Combinatorial Expansions for Paths, Chung-Feller Theorem and Hankel Matrix Speaker: Yeong-Nan Yeh Institute of mathemetics, Academia sinica Online Part I. Functions of uniform-partition type Part II. Combinatorial interpretations for a class of function equations Part III. Lattice paths and Fluctuation theory Part IV Paths with some avoiding sets shift equivalence Part V. Addition formulas of polynomials and Hankel determinants 2 Part I. Functions of uniform-partition type 3 Catalan paths An n-Catalan path is a lattice path in the first quadrant starting at (0,0) and ending at (2n,0) with only two kinds of steps---up-step: U=(1,1) and down- step: D=(1,-1). 4 Catanlan number 1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, … , 5 Catanlan number The Catalan sequence was first described in the 18th century by Leonhard Euler, who was interested in the number of different ways of dividing a polygon into triangles. 6 Eugène Charles Catalan (May 30, 1814 – February 14, 1894) was a French and Belgian mathematician. The sequence is named after Eugène Charles Catalan, who discovered the connection to parenthesized expressions during his exploration of the Towers of Hanoi puzzle. ((ab)c)d (ab)(cd) (a(bc))d a((bc)d) (ab)(cd) E.C. Catalan, Note surune equation aux differences finies, J. Math.Pures Appl. 3(1838), 508-515. 7 Catanlan number The counting trick for Catalan words was found by D. André in 1887 D. André, Solution directe du problème résolu par M. Bertrand, Comptes Rendus de l’Académie des Sciences, Paris 105 (1887) 436–437. 8 Chung-Feller Theorem (The number of Dyck path of semi-length n with m nonpositive up-steps is the n-th Catalan number and independent on m.) We say ChungFeller theorem is an uniform partition of up- down type. K.L. Chung, W. Feller, On fluctuations in-coin tossing, Proc. Natl. Acad. Sci. USA 35 (1949) 605-608 9 The classical Chung-Feller theorem was proved by Macmahon. MacMahon, P. A. Memoir on the theory of the partitions of numbers, Philos. Trans. Roy. Soc. London, Ser. A, 209 (1909), 153-175. Chung and Feller reproved the theorem by analytic method. Chung, K. L. and Feller, W. On fluctuations in-coin tossing, Proc. Natl. Acad. Sci. USA 35 (1949) 605-608. A combinatorial proof. Narayana, T. V. Cyclic permutation of lattice paths and the Chung-Feller theorem, Skand. Aktuarietidskr. (1967) 23-30 Eu, Liu and Yeh proved the Chung-Feller theorem by using the Taylor expansions of generating functions. Eu, S. P. Liu, S. C. and Yeh, Y. N. Taylor expansions for Catalan and Motzkin numbers, Adv. Appl. Math. 29 (2002) 345-357 Eu, Fu and Yeh gave a strengthening of the Chung-Feller Theorem and a weighted version for schroder paths. Eu, S. P. Fu, T. S. and Yeh, Y. N. Refined Chung-Feller theorems for lattice paths, J. Combin. Theory Ser. A 112 (2005) 143-162 10 Bijection proofs. D. Callan, Pair them up! A visual approach to the Chung-Feller theorem, Coll. Math. J. 26(1995)196-198. R.I. Jewett, K. A. Ross, Random walk on Z, Coll. Math. J. 26(1995)196-198. Mohanty’s book devotes an entire section to exploring the Chung-Feller theorem. Mohanty, S. G. Lattice path counting and applications, NewYork : Academic Press, 1979. Narayana's book introduced a refinement of this theorem. T.V. Narayana, Lattice path combinatorics, with statistical applications,Toronto;Buffalo : University of Toronto Press, c1979. Callan reviewed and compared combinatorial interpretations of three different expressions for the Catalan number by cycle method. D. Callan, Why are these equal? http://www.stat.wisc.edu/~callan/notes/ Huq developed generalized versions of this theorem for lattice paths. A. Huq, Generalized Chung-Feller Theorems for Lattice Paths(Thesis), http://arxiv.org/abs/0907.3254 11 Another uniform partition for Dyck paths The number of up-steps at the left of the rightmost lowest point of a dyck path We say this uniform partition is of left-right type. W.J. Woan, Uniform partitions of lattice paths and Chung-Feller Generalizations, Amer. Math. Monthly 108(2001) 556-559. 12 Motzkin paths An n-Motizkin path is a lattice path in the first quadrant starting at (0,0) and ending at (n,0) with only two kinds of steps---level-step: (1,0), up-step: U=(1,1) and down- step: D=(1,-1). 13 An uniform partition for Motzkin paths Shapiro found an uniform partition for Motzkin path. L. Shapiro, Some open questions about random walks, involutions, limiting distributions, and generating functions, Advances in Applied Math. 27 (2001), 585596. The number of steps at the left of the rightmost lowest point of a lattice path This uniform partition is of left- right type. Eu, Liu and Yeh proved this proposition. Eu, S. P. Liu, S. C. and Yeh, Y. N. Taylor expansions for Catalan and Motzkin numbers, Adv. Appl. Math. 29 (2002) 345-357 14 Another uniform partition of up-down type for Motzkin paths. The number of steps touching x-axis and under x-axis 15 Our main results 1. Eu, Liu and Yeh proved the Chung-Feller theorem by using the Taylor expansions of generating functions. Eu, S. P. Liu, S. C. and Yeh, Y. N. Taylor expansions for Catalan and Motzkin numbers, Adv. Appl. Math. 29 (2002) 345-357 2. Eu, Fu and Yeh gave a strengthening of the Chung-Feller Theorem and a weighted version for schroder paths. Eu, S. P. Fu, T. S. and Yeh, Y. N. Refined Chung-Feller theorems for lattice paths, J. Combin. Theory Ser. A 112 (2005) 143-162 3. Ma and Yeh gave a generalizations of Chung-Feller theorems J. Ma, Y.N. Yeh, Generalizations of Chung-Feller theorems, Bull. Inst. Math., Acad. Sin.(N.S.)4(2009) 299-332. 第16页 Our main results 4. Ma and Yeh gave a characterization for uniform partitions of cyclic permutations of a sequence of real number 5. J. Ma, Y.N. Yeh, Cyclic permutations ofsequences and uniform partitions, The electronic journal ofcombinatorics 17 (2010), #R117. Liu, Wang, Yeh gave the concepts of functions of Chung-Feller type 6. S.C. Liu, Y. Wang, Y.N. Yeh, Chung-Feller Property in View of Generating Functions, Electron. J. Comb. 18(2011), #P104. Ma and Yeh gave a refinement of Chung-Feller theorems J. Ma, Y.N. Yeh, Refinements of (n,m)-Dyck paths, European. J. Combin. 32(2011) 92-99. 第17页 Our main results 7. Ma and Yeh generalized the cycle lemma. 8. J. Ma, Y.N. Yeh, Generalizations of the cycle lemma, (Accepted 2014). Ma and Yeh gave a characterization for uniform partitions of cyclic permutations of a sequence of real number 9. J. Ma, Y.N. Yeh, Rooted cyclic permutations of a lattice paths and uniform partitions, submitted. Ma and Yeh studied a class of generating functions and their functions of Chung-Feller type J.Ma, Y.N.Yeh, Combinatorial interpretations for a class of functions of Chung-Feller theorem. submitted 第18页 Part II. Combinatorial interpretations for a class of function equations 19 Uniform-partition Extension (, ) 20 Let S ( z ) sn z n . Suppose f n ,k sn for any 0 k n. n 0 We consider the generatingfunction n n n CS( y , z ) f n ,k y k z n sn y k z n sn y k z n n 0 k 0 n 0 k 0 n 0 n n s z y s ( yz ) n n 1 y n 1 n n 0 n 0 sn z 1 y 1 y n 0 yS ( yz) S ( z ) yzS ( yz) zS ( z ) T hen CS ( y , z ) . y 1 yz z the function of uniform-partition type for k 0 . : Liu, S. C. Wang, Y. and Yeh, Y. N. The function of uniform-partition type, submitted 21 An example for catalan sequence (up-down type) 22 An example for Motzkin sequence (left-right M 1 zM z 2 M 2 zM z z 2 M z[zM ]2 type) A z 1 A A2 A* w 1 A* [ A* ]2 rightmost lowest point 23 • In general, given a combinatorial structure, let f(z) be a generating function correspoding with this combinatorial structure. We can obtain a functional equation which f(z) satisfies . 第24页 Combinatorial structure Generating Functional equation which function f(z) satisfies f(z) Catalan path:(1,1),(1,-1) in the first quadrant C(z) Motzkin path:(1,1),(1,-1),(1,0) in the first quadrant M(z) Schroder path:(1,1),(1,-1),(2,0) in the first quadrant S(z) 第25页 Combinatorial structure Generating Functional equation which function f(z) satisfies f(z) Catalan path:(1,1),(1,-1) in the first quadrant C(z) C(z)=1+z[C(z)]2 Motzkin path:(1,1),(1,-1),(1,0) in the first quadrant M(z) M(z)=1+zM(z)+z2[M(z)] 2 Schroder path:(1,1),(1,-1),(2,0) in the first quadrant S(z) S(z)=1+zS(z)+z[S(z)] 2 第26页 Combinatorial structure Generating Functional equation which function f(z) satisfies f(z) Catalan path:(1,1),(1,-1) in the first quadrant C(z) C(z)=1+z[C(z)]2 Motzkin path:(1,1),(1,-1),(1,0) in the first quadrant M(z) M(z)=1+zM(z)+z2[M(z)] 2 Schroder path:(1,1),(1,-1),(2,0) in the first quadrant S(z) S(z)=1+zS(z)+z[S(z)] 2 Given a functional equation , how to find a combinatorial structure and its corresponding generating function f(z) such that f(z) satisfies this functional equation? 第27页 Combinatorial structure Generating Functional equation which function f(z) satisfies f(z) Catalan path:(1,1),(1,-1) in the first quadrant C(z) C(z)=1+z[C(z)]2 Motzkin path:(1,1),(1,-1),(1,0) in the first quadrant M(z) M(z)=1+zM(z)+z2[M(z)] 2 Schroder path:(1,1),(1,-1),(2,0) in the first quadrant S(z) S(z)=1+zS(z)+z[S(z)] 2 Given a functional equation , how to find a combinatorial structure and its corresponding generating function f(z) such that f(z) satisfies this functional equation? 第28页 Combinatorial structure Catalan path:(1,1),(1,-1) in the first quadrant C(z) C(z)=1+z[C(z)]2 Motzkin path:(1,1),(1,-1),(1,0) in the first quadrant M(z) M(z)=1+zM(z)+z2[M(z)] 2 Schroder path:(1,1),(1,-1),(2,0) in the first quadrant S(z) S(z)=1+zS(z)+z[S(z)] 2 ??? Generating Functional equation which function f(z) satisfies f(z) ??f(z) Given a functional equation , how to find a combinatorial structure and its corresponding generating function f(z) such that f(z) satisfies this functional equation? 第29页 Let . The recurrence relation which the sequence satisfies is independent on a0(z) . Hence, let a0(z) =1. . We focus on the following functional equation. Let S be a set of vector in the plane Z×Z. We also call the set S step set and vectors in S steps. Let L be a function from S to N, where N is the set of nonnegative integers . We call L a step-length function of the set S and L(s) the step length of the step s in the set S repectively. Let W be a function from S to R, where R is the set of real numbers. We call W a weight function of the set S and W(s) the weight of the setp s in the set S respectively, 第30页 Let P be a sequence of vectors (x1,y1)…(xn,yn) in the set S such that y1+…yn=0, y1+…yi≥0 for all i. We call P an S-path. Let Ω(S) be the set of all S-paths. Define the L-length of a S-path P= (x1,y1)…(xn,yn) , denoted by L(P), as L(P)=L(x1,y1)+…L(xn,yn). Define the W-length of a S-path P= (x1,y1)…(xn,yn) , denoted by W(P), as W(P)=W(x1,y1)…W(xn,yn). Define a generating function f(z) as 第31页 T heorem: Let S {(0,1)} {( j ,i 1) i 1,2, r , j 1,2, , m)} w( 0 ,1 ) 1,w(j,-i 1 ) aij . Let f 0 1 and f n be thesum of weights of S - paths from (0,0) to (n,0). T hen thegeneratingfunction f(z) f n z n n0 satisfies thefunctionalequation r m f(z) 1 aij z j [f(z)] i . i 1 j 1 第32页 • A decomposition of a S-path. P=(0,1)P1(0,1)P2(0,1)P3…Pi-1(j,-i+1)Pi W(1,1)=1,W(j,-i+1)=ai,j 第33页 Part III. Lattice paths and Fluctuation theory 第34页 • Fluctuation theory is the name given to that part of probability theory which deals with the fluctuations of the partial sums sn=x1+...+xn of a sequence of random variables x1,…,xn. 35 • Consider x=(r1,…rn). Let s0=0,si=r1+…+ri • Let p(x) be the number of positive sums si • Let m(x) be the index where the maximum is attained for the first time. 36 r1=3,r2=1,r3=-2 (1,2,3) (1,3,2) (2,1,3) (3,1,2) (2,3,1) (3,2,1) x (3,1,-2) (3,-2,1) (1,3,-2) (-2,3,1) (1,-2,3) (-2,1,3) partial sum p(x) m(x) (3,4,2) (3,1,2) (1,4,2) (-2,1,2) (1,-1,2) (-2,-1,2) 3 3 3 2 2 1 2 1 2 3 3 3 37 (1,2,3) (1,3,2) (2,1,3) (3,1,2) (2,3,1) (3,2,1) x (1,2,-2) (1,-2,2) (2,1,-2) (-2,1,2) (2,-2,1) (-2,2,1) partial sum p(x) m(x) (1,3,1) (1,-1,1) (2,3,1) (-2,-1,1) (2,0,1) (-2,0,1) 3 2 3 1 2 1 2 1 2 3 1 3 38 • Fix X=(r1,…rn). • Let Xi=(ri,…rn,r1,…,ri-1) (cyclic permutations.) • Let P(X)={p(Xi)| i=1,2,…,n} M(X)={m(Xi)| i=1,2,…,n} 39 • F. Spitzer, (1956) • Let X be a sequence of real numbers of length n such that sn=0 and no other partial sum of distinct elements vanishes. Then P(X)=M(X)=[0,n-1]. 40 Remark • Fix X=(r1,…rn). Suppose r1+…+rn=m. • Let m=0. The conditions in the results of Spitzer are necessary and sufficient conditions for P(X)=[0,n-1] The conditions in the results of Spitzer are not necessary for M(X)=[0,n-1]. 41 • T.V. Narayana, (1967) • Let n be a positive integer and X be a sequence of integers with -n<ri< 2 for all i=1,2,…,n such that sn=1. Then P(X)=[n]. 42 • J. Ma, Y.N. Yeh, Generalizations of The Chung-Feller Theorem II, submitted. • Let n be a positive integer and X be a sequence of integers with -n<ri< 2 for all i=1,2,…,n such that sn=1. Then M(X)=[n]. 43 Two natural problems • What are necessary and sufficient conditions for M(X)=[n] and P(X)=[n] if m>0? • What are necessary and sufficient conditions for M(X)=[0,n-1] and P(X)=[0,n-1] if m<=0? 44 • Fix X=(r1,…rn). Given an index j=1,…,n, define LP(X;j)={i|sj>si,i=1,…,j-1} and RP(X;j)={i|sj>=si i=j+1,…,n} 45 • Let m>0. The necessary and sufficient conditions for M(X)=[n] are sm(X)-si>=m for all i in LP(X;m(X)) The necessary and sufficient conditions for P(X)=[n] are sj-si>=m for any j in [n] and any all i in [0,j-1]\LP(X;j) 46 • Let m<=0. The necessary and sufficient conditions for M(X)=[0,n-1] are si -sm(X)<m for all i in RP(X;m(X)) The necessary and sufficient conditions for P(X)=[0,n-1] are sj-si<m for any j in [n] and any all i in [0,j-1]\LP(X;j) 47 Part IV. Paths with some avoiding sets shift equivalence 48 Let M be a Motzkin path. LM: the set of the height of the level steps LM={0,3} PM: the set of the height of the peaks PM={2,1} VM: the set of the height of the valleys VM={0,1} 49 Motzkin paths from (0,0) to (2(n-1),0) without level of height larger than 0 50 Peaks-, Valleys- and Level-avoiding Sets Given the sets A , B P, C we consider the Motzkin path such that (1): LM A , PM B ,VM C A: level-avoiding set B: peak-avoiding sest C: valley-avoiding set (2): LM A, PM B,VM C A: level-restricting set B: peak-restricting set C: valley-restricting set 51 Generating Functions mn,k ,l ,s; A, B,C the number of the Motzkin path of length n with k levles, l peaks and s valleys M A, B,C( x, y, z, q) mn,k ,l , s; A, B,C z n x k y l q s 52 M A , B ,C 1 (0 A) xzM A, B ,C z 2 {M A1, B 1,C 1 1 (1 B ) y}{1 (0 A) xzM A, B ,C [ M A, B ,C 1 (0 A) xzM A, B ,C ]q (0 C )} 53 Some results E. Deutsch, Dyck path enumeration, Discrete Math. 204 (1999), 167--202. mn;P,{1}, the n-th Fine number P. Peart and W-J. Woan, Dyck paths with no peaks at height k, J. Integer Seq. 4 (2001), Article 01.1.3. mn;P,{h}, mn;P,{2}, the (n-1)-th Catalan number 54 S.-P. Eu, S.-C. Liu, and Y.-N. Yeh, Dyck Paths with Peaks Avoiding or Restricted to a Given Set, Stud. Appl. Math. 111 Iss 4 (2003), 453--465. mn,k ;P, B, Shu-Chung Liu, Jun Ma, Yeong-Nan Yeh, Dyck Paths with Peak- and Valley-Avoiding Sets, Stud. Appl. Math. 121:263-289 mn,k ;P, B,C • Continued fractions • Close forms • Shift equivalence 55 an s bn If there exist non-negative integers p and q such that a p n bq n Suppose H ( z ) n a z n n 0 Then and G( z) bn z n n 0 an s bn if and only if there is a positive integer m such that H ( z) z mG( z) is a polynomial. 56 Some interesting shift equivalence mn; N , , s mn; N , ,{0} s mn; N , ,{0,1} s ... s mn; N ,{2}, s mn; N ,{2,3}, s ... s Catalan num ber mn; N ,E ,O s mn; N ,O, E s ... s Generalized Catalan num ber mn; N , ,{1,2} s mn; N , ,{0,1} 3 3 s mn; N , ,{0,2} 3 s ... s mn; N ,{1}3 \{1},{1}3 s mn; N ,{2}3 ,{2}3 s mn; N ,{0}3 ,{0}3 s ... s mn; N ,{1}3 ,{1}3 s ... Generalized Catalan num ber 57 Some interesting shift equivalence mn; N , ,O s mn; N ,{2},O1 s mn; N ,{2,3},O 2 s ... s mn; N , ,E s mn; N ,{2},E 1 s mn; N ,O \{1}, s mn; N ,E , s ... s ... s Motzkin num ber mn; N ,O,O s mn; N ,E , E s mn; N ,( E 1){2}, E 1 s ... s A126120 in Sloane mn; N ,{0}3 ,{2}3 s mn; N ,{1}3 ,{0}3 s ... s A025265 in Sloane mn; N ,{1}3 ,{2}3 s mn; N ,{2}3 ,{0}3 s ... s A127389 in Sloane 58 Continued fractions It is difficult to represent M A, B,C( x, y, z, q) as a continued fractions 59 Close form • Matrix methods We just consider FA, B ( x, y, z) M A, B, ( x, y, z,1) 60 FA, B ( x, y, z ) 1 { (0 A) xz z [ M A1, B 1,C 1 1 (1 B) y ]}FA, B ( x, y, z ) 2 61 FA, B ( x, y, z) | A | and | B | F , ( x, y , z ) | A | and | B | F , P ( x, y , z ) | A | and | B | FN , ( x, y , z ) | A | and | B | F , P ( x, y , z ) 62 Let A,B,i 1 z 2 xz (i 1 A) z 2 y (i B) | A | and | B | m max{maxA, max B} (max A max B) 1 S x x Then where 0 1 0 Tx 2 z 1 x am bm z F , ( x, y, z ) 2 FA, B ( x, y, z ) am cm cm d m z 2 F , ( x, y, z ) bm m 1 T A,B ,i S A,B ,m d m i 1 63 am cm bm 2 ( m 1) z dm 64 (A,B) being Congruence classes Define the congruence classes I k : {n | n j (modk ) for some j I } Let ( A, B) ( I k , J k ) Then where FI a c k ,Jk ( x, y , z ) a bz2 FI k , J k ( x, y, z ) c dz2 FI k ,Jk ( x, y , z ) k 1 b T I , J ,i S I , J ,k d i 1 65 If F(x) is a solution of an equation ad ( x) y d ad 1 ( x) y d 1 ... a0 ( x) 0 then we say that F(x) is algebraic The algebraic degree of F(x): ad (F ) 66 | A | and | B | | A | and | B | FA, B ( x, y, z) | A | and | B | | A | and | B | A I k and B J k FA, B ( x, y, z) is algebraic since it is a solution of a quadratic Equations ad(FA,B ( x, y, z)) 2 67 Problem I • Characterize the set ( A, B) | ad(F ( x , y , z )) d A, B 68 Problem II • Given a sequence a1,a2,…,an,…, find a pair (A,B) of the sets such that [ z ]FA, B ( z ) an n 69 We consider the coefficients in mn; A, B FA,B ( z) FA,B ( x, y, z) |x1, y1 70 Let | Ai | and| Bi | ri max{maxAi , max Bi } (maxBi max Ai ) 1 Suppose FA , B ( z ) i Then (1) (2) (3) i ai bi z 2 F , ( z ) ci d i z 2 F , ( z ) mn; A1 , B1 s mn; A2 , B2 if and only if k Such that 2 r1 2 2 2 z ( c c d ( 1 z ) d k 2 2 2 2z ) z 2 r2 2 z (c1 c1d1 (1 z ) d12 z 2 ) c1c2 d1d2 z 2 | a1c2 b1d2 z 2 z k [a2c1 b2d1 z 2 ] c1d 2 c 2 d1 d1d 2 (1 z ) a1d 2 b1c2 b1d 2 (1 z ) z k [a2 d1 b2c1 b2 d1 (1 z )] 71 Bijection methods Suppose Then 1 B mn; A, B mn2;( A1){0},{B1}{2} In fact, if 1 B F( A1){0},( B1){2} ( z ) z 2 FA, B ( z ) 1 72 Problem III If the sequences mn; A , B 1 1 s mn; A2 , B2 then we say that (A1,B1) and (A2,B2) shift equivalent, denoted by ( A1 , B1 ) s ( A2 , B2 ) • Give a characterization of ( A1 , B1 ) s ( A2 , B2 ) 73 mn;{i},{1} is shift equivalent to the Fibonacci numbers The sequence mn+2i;{i},{1} has Chung-Feller property, i.e., mn+2i;{i},{1} =Fn is independent on i, where Fn is the n-th Fibonacci number. 74 mn;{i},[2,i 1] is shift equivalent with the Central binomial coefficients (i=1) Replace valleys(DU) of height 0 and level into peak DU and U respectively Remove the first and final steps Left factor of Dyck 75 path Addition formulas and Hankel matrix For a sequence{an }n0 and k 0, the problem to evaluatethedeterminants of theHankelmatrices of thesequence, ai j k 0i, jn1 , has been widely studied. 第76页 1 2n , For theCatalan numberscn n 1 n det ci j 0i , j n 1 1 det ci j 1 0i , j n 1 1 2n i j det ci j m 0i , j n 1 i j 0i j m 1 第77页 For theMotzkinnumbersmn count the number of Motzkinpathsof length n, det mi j 0i , j n 1 1 1 if n 0, 1 (mod6) det mi j 1 0i , j n 1 0 if n 2, 5 (mod6) 1 if n 3, 4 (mod6) 第78页 For thelarge Schroder numbersrn n0 count the number of large Schroder paths, det ri j 0i , j n 1 2 n 2 det ri j 1 0i , j n 1 2 n 1 2 . Noticethatdet ai j m 0i,jn 1 can be evaluatedby det ai j 0i,jn 1 and det ai j 1 0i,jn 1 by Desnanot- Jacobi identity. 第79页 Our results Let Pn (0,k )denotetheset of weightedlatticepathsfrom (0,0) to (n,k) with step set S {(1,1) ,(1, -1) ,(1, 0)} thatare neverbelow the x - axis. Let w((1,1)) 1, w((1, 0)) u, w((1, 1)) v. Denotean,k w( Pn (0,k )). Define thegeneratingfunctionof n P n (0,i) by i r n an,r ( x) an,k x k r . k r 第80页 Theorem For m ,n 0, am n,0 ( x) min(m , n ) r v am,r ( x)an,r ( x). r 0 Sketch of theproof : Since am n,k [ x k ]am n,0 ( x) w( Pm n (0,k)) , we classify the pathsof Pm n (0,k) accordingto thedifferencer between the end height of themth step and thelowest height of thelast n steps.T hen count such pathswith differencer by a bijection. 第81页 第82页 Example. Let u v 1, 1 1 1 2 2 1 ai,j 4 5 3 1 9 12 9 4 1 21 30 25 14 5 1 1 1 x 2 2x x2 ai,j ( x) 4 5x 3x 2 x3 9 12x 9 x 2 4 x 3 x 4 21 30x 25x 2 14x 3 5 x 4 x 5 1 2 x 5 3x x 2 12 9 x 4 x 2 x 3 1 3 x 9 4x x2 1 4 x 1 1 T henfor (m ,n) {( 0,5 ),(1,4 ),( 2,3 ),( 3,2 ),( 4,1 ),( 5,0 )}, we have a5, 0 ( x) 21 30x 25x 14x 5 x x 2 3 4 5 min(m , n ) a r 0 m,r ( x)an ,r ( x). 第83页 Corollary det(ai j ,0 ( x))0i , j n1 v n ( n 1) 2 . Proof: min(i , j ) r ai j ,0 ( x)0i, jn1 v ai,0 ( x)a j ,0 ( x) diag(1, v, v 2 ,, v n 1 ) AAT , r 0 0i , j n 1 a0,0 ( x) a1,1 ( x) a1, 0 ( x) . a2,1 ( x) a2 , 2 ( x ) where A a2,0 ( x) a ( x) a ( x) a ( x) a ( x ) n 1 , 0 n 1 , 1 n 1 , 2 n 1 , n 1 since ai ,i ( x) ai ,i 1, then det ai j ,0 ( x) 0i , j n 1 v n ( n 1) 2 A v 2 n ( n 1) 2 . Remark : Let x 0 in the corollary, we have det ai j 0i, j n -1 v n ( n 1) 2 . 第84页 Special cases For thecase u 0, v 1, we havecn ,k a2 n,2 k , where cn,k is thenumber of pathfrom(0,0) to (2n,2k ) using steps(1,1)and (1,-1)thatneverpass n below x - axis. Denotecn,r ( x) cn,i x i r . i r Since an,k 0 if n and k havedifferentparity, by thetheoremwe have det ci j , 0 (x)0i , j n 1 1 x n 1 let x 0, noticethat thecatalannumber cn cn,0 a2 n,0 , then det ci j 0i , j n 1 1. For thecase u 1 and v 1, themotzkinnumber mn an,0 , we have det m i j,0 ( x) 0i , j n 1 1, det m i j 0i , j n 1 1, 第85页 LargeSchroder paths Let S 1,1, (1,1), (2,0), theweightsof stepsare1,v and u. rn,k and rn,r ( x) are definedsimilar toan,k and an,r ( x) . Theorem For m ,n 1, rm n,0 ( x) u min(m 1, n 1) r min(m , n ) r 0 r 0 v am1,r ( x)an1,r ( x) r v am,r ( x)an,r ( x). 第86页 第87页 Thank you for your attention! 第88页