A Generalization of the k-bonacci Sequence from Riordan Arrays

Transcription

1 A Generalization of the k-bonacci Sequence from Riordan Arrays José L Ramírez Departamento de Matemáticas Universidad Sergio Arboleda Bogotá, Colombia joselramirez@imausergioarboledaeduco Víctor F Sirvent Departamento de Matemáticas Universidad Simón Bolívar Caracas, Venezuela vsirvent@usbve Submitted: Aug 2, 204; Accepted: Jan 5, 205; Published: Feb 6, 205 Mathematics Subject Classifications: B39, 05A9, 5A24 Abstract In this article, we introduce a family of weighted lattice paths, whose step set is {H (,0,V (0,,D (,,,D m (,m } Using these lattice paths, we define a family of Riordan arrays whose sum on the rising diagonal is the k-bonacci sequence This construction generalizes the Pascal and Delannoy Riordan arrays, whose sum on the rising diagonal is the Fibonacci and tribonacci sequence, respectively From this family of Riordan arrays we introduce a generalized k-bonacci polynomial sequence, and we give a lattice path combinatorial interpretation of these polynomials In particular, we find a combinatorial interpretation of tribonacci and tribonacci-lucas polynomials Keywords: Riordan arrays, k-bonacci sequence, lattice paths Introduction A lattice path Γ in the xy-plane with steps in a given set S Z 2 is a concatenation of directed steps of S, ie, Γ s s 2 s l, where s i S for i l For any positive integer n, let M (n,k denote the set of lattice paths from the point (0,0 to the point (k,n, with step set S {H (,0,V (0,} Let D (n,k be the number of lattice paths of M (n,k, ie, D (n,k M (n,k It is well known that ( ( n+k n+k D (n,k n k The author was partially supported by Universidad Sergio Arboleda under grant no DII-262 the electronic journal of combinatorics 22( (205, #P38

2 [ Let H be the infinite lower triangular array defined by H { n,k D (n k,k ( n k, if n k; 0, if n < k h ( h ( n,k ] n,k N, where Therefore, H is the Pascal matrix This matrix has a lot of interesting properties, (see for instance [2, 5, 6, 7, 2, 23, 24], one of them is the sum of the elements on the rising diagonal is the Fibonacci sequence F n, (sequence A If we add a third step, D (,, we obtain Delannoy paths; among other references see [, 9, 0, 3, 22] We denote by M 2 (n,k the set of lattice paths from the point (0,0 to the point (k,n, with step set S 2 {H (,0,V (0,,D (,} If D 2 (n,k M 2 (n,k, then it is known that D 2 (n,k k ( n i ( n+k i The sequence D 2 (n,n is called central Delannoy numbers [ ](sequence A00850 Let H 2 be the infinite lower triangular array defined by H 2, where h (2 n { n,k D 2 (n k,k, if n k; 0, if n < k h (2 n,k n,k N This array is called Delannoy or tribonacci matrix It satisfies that the sum of the elements on the rising diagonal is the tribonacci sequence t n (sequence A A generalized Delannoy number D 2(n,k is the number of weighted lattice paths such that the steps H,V and D are labelled with weights a,b and c, respectively This kind of paths is called (a, b, c-weighted paths [9] The generalized Delannoy numbers are given by the following formula (cf [5]: D 2(n,k k ( k i ( n+k i k a k i b n i c i ( Cheon et al [9] defined the generalized Delannoy triangle H 2 (a,b,c [d n,k ] n,k N, where { D d n,k 2(n k,k, if n k; 0, if n < k; Many integer sequences and their properties are expounded on The On-Line Encyclopedia of Integer Sequences [20] the electronic journal of combinatorics 22( (205, #P38 2

3 and, they studied this array by means of Riordan arrays The first few terms of this triangle are b a b 2 c+2ab a H 2 (a,b,c b 3 b(2c+3ab a(2c+3ab a 3 0 b 4 b 2 (3c+4ab c 2 +6abc+6a 2 b 2 a 2 (3c+4ab a 4 Additionally, they introduced a generalized Lucas polynomial sequence, and they gave a lattice path combinatorial interpretation for these polynomials In particular, they found a combinatorial interpretation of the polynomials of Fibonacci, Lucas, Pell, Pell- Lucas Chebyschev of first and second kind, among others The study of lattice paths through Riordan arrays is not new, see [8, 2] A natural questions is: What family of steps are required to find a Riordan array, such that the sum of the elements on its rising diagonal is the k-bonacci numbers F n (k? The k-bonacci numbers are defined by the recurrence F n (k F (k n +F (k n 2 + +F (k n k, n, with initial values F (k F (k 2 F (k (k 0 and F(k 0 In particular, we obtain the Fibonacci sequence and tribonacci sequence, in the case k 2, 3, respectively In this paper, we introduce a family of weighted lattice paths whose step set is S m {H (,0,V (0,,D (,,,D m (,m }, where each step is labelled with weights a,a 2,,a m+, respectively From the number of these weighted lattice paths, we obtain a new family of Riordan arrays, such that the sum of the elements on its rising diagonal is the k-bonacci sequence Moreover, we obtain a generalized k-bonacci polynomial sequence The new family of weighted lattice paths leads us to a combinatorial interpretation for the generalized k-bonacci polynomial sequence In particular, we study a generalization of the tribonacci polynomials and tribonacci-lucas polynomials We also show three combinatorial interpretations of the central tetrabonacci numbers 2 Riordan arrays A Riordan matrix L [l n,k ] n,k N is defined by a pair of generating functions g(z +g z+g 2 z 2 + and f(z f z+f 2 z 2 +, where f 0, such that the k-th column satisfies l n,k z n g(z(f(z k n 0 the electronic journal of combinatorics 22( (205, #P38 3

4 The first column being indexed by 0 From the definition, follows l n,k [z n ]g(z(f(z k, where [z n ] is the coefficient operator The matrix corresponding to the pair f(z,g(z is denoted by R(g(z,f(z or (g(z,f(z The product of two Riordan arrays (g(z,f(z and (h(z,l(z is defined by: (g(z,f(z (h(z,l(z (g(zh(f(z,l(f(z The set of all Riordan matrices is a group under the operator, (cf [9] The identity elementisi (,z,andtheinverseof(g(z,f(zis(g(z,f(z ( / ( g f (z,f(z, where f(z is the compositional inverse of f(z Example ThePascalmatrixH andthedelannoymatrixh 2 aregivenbythefollowing Riordan arrays, respectively ( ( H z, z, H 2 z z, z(+z z Example 2 ([9] The generalized Delannoy array H 2 (a,b,c has a Riordan array expression given by ( H 2 (a,b,c bz,za+cz bz The following theorem is known as the fundamental theorem of Riordan arrays or summation property (cf [2] Theorem 3 If [l n,k ] n,k N (g(z,f(z is a Riordan matrix Then for any sequence {h k } k N n l n,k h k [z n ]g(zh(f(z, k0 where h(z is the generating function of the sequence {h k } k N Let [l n,k ] n,k N (g(z,f(z be a Riordan array, then the elements {l n sk : k 0} are called the d-diagonal of (g(z, f(z (cf [2] 3 Generalized Tribonacci and Tribonacci-Lucas polynomial sequence In this section, we defined a new kind of lattice paths: (a,b,c,d-weighted paths From the number of (a,b,c,d-weighted lattice paths, we introduce a new infinite lower triangular array Then, we study its properties and we obtain the generalized tribonacci and the electronic journal of combinatorics 22( (205, #P38 4

5 tribonacci-lucas polynomial sequence Let M 3 (n,k denote the set of (a,b,c,d-weighted paths from the point (0,0 to the point (k,n, with step set S 3 {H (,0,V (0,,D (,,D 2 (,2}, where each step is labelled with weights a,b,c and d, respectively The weight of a weighted path is the product of the weights of all its steps in the weighted path and the length of a weighted path is the number of steps making up the path For example, in Figure, we show a (a,b,c,d-lattice path of length 5 and weight a 2 bcd b a a d c Figure : Example of (a,b,c,d-weighted path ThelaststepofanypathfromM 3 (n,kisoneofs 3 Therefore,thenumberD 3 (n,k : M 3 (n,k satisfies the following four-term recurrence relation: D 3 (n,k ad 3 (n,k+bd 3 (n,k +cd 3 (n,k +dd 3 (n,k 2, (2 with k 2,n and initial conditions D 3 (0,k b k and D 3 (n,0 a n Theorem 4 The number of (a,b,c,d-lattice paths is given by D 3 (n,k Proof For n, let n j0 j0 j l0 l0 ( n j ( j l n j n ( ( n n j j l Then by Equation (2, we obtain ( n+k 2j +l a n j c l d j l b k 2j+l k 2j +l W (3 n (z : ( k n+2j +l n D 3 (n,iz i a j c l d n j l b k 2n+2j+l W (3 n (z aw (3 n (z+bzw (3 n (z+czw (3 n (z+dz 2 W (3 n (z the electronic journal of combinatorics 22( (205, #P38 5

6 Then, W (3 n (z ( a+cz +dz 2 bz ( a+cz +dz 2 bz W (3 ( a+cz +dz 2 n (z bz n bz (a+cz +dz2 n ( bz n+ Therefore, by the binomial theorem [ ] z k W n (3 [ z k]( (a+cz +dz 2 n ( bz (n+ [ z k]( n j0 [ z k]( n n j0 j l0 j0 ( n j j l0 j l0 ( j l ( n j ( j l ( n j [ Definition 5 Let H 3 : H 3 (a,b,c,d : d (3 a n j (cz l (dz 2 j l ( j l ( n+k 2j +l k 2j +l d (3 n,k n W (3 0 (z ( k +i (bz i i ( k +i a n j c l d j l b i z 2j l+i ] i n,k N a n j c l d j l b k 2j+l, where { n,k D 3 (n k,k, if n k; 0, if n < k The first few terms of this triangle array are b a b 2 2ab+c a H 3 (a,b,c,d b 3 3ab 2 +2cb+d 3ba 2 +2ca a 3 0 b 4 4ab 3 +3cb 2 +2db 6a 2 b 2 +6acb+c 2 +2ad 4ba 3 +3ca 2 a 4 Theorem 6 The infinite triangular array H 3 (a,b,c,d has a Riordan array expression given by ( +dz2 H 3 H 3 (a,b,c,d,za+cz bz bz Proof The proof runs like in Theorem 4, because [z n ]H 3 (a,b,c,d [z n ] ( z k (a+cz +dz 2 k ( bz (k+ [ z n k]( (a+cz +dz 2 k ( bz (k+ the electronic journal of combinatorics 22( (205, #P38 6

7 Proposition 7 Let A 3 (z be the generating function for the rows sums of the Riordan array H 3 Then A 3 (z (a+bz cz 2 dz3 (3 Proof By applying Theorem 3 to the Riordan array H 3 with the generating function h(z ( z, we have ( ( A 3 (z bz ( z a+cz+dz2 (a+bz cz 2 dz 3 bz Each step weight a,b,c,d can be considered as weighted functions a a(x,b b(x,c c(x and d d(x Let a+b p(x,c q(x,d r(x Let F n (3 (x be the n-th row sum of H 3, then from above proposition we obtain the generating function for F n (3 (x: A 3 (z F (3 i (xz i p(xz q(xz 2 r(xz3 (4 The polynomials F (3 n (x are called generalized tribonacci polynomials The first few terms are, p, p 2 +q, p 3 +2pq +r, p 4 +3p 2 q +2pr +q 2, p 5 +4p 3 q +3p 2 r+3pq 2 +2qr,, where q q(x,p p(x,r r(x If p x 2,q x and r, we get the tribonacci polynomials (cf [6] Theorem 8 For all integer n 0, and for any polynomials p(x,q(x,r(x n 2 F n (3 (x i ( ( n i j i q i j (xr j (xp n 2i j (x i j j0 Proof The generating function of the sum of elements on the rising diagonal lines (- diagonal in the generalized Delannoy triangle H 2 (q(x,p(x,r(x is ( ( p(xz p(xz q(xz 2 r(xz A 3(z 3 Then, by Equation ( F (3 n (x n 2 z 2q(x+r(xz p(xz D 2(n i,i n 2 i j0 ( i j ( n i j i q i j (xr j (xp n 2i j (x the electronic journal of combinatorics 22( (205, #P38 7

8 where We give an alternative proof of Theorem 8 Observe the equality A 3 (z p(xz q(xz 2 r(xz 3 q(xz 2 w, w p(xz r(xz3 q(xz 2 Therefore by Theorem 3, polynomials F n (3 (x can be expressed by n-th sum of the rows of the following Riordan array p ( p(xz r(xz3 q(xz2, q(xz 2 q 0 p pq r 0 p q 2 0 3p 2 q 2pr 0 p pq 2 2qr 0 4p 3 q 3p 2 r 0 p 5 where q q(x,p p(x,r r(x So, the identity is clear Following the same ideas of Theorem 8 in [7], we obtain the following identity: Proposition 9 For all integer n 0, and for any polynomials p(x,q(x,r(x F n (3 (x ( ( n i i j q i 2j (xr j (xp n 2i+j (x i j j 0 j i n Theorem 0 For all integer n 3, the polynomials F (3 n (x satisfy the following recurrence:, F (3 n (x p(xf (3 n (x+q(xf (3 n 2(x+r(xF (3 n 3(x, (5 where F (3 0 (x,f (3 (x p(x,f (3 2 (x p 2 (x+q(x Proof The result follows from the equality A 3 (z p(xza 3 (z+q(xz 2 A 3 (z+r(xz 3 A 3 (z We can write the polynomials F (3 n (x as a Binet-like formula, ie, F (3 n (x α(x n+2 (α(x β(x(α(x γ(x β(x n+2 + (β(x α(x(β(x γ(x + γ(x n+2 (γ(x α(x(γ(x β(x, where α(x,β(x and γ(x are the roots of the characteristic equation of (5 the electronic journal of combinatorics 22( (205, #P38 8

9 Definition The generalized tribonacci-lucas polynomials are defined by Note that L (3 n (x α(x n +β(x n +γ(x n, n 0 L (3 n (x p(xf n (x+2q(xf (3 n 2(x+3r(xF (3 n 3(x (3 (6 F n (3 (x+q(xf n 2(x+2r(xF (3 n 3(x, (3 n 3, (7 n 2 i j0 The first few terms are ( n i n i j j ( n i j i q i j (xr j (xp n 2i j (x, n > i+j (8 3, p, p 2 +2q, p 3 +3pq +3r, p 4 +4p 2 q +4pr +2q 2, p 5 +5p 3 q +5p 2 r +5pq 2 +5qr,, where q q(x,p p(x,r r(x If p q r, we obtain the tribonacci-lucas sequence, (sequence A00644 Proposition 2 The generating function of the tribonacci-lucas polynomials is L 3 (z : L (3 i (xz i 3 2p(xz q(xz 2 p(xz q(xz 2 r(xz 3 Proof From Equation (6, we have the following matrix equation: R(+q(xz 2 +2r(xz 3,z [ ] T [ T F (3 0 (x,f (3 (x,,l (3 (x,l (3 2 (x,], where A T is the transpose of the matrix A Then by applying Theorem 3 to the Riordan array R(+q(xz 2 +2r(xz 3,z with the generating function A 3 (z, we get the generating function: Therefore, L 3 (z 2+ +q(xz 2 +2r(xz 3 p(xz q(xz 2 r(xz 3 +q(xz 2 +2r(xz 3 p(xz q(xz 2 r(xz 3 3 2p(xz q(xz 2 p(xz q(xz 2 r(xz 3 Theorem 3 Let p(x ax+b,q(x cx, and r(x dx Then ( n k j ( ( ( k j n 2j +l F n (3 (x a k j b n k 2j+l c l d j l x k j l k k0 j0 l0 the electronic journal of combinatorics 22( (205, #P38 9

10 Proof By applying Theorem 3 to the Riordan array H 3 with the generating function h(z /( xz, we have n ( ( D 3 (n k,kx k [z n ] bz xz ( a+cz+dz 2 k0 bz ( [z n ] (b+axz cxz 2 dxz ( 3 [z n ] p(xz q(xz 2 r(xz 3 The coefficients of the above polynomial are the numbers of (a, b, c, d-lattice paths 3 Combinatorial Interpretations The following theorem shows a combinatorial interpretation for the generalized tribonacci and tribonacci-lucas polynomial sequences Theorem 4 For any polynomial F n (3 (x and L (3 n (x, we have n (i F (3 n (x k0 n (ii L (3 n (x p(x ω (3 n,k (x, n 0, k0 n 2 ω (3 n,k (x+2q(x k0 n 3 ω (3 n 2,k (x+3r(x k0 ω (3 n 3,k (x, n 3, where ω (3 n,k (x : ω(3 n,k h(3 n,k (x D 3(n k,k is the sum of weights of (a(x,b(x,c(x,d(x-weighted paths from (0,0 to (k,n k with step set S 3 {H (0,0,V (0,,D (,,D 2 (,2}, such that a(x+b(x p(x,c(x q(x and d(x r(x Proof By definition, F (3 n (x is the n-th row sum of the Riordan array H 3 Then (i follows from Definition 5 Identity (ii follows from Equation (6 Example 5 The tribonacci numbers are defined by the recurrence relation: t 0, t, t 2 2, t n+3 t n+2 +t n+ +t n, for n 0 Thefirstfewtermsofthetribonaccinumberare,,2,4,7,3,24,44,8,49,274,, (sequence A Hoggatt and Bicknell [6] introduced tribonacci polynomials The tribonacci polynomials T n (x are defined by the recurrence relation T 0 (x, T (x x 2, T 2 (x x 4 +x, T n+3 (x x 2 T n+2 (x+xt n+ (x+t n (x, for n 0 the electronic journal of combinatorics 22( (205, #P38 0

11 Note that T n ( t n for all integer positive n The first few tribonacci polynomials are T 0 (x, T 4 (x x 8 +3x 5 +3x 2, T (x x 2, T 5 (x x 0 +4x 7 +6x 4 +2x, T 2 (x x 4 +x, T 6 (x x 2 +5x 9 +0x 6 +7x 3 +, T 3 (x x 6 +2x 3 +, T 7 (x x 4 +6x +5x 8 +6x 5 +6x 2 The generating function of the tribonacci polynomials is T n (xz n n0 x 2 z xz 2 z 3 If a(x x 2,b(x 0,c(x x and d(x in (3, we obtain the tribonacci polynomials from the row sums of the Riordan array H 3 (x 2,0,x,: T 0 (x 0 x T (x 0 x x T 2 (x 0 2x 3 x T 3 (x 0 0 3x 2 3x 5 x T 4 (x 0 0 2x 6x 4 4x 7 x 0 0 T 5 (x 0 0 7x 3 0x 6 5x 9 x 2 T 6 (x From Theorem 4-(i, we obtain a combinatorial interpretation of T n (x For example, T 4 (x x 8 + 3x 5 + 3x 2 4 ω(3 4,i Note that, ω(3 4,0 ω (3 4, 0,ω (3 4,2 3x 2,ω (3 4,3 3x 5 and ω (3 4,4 x 8 In Figure 2, we show the corresponding weighted lattice paths Example 6 The tribonacci-lucas numbers are defined by the recurrence relation: k 0 3, k, k 2 3, k n+3 k n+2 +k n+ +k n, for n 0 The first few terms of the tribonacci-lucas numbers are 3,, 3, 7,, 2, 39, 7, 3, 24, 443,, (sequence A00644 The tribonacci-lucas polynomials K n (x are defined by the recurrence relation K 0 (x 3, K (x x 2, K 2 (x x 4 +2x, K n+3 (x x 2 K n+2 (x+xk n+ (x+k n (x, n 0 NotethatK n ( k n forallintegerpositiven Thefirstfewtribonacci-Lucaspolynomials are K 0 (x 3, K 4 (x x 8 +4x 5 +6x 2, K (x x 2, K 5 (x x 0 +5x 7 +0x 4 +5x, K 2 (x x 4 +2x, K 6 (x x 2 +6x 9 +5x 6 +4x 3 +3, K 3 (x x 6 +3x 3 +3, K 7 (x x 4 +7x +2x 8 +28x 5 +4x 2 the electronic journal of combinatorics 22( (205, #P38

12 x x } {{ } 3x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 8 x x x x 2 x 2 x 2 x 2 x 2 x 2 } {{ } 3x 5 Figure 2: (x 2,0,x,-weighted paths and tribonacci polynomials Using standard techniques it can be shown that the generating function of the tribonacci- Lucas polynomials is K n (xz n 3 2zx2 z 2 x x 2 z xz 2 z 3 n0 Since p(x a(x + b(x x 2,q(x c(x x and r(x d(x, we may choose a(x x 2,b(x 0,c(x x and d(x From Theorem 4-(ii, we obtain a combinatorial interpretation of K n (x For example, K 3 (x x 6 +3x 3 +3 x 2 2 ω (3 2,i +2x ω (3,i +3ω(3 0,0 Note that ω (3 2,0 0,ω (3 2, x,ω (3 2,0 x 4,ω (3,0 0,ω (3, x 2 and ω (3 0,0 In Figure 3, we show the corresponding weighted lattice paths x x 2 x 2 x 2 }{{} x 2 (x 4 +x x 6 +x 3 2x(x 2 2x 3 3( 3 Figure 3: (x 2,0,x,-weighted paths, and tribonacc-lucas polynomials the electronic journal of combinatorics 22( (205, #P38 2

13 32 The diagonal of D 3 Let the Riordan array H 3 (,,, The first few terms of this array are H 3 (,,, { From } Theorem 6 of [3], we have that the generating function of the central sequence d (3 2n,n {,3,5,8,459,2673,} is n N n0 d (3 2n,nz n D 3 (n,nz n n0 6z 3z 2 On the other hand, a Motzkin path of length n is a lattice path of Z Z running from (0,0 to (n,0 that never passes below the x-axis and whose permitted steps are the up diagonal step U (,, the down diagonal step D (, and the horizontal step H (, 0, called rise, fall and level step, respectively The number of Motzkin paths of length n is the n-th Motzkin number m n, (sequence A00006 Many other examples of bijections between Motzkin numbers and others combinatorial objects can be found in [4] A Grand Motzkin path of length n is a Motzkin path without the condition that never passes below the x-axis The number of Grand Motzkin paths of length n is the n-th Grand Motzkin number g n, sequence A A a H b U -Motzkin path is a Motzkin path such that each horizontal step is colored with one of a specific colors and each up diagonal step is colored with one of b specific colors The number of a H b U -Motzkin paths of length n is the n-th a H b U -Motzkin number m (a,b n Analogously, we have a H b U -Grand Motzkin paths, the number of a H b U -Grand Motzkin paths of length n is denoted by g n (a,b From Theorem of [8], we have n0 g (3,3 n z n Then, we get the following corollary 6z 3z 2 Corollary 7 The number of 3 H 3 U -Grand Motzkin path is equal to the number of (,,,-weighted paths from the point (0,0 to the point (n,n, ie, g (3,3 n D 3 (n,n the electronic journal of combinatorics 22( (205, #P38 3

14 Let S(n,k be the number of lattice paths from (0,0 to (n,k in the plane Z N with set of steps S 3 {H (,0,V (0,,D (,,L (,} In [], Dziemiańczuk shows that S(0,nz n 6z 3z 2 Therefore Moreover, n0 then S(i 2j,j D 3 (i j,j S(0,n g (3,3 n D 3 (n,n (S(i 2j,j i,j 0 H 3 (,,,, 4 Riordan Arrays and Generalized k-bonacci numbers In this section, we generalize the results of the above section We introduce a new family of weighted lattice paths, the A s -weighted paths Let A s (a,a 2,,a s be a vector of weights Then, we denote by M m (n,k the set of A m+ -weighted paths from the point (0,0 to the point (k,n, with step set S m {H (,0,V (0,,D (,,,D m (,m }, where each step is labelled with weights a,a 2,,a m+, respectively Let D m (n,k M m (n,k Note that if m 2,3 we obtain the generalized Delannoy paths and the (a, b, c, d-weighted paths, respectively Lemma 8 The numbers D m (n,k satisfy the following (m+-term recurrence relation D m (n,k a D m (n,k+a 2 D m (n,k + m j a j+2 D m (n,k j, (9 with k m,n and initial conditions D m (0,k a k 2 and D m (n,0 a n Theorem 9 The number of A m+ -lattice paths is given by D m (n,k n n j j 0 j 2 0 n m 2 j j i j m 0 ( ( n n j j j 2 ( n m 2 j j i j m ( n+k u a j n a j 2 3 a j m m a n m i j i m+ a k u 2, where m u (m (n j + (i mj i i2 the electronic journal of combinatorics 22( (205, #P38 4

15 Proof For n, let Then from Equation (9, we have Then, W (m n (z : D m (n,iz i m W n (m (z a W n (z+a (m 2 zw n (m (z+ a j+2 z j W n (z (m ( a + m W n (m j (z a j+2z j a 2 z ( a + m j a j+2z j a 2 z W (m n (z n a 2 z Therefore, by the binomial theorem [ ] z k W n (m [ z k](( m n a + a j+2 z j ( a 2 z (n+ [ z k] n n j j 0 j 2 0 j n m i j i j m 0 ( ( n n j j j ( a + m j a j+2z j a j (a 3 z j2 (a m z m 2 j m 2 (a m+ z m n m i j i [ z k] a j a j 2 n j 0 j 2 0 a j a j 2 n n j j 0 j a j m 2 n j m 3 a j m 2 m n m i j i j m 0 j 2 ( ( n n j l0 a n m i j i m+ a l 2z u+l n m i j i j m 0 j ( ( n n j j a n m i j i m+ a k u 2 [ Definition 20 Let H m : H m (a,a 2,,a m+ : j 2 a 2 z ( a + n m j a j+2z j ( a 2 z n+ ( n m 2 i j i j m j 2 n W (m 0 (z ( n+l (a 2 z l l0 ( n m 2 i j i j m d (m n,k ] n,k N { n,k D m (n k,k, if n k; 0, if n < k d (m n ( n m 2 i j i j m ( n+k u, where i ( n+l n the electronic journal of combinatorics 22( (205, #P38 5

16 Theorem 2 The infinite triangular array H m has a Riordan array expression given by ( H m a 2 z,za +a 3 z +a 4 z 2 + +a m+ z m a 2 z Proof It is clear from Theorem 9 For example, the first few terms of the Riordan array H 4 (a,b,c,d,e are b a b 2 2ab+c a H 4 b 3 3ab 2 +2cb+d 3ba 2 +2ca a 3 0 b 4 4ab 3 +3cb 2 +2db+e 6a 2 b 2 +6acb+c 2 +2ad 4ba 3 +3ca 2 a 4 Proposition 22 Let A m (z be the generating function for the rows sums of the Riordan array H m Then A m (z Proof The proof runs like in Proposition 7 (a +a 2 z a 3 z 2 a m+ zm (0 Let F n (k (x be the n-th row sum of H k, then from above proposition we obtain the generating function for F n (k (x: A k (z F (k i (xz i p (xz p 2 (xz 2 p k (xzk, ( where p (x a +a 2, and p j (x a j+,j 2,,k The polynomials F (k n (x are called generalized k-bonacci polynomials For example, if k 4 we obtain the tetrabonacci polynomials The first few terms of the tetrabonacci polynomials sequence are, p, p 2 +p 2, p 3 +2p 2 p +p 3, p 4 +3p 2 p 2 +2p 3 p +p 2 2 +p 4,, where p i p i (x,i,2,3,4 The sum on the rising diagonal in the Pascal triangle and in the tribonacci triangle are the Fibonacci numbers and tribonacci numbers, respectively In general, we have the following proposition Proposition 23 The k-bonacci triangle T k defined by the following Riordan array ( + +zk T k,z+z, z z satisfies that the sum on the rising diagonal is the k-bonacci sequence F n (k, where F (k F (k n +F (k n 2 + +F (k n k n the electronic journal of combinatorics 22( (205, #P38 6

17 From Theorem 3 and Theorem 9, we have the following identity Theorem 24 For all integern 0, andfor anypolynomialsp i (x p(x,i,2,,k+, F (k n (x where n 2 D k (n s,s n 2 s s j j 0 j 2 0 s k 2 j j i j k 0 ( ( s s j j j 2 ( n u+ s ( s k 2 j j i j k p j p j 2 k u (k (s j + (i kj i i2 3 p j k k p s k i j i k+ p2 n u s, Theorem 25 The polynomials F (k n (x satisfy the following recurrence for n k F n (k (x p (xf n (x+p (k 2 (xf n 2(x+ +p (k k (xf (k n k (x, (2 where F (k 0 (x,f (k i (x 0 for i, 2,, (k Proof The identity follows from Equation ( We can write the polynomials F (k n (x as a Binet-like formula [4], ie, F (k n (x k i r i,k (x n+k k l,l i (r i,k(x r j,k (x, where r i,k (x are the roots of the characteristic equation of (2 Definition 26 The generalized m-bonacci-lucas polynomials are defined by Note that L (m n (x r,m (x n +r 2,m (x n + +r m,m (x n L (m n (x p (xf (m n (x+2p 2 (xf (m n 2(x+ +mp m (xf (m n m(x, F (m n (x+p 2 (xf (m n 2(x+ +(m p m (xf (m n m(x For example, if m 4 we obtain the tetrabonacci-lucas polynomials The first few terms of this sequence are 4, p, p 2 +2p 2, p 3 +3p 2 p +3p 3, p 4 +4p 2 p 2 +4p 3 p +2p p 4,, where p i p i (x,i,2,3,4 If p i,i,2,3,4, we obtain the sequence A07387 the electronic journal of combinatorics 22( (205, #P38 7

18 Proposition 27 The generating function of the m-bonacci-lucas polynomials is L m (z : L (m i (xz i m (m p (xz (m 2p 2 (xz 2 p m (xz m p (xz p 2 (xz 2 p m (xz m Proof The proof runs like in Proposition 4 4 Combinatorial Interpretation Theorem 28 For any polynomial F (m n (x and L (m n (x, we have (i F (m n (x n k0 ω(m n,k (x for n 0, (ii For n m, n n (x p (x Ls (m k0 n,k (x+2p n 2 2(x ω (m k0 n m ω (m n 2,k (x+ +mp m(x k0 ω (m n m,k (x where ω (m n,k (x : ω(m n,k h(m n,k D m(n k,k is the sum of weights of Am+ -weighted paths from (0,0 to (k,n k with step set S m {H (0,0,V (0,,D (,,D 2 (,2,,D m (,m }, such that a (x+a 2 (x p (x, and a i (x p i (x for i,,m+ Example 29 The tetrabonacci numbers are defined by the recurrence relation: l 0, l, l 2 2, l 3 4 l n+4 l n+3 +l n+2 +l n+ +l n, for n 0 The first few terms of the tetrabonacci number are,, 2, 4, 8, 5, 29, 56, 08, 208, 40,, (sequence A The tetrabonacci polynomials R n (x are defined by the recurrence relation R 0 (x, R (x x 3, R 2 (x x 6 +x 2, R 3 (x x 9 +2x 5 +x, R n+4 (x x 3 R n+3 (x+x 2 R n+2 (x+xr n+ (x+r n (x, for n 0 The generating function of the tetrabonacci polynomials is R n (xz n x 3 z x 2 z 2 xz 3 z 4 n0 If a (x x 3,a 2 (x 0,a 3 (x x 2,a 4 (x x and a 5 (x in (3, we obtain the tetrabonacci polynomials from the row sums of the Riordan array H 4 (x 3,0,x 2,x, FromTheorem4-(i, weobtainacombinatorialinterpretationofr n (x Forexample, R 3 (x x 9 +2x 5 +x 3 k0 ω (4 3,k 0+x+2x5 +x 9 In Figure 4, we show the corresponding weighted lattice paths the electronic journal of combinatorics 22( (205, #P38 8

19 x x 3 x 3 x 2 x 2 x 3 x 3 x 3 }{{} 2x 5 x 9 Figure 4: (x 3,x 2,0,x,-weighted paths and tretranacci polynomials Acknowledgements We would like to thank the anonymous referee for valuable comments to improve the final version The second author thanks the invitation to Bogotá where the mayor part of this paper was done References [] C Banderier and S Schwer Why Delannoy numbers? J Statist Plann Inference, 35:40 50, 2005 [2] P Barry On integer-sequence-based constructions of generalized Pascal triangles J Integer Seq, 9 Article 0624, 2006 [3] P Barry On the central coefficients of Riordan matrices J Integer Seq, 6 Article 35, 203 [4] F Bernhart Catalan, Motzkin, and Riordan numbers Discrete Math, 204(-3: 73 2, 999 [5] B Birregah, P K Dohb and K H Adjallah A systematic approach to matrix forms of the Pascal triangle: The twelve triangular matrix forms and relations European J Combin, 3:205 26, 200 [6] R Brawer and M Pirovino The linear algebra of the Pascal matrix Linear Algebra Appl, 74:3 23, 992 [7] G S Call and D J Velleman Pascal s matrices Amer Math Monthly, 00: , 993 [8] D Callan On generating functions involving the square root of a quadratic polynomial J Integer Seq, 0 Article 0752, 2007 [9] G-S Cheon, H Kim and L W Shapiro A generalization of Lucas polynomial sequence Discrete Appl Math 57: , 2009 [0] L Comtet Advanced combinatorics D Reidel Publishing Company, 970 [] M Dziemiańczuk Counting lattice paths with four types of steps Graphs Combin, 30: , 204 the electronic journal of combinatorics 22( (205, #P38 9

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