s = t j + j (n + 1. In other words, we have removed the ball B thrown at tme j and thus throw all balls after tme j one tme unt sooner. Then at tme n we throw the ball B so that t lands one tme unt sooner than t would have orgnally landed. Then we repeat ths procedure to all the balls thrown before tme j. 3 The polynomals B n,k (y In ths secton we wll characterse the polynomals B n,k (y. Ths s done by frst fndng a recurrence for the polynomals and then solvng the recurrence by explotng some aspects of ther assocated char-

5 Descent polynomals for permutatons wth bounded drop 119 acterstc polynomals. The latter step s qute nvolved and so we present the specal case dealng wth B n,4 (y frst. 3.1 Dervng the recurrence for B n,k (y We wll derve the followng recurrence for B n,k (y. Theorem 1 For n 0, wth the ntal condtons k+1 ( k + 1 B n+k+1,k (y = (y 1 1 B n+k+1,k (y (5 =1 B,k (y = A (y, 0 k. We use the notaton [a, b] = { Z : a b} and [b] = [1, b]. Let A = {a 1,..., a n } wth a 1 < < a n be any fnte subset of N. The standardzaton of a permutaton π on A s the permutaton st(π on [n] obtaned from π by replacng the nteger a wth the nteger. Thus π and st(π are order somorphc. For example, st(19452 = If the set A s fxed, the nverse of the standardzaton map s well defned, and we denote t by st 1 A (σ; for nstance, wth A = {1, 2, 4, 5, 9}, we have st 1 A (15342 = Note that st and st 1 A each preserve the descent set. For any set S [n 1] we defne A n,k (S = {π B n,k : Des(π S} and t n (S = max{ N : [n, n 1] S}. Note that t n (S = 0 n the case that n 1 s not a member of S. Now, for any permutaton π = π 1... π n n A n,k (S defne f(π = (σ, X, where σ = st(π 1... π n 1, X = {π n,..., π n } and = t n (S. Example 1 Let S = {3, 7, 8}, and choose the permutaton π = n A 9,3 (S. Notce that Des(π = {3, 4, 7, 8} S. Now t 9 (S = 2. Ths gves f(π = (σ, X where σ = st( = and X = {π 7, π 8, π 9 } = {6, 7, 9}. Hence f( = (136425, {6, 7, 9}. Lemma 2 For any π n A n,k (S, the mage f(π s n the Cartesan product ( [n k, n] A n 1,k (S [n t n (S 2], t n (S + 1 where ( X m denotes that set of all m-element subsets of the set X. Proof: Gven π A n,k (S, let f(π = (σ, X. Suppose = t n (S. Then there are descents at postons n,..., n 1 (ths s an empty sequence n case = 0. Thus n π n > π n +1 > > π n 1 > π n n k, where the last nequalty follows from the assumpton that maxdrop(π k. Hence X s an ( + 1- element subset of [n k, n], as clamed. Clearly σ S n 1.

6 120 Fan Chung, Anders Claesson, Mark Dukes and Ronald Graham Next we shall show that σ s n A n 1,k. Notce that the entres of (π 1,..., π n 1 that do not change under standardzaton are those π l whch are less than π n. Snce these values reman unchanged, the values l π l are also unchanged and are thus at most k. Let (π a(1,..., π a(m be the subsequence of values whch are greater than π n. The smallest value that any of these may take after standardzaton s π n n k. So σ a(j π n n k for all j [1, m]. Thus a(j σ a(j a(j (n k = k (n a(j k for all j [1, m]. Therefore l σ l k for all l [1, n 1] and so σ A n 1,k. The descent set s preserved under standardzaton, and consequently σ s n A n 1,k (S [n 2], as clamed. We now defne a functon g whch wll be shown to be the nverse of f. Let π be a permutaton n A m,k (T, where T s a subset of [m 1]. We wll add + 1 elements to π to yeld a new permutaton σ n A m+,k (T [m+1, m+]. Choose any (+1-element subset X of the nterval [m++1 k, m++1], and let us wrte X = {x 1,..., x +1 }, where x 1 x +1. Defne g(π, X = st 1 V (π 1... π m x +1 x... x 1, where V = [m + + 1] \ X. Example 2 Let T = {1}, and choose the permutaton π = 3142 n A 4,3 (T. Notce that Des(π = {1, 3} T. Choose = 2 and select a subset X from [ , ] = {4, 5, 6, 7} of sze + 1 = 3. Let us select X = {4, 6, 7}. Now we have g(π, X = st 1 V ( = , where V s the set [ ] \ {4, 6, 7} = {1, 2, 3, 5}. Lemma 3 If (π, X s n the Cartesan product ( [m k, m + + 1] A m,k (T + 1 for some > 0 then g(π, X s n A m++1,k (T [m + 1, m + ]. Proof: Let σ = g(π, X. For the frst m elements of σ, snce σ j π j for all 1 j m, we have j σ j j π j whch gves max{j σ j : j [m]} max{j π j : j [m]} k. The fnal + 1 elements of σ are decreasng so the maxdrop of these elements wll be the maxdrop of the fnal element, m σ m++1 = m x 1 m (m k = k. Thus maxdrop(σ k and so σ B m++1,k. The descents of σ wll be n the set T [m + 1, m + ] snce descents are preserved under standardzaton and the fnal +1 elements of σ are lsted n decreasng order. Hence σ A m++1,k (T [m + 1, m + ], as clamed. We omt the straghtforward, but a bt tedous, proof of the followng mportant Lemma. Lemma 4 The functon f s a bjecton, and g s ts nverse.

7 Descent polynomals for permutatons wth bounded drop 121 Corollary 1 Let a n,k (S = A n,k (S and = t n (S. Then Proposton 1 For all n 0, a n,k (S = ( k + 1 a n (+1,k (S [1, n ( + 1]. + 1 k+1 ( k + 1 B n,k (y + 1 = y 1 B n,k (y + 1. =1 Proof: Notce that B n,k (y + 1 = π B n,k (y + 1 des(π = π B n,k = des(π =0 π B n,k S Des(π = S [n 1] y S ( des(π y S π A n,k (S y 1 = S [n 1] y S a n,k (S. From Corollary 1, multply both sdes by y S and sum over all S [n 1]. We have B n,k (y + 1 = ( k + 1 y S a n (t t n (S + 1 n(s+1,k(s [n (t n (S + 2] S [n 1] = ( k + 1 y y S a n (+1,k (S [n ( + 2] S [n 1] tn(s= = ( k = ( k = ( k = ( k y y S [n 1] tn(s= S [n (+1] y B n (+1,k (y + 1 y 1 B n,k (y + 1. a n (+1,k (S [n ( + 2]y S a n (+1,k (Sy S

8 122 Fan Chung, Anders Claesson, Mark Dukes and Ronald Graham Proof of Theorem 1: Replacng n and y by n + k + 1 and y 1, respectvely, n Proposton 1 yelds the recurrence (5: k+1 ( k + 1 B n+k+1,k (y = (y 1 1 B n+k+1,k (y for n 0, wth the ntal condtons B,k (y = A (y for 0 k. =1 By multplyng the above recurrence by z n and summng over all n 0, we have the generatng functon B k (z, y gven n equaton ( Solvng the recurrence for B n,4 (y. Before we proceed to solve the recurrence for B n,k, we frst examne the specal case of k = 4 whch s qute llumnatng. We note that the characterstc polynomal for the recurrence for B n,4 s h(z = z 5 5z (1 yz 3 10(1 y 2 z 2 + 5(1 y 3 z (1 y 4 = (z 1 + y5 yz 5. 1 y Substtutng y = t 5 n the expresson above, we see that the roots of h(z are just ρ j (t = 1 t5 1 ω j t, 0 j 4, where ω = exp( 2π 5 s a prmtve 5th root of unty. Hence, the general term for B n,4(t can wrtten as B n,4 (t = 4 α (tρ n (t =0 where the α (t are approprately chosen coeffcents (polynomals n t. To determne the α (t we need to solve the followng system of lnear equatons: 4 α (tρ j (t = B j,4(t = A j (t 5, 0 j 4. =0 Thus, α (t can be expressed as the rato N 4,+1 (t/d 4 (t of two determnants. The denomnator D 4 (t s just a standard Vandermonde determnant whose ( + 1, j + 1 entry s ρ j (t. The numerator N 4,+1(t s formed from D 4 (t by replacng the elements ρ j (t n the (+1st row by A j(t 5. A quck computaton (usng the symbolc computaton package Maple gves: D 4 (t = 25 5 (1 t 5 6 t 10 ; N 4,1 (t = 5 5 (t 12 + t t 9 + 4t 8 + 4t 7 + 4t 5 + 4t 4 + 2t 3 + t 2 + 1(1 t 5 3 (1 t 3 t 10 and, n general, N 4,+1 (t = N 4,1 (ω t. Substtutng the value α 0 (t = N 4,1 (t/d 4 (t nto the frst term n the expanson of B n,4, we get α 0 (t(1 + t + t 2 + t 3 + t 4 n = 1 5 (t12 + t t 9 + 4t 8 + 4t 7 + 4t 5 + 4t 4 + 2t 3 + t 2 + 1(1 + t + t 2 + t 3 + t 4 n 3.

9 Descent polynomals for permutatons wth bounded drop 123 Now, snce the other four terms α (t(1 + t + t 2 + t 3 + t 4 n arse by replacng t by ω t then n the sum of all fve terms, the only powers of t that survve are those whch have powers whch are multples of 5. Thus, we can conclude that f we wrte (t 12 + t t 9 + 4t 8 + 4t 7 + 4t 5 + 4t 4 + 2t 3 + t 2 + 1(1 + t + t 2 + t 3 + t 4 n 3 = r β(rt r then b n,4 (d = β(5d. In other words, the number of permutatons π B n,4 wth d descents s gven by the coeffcent of t 5d n the expanson of the above polynomal. Incdentally, the correspondng results for the earler B n, are as follows: b n,1 (d = β(2d n the expanson of (1 + t n = r β(rt r, so b n,1 (d = ( n 2d ; bn,2 (d = β(3d n the expanson of (1 + t 2 (1 + t + t 2 n 1 = r β(rt r ; and b n,3 (d = β(4d n the expanson of (1 + t 2 + 2t 3 + t 4 + t 6 (1 + t + t 2 + t 3 n 2 = r β(rt r. The precedng arguments have now set the stage for dealng wth the general case of B n,k. Of course, the arguments wll be somewhat more nvolved but t s hoped that treatng the above specal case wll be a useful gude for the reader. 3.3 Solvng the recurrence for B n,k (y Theorem 2 We have B n,k (y = d β k( (k + 1d y d, where β k (ju j = P k (u ( 1 + u + + u k n k and j P k (u = k A k j (u k+1 (u k+1 1 j j=0 The frst few values of P k (u are shown below. k P k (u u u + 2u 2 + u 3 + u u + 2u 2 + 4u 3 + 4u 4 + 4u 5 + 4u 6 + 2u 7 + u 8 + u u + 2u 2 + 4u 3 + 8u u u u u u u u u u u 14 + u 15 + u 16 k =j ( u. (6 j There s clearly a lot of structure n the polynomals P k (u whch wll be dscussed n the next secton.

10 124 Fan Chung, Anders Claesson, Mark Dukes and Ronald Graham 4 The structure of P k (u Consder Equaton 6 of Theorem 2. We wrte P k (u = k 2 =0 α u and defne the stretch of P k (u to be k k 2 P P k (u = α 0 + α k 2u k2 +k + α 1++(k+1j u 2++(k+1j+j. =0 j=0 What ths does to P k (u s to nsert 0 coeffcents at every (k + 1st term, startng after α 0. Thus, the stretched polynomals correspondng to the values of P k (u gven n the array above are: k P P k (u u u 2 + 2u 3 + u 4 + u u 2 + 2u 3 + 4u 4 + 4u 5 + 4u 7 + 4u 8 + 2u 9 + u 10 + u u 2 + 2u 3 + 4u 4 + 8u u u u u u u u u u u 17 + u 18 + u 20 Note that f P k (u has degree k 2 then P P k (u has degree k 2 + k. Theorem 3 For all k 1, P k+1 (u = P P k (u (1 + u + u u k+1. Theorem 4 The coeffcents of P k (u are symmetrc and unmodal. Proof: It follows from Theorem 3 that we can construct the coeffcent sequence for P k+1 (u from that of P k (u by the followng rule (where we assume that all coeffcents of u t n P k (u are 0 f t < 0 or t > k 2. Namely, suppose we wrte P k (u = k 2 =0 α u so that we have the coeffcent sequence A k = (α 0, α 1,..., α k 2. Now form the new sequence B k = (β 0, β 1,... β k2 +k by the rule β = α j, 0 k 2 + k. j= k Fnally, startng wth β 0, nsert duplcate values for the coeffcents Thus, ths wll generate the sequence β 0, β k+1, β 2(k+1,..., β t(k+1,..., β (k 1(k+1 and β k(k+1. (β 0, β 0, β 1, β 2,..., β k, β k+1, β k+1, β k+2,..., β k 2 +k 1, β k 2 +k, β k 2 +k. Ths new sequence wll n fact just be the coeffcent sequence A k+1 for P k+1 (u. For example, startng wth P 1 (u = 1 + u, we have A 1 = (1, 1 and so B 1 = (1, 2, 1. Now, nsertng the duplcate values for β 0 = 1 and β 2 = 1, we get the coeffcent sequence A 2 = (1, 1, 2, 1, 1 for P 2 (u = 1 + u + 2u 2 + u 3 + u 4. Repeatng ths process for A 2, we sum blocks of length 3 to get B 2 = (1, 2, 4, 4, 4, 2, 1. Insertng duplcates for entres at postons 0, 3 and 6 gves us the new coeffcent sequence A 3 = (1, 1, 2, 4, 4, 4, 4, 2, 1, 1 of P 3 = 1 + u + 2u 2 + 4u 3 + 4u 4 + 4u 5 + 4u 6 + 2u 7 + u 8 + u 9, etc. It s also clear from ths procedure that f A k s symmetrc and unmodal, then so s B k, and consequently, so s A k+1. Ths s what we clamed.

11 Descent polynomals for permutatons wth bounded drop An Euleran dentty Note that snce P k (u s symmetrc and has degree u k2, we have P k (u = u k2 P k ( 1 u. Replacng P k(u by ts expresson n (6, we obtan (wth some calculaton the nterestng dentty a+b ( a a+b ( ( b b ( 1 j (1 x j A a+b j (x = x (1 x j A a+b j (x + (1 x a+b+1 j j a + b j=0 j=0 for all ntegers a and b provded that a + b 0. References [1] J. Buhler, D. Esenberg, R. Graham and C. Wrght, Jugglng drops and descents, Amer. Math. Monthly 101 (1994, [2] F. Chung, and R. L. Graham, Prmtve jugglng sequences, Amer. Math. Monthly 115 (2008, [3] L. Euler, Methodus unversals seres summand ulterus promota, Commentar academae scentarum mperals Petropoltanae 8 (1736, Reprnted n hs Pera Omna, seres 1, volume 14, [4] D. Foata and G. Han, q-seres n Combnatorcs; permutaton statstcs (Lecture Notes, prelmnary edton, [5] R. L. Graham, D. E. Knuth and O. Patashnk, Concrete Mathematcs, Addson-Wesley, [6] D. E. Knuth: The Art of Computer Programmng, Vol. 1, Fundamental algorthms. Addson-Wesley, Readng, [7] D. E. Knuth, The Art of Computer Programmng, Vol. 3, Sortng and Searchng, Addson-Wesley, Readng, 2nd ed., [8] P. A. MacMahon, Combnatory Analyss, 2 volumes, Cambrdge Unversty Press, London, Reprnted by Chelsea, New York, [9] O. Rodrgues, Note sur les nversons, ou dérangements produts dans les permutatons, J. de Math. 4 (1839, [10] J. Shareshan and M. L. Wachs, q-euleran polynomals: excedance number and major ndex, Electron. Res. Announc. Amer. Math. Soc. 13 (2007, [11] J. Shareshan and M. L. Wachs, Euleran quassymmetrc functons, preprnt 2009.

12 126 Fan Chung, Anders Claesson, Mark Dukes and Ronald Graham