Download Combinatorial Expansions for Paths, Chung

Combinatorial Expansions for Paths,
Chung-Feller Theorem
and Hankel Matrix
Speaker:
Yeong-Nan Yeh
Institute of mathemetics, Academia sinica
Online
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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 t he generating function
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 )
Then 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
Theorem :
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 the sum of weights of S - paths
from (0,0) to (n,0) .
Then the generating function f(z)   f n z n
n0
satisfies the functional equation
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 A1, 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 ) 
a z
n0
Then
n
n
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 number
mn; N ,E ,O  s mn; N ,O, E  s ...  s Generalize d Catalan number
mn; N , ,{1,2}  s mn; N , ,{0,1}
3
 s mn; N , ,{0,2}
3
3
 s ...
 s mn; N ,{1} \{1},{1}
 s mn; N ,{2} ,{2}  s mn; N ,{0} ,{0}  s ...
 s mn; N ,{1}3 ,{1}3
 s ... Generalize d Catalan number
3
3
3
3
3
3
57
Some interesting shift equivalence
mn; N , ,O  s mn; N ,{2},O 1  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 number
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 A1, 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{max A, 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
bm
cm
dm
z
2 ( m  1)
64
(A,B) being Congruence classes
Define the congruence classes
I k : {n   | n  j (mod k ) for some j  I }
Let
( A, B)  ( I k , J k )
Then
where
FI
a

c

k
,Jk
( x, y , z ) 
a  bz 2 FI
k
,Jk
( x, y , z )
c  dz 2 FI k , J k ( 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) |x1, y 1
70
Let | Ai |  and | Bi | 
ri  max{max Ai , max Bi }   (max Bi  max Ai )  1
Suppose
Then
(1)
(2)
(3)
FAi , Bi ( z ) 
ai  bi z 2 F , ( z )
ci  di 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  d1d 2 z 2 | a1c2  b1d 2 z 2  z k [a2c1  b2 d1 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  mn2;( A1){0},{B1}{2}
In fact, if
1 B
F( A1){0},( B 1){2} ( z )  z FA, B ( z )  1
2
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 }n0 and k  0, the problem to
evaluate the determinan ts of the Hankel matrices
of the sequence,
a

i  j  k 0i , j  n 1
,
has been widel y studied.
第76页
1  2n 
 ,
For the Catalan numbers cn 
n 1  n 
det ci  j 0i , j  n 1  1
det ci  j 1 0i , j  n 1  1
2n  i  j
det ci  j  m 0i , j  n 1  
i j
0i  j  m 1
第77页
For the Motzkin numbers mn count the
number of Motzkin paths of length n,
det mi  j 0i , j  n 1  1
 1 if n  0, 1 (mod 6)

det mi  j 1 0i , j  n 1   0 if n  2, 5 (mod 6)
 1 if n  3, 4 (mod 6)

第78页
For the large Schr oder numbers rn n0 count the
number of large Schr oder paths,
det ri  j 0i , j  n 1  2
n
 
 2
det ri  j 1 0i , j  n 1  2
 n 1 


2


.
Notice that det ai  j  m 0i,j  n 1 can be evaluated by
det ai  j 0i,j  n 1 and det ai  j 1 0i,j  n 1 by
Desnanot - Jacobi identity.
第79页
Our results
 Let Pn (0 ,k )denote the set of weighted lattice paths from
(0 ,0) to (n,k ) with step set S  {(1,1) ,(1, -1) ,(1, 0)} that are
never below the x - axis.
 Let w((1,1))  1, w((1, 0))  u , w((1,  1))  v.
 Denote an,k  w( Pn (0 ,k )).
 Define the generating function of
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 the proof : Since
am n,k  [ x k ]am  n,0 ( x)  w( Pm n (0, k)) ,
we classify t he paths of Pm n (0, k) according to
the difference r between th e end height of the mth
step and the lowest height of the last n steps. Then
count such paths with difference r 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  5 x  3x 2  x3
 9  12 x  9 x 2  4 x 3  x 4

 21  30 x  25 x 2  14 x 3  5 x 4  x 5


1
2 x
1
5  3x  x 2
3 x
12  9 x  4 x 2  x 3 9  4 x  x 2









1


4 x 1



 1


  
Then for (m,n)  {( 0 ,5 ),( 1,4 ),( 2 ,3 ),( 3,2 ),( 4 ,1 ),( 5,0 )}, we have
a5, 0 ( x)  21  30 x  25 x  14 x  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)) 0i , j  n 1  v
n ( n 1)
2
.
Proof :
 min( i , j ) r
ai  j ,0 ( x)0i, j n1    v ai,0 ( x)a j ,0 ( x) 
 diag (1, v, v 2 ,  , v n 1 ) AAT ,
 r 0
 0i , 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 1,1
n 1, 2
n 1, n 1
 n 1, 0

since ai ,i ( x)  ai ,i  1, then det ai  j , 0 ( x) 0i , 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 0i, jn -1  v
n ( n 1)
2
.
第84页
Special cases
 For the case u  0 , v  1, we have cn ,k  a2 n,2 k , where cn,k is the number
of path from (0 ,0) to (2n,2k ) using steps (1,1) and (1,-1) that never pass
n
below x - axis. Denote cn,r ( x)   cn,i x i  r .
i r
Since an,k  0 if n and k have different parity , by the theorem we have
det ci  j , 0 (x) 0i , j  n 1  1  x 
n 1
let x  0, notice that the catalan number cn  cn,0  a2 n,0 , then
det ci  j 0i , j  n 1  1.
 For the case u  1 and v  1, the motzkin number mn  an,0 , we have
det m i  j,0 ( x) 0i , j  n 1  1,
det m i  j 0i , j  n 1  1,
第85页
Large Schr oder paths
 Let S  1,1, (1,1), (2,0), the weights of steps are 1,v and u.
 rn,k and rn,r ( x) are defined similar to an,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 am1,r ( x)an1,r ( x) 
r
v
 am,r ( x)an,r ( x).
第86页
第87页
Thank you
for your attention!
第88页