Generalized Eulerian Sums

Transcription

1 Generalzed Eleran Sms Fan hng Ron Graham Dedcated to Adrano Garsa on the occason of hs 84 th brthday Abstract In ths note, we dere a nmber of symmetrcal sms nolng Eleran nmbers and some of ther generalzatons These extend earler denttes of Don Knth and the athors, and also nclde seeral -nomal sms nspred by recent wor of Shareshan and Wachs on the jont dstrbton of aros permtaton statstcs, sch as the nmber of excedances, the major ndex and the nmber of fxed ponts of a permtaton We also prodce symmetrcal sms nolng restrcted Eleran nmbers whch cont permtatons π on {1, 2,, n} wth a gen nmber of descents and whch, n addton, hae the ale of π 1 (n specfed 1 Introdcton The classcal Eleran nmbers, whch we wll denote by n, occr n a arety of contexts n combnatorcs, nmber theory and compter scence (see [4] These nmbers were ntrodced by Eler [3] n 1736 and hae many nterestng propertes We lst some small ales n Table n Some Eleran nmbers n Table 1 In partclar, n enmerates the nmber of permtatons π on [n] : {1, 2,, n} whch hae descents (e, n1 wth π( > π(+1 as well as the nmber Unersty of alforna, San Dego 1

2 of permtatons π on [n] whch hae excedances (e, < n wth π( > Eleran nmbers satsfy the reflecton property n n, n > 0, (1 n 1 and the recrrence n n 1 n 1 ( (n 1 They also hae the explct representaton E n (t n ( n + 1 (1 ( + 1 n 0 and hae the followng generatng fncton (see [4]: n E(t, x t, n > 0, (2 (1 tex e tx te x 1 + n>0, 0 n t xn n! 1 + E n (t xn n! (3 n>0 Eleran nmbers grow rapdly n sze wth ncreasng n, and one doesn t expect as many combnatoral denttes to exst for them compared wth other seences sch as bnomal coeffcents or Strlng nmbers Neertheless, t was shown n [1] that the followng symmetrcal dentty holds: ( ( a + b a + b 0 a 1 0 b 1 for a, b > 0 (where nless stated otherwse, we se the conenton that In Secton 2, we proe seeral natral generalzatons of ths dentty In Secton 3, we consder a arant of the sal Eleran nmbers n whch we cont the nmber of permtatons π on [n] hang a gen nmber of descents wth the addtonal constrant that π(n 1 s specfed In Sectons 4 and 5, we deal wth -coeffcents arsng n the jont dstrbton of seeral permtaton statstcs stded by Shareshan and Wachs [7, 8] Fnally, n Secton 7, we close wth a nmber of open problems 2 An alternatng Eleran sm dentty Theorem 1 ( a + b (1 a for a, b > 0 0 ( a + b (1 0 b (4 2

3 Proof The generatng fncton for or modfed Eleran nmbers (e, wth s obtaned by sbtractng 1 from (1te x n (3 Ths ges Ths, ( 1 e x e x e tx e tx te x t e tx n, n t xn n! n, 0 e tx te x n t xn n! (5 ( 1 e x t ( e x e tx e tx e tx te x 1 e tx 1 e x Expressng the exponentals as sms mples (1 x! t (1 t x n t x n! n! 0 0 n, n t (1 x n+!n!,n, n (1,n, Shftng the ndces of smmaton yelds (1 n tn x n (1 n xn n! n!, t ++1 x n+!n! (1 n (tn 1x n n! ( n n t (1 x n n!,n, ( n n t (1 x n 1 n!,n, (1 n (tn 1x n n! (6 Identfyng the coeffcents of x n ges ( n n (1 t n (1 1,, ( n t (1 n (t n 1 Assmng 0, n and dentfyng the coeffcents of t ges ( n n (1 ( n n (

4 Fnally, sng the reflecton propertes of the Eleran nmbers n (1, replacng by n, and sng the reflecton propertes of the bnomal coeffcents, we obtan ( n (1 (1 ( n, n for 0 < < n Therefore, f we set n a + b, a, then we hae ( a + b (1 ( a + b (1 a b 0 0 (7 for a, b > 0 Ths proes the theorem Note that (7 can be consdered as a companon to (4 We can rewrte (4 n the form ( n n ( n n, (8 1 for 0 < < n As ponted ot n [1], ths s the frst n an nfnte seence of (ncreasngly complex sms of ths type The next two are: ( n n 2 ( n n 2 2 ( ( n n, for 1 < < n (9 1 ( n n 3 ( n n 3 3 ( ( n n n+1 1 ( ( n n n for 2 < < n (10 The companon sm (7 s also the frst of an nfnte seence of smlar sms, 4

5 the next few beng: ( n (1 2 n ( n (1 3 n ( n (1 2 n n + 1 ( ( n n, for 1 n (11 1 ( n (1 3 n n + 2 ( ( n n 2 n 2 n ( n ( n 2 for 1 n (12 The proofs of these follow the same lnes as that of (7 and are omtted 3 Generalzed Eleran denttes for restrcted descent polynomals In ths secton we consder a restrcted erson of Eleran nmbers We defne the restrcted Eleran nmber b(n,,, for 1 n, 0 < n, to be the nmber of π S n wth des(π and wth π( n Ths restrcton s smlar to one sed by the athors for nestgatng the jont statstcs of permtatons hang a gen nmber of descents and a gen bond on drop sze [2] (A permtaton π has a drop at f π( < and the drop sze s π( From the defnton of b(n,,, t follows that n b(n,,, 1 b(,,, b(n,, 0, f n < or n or < 0 The anttes b(n,, satsfy the followng reflecton property whch wll be needed later Lemma 1 For 1 < < n,, we hae b(n,, b(n,, n 1 (13 Proof The proof follows by consderng the bjecton whch maps π S n wth π( n to π S n wth π ( n defned by π (π( 1,, π(1, n, π(n,, π( + 1 5

6 We consder the followng descent polynomal B n, (t defned by: B n, (t 0 b(n,, t We wll need the generatng fncton for ths polynomal Lemma 2 The descent polynomal B n, (t has the followng generatng fncton: B (t, x n B n, (t xn1 (n 1! E x 1 (e tx te tx 1(t ( 1!(e tx te x (14 Proof We frst consder B n, (t for the case that < n Let S m,l denote the set of permtatons π S m whch hae l descents For a fxed ( 1-sbset of {1,, n}, there s a straghtforward bjecton from the set of permtatons π S n wth π( n and hang descents to the followng set: ( S1,j S n,j1 Ths, for < n, we hae 0 j< B,n (t b(n,, t ( n 1 1 n t t j1 t j2 1 By mltplyng both sdes by fxed, b(n,, t x n1 (n 1! n> j 1 j 1 j 2 xn1 (n1! and smmng oer all n >, we hae for t ( 1! E 1(t E n (tx n1 (n! n> tx1 E 1 (t E n (tx n ( 1! (n! n> ( tx1 E 1 (t (1 te x ( 1! e tx te x 1 ( e x e tx tx1 E 1 (t ( 1! j 2 e tx te x 6

7 Addng to ths sm a term for the case n, we obtan B (t, x b(n,, t x n1 (n 1! n ( tx1 E 1 (t e x e tx ( 1! e tx te x ( tx1 E 1 (t e x e tx ( 1! e tx te x x1 E 1 (t ( 1! x1 E 1 (t ( 1! ( t(e x e tx e tx te x + 1 ( e tx te tx e tx te x + + x1 E 1 (t ( 1! x 1 b(,, t ( 1! Now we are ready to proe the followng generalzed ealty for or restrcted descent polynomals Theorem 2 For > 1, r, s > 0, the restrcted Eleran nmbers b(n,, satsfy ( r + s + 1 b(,, r ( r + s + 1 b(,, s ( Proof We start wth the generatng fncton gen n Lemma 2 By crossmltplyng, we hae b(n,, t x n1 (e tx te x x1 E 1 (t (e tx te tx (n 1! ( 1! n, Expandng the exponental fnctons, we hae n,,j b(n,, t+j x n+j1 j!(n 1! n,,j b(n,, t+1 x n+j1 j!(n 1! x1 E 1 (t ( 1! (1 tt n x n n! Identfyng the coeffcent of x n1 ges t +j b(n j,, j!(n j 1!,j,j E 1(t ( 1! t +1 b(n j,, j!(n j 1! ( t n (n! tn+1 (n! 7

8 We frther dentfy the coeffcents of t l to get b(n l +,, (l!(n l + 1! j b(n j,, l 1 j!(n j 1! ( 1 l + n ( 1 n l 2 1 ( 1!(n! 1 ( 1!(n! 1 l n n l 1 Mltplyng both sdes by (n 1!, we obtan ( n 1 b(n l +,, ( n 1 b(n j,, l 1 l j j ( ( n n l 2 n l 1 (16 We wll deal wth each term n (16 separately For the frst sm on the left sde of (16, we frst change arables by settng l and then n as follows: ( n 1 b(n l +,, ( n 1 b(n,, l l ( n 1 b(,, l n + n Now, sng the reflecton property n Lemma 1, we hae ( n 1 b(,, l n + n ( ( n 1 1 n 1 b(,, n l l n + 1 ( ( n 1 1 n 1 b(,, n l n l 2 1 ( ( r + s n 1 b(,, r + 1 r 1 1 (17 by settng l s + 1 and n r + s + 2 8

9 The second sm of the left sde of (16 can be treated as follows: ( n 1 b(n j,, l 1 ( n 1 b(,, l 1 j n j ( n 1 b(,, l 1 1 ( r + s + 1 b(,, s (18 1 The rght sde of (16 ges ( ( 1 1 n 1 n l 2 n l 1 1 ( ( 1 1 n 1 r 1 r 1 ( ( 1 n 1 n 1 b(,, r (19 r Sbstttng (17, (18 and (19 nto (16, we hae ( r + s + 1 b(,, r ( r + s + 1 b(,, s 1 1 as clamed The proof of Theorem 2 s complete The only case that s left ot n Theorem 2 s the case of 1 Howeer, for ths case, b(n, 1, n1 1, and ths case redces to the orgnal Eleran ealty (4 4 Some -nomal generalzatons Recently, Shareshan and Wachs [7, 8] fond an elegant expresson for the jont dstrbton of the excedance and the major ndex of a permtaton n S n Frst, we ge some standard defntons (see [8] For π S n, defne Also set EX(π { : π( > }, exc(π EX(π, DES(π { : π( > π( + 1}, des(π DES(π, maj(π DES(π A maj,exc n (, t maj(π t exc(π π S n a n (, j t j (20,j 0 9

10 We wll defne the polynomal n (j n (j( by n (j : a n (, j j (21 so that we can wrte An maj,exc (, t j n (j(t j As sal, defne n n1, n! [n] [n 1] [1], a [a]! b [b]![a b]! Theorem 3 ([7, 8] An maj,exc (, t zn [n]! (1 t exp (z exp (tz t exp (z (22 where An maj,exc (, 0 1 and exp (z z n [n]! From ths reslt, we wll show how to dere -nomal generalzatons of (4 and (7 Theorem 4 For a, b > 0, the polynomals defned n (21 and (20 satsfy: a + b (a 1 a + b (b 1 (23 Note that (21 generalzes (4 snce n (j 1 0 a n (, j n j Proof We wll se the conenton that A maj,exc 0 (, t 0 (rather than 1 In ths case, sbtractng 1 from the expresson on (22, we hae exp (z exp (tz exp (tz t exp (z n,j 0 zn n (j(t j [n]! (24 10

11 Mltplyng by exp (tgz t exp (z and expandng, we obtan (tz t z n (j(t j zn []! []! [n]! z n [n]! 0 0 n,j 0 Ths,,n,j 0 n (j(t j+ z n+ []![n]!,n,j 0 Shftng the smmaton ndex n ges n (j(t j+ n z n [n],n,j 0 n (j(t j+1 z n+ []![n]!,n,j 0 (tz n [n]! (25 (1 (t n z n [n]! n (j(t j+1 [ n Identfyng the coeffcents of z n ges n (j(t j+ n n (j(t j+1 n,j 0,j 0 Now, shftng the smmaton ndex j, we hae n (j (t j n n (j 1(t j n,j 0,j 0 Identfyng coeffcents of t j then ges [ n n (j ] [ n n (j 1 ] ] z n [n] (1 (t n z n (26 [n]! 1 (t n 1 (t n 1 f j 0, 1 f j n, 0 f 0 < j < n It was shown n [8] that the n (j enjoy the symmetry property: Hence, we can rewrte (27 as [ n n (n 1 j + ] [ n (n 1 j ] (27 n (r n (n 1 r (28 [ n n (j 1 ] [ n (j 1 ] 1 f j 0, 1 f j n, 0 f 0 < j < n 11

12 Settng n a + b and j b yelds a + b (a 1 a + b (b 1, for a, b > 0 Ths proes the theorem The same technes can be sed to proe the followng companon sm for (22 Theorem 5 For a, b > 0, (1 a + b (a+b 2 (a Ths s a -nomal generalzaton of (7 5 Frther -nomal generalzatons (1 a + b (a+b 2 (b (29 In [7, 8], Shareshan and Wachs proe the followng more general erson of (22: Theorem 6 ([7, 8] A maj,exc,fx n (, t, r zn [n]! (1 t exp (rz exp (tz t exp (z (30 where A maj,exc,fx n (, 0, 0 1 and fx(π denotes the nmber of fxed ponts of π S n, e, the nmber of sch that π( Let s wrte (1 t exp (rz exp (zt t exp (z n,,j, and defne the polynomals n (j, n (j, ( by zn a n (, j, t j r [n]! n (j, a n (, j, j Theorem 7 For a, b, 0, the polynomals ( as defned n (31 satsfy a + b (a, a + b (b, (31 Proof ross-mltplyng, sbstttng and expandng the exponentals n (30, we obtan (tz n (j, (t j r zn []! [n]! t z n (, j, (t j r zn []! [n]! 0 n,j, 0 n,j, (rz n [n]! (1 t 12

13 Now, shftng ndces and dentfyng the coeffcents of z n (as before yelds n (j, (t j r n n (j 1, (t j r n (1 tr n j,, j,, Hence, for < n, we can dentfy the coeffcents of r and t and obtan n n (j, n n (j 1, (32 We next note the followng nce symmetry property of the n (j, Fact for n, j, 0 n (j, n (n j, (33 Proof Frst, a straghtforward comptaton from (30 erfes that A maj,exc,fx n (, t, r zn [n]! ( A maj,exc,fx 1 n, 2 t, r (tz n t [n]! (34 Ths mples a n (, j, t j r zn [n]! a n (, j, 2j+n t j+n r zn [n]! (35 n,,j, n,,j, Identfyng coeffcents of z n and r then ges a n (, j, t j a n (, j, 2j+n t j+n (36,j,j Shftng the ndces of smmaton n the second sm by 2j + n and j j + n yelds a n (, j, t j a n ( 2j + n, j + n, t j (37,j,j Now, dentfyng the coeffcents of and t j ges s a n (, j, a n (n + 2j, n j, (38 for, j,, n 0 Fnally, smmng oer ges a n (, j, j a n (n + 2j, n j, j a n (n + 2j, n j, (n+2j(nj a n (, n j, n+j+ 13

14 e, n (j, n (n j,, and the Fact s proed Now, contnng the proof of (31, we wll apply eaton (33 to (32 Ths, [ n n (j, ] [ n n (n j, ] [ n (j 1, ] Fnally, lettng n a + b + + 1, j b + 1, we get a + b (a, a + b (b, as rered Obsere that settng 1 n Theorem 6 ges s symmetrcal sms for the anttes c n (j, n (j, 1, whch cont the nmber of π n S n wth exc(π j and fx(π These sms can be wrtten as follows: Theorem 8 For a, b, > 0, the c defned aboe satsfy ( a + b c (a, ( a + b c (b, (39 6 oncldng remars In general, we wold le to fnd symmetrcal denttes smlar n form to (4, (15, or (23 whch depend on coeffcents arsng n the enmeraton of aros jont statstcs on permtatons For example, we cold try to proe denttes of the form a + b P (, a 1 a + b P (, b 1 (40 for some approprate polynomals P (, j( A natral place to loo s at the polynomals S(n, j s(n,, j where s(n,, j s defned to be the nmber of π S n sch that n(π and des(π j Unfortnately, we ddn t fnd any araton of (40 whch was ald We fnd ths to be slghtly srprsng snce a classc reslt of Stanley [6], namely, n A n(π,des(π (, t zn [n]! 1 t exp (tz t n zn S(n, jt j [n]!, wold seem to be n the approprate form for or analyss Howeer, we sspect that sch symmetrcal sms do exst 14

15 In another drecton, the precedng technes can be appled to a arety of other expressons whch hae an approprate generatng fncton For example, for π S n, we can defne D n (, t : maj(π t des(π d(n,, j t j, π S n,j Dn(, t : d(n,,, j t j π S n,π(n maj(π t des(π,j We can then apply the precedng argments to obtan sms smlar to (15 and (29 for those coeffcents It wold be nterestng to fnd bjecte proofs for some of these denttes sch as (7, (9, (10, (11, (15, (23 or (31 A beatfl (and non-tral bjecte proof of (4 was dscoered by Don Knth [1] bt no one has yet fond a correspondng bjecte proof for ts companon (7 7 Acnowledgements The athors wsh to than Adrano Garsa, Jeff Remmel and an anonymos referee for ther helpfl comments whle we were preparng ths note References [1] F hng, R Graham and D Knth, A symmetrcal Eleran dentty, Jornal of ombnatorcs, 1 (2010, [2] F hng and R Graham, Restrcted Eleran nmbers and descent polynomals, preprnt [3] L Eler, Methods nersals seres smmand lters promota, ommentar academae scentarm mperals Petropoltanae, 8 (1736, Reprnted n hs Pera Omna, seres 1, olme 14, [4] R Graham, D Knth and O Patashn, oncrete Mathematcs, A Fondaton for ompter Scence, Addson-Wesley, Boston, 1994, x, 657p [5] Jeff Remmel, (personal commncaton [6] R P Stanley, Bnomal posets, Möbs nerson, and permtaton enmeraton, J ombnatoral Theory Ser A 20 (1976, [7] J Shareshan and M Wachs, -Eleran polynomals, excedance nmber and major ndex, Electronc Res Annonc, Amer Math Soc 13 (2007, [8] J Shareshan and M Wachs, Ecleran assymmetrc fnctons, Ad n Math, 225 (2010,