THE NUMBER OF PATHS IN BOUNDARY RESTRICTED PASCAL TRIANGLE

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1 THE NUMBER OF PATHS IN BOUNDARY RESTRICTED PASCAL TRIANGLE KINGO KOBAYASHI, HAJIME SATO, AND MAMORU HOSHI Abstract In this tal, we provide several forms of the generator for counting the number of paths in the boundary restricted Pascal triangle In the section 1, we consider the problem of single boundary In the section, we proceed the discussion into the problem of two boundaries Specifically, we give three seemingly different forms to count the number of paths If time permits, we will give the proof of the equivalence between them The study treated in the paper is a continuation of papers [, 3] 1 Paths in Pascal triangle with single boundary At first, let us consider the random wal by a particle with an upper boundary at s 0 starting from 0 at time 0 The particle can move up or down in one step except at the boundary where it should move down only Following the idea in [1], the paths starting from 0 and ending at the boundary at s are symbolically represented as A zd s D 11 where D is the class of Dyc paths see Figure 1 Therefore, we can obtain the generating function Az, s for counting the number of paths starting from 0 ending at the boundary s as Az, s z s Dz s+1 1 Note that the Dyc path always remains in the non-negative region and ends at the starting point 0 The generating function Dz is expressed as Dz 1 1 4z 1 m + 1 z z m 13 m + 1 m m0 The coefficient of z m is nown as Catalan number: [z m ]Dz 1 m 1 m m + 1 m m + 1 m More generally, we have [z m ]Dz m + m + m 15 Key words and phrases random wal, symbolic method, Pascal triangle, Dyc path, Fibonacci polynomial, Chebyshev polynomial of the second ind 1

Figure 1 A path of random wal ending at the upper boundary For convenience, we transform the position coordinate from the variable u s, s

the transformed variable as Âz, a Az, a 1 z a 1 Dz a 16 We can represent the generator Âz, a, b starting from a and ending at b by the

+ 1 + 1 if b a In other words, Âz, a, b { 1 z 1 z Âz, a, b 1 Âz, a, b if b a + 1 Âz, a, b 1 1 Âz, a, b if b a + 1 Figure The generators

2 Figure 1 A path of random wal ending at the upper boundary For convenience, we transform the position coordinate from the variable u s, s 1, to v 1,,, ie, v s u+1 Then, the correspondence is recognized as s 1 and 0 a, that is, s a 1 Thus, the generator can be represented for the transformed variable as Âz, a Az, a 1 z a 1 Dz a 16 We can represent the generator Âz, a, b starting from a and ending at b by the following recursion see Fgure : Âz, a, 0 0 Âz, a, 1 Âz, a za 1 Dz a { z Âz, a, b 1 + Âz, a, b + 1 if b a Âz, a, b z Âz, a, b 1 + Âz, a, b if b a In other words, Âz, a, b { 1 z 1 z Âz, a, b 1 Âz, a, b if b a + 1 Âz, a, b 1 1 Âz, a, b if b a + 1 Figure The generators of random wal from a to b From this recursion, we can obtain the explicit form of the generator as, Âz, a, b z a b fz, b 1Dz a fz, b a 1 17

3 where fz, t is the Fibonacci polynomial defined as, t t i fz, t 1 i z i 18 i i0 and note that fz, t is an even polynomial of order t, and if b a, fz, b a 1 is zero Usually, f z, t is called as the Fibonacci polynomial, and f 1, t is Fibonacci number Actually, fz, t is satisfied the recursion, fz, t fz, t 1 z fz, t Furthermore, by using the Chebyshev polynomial of the second ind, we can express fz, t for t 0 as, 1 fz, t z t U t 19 z As the Chebyshev polynomial of the second ind U t z satisfies sint + 1θ U t cos θ, 110 sin θ we have another expression for fz, t for t 1 as, t fz, t 1 z cos π 111 t Now, we determine the coefficient of z n in the generator Âz, a, b: c n,a,b [z n ]Âz, a, b [zn+b a ]fz, b 1 Dz a Taing into account of the fact that both of Dz and fz, t are the even polynomials, b 1 c n,a,b 1 a m + a m + a m m+n+b a b b 1 where we use 15 and from 18, a n + b [z ] fz, b 1 1 b 1 n + b n+b a 11 On the other hand, we can have another formula for the number of paths in Pascal triangle with single boundary by the mirror image method Now, we come bac to the original coordinate u In order to cancel the paths starting from 0 and arriving to the position s + 1, we have to subtract the number of paths starting from the image point s + of the origin 0 with respect to s + 1, from the number of paths from 0 Thus, the number of paths starting from 0 and arriving to the position l n l s after n

steps without visiting the position s + 1, is given by the formula see Figure 3, n n c n,s,l n+l n+l if n + l is even, s + 1 113 0 otherwise, Figure 3 The mirror image method for single boundary When

: n n c n,a,b n+b a n+b+a if n + b a is even, 114 0 otherwise, Therefore, it should hold an identity for the case when n + b a is even, b 1 0 1 b 1 a n + b We omit the proof for this identity n + b

4 steps without visiting the position s + 1, is given by the formula see Figure 3, n n c n,s,l n+l n+l if n + l is even, s otherwise, Figure 3 The mirror image method for single boundary When we consider the situation in the coordinate system {v 1,,, a, } instead of {u s, s 1,, 0, }, the count function c n,s,l should become c n,a,b for the number of paths from a to b in n steps as follows : n n c n,a,b n+b a n+b+a if n + b a is even, otherwise, Therefore, it should hold an identity for the case when n + b a is even, b b 1 a n + b We omit the proof for this identity n + b n+b a n n+b a n n+b+a 115 Three formulas for the number of paths in Pascal triangle with two boundaries 1 The height restricted Dyc path Here, we study the number of Dyc paths, the height of which is restricted to be up to s The generating function is recursively defined as 1 Dz, s 1 z 1 Dz, s 1

5 with the initial condition, Dz, 0 1 For the small s, we can show 1 Dz, 1 1 z, Dz, 1 z 1 z, Dz, 3 1 z 1 3z + z 4, Dz, 4 1 3z + z 4 1 4z + 3z 4, 1 4z + 3z 4 Dz, 5 1 5z + 6z 4 z 6 In general, we have fz, s Dz, s fz, s + 1, where fz, s is Fibinacci polynomial defined in 18 The generator for the number of paths between finite width By using the generator Dz, s, we can express the generator Bz, s, t that gives the number of paths restricted between the upper boundary s and the lower boundary t starting from the origin 0 and ending at the upper boundary s s, t 0 as follows see Figure 4 : Bz, s, t Dz, tzdz, t + 1zDz, t + z Dz, t + s s z s Dz, t + i i0 z s s i0 fz, t + i fz, t + i + 1 z s fz, t fz, t + s It should be mentioned that after the first touch of the upper boundary, the path continues as a Dyc path with height restriction s + t + 1 s t Figure 4 Path ending at the upper boundary s restricted between s and t

6 As before, we proceed the coordinate transform from u s, s 1,, 0,, t to v 1,,, a,, w, ie, s a 1, t w a, s + t + 1 w It means that the state space is defined to be S {1,,, w}, and the starting state is a S Then, the generator of number of paths starting from a and ending at 1 is obtained by transforming 3 as, a 1 fz, w a ˆBz, w, a z fz, w 4 Furthermore, we can represent the generator starting from a ending at b by the following recursion cf also Figure : ˆBz, w, a, 0 0 ˆBz, w, a, 1 ˆBz, a 1 fz, w a w, a z fz, w ˆBz, w, a, b In other words, z ˆBz, w, a, b 1 + ˆBz, w, a, b + 1 if b a z ˆBz, w, a, a 1 + ˆBz, w, a, a if b a ˆBz, w, a, b { 1 z ˆBz, w, a, b 1 ˆBz, w, a, b if b a z ˆBz, w, a, a 1 ˆBz, w, a, a 1 if b a + 1 From the above recursion, we can obtain the explicit form of the generator as, fz, b 1 fz, w a ˆBz, w, a, b z a b fz, b a 1 5 fz, w Here, we note that if a b, then fz, b a 1 0 By applying 111, we also have b 1 1 z cos i π w a b ˆBz, w, a, b z a b i1 j1 w 1 z cos b a 1 i1 1 1 z cos π j π w a + 1 i π 1 z cos 6 b a 3 The path counting formula by the finite state model By changing the viewpoint, this problem can also be modeled by the finite state

7 transition matrix, T Let introduce an 0-1 vector determining the starting state as, s a s i i1,,w 0,, 0, 1, 0,, 0, 7 where s a 1 and s i 0 for i a Then, the numbers of paths of random wal between 1 and w starting from the state a are given by the vector, s a T n The characteristic function h w x def T xi of T is satisfies the recursion by the expansion of determinant as, h w x xh w 1 x h w x 8 On the other hand, the Chebyshev polynomial of the second ind satisfies the following recursion, U n x xu n 1 x U n x 9 By comparing 8 and 9, we have x h w x 1 w U w 10 Taing into account of 110, the eigenvalues of T, that is, the zeros of h w x are given as iπ µ i cos, for i 1,, w 11 Now, let us determine the eigenvector v i row vector with respect to the eigenvalue µ i Thus, by setting from the characteristic equation it holds the relation v i v i,1, v i,,, v i,w, v i T µ i v i, v i,j µ i v i,j 1 v j Under the initial condition v i,0 0, v i,1 1, v i, µ i, we can solve the above recursive equation by using also the Chebyshev polynomial of the second ind as, µi v i,j U j 1 sin j i π w+1 sin i π w+1 1

8 Here, we will redefine v i,j sin i π w+1 as v i,j, the eigenvector v i with respect to the eigenvalue µ i is determined as v i sin i π, sin i π,, sin w i π 13 Now, let define two w w matrices M and V M is the diagonal matrix having the eigenvalues µ i cos iπ w+1 for i, i-elements i 1,, w, ie, µ µ 0 0 M 0 µ w µ w Next, the matrix V is defined as v 1 v Then, we have V In other words, it holds v w Furthermore, we have the identity, w where j1 sin i j π i,j1,,w V T M V, T V 1 M V 14 sin i j π sin j π δ i, δ i, { 1 if i, 0 otherwise, Therefore, the inverse of V is propotional to V itself Finally, we obtain T n V 1 M n V V M n V V 1 V 15 w µ n i v T i v i 16 As the number of paths starting from a to b in n steps is given by c w,a,b,n s a T n s T b, i1

9 we calculate the a, b-element in the matrix T n in 16 and give the value explicitly for 1 a, b w, as, c w,a,b,n w i1 i π n cos sin a i π sin b i π if n a + b is even, 0 otherwise Remar 1 Feller gave a similar form for the duration of banrupt at page 353, 57 of the boo [4] from the probabilistic argument He treated the situation of absorbing barrier On the other hand, we consider that of reflecting barrier, and have studied from the combinatorial point of view It is quite interest to recognize that the right sum of first case of 17 is a positive integer From this result, we obtain another form of the generator, ˆBz, w, a, b c w,a,b,n z n n0 w i1 i1 v a,i v i,b 1 µ i z w sin a i π w+1 sin b i π w+1 1 z cos i π, 18 w+1 that should be identical to the form of 6 Here, we have obtained two forms of the generator ˆBz, w, a, b, that is, 6 and 18 We can prove the equivalence of both forms by showing that 18 is the partial fractional expansion of 6 The detail of the proof is omitted 4 The path counting formula by the mirror image method By applying the mirror image method for Pascal triangle with two boundaries, we can conclude that the number of paths starting from a and ending at b is obtained for 1 a, b w, as, c w,a,b,n 0 n, n a+b mod w+1 n 0 n, n+a+b mod w+1 n 17 if n a + b is even, 0 otherwise The detail of derivation for the formula 19 is omitted 19

10 Now, we have obtained two formulas for the number of paths starting from a and ending at b in n steps, that is, 17 and 19 We can show the equivalence of these two formulas, but the proof is omitted acnowledge The authors wish to than Prof H Morita and Prof T Naata for useful information and discussions on the series of related topics References [1] P Flajolet and R Sedgewic, Analytic Combinatorics, Cambridge University Press, 009 [] K Kobayashi, M Hoshi and H Morita, The distributions of some characteristics of random wals and related combinatorial identities, in Proceedings of ITA, 015 [3] K Kobayashi, H Sato, M Hoshi and H Morita, A wal on random wals through the symbolic method, Proceedings of STW, Sep 4-6, 015 [4] W Feller, An Introduction to Probability Theory and Its Applications, vol 1, 3rd Edition, Johm Wiley & Sons, Inc, 1068 The University of Electro-Communications, Toyo, JAPAN address: ingo@ieeeorg Senshu University, Toyo, JAPAN address: h-sato@iscsenshu-uacjp The University of Electro-Communications, Toyo, JAPAN address: mmrhoshi@gmailcom

Yi Wang Department of Applied Mathematics, Dalian University of Technology, Dalian , China (Submitted June 2002)

SELF-INVERSE SEQUENCES RELATED TO A BINOMIAL INVERSE PAIR Yi Wang Department of Applied Mathematics, Dalian University of Technology, Dalian 116024, China (Submitted June 2002) 1 INTRODUCTION Pairs of