COUNTING DESCENTS, RISES, AND LEVELS, WITH PRESCRIBED FIRST ELEMENT, IN WORDS

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1 COUNTING DESCENTS, RISES, AND LEVELS, WITH PRESCRIBED FIRST ELEMENT, IN WORDS Sergey Kitev The Mthemtic Intitute, Reykvik Univerity, IS-03 Reykvik, Icelnd Toufik Mnour Deprtment of Mthemtic, Univerity of Hif, 3905 Hif, Irel Jeffrey B Remmel 2 Deprtment of Mthemtic, Univerity of Cliforni, Sn Diego, L Joll, CA 92093, USA remmel@ucdedu Abtrct Recently, Kitev nd Remmel [9] refined the well-known permuttion ttitic decent by fixing prity of one of the decent number Reult in [9] were extended nd generlized in everl wy in [7, 0,, 2] In thi pper, we hll fix et prtition of the nturl number N, N,, N, nd we tudy the ditribution of decent, level, nd rie ccording to whether the firt letter of the decent, rie, or level lie in N i over the et of word over the lphbet [k] {,, k} In prticulr, we refine nd generlize ome of the reult in [4] Introduction The decent et of permuttion π π π n S n i the et of indice i for which π i > π i+ Thi ttitic w firt tudied by McMhon [3] lmot hundred yer go nd it till ply n importnt role in the field of permuttion ttitic The number of permuttion of length n with exctly m decent i counted by the Eulerin number A m n The Eulerin number re the coefficient of the Eulerin polynomil A n t π S n t +deπ It i well-known tht the Eulerin polynomil tify the identity m 0 mn t m A nt t For more propertie of the Eulerin polynomil ee [5] n+ Recently, Kitev nd Remmel [9] tudied the ditribution of refined decent ttitic on the et of permuttion by fixing prity of exctly one of the decent number For exmple, they howed tht the number of permuttion in S 2n rep S 2n+ with exctly k decent uch tht the firt entry of the decent i n even number i given by n 2n! 2 k rep n 2n k+ k +! 2 In [0], the uthor generlized reult of [9] by tudying decent ccording to whether the firt or the econd element in decent pir i congruent to 0 mod k 2 The work preented here w upported by grnt no /3 from the Icelndic Reerch Fund 2 The work preented here w prtilly upported by NSF grnt DMS

2 2 COUNTING DESCENTS, RISES, AND LEVELS, WITH PRESCRIBED FIRST ELEMENT, IN WORDS Subequently, Hll nd Remmel [7] generlized reult of [0] by conidering X, Y -decent, which re decent whoe top firt element i in X nd whoe bottom econd element i in Y where X nd Y re ny ubet of the nturl number N In prticulr, Hll nd Remmel [7] howed tht one cn reduce the problem of counting the number of permuttion σ with k X, Y -decent to the problem of computing the k-th hit number of Ferrer bord in mny ce Liee [] lo conidered the itution of fixing equivlence cle of both decent number imultneouly Alo, pper [6] nd [2] dicu q-nlogue of ome of the reult in [7, 9, 0, ] Hll nd Remmel [7] extended their reult on counting permuttion with given number of X, Y -decent to word Tht i, let Rρ be the rerrngement cl of the word ρ 2 ρ2 m ρ m ie, ρ copie of, ρ 2 copie of 2, etc where ρ + + ρ m n For ny et X N nd ny et [m] {, 2,, m}, we let X m X [m] nd X c m [m] X Then given X, Y N nd word w w w n Rρ, define De X,Y w {i : w i > w i+ & w i X & w i+ Y }, de X,Y w De X,Y w, nd P X,Y ρ, {w Rρ : de X,Y w } Hll nd Remmel [7] proved the following theorem by purely combintoril men Theorem P X,Y ρ, where X c m {v, v 2,, v b }, + r n + r ρx + r + α X,ρ,x + β Y,ρ,x ρ v, ρ v2,, ρ vb r r b i r0 ρ vi, nd for ny x X m, α X,ρ,x β Y,ρ,x z / X x < z m z / Y z < x x X ρ z, nd In thi pper, we hll tudy imilr ttitic over the et [k] n of n-letter word over fixed finite lphbet [k] {, 2,, k} In wht follow, E {2, 4, 6, } nd O {, 3, 5, } re the et of even nd odd number repectively Alo, we let x[t] x,, x t Then given word π π π 2 π n [k] n nd et X N, we define the following ttitic: De X π {i : π i > π i+ nd π i X} nd de X π De X π, Ri X π {i : π i < π i+ nd π i X} nd ri X π Ri X π, Lev X π {i : π i π i+ nd π i X} nd lev X π Lev X π Let N,, N be et prtition of the nturl number N, ie N N N 2 N, N i for ll i, nd N i N for i Then the min gol of thi pper i to tudy the following multivrite generting function MGF 2 A k A k x[]; y[]; z[]; q[] π i ρ z xi de Ni π yi ri Ni π ρ x z lev N i π i q iπ i,

3 COUNTING DESCENTS, RISES, AND LEVELS, WITH PRESCRIBED FIRST ELEMENT, IN WORDS 3 where iπ i the number of letter from N i in π nd the um i over ll word over [k] The outline of thi pper i follow In Section 2, we hll develop ome generl method to compute 2 In Section 3, we hll concentrte on computing generting function for the ditribution of the number of level Tht i, we hll tudy A k where et x i y i for ll i In Section 4, we hll find formul for the number of word in [k] n tht hve decent tht trt with n element le thn or equl to t greter thn t for ny t k Note tht if we replce word w w w n [k] n by it complement w c k + w k + w n, then it i ey to ee tht de [t] w ri {k+ t,,k} w nd de {t+,,k} w ri [k t] w Thu we will lo obtin formul for the number of word in [k] n tht hve rie tht trt with n element le thn or equl to t greter thn t for ny t k In Section 5, we hll pply our reult to tudy the problem of counting the number of word in [k] n with p decent rie tht trt with n element which i congruent to i mod for ny 2 nd i,, In prticulr, if 2 nd N,, N i the et prtition of N where N i {x N : x i mod } for i,,, then we hll tudy the generting function 3 A k x[]; y[]; z[]; q[] de Ni π ri Ni π xi yi z lev N i π i q iπ i π nd 4 A k x[]; y[]; z[]; q π i q π i xi de Ni π yi ri Ni π z lev N i π i Our generl reult in Section 2 llow u to derive n explicit formul for A k x[]; y[]; z[]; q[] depending on the equivlence cl of k mod For exmple, in the ce where 2, our generl reult implie tht 5 A 2 2k q, q 2 A k x, x 2, y, y 2, z, z 2, q, q 2 π x de O π de x E π 2 y ri O π y ri E π 2 z lev Oπ z lev Eπ 2 q oddπ q evenπ 2 + λ µ 2 + λ 2 µ k µk 2 µ µ 2 ν µ 2 + ν 2 µk µk 2 µ µ 2 where the um i over ll word over [2k], evenπ rep oddπ i the number of even rep odd number in π, λ q y q, µ z y i q iz i x i q, nd ν q iz i y i y q z y for, 2 Then by pecilizing the vrible ppropritely, we will find explicit formul for the number of word w [2k] n uch tht de E π p, de O π p, ri E π p, ri O π p, etc For exmple, we prove tht the number of n-letter word π on [2k] hving de O π p rep ri E π p i given by n p 0 i0 i n n+p+i 2 ki n p In fct, we hll how tht imilr formul hold for the number of word π [k] n with p decent rie, level whoe firt element i congruent to t mod for ny 2 nd 0 t Our reult refine nd generlize the reult in [4] relted to the ditribution of decent, level, nd rie in word Finlly, in Section 6, we hll dicu ome open quetion nd further reerch

4 4 COUNTING DESCENTS, RISES, AND LEVELS, WITH PRESCRIBED FIRST ELEMENT, IN WORDS We need the following nottion: 2 The generl ce A k i,, i m A k x[t]; y[t]; z[t]; q[t]; i[m] π t i xi de Ni π yi ri Ni π z lev N i π i q iπ i where the um i tken over ll word π π π 2 over [k] uch tht π π m i i m From our definition, we hve tht 2 A k + k A k i Thu, to find formul for A k, it i ufficient to find formul for A k i for ech i, 2,, k Firt let u find recurrence reltion for the generting function A k i Lemm 2 For ech N i, k nd i t, we hve 22 A k Proof From the definition we hve tht i q i y i q i z i y i A k + q i y i q i z i y i + q ix i y i A k q i z i y i A k q i + k A k, q i + A k, + A k, + k + A k, Let π be ny n-letter word over [k] where n 2 nd π > π 2 If we let π π 2 π 3 π n, then it i ey to ee tht de Ni π + de Ni π, iπ + iπ It i lo ey to ee tht remining 4t 2 ttitic of interet tke the me vlue on π nd π Thi implie tht A k, q i x i A k for ech < Similrly, A k, q i z i A k nd A k, q i y i A k for < k Therefore, k A k q i + q i x i A k + q i z i A k + q i y i A k + Uing 2, we hve k + A k A k A k A k, nd thu deired A k Lemm 22 For ech k nd [k], 23 A k where, for i N m nd i, γ i q i y i q i z i y i A k + q i y i q i z i y i + q ix i y i A k, q i z i y i γ i+ α i q m y m q mz m y m A k + q m y m q mz m y m nd α i q my m x m q mz m y m

5 COUNTING DESCENTS, RISES, AND LEVELS, WITH PRESCRIBED FIRST ELEMENT, IN WORDS 5 Proof We proceed by induction on Note, tht given our definition of γ i nd α i, we cn rewrite A k γ α A k It follow tht A k γ o tht 23 hold for Thu the be ce of our induction hold Now ume tht 23 hold for where < k Then uing our induction hypothei nd 24, it follow tht A k + + A k + A k + γ α i + γ + α + γ + i+ + + γ i+ + γ i+ α i α i γ i+ α i Thu the induction tep lo hold o tht 23 mut hold in generl Lemm 2 give tht the A k i, for i k, re the olution to the following mtrix eqution α A k γ α 3 α A k 2 25 α 4 α 4 α γ 2, A k k γ k α k α k α k α k α k where, for i N m nd i, γ i q my m q m z m y m A k + qm ym q m z m y m, nd, for i N m nd i 2, α i q m y m x m q mz m y m Notice tht α i α nd γ i γ whenever i nd re from the me et N m for ome m In fct, it i ey to ee tht 22 nd 23 imply tht i 26 A k i γ i α i γ i i+ α i hold for i,, k o tht 25 h n explicit olution By combining 2 nd 26, we cn obtin the following reult Theorem 23 For α i nd γ i bove defined in Lemm 22, we hve A k + k γ k i+ α i

6 6 COUNTING DESCENTS, RISES, AND LEVELS, WITH PRESCRIBED FIRST ELEMENT, IN WORDS olving which for A k give 27 A k + k k q y k q iz i x i q z y i+ q i z i y i q y k q iz i x i q z y i+ q i z i y i where for ech vrible {x, y, z, q} we hve i m if i N m Even though we tte Theorem 23 the min theorem in thi pper, it ttement cn be eily generlized if one conider compoition inted of word ee [8] Indeed, let B k B k x[t]; y[t]; z[t]; q[t]; v t v π de Ni π ri Ni π xi yi z lev N i π i q iπ i π i where the um i tken over ll compoition π π π 2 with prt in [k] nd π π + π 2 + i the weight of the compoition π Alo, we let B k i,, i m B k x[t]; y[t]; z[t]; q[t]; i[m]; v t v π de Ni π ri Ni π xi yi z lev N i π i q iπ i π where gin the um i tken over ll compoition π π π 2 with prt in [k] Next, one cn copy the rgument of Lemm 2 ubtituting q i by v q i to obtin the following generliztion of Lemm 2: B k v q i y i q i z i y i B k + v q i y i q i z i y i + v q i x i y i B k q i z i y i One cn then prove the obviou nlogue of Lemm 2 by induction nd pply it to prove the following theorem Theorem 24 We hve where γ i B k + k k γ i+ i α i v i q my m q m z m y m B k + vi q m y m q m z m y m, nd α i vi q my m x m q m z m y m if i belong to N m Thu, B k + k k v q y k q i z i y i +v i y i x i q z y i+ q iz i y i v q y k q i z i y i +v i y i x i q z y i+ q iz i y i where for ech vrible {x, y, z, q} we hve i m if i N m Theorem 24 cn be viewed q-nlogue of Theorem 23 Set v in Theorem 24 to get Theorem 23 3 Counting word by the type of level Suppoe we re given et prtition N N N 2 N Firt oberve tht for ny fixed i, if we wnt the ditribution of word in [k] n ccording to the number of level which involve element in N i, then it i ey to ee by ymmetry tht the ditribution will depend only on the crdinlity of N i [k] Thu we only need to conider the ce where 2 nd N {,, t} for ome t k

7 COUNTING DESCENTS, RISES, AND LEVELS, WITH PRESCRIBED FIRST ELEMENT, IN WORDS 7 Let λ ν q y q z y, q y q z y, nd µ i q iz i x i q i z i y i Then we cn rewrite 27 for ny rbitrry et prtition N N N 2 N 34 A k + k λ k i+ µ i k ν k i+ µ i where for ech vrible {x, y, z, q}, we hve i m if i N m Suppoe we et x x 2 y y 2 z 2 nd q q 2 q in 34 in the pecil ce where 2 nd q N [t] for ome t k Then λ λ 2 0, ν qz, ν 2 q, nd µ µ 2 It follow in thi ce tht A k tq qz + qk t m q m t + k t qz m 0 m i m q k m t m i t i i qz m 0 i0 Since 35 it follow tht 36 A k n 0 i i qz! q z 0 i + q z, 0 n m m0 i0 i i + n m n m k t m i t i z n m Thu tking the coefficient of z on both ide of 36, we obtin the following reult Theorem 3 Let N N N 2 where N [t] nd N 2 N N Then if t k, the number of word in [k] n with level tht trt with element in N i n m m i + n m n m 37 n m k t m i t i i n m m0 i0 Going bck to the generl et prtition N N N 2 N, we cn obtin generl formul for the number of word in [k] n for which there re t i level which trt with n element of N i for i,,

8 8 COUNTING DESCENTS, RISES, AND LEVELS, WITH PRESCRIBED FIRST ELEMENT, IN WORDS follow Let n i N i [k] for i,, n Then if et x y nd q q for ll, then it will be the ce tht λ 0 nd µ nd ν for ll It ey follow tht in thi ce, A k Then uing 35, we ee tht 38 A k m 0 n 0 q m q qz i n i q qz i q m qz m 0 i i q m m,, m m 0 i + +m,, 0 n m0 + +m,, 0 + m,, 0 n i m m n,, n m b + +bn m b,,b 0 m,, m i b i 0 n i qz i i i bi q bi z bi b i! n n i + b i z i b i Tking the coefficient of z t zt on both ide of 38, we obtin the following reult Theorem 32 Let N N N be et prtition of N Let n i N i [k] for i,, Then the number of word in [k] n with t i level tht trt with element in N i for i,, i n m 39 n,, i + b i bi n m b i t i m0 + +m,, 0 b + +bn m b,,b 0 i i b i 4 Clifying word by the number of decent tht trt with element t t + In thi ection, we hll conider the et prtition N N N 2 where N [t] Now if t k, then it i ey to ee tht we cn rewrite 27 4 A k + k λ k i+ µ i k ν k i+ µ i where nd λ q y q z y, ν q y q z y, nd µ q z x q z y if t λ q 2 y 2 q 2 z 2 y 2, ν q 2 y 2 q 2 z 2 y 2, nd µ q 2z 2 x 2 q 2 z 2 y 2 if > t Now if we wnt to find formul for the number of word in [k] n with decent tht trt with n element le thn or equl to t, then we need to et x 2 y y 2 z z 2 nd q q 2 q in 4 In tht

9 COUNTING DESCENTS, RISES, AND LEVELS, WITH PRESCRIBED FIRST ELEMENT, IN WORDS 9 ce, we will hve λ 0 nd ν q for ll, µ + qx for t, nd µ for > t It follow tht A k k t+ q + t q t i+ + qx k tq + q +qx t +qx x k tqx + + qx t k tqx x m + + qx t m m 0 m 0 m 0 x m x m m 0 m m k tqx m + qx t m t 0 b0 c0 b t m b q b k t b x b q c x c c If we wnt to tke the coefficient of, then we mut hve b + c n or c n b Thu 42 A k q m m m m t n m b k t b x n m b n b n 0 m 0 0 b0 Tking the coefficient of of both ide of 42, we ee tht 43 de x [t] π π [k] n m 0 m m 0 b0 b t m b k t b x n m n b for ll n However, if we replce x by z + in 43, we ee tht the polynomil de z + [t] π π [k] n h the Lurent expnion m m m 0 0 b0 m t m b k t b z n m b n b It follow tht it mut be the ce tht m m m m t m b k t b z n m 0, b n b o tht 44 A k n 0 m n+ 0 b0 n m m m m b b m0 0 b0 t n b k t b x n m

10 0 COUNTING DESCENTS, RISES, AND LEVELS, WITH PRESCRIBED FIRST ELEMENT, IN WORDS Thu if we tke the coefficient of x on both ide of 44 nd we ue the remrk in the introduction tht de [t] w ri {k+ t,,k} w for ll w [k] n, then we ee tht the number of word w [k] n uch tht de [t] w ri {k+ t,,k} w i equl to 45 m m m n b b m0 0 b0 t n b n m k t b Reordering the ummnd, we ee tht 45 i equl to 46 t k t b n b n b b0 b m+b m b n m However it i the ce tht 47 m+b m m n m b n + + b b Thi i ey to ee combintorilly Tht i, we cn interpret the RHS of 47 the number of wy of chooing + b + + point, x < x 2 < < x +b++ n +, from the et [n + ] nd then circling of the point from x < < x +b However if we clify our choice by the vlue m + of x +b+, then we ee tht m cn be viewed the number of wy to pick the circled point from [m], m b cn be viewed the number of wy to pick the non-circled choen point from [m], nd the binomil coefficient n m cn be viewed the number of wy to pick the point x+b+2 < < x +b++ from the intervl {m + 2,, n + } Thu we hve proved the following theorem Theorem 4 If t k, then the number of word w [k] n uch tht de [t] w ri {k+ t,,k} w i equl to 48 t k t b n b n + + b n b + b + + b0 b If we wnt to find formul for the number of word in [k] n with decent tht trt with n element greter thn t, then we need to et x y y 2 z z 2 nd q q 2 q in 4 In tht ce, we will hve

11 COUNTING DESCENTS, RISES, AND LEVELS, WITH PRESCRIBED FIRST ELEMENT, IN WORDS λ 0 nd ν q for ll, µ + qx 2 for > t, nd µ for t It follow tht A k t 0 q + qx 2 k t + k t+ q k i+ + qx 2 qt + qx 2 k t + q +qx2 k t +qx 2 x 2 qtx 2 + qx 2 k t + + qx k t x 2 qtx qx 2 k t qtx2 x 2 m + + qx 2 k t m m 0 m 0 m 0 x 2 m x 2 m m 0 m 0 b0 m m qtx qx 2 k t k t c0 m b k t c q b t b x 2 b q c x 2 c Agin, if we wnt to tke the coefficient of, then we mut hve b + c n or c n b Thu 49 A k q m m k t n m t b x 2 n b n b n 0 m 0 0 b0 Tking the coefficient of of both ide of 49, we ee tht de 40 {t+,k} π 2 m m k t m t b x 2 n m b n b m 0 0 π [k] n x for ll n However, if we replce x 2 by z + in 43, we ee tht the polynomil de z + {t+,,k} π π [k] n h the Lurent expnion m m 0 0 b0 b0 m It follow tht it mut be the ce tht m m o tht 4 A k n 0 m n+ 0 b0 n m m0 0 b0 b b m k t n b k t n b b k t n b t b z n m t b z n m 0, t b x n m

12 2 COUNTING DESCENTS, RISES, AND LEVELS, WITH PRESCRIBED FIRST ELEMENT, IN WORDS Thu if we tke the coefficient of x 2 on both ide of 4 nd we ue the remrk in the introduction tht de {t+,,k} w ri [k t] w for ll w [k] n, then we hve tht the number of word w [k] n uch tht de {t+,,k} w ri [k t] w i equl to 42 m m0 0 b0 b0 n b Reordering the ummnd, we ee tht 42 i equl to k t 43 t b n b n b b k t n m n b m t b n m However it i ey to ee tht 44 m n m n becue the LHS of 44 i the reult of clifying the wy to pick + + point from [n + ] by the vlue m + of the + -t point reding from left to right Thu we hve obtin the following theorem Theorem 42 If t k, then the number of word w [k] n uch tht de {t+,,k} w ri [k t] w i equl to k t n + 45 t b n b n b + + b0 b In ome pecil ce, we cn give direct combintoril proof of Theorem 4 nd 42 For exmple, in the pecil ce of Theorem 42 where t k, one cn ue the Pfff-Slchütz Theorem to how tht 46 o tht 45 reduce to 47 b n b n + b n b + + b b n b k b b0 It i ey to ee tht 47 i the reult of clifying the word uch tht de {k} w by the number of letter which re not equl to k in the word Tht i, if there re b uch letter, then we cn pick word of length b in the lphbet {,, k } in k b wy Next we inert k directly in front of different letter in u in b wy to crete decent tht trt with k Finlly we cn plce the remining n b k either in block with one of the k tht trt decent or t the end of u The number of wy to plce the remining k i the number of non-negtive integer vlued olution to x + + x + n b or, equivlently, the number of poitive integer vlued olution to y + + y + n b + which i clerly We cn give imilr combintoril proof of Theorem 4 in the pecil ce when t 2 n b

13 COUNTING DESCENTS, RISES, AND LEVELS, WITH PRESCRIBED FIRST ELEMENT, IN WORDS 3 5 Clifying decent nd rie by their equivlence cle mod for 2 In thi ection we tudy the et prtition N N N 2 N where > 2 nd N i { i mod } for i,, In thi ce, we hll denote N i N + i for i,, nd N N Recll tht we cn rewrite 27 5 A k + k λ k i+ µ i k ν k i+ µ i where λ q y q z y, µ i qizi xi q i z i y i, nd ν q y q z y We let A k denote A k under the ubtitution λ i+ λ, µ i+ µ, nd ν i+ ν for ll i nd,, Then it i ey to ee tht for k, + A k + λ i+ µ k i r0 µ µ 2 µ r ν i+ µ k i r0 µ µ 2 µ r λ i+ µ µ µ 2 µ k i µ µ 2 µ ν i+ µ µµ 2 µ k i µ µ 2 µ Similrly for t, 52 A k+t + t λ t i+ µ i t ν t i+ µ i + t λ t i+ µ i t ν t i+ µ i k k r0 µ µ r + µ µ 2 µ t t+ λ i+ µ i r0 µ µ r µ µ 2 µ t t+ ν i+ µ i + µ µ 2 µ t t+ λ i+ µ i µ µ 2 µ t t+ ν i+ µ i µ µ 2 µ k+ µ µ 2 µ µµ 2 µ k+ µ µ 2 µ k k r0 µ µ r r0 µ µ r µ µ 2 µ k µ µ 2 µ µµ 2 µ k µ µ 2 µ 5 The ce where k i equl to 0 mod Firt we hll conider formul for the number of word in [k] n with p decent whoe firt element i congruent to r mod where r Note tht if we conider the complement mp comp k : [k] n [k] n given by compπ π n k + π k + π n, then it i ey to ee tht de N+r π ri N++ r comp k π for r,, Thu the problem of counting the number of word in [k] n with p decent whoe firt element i congruent to r mod i the me counting the number of word in [k] n with p rie whoe firt element i congruent to + r mod Now conider the ce where z i y i nd q i q for i,, nd x i for i r In thi ce, A k n 0 de x N+r π r π [k] n

14 4 COUNTING DESCENTS, RISES, AND LEVELS, WITH PRESCRIBED FIRST ELEMENT, IN WORDS Subtituting into our formul for A k, we ee tht in thi ce λ i 0 nd ν i q for i,, nd µ i for i r nd µ r + qx r Thu under thi ubtitution, 52 become A k r qµ r + r + q µk r qx r x + r qx r r µ k r x r + r qx r µ k r x r x r i,i 20 i,i 20 i i i r i q i x r i i2 i2 µ ki2 r i r i q i x r i Tking the coefficient of in 53, we ee tht n t + i o tht 53 A k Thu we mut hve 54 n 0 de x N+r π r π [k] n 0 i,i i,i 2 0 i 2 i r i i 2 ki2 i ki2 2 q t x r t t ki2 i i 2 n i ki2 i2 i r i i i 2 n i for ll n However, if we replce x r by z + in 54, we ee tht the polynomil de z + N+r π π [k] n h the Lurent expnion 0 i,i 2 0 ki2 i2 i r i i i 2 n i It follow tht it mut be the ce tht i 2 i r i ki2 i i 2 n i o tht 55 A k n+ i,i 20 n 0 n 0 i,i 2 0 t0 x r n z n ki2 i2 i r i i i 2 n i z n 0 x r n x r n Thu we hve the following theorem by tking the coefficient of x p r on both ide of 55

15 COUNTING DESCENTS, RISES, AND LEVELS, WITH PRESCRIBED FIRST ELEMENT, IN WORDS 5 Theorem 5 The number of word π [k] n with de N+r π p ri N++ r π p i 56 n p 0 i,i 2 0 ki2 n n+p+i2 i r i i i 2 n i p In the ce 2, our formul implify omewht For exmple, putting 2 nd r 2 in Theorem 5, we obtin the following Corollry 52 The number of n-letter word π on [2k] hving de E π p rep ri O π p i given by 0 i,i 20 n+p+i 2 2 i ki2 i i 2 n i n Similrly, putting 2 nd r in Theorem 5, we obtin the following Corollry 53 The number of n-letter word π on [2k] hving de O π p rep ri E π p i given by n p ki n n+p+i 2 i n p 0 i0 52 The ce where k i equl to t mod for t,, Fix t where t Firt we hll conider formul for the number of word in [k + t] n with p decent whoe firt element i congruent to r mod where r We hll ee tht we hve to divide thi problem into two ce depending on whether r t or r > t Note tht if we conider the complement mp comp k+t : [k + t] n [k + t] n given by compπ π n k + t + π k + t + π n, then it i ey to ee tht de N+r π ri N+t+ r comp k π for r,, t nd de N+r π ri N++r t comp k π for r t +,, Firt conider the ce where y i z i for i,, nd x i for i r where r > t In thi ce, A k+t de N+r π r n 0 π [k+t] n x Subtituting into our formul for A k+t, we ee tht in thi ce λ i 0 nd ν i q for i,, nd µ i for i r nd µ r + qx r Thu under thi ubtitution, 52 become A k+t qt µk+ r qx r r tqµ r + r + q µk r qx r x r [tµk+ r + t + r tqx r µ k r ] x r [µk r[ + r qx r ] [ + r tqx r ]] x r m [µk r[ + r qx r ] [ + r tqx r ]] m m0 m m0 0 x r m m + r tqx r + r qx r µ k r p

16 6 COUNTING DESCENTS, RISES, AND LEVELS, WITH PRESCRIBED FIRST ELEMENT, IN WORDS Uing the expnion + r tqx r + r qx r µ k r i 0 i 2 0 k i 30 nd etting i + i 2 + i 3 n, we ee tht 57 become A k+t 57 Thu we mut hve 58 n 0 m m0 0 i 0 i 20 i 2 k i 3 i i r t i q i x r i, i 2 r i 2 q i 2 x r i 2, nd q i 3 x r i 3, m i i 2 r t i r i 2 m k i i 2 n i i 2 de x N+r π r π [k t ] n m m0 0 i 0 i 2 0 x r n m m i i 2 r t i r i 2 m k i i 2 n i i 2 x r n m for ll n However, if we replce x r by z + in 58, we ee tht the polynomil de z + N+r π π [k+t] n h the Lurent expnion m m i i 2 r t i r i 2 m0 0 i 0 i 20 It follow tht m mn+ 0 i 0 i 20 o tht 59 n A k+t m m0 0 i 0 i 2 0 m m i i 2 r t i r i 2 m i i 2 r t i r i 2 m k i i 2 n i i 2 m k i i 2 n i i 2 m k i i 2 n i i 2 Thu we hve the following theorem by tking the coefficient of x p r on both ide of 59 z n m z n m 0 x r n m

17 COUNTING DESCENTS, RISES, AND LEVELS, WITH PRESCRIBED FIRST ELEMENT, IN WORDS 7 Theorem 54 If t,, nd t < r, then the number of word π [k + t] n with de N+r π p ri N++r t π p i 50 n p m m0 0 i 0 i 20 n+p+ m i i 2 r t i r i 2 k n m i i 2 n i i 2 p In the ce 2, our formul implify omewht For exmple, putting 2, r 2 nd t in Theorem 54, we obtin the following Corollry 55 The number of n-letter word π over [2k + ] hving de E π p rep given by n p m m k n m n+p+ 2 m i i n i p m0 0 i0 ri E π p i Next conider the ce where y i z i for i,, nd x i for i r where r t In thi ce, A k+t de N+r π r n 0 π [k+t] n x Subtituting into our formul for A k+t, we ee tht in thi ce λ i 0 nd ν i q for i,, nd µ i for i r nd µ r + qx r Thu, under thi ubtitution, 52 become A k+t r qµ r + qt r + µk+ r qx r tqµ r qx r x [t + r qx r r µ k+ r + tµ r µ k r ] x r [µk+ r [ + r qx r ] [ + t + r qx r ]] x r m [µk+ r [ + r qx r ] [ + t + r qx r ]] m m0 m m0 0 Uing the expnion x r m + t + r qx r µk r m + t + r qx r + r qx r µ k+ r + r µ r µ k+ r i 0 i 2 0 k+ i 30 i 2 i i t + r i q i x r i, i 2 r i 2 q i 2 x r i 2, nd k + i 3 q i3 x r i3,

18 8 COUNTING DESCENTS, RISES, AND LEVELS, WITH PRESCRIBED FIRST ELEMENT, IN WORDS nd etting i + i 2 + i 3 n, we ee tht 5 become 5 Thu we mut hve 52 n 0 A k+t m m0 0 i 0 i 2 0 m i i 2 t + r i r i 2 k + i i 2 de x N+r π r π [k t ] n m m0 0 i 0 i 20 n i i 2 x r n m m i i 2 t + r i r i 2 k + i i 2 n i i 2 x r n m for ll n However, if we replce x r by z + in 52, we tht the polynomil de z + N+r π π [k+t] n h the Lurent expnion m m k + m m i i2 t + r i r i2 i i 2 n i i 2 m0 0 i 0 i 20 It follow tht it mut be the ce tht m m i i 2 t + r i r i 2 mn+ 0 i 0 i 2 0 o tht n A k+t m m0 0 i 0 i 2 0 m i i 2 t + r i r i 2 k + i i 2 n i i 2 k + i i 2 n i i 2 Thu we hve the following theorem by tking the coefficient of x p r on both ide of 59 z n m z n m 0 x r n m Theorem 56 If k 0, 2, t,,, nd t < r, then the number of word π [k + t] n with de N+r π p ri N++r t π p i n p m n+p+ m i i 2 t + r i r i 2 m m k + n m i i 2 n i i 2 p m0 0 i 0 i 2 0 In the ce 2, our formul implify omewht For exmple, putting 2, r nd t in Theorem 54, we obtin the following

19 COUNTING DESCENTS, RISES, AND LEVELS, WITH PRESCRIBED FIRST ELEMENT, IN WORDS 9 Corollry 57 The number of n-letter word π over [2k + ] hving de O π p rep given by n p m m m k + n m n+p+ 2 m i i n i p m0 0 i0 ri O π p i 6 Concluding remrk A prticulr ce of the reult obtined by Burtein nd Mnour in [4] i the ditribution of decent rep level, rie, which cn be viewed occurrence of o clled generlized pttern 2 rep, 2 in word To get thee ditribution from our reult, we proceed follow we explin only the ce of decent; rie nd level cn be conidered imilrly Set x x 2 x, y y 2 z z 2, nd q q 2 q in A 2 2k nd A 2 2k+ to get the ditribution in [4, Theorem 22] for l 2 the ce of decent/rie Thu, our reult refine nd generlize the known ditribution of decent, level, nd rie in word It i intereting to compre our formul with formul of Hll nd Remmel [7] For exmple, uppoe tht X E nd Y N nd ρ ρ,, ρ 2k i compoition of n Then Theorem tell tht the number of word π of [2k] n uch tht de E π p i p + r n + k 6 p r ρ2i + r + ρ 2i+ + ρ 2i ρ 2k, ρ 2, ρ 4,, ρ 2k r p r ρ 2i r0 where ρ 2 + ρ ρ 2k Thi how tht once we re given the ditribution of the letter for word in [2k] n, we cn find n expreion for the number of word π uch tht de E π p with ingle lternting um of product of binomil coefficient Thi contrt with Corollry 52 where we require triple lternting um of product of binomil coefficient to get n expreion for the number of word of [2k] n uch tht de E π p Of coure, we cn get imilr expreion for the number of word of [2k] n uch tht de E π p by umming the formul in 6 over ll n+k k compoition of n into k prt but tht h the didvntge of hving the outide um hve lrge rnge n nd k get lrge Neverthele, we note tht for 6 Hll nd Remmel hve direct combintoril proof vi ign-revering involution It i therefore nturl to k whether one cn find imilr proof for our formul in Section 3 nd 4 There re everl wy in which one could extend our reerch For exmple, one cn tudy our refined ttitic De X π, Ri X π, Lev X π on the et of ll word voiding fixed pttern or et of pttern ee [, 2, 3, 4] for definition of pttern in word nd reult on them More generlly, inted of conidering the et of ll word, one cn conider ubet of it defined in ome wy, nd then to tudy the refined ttitic on the ubet Alo, inted of conidering refined decent, level, nd rie pttern of length 2, one cn conider pttern of length 3 nd more in which the equivlence cl of the firt letter i fixed, or, more generlly, in which the equivlence cle of more thn one letter poibly ll letter re fixed Once uch pttern or et of pttern i given, the quetion on voidnce or the ditribution of occurrence of the pttern in word over [k] cn be ried i Reference [] A Burtein, Enumertion of word with forbidden pttern, PhD thei, Univerity of Pennylvni, 998 [2] A Burtein nd T Mnour, Word retricted by pttern with t mot 2 ditinct letter, Electron J Combin 9, no 2, #R3 2002

20 20 COUNTING DESCENTS, RISES, AND LEVELS, WITH PRESCRIBED FIRST ELEMENT, IN WORDS [3] A Burtein nd T Mnour, Word retricted by 3-letter generlized multipermuttion pttern, Ann Combin , 4 [4] A Burtein nd T Mnour, Counting occurrence of ome ubword pttern, Dicr Mth Theor Comp Sci , [5] L Comtet, Advnced Combintoric, D Reidel Publihing Co, Dordrecht, 974 [6] J Hll, J Liee, nd J Remmel, q-nlogue of formul counting decent pir with precribed top nd bottom, in preprtion [7] J Hll nd J Remmel, Counting decent pir with precribed top nd bottom, J Combin Theory, Serie A 5, Iue 5, 2008, [8] S Heubch nd T Mnour, Counting rie, level, nd drop in compoition, Integer: Elect J Combin Number Theory , Article A [9] S Kitev nd J Remmel, Clifying decent ccording to prity, Ann Combin 2007, [0] S Kitev nd J Remmel, Clifying Decent According to Equivlence mod k, Elect J Combin , #R64 [] J Liee, Clifying cent nd decent with pecified equivlence mod k, Proceeding of 8-th Interntionl Conference on Forml Power Serie nd Algebr Combintoric, Sn Diego, CA 2006 [2] J Liee nd J Remmel, q-nlogue of formul f or the number of cent nd decent with pecified equivlence mod K, Permuttion Pttern Conference, 2006 [3] P A McMhon, Combintory Anlyi, Vol nd 2, Cmbridge Univ Pre, Cmbridge, 95 reprinted by Chele, New York, 955

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