Lattice Paths, k-bonacci Numbers and Riordan Arrays

Transcription

1 Lattice Paths, k-bonacci Numbers and Riordan Arrays José L. Ramírez Departamento de Matemáticas Universidad Nacional de Colombia Joint work: Victor Sirvent; Universidad Simón Bolivar. 4th International Symposium on Riordan Arrays and Related Topics Universidad Complutense de Madrid Madrid - Spain July

2 Outline 1. Lattice Path - Pascal Matrix - Delannoy Matrix 2. (a, b, c, d) weighted paths 3. The General Case 4. Schröder Matrix 5. Some additional comments.

3 Lattice Path Riordan Arrays D. Merlini, D. G. Rogers, R. Sprugnoli and M. Cecilia Verri. Underdiagonal lattice paths with unrestricted steps. Discrete Appl. Math., 91(1-3): , 1999.

4 Lattice Path Riordan Arrays D. Merlini, D. G. Rogers, R. Sprugnoli and M. Cecilia Verri. Underdiagonal lattice paths with unrestricted steps. Discrete Appl. Math., 91(1-3): , 1999.

5 Basic Problem How many lattice paths are there from (0,0) to (n,m)? x (n,m) (0,0) y

6 Basic Problem How many lattice paths are there from (0,0) to (n,m)? x 1 (n,m) (0,0) y ( ) ( ) n+m n+m D 1 (n,k) = = n m

7 Pascal Matrix - Lattice Paths [ H 1 := h (1) n,k H 1 = ] n,k N, where h (1) n,k = { D 1 (n k,k) = ( n k), if n k; 0, if n < k. ( ) 1 1 z, z = 1 z H 1 is the Pascal matrix

8 Pascal Matrix - Lattice Paths F n = n 1 2 ( n l 1 ) l=0 l := is the number of lattice path from (0,0) to (n 1 2k,k), for k = 0,1,..., (n 1)/2.

9 Problem 2 How many (a, b, c)-weighted lattice paths are there from (0,0) to (n,m)? b c a c a a c b b a (n,m) b (0,0) D 2 (n,k) = k i=0 ( k i )( n+k i k ) a k i b n i c i

10 Generalized Delannoy number Generalized Delannoy matrix: H 2 (a,b,c) = [d n,k ] n,k N, where d n,k = { D2 (n k,k), if n k; 0, if n < k. ( ) 1 H 2 (a,b,c) = 1 bz,za+cz 1 bz b a b 2 c +2ab a = 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. GF Row Sum: 1 1 (a+b)z cz 2 G.-S. Cheon, H. Kim and L. W. Shapiro. A generalization of Lucas polynomial sequence. Discrete Appl. Math. 157(5),

11 Delannoy Matrix Delannoy Matrix: ( 1 H 2 (1,1,1) = 1 z, z(1+z) ) 1 z = GF Row Sum: 1 1 2z z 2

12 Delannoy Matrix The sum of the elements on the rising diagonal is the tribonacci sequence t n

13 Delannoy Matrix T (a,b,c) n+1,q,r,p := the sum on the diagonal of direction (r,q), i.e., T (a,b,c) n+1,q,r,p = n p q+r k=0 d(n qk,p +rk). Direction (1, q). Theorem The sequence T n,q := T (a,b,c) n,q,1,0 recurrence relation satisfies the following linear T n,q = bt n 1,q +at n q 1,q +ct n q 2,q, with the initial values T 1,q = 1, and T 0,q = T 1,q = = T n q 2,q = 0.

14 Delannoy Matrix T (a,b,c) n+1,q,r,p := the sum on the diagonal of direction (r,q), i.e., T (a,b,c) n+1,q,r,p = n p q+r k=0 d(n qk,p +rk). Direction (1, q). Theorem The sequence T n,q := T (a,b,c) n,q,1,0 recurrence relation satisfies the following linear T n,q = bt n 1,q +at n q 1,q +ct n q 2,q, with the initial values T 1,q = 1, and T 0,q = T 1,q = = T n q 2,q = 0.

15 Delannoy Matrix Proof: A0 b B0 D1 C1 a A1 c b B1 D2 C2 a A2 c b B2 (0,0) Corollary The generating function of the sequence {T n+1,q } n 0 is T n+1,q z n = n bz az q+1 +cz q+2.

16 Delannoy Matrix Proof: A0 b B0 D1 C1 a A1 c b B1 D2 C2 a A2 c b B2 (0,0) Corollary The generating function of the sequence {T n+1,q } n 0 is T n+1,q z n = n bz az q+1 +cz q+2.

17 Delannoy Matrix General Case: this corresponds to the finite sequences lying over finite transversals of Delannoy matrix. Theorem The sequence T n,q,r,p := T n,q,r,p (a,b,c) satisfies the following linear recurrence relation r ( ) r r ( ) r ( b) s T n s,q,r,p = a r s c s T n r q s,q,r,p. s s s=0 s=0 Theorem The generating function of the sequence {T n+1,q,r,p } n 0 is given by n 0 T n+1,q,r,p z n = (1 bz)r p 1 (a+cz) p z p (1 bz) r (a+cz) r z q+r.

18 Delannoy Matrix General Case: this corresponds to the finite sequences lying over finite transversals of Delannoy matrix. Theorem The sequence T n,q,r,p := T n,q,r,p (a,b,c) satisfies the following linear recurrence relation r ( ) r r ( ) r ( b) s T n s,q,r,p = a r s c s T n r q s,q,r,p. s s s=0 s=0 Theorem The generating function of the sequence {T n+1,q,r,p } n 0 is given by n 0 T n+1,q,r,p z n = (1 bz)r p 1 (a+cz) p z p (1 bz) r (a+cz) r z q+r.

19 Delannoy Matrix General Case: this corresponds to the finite sequences lying over finite transversals of Delannoy matrix. Theorem The sequence T n,q,r,p := T n,q,r,p (a,b,c) satisfies the following linear recurrence relation r ( ) r r ( ) r ( b) s T n s,q,r,p = a r s c s T n r q s,q,r,p. s s s=0 s=0 Theorem The generating function of the sequence {T n+1,q,r,p } n 0 is given by n 0 T n+1,q,r,p z n = (1 bz)r p 1 (a+cz) p z p (1 bz) r (a+cz) r z q+r.

20 Delannoy Matrix - Tribonacci case Example For r = q = 1 and p = 0, we have the tribonacci sequence (T n ) n 0 = (0,1,1,2,4,7,13,24,...). Therefore, we have the explicit relation T n+1 = n/2 k=0 k j=0 ( j k )( n j k k ).

21 Particular Case H 2 (1,0,1) = Row Sum = Fibonacci Sequence GF Row Sum: 1 1 z z 2

22 Combinatorial Interpretation - Fibonacci Numbers F n+1 = n k=0 ω k,n k ω k,n k := is the sum of weights of paths from (0,0) to (k,n k) using the steps: (1,0), (1,1) G.-S. Cheon, H. Kim and L. W. Shapiro. A generalization of Lucas polynomial sequence. Discrete Appl. Math. 157(5), 2009.

23 Combinatorial Interpretation - Fibonacci Numbers Example F 5 = 5 (3,1) (4,0) (2,2) (3,1) (3,1)

24 Summarizing Pascal Matrix Sum of the elements on the rising diagonal = Fibonacci numbers. Delannoy Matrix Sum of the elements on the rising diagonal = Tribonacci numbers. XXXX Matrix (Riordan array) Sum of the elements on the rising diagonal = k-bonacci numbers. What family of steps are required to find a Riordan array, such that the sum of the elements on its main diagonal (or on its row) is the k-bonacci sequence F (k) n?

25 Summarizing Pascal Matrix Sum of the elements on the rising diagonal = Fibonacci numbers. Delannoy Matrix Sum of the elements on the rising diagonal = Tribonacci numbers. XXXX Matrix (Riordan array) Sum of the elements on the rising diagonal = k-bonacci numbers. What family of steps are required to find a Riordan array, such that the sum of the elements on its main diagonal (or on its row) is the k-bonacci sequence F (k) n?

26 Summarizing Pascal Matrix Sum of the elements on the rising diagonal = Fibonacci numbers. Delannoy Matrix Sum of the elements on the rising diagonal = Tribonacci numbers. XXXX Matrix (Riordan array) Sum of the elements on the rising diagonal = k-bonacci numbers. What family of steps are required to find a Riordan array, such that the sum of the elements on its main diagonal (or on its row) is the k-bonacci sequence F (k) n?

27 Summarizing Pascal Matrix Sum of the elements on the rising diagonal = Fibonacci numbers. Delannoy Matrix Sum of the elements on the rising diagonal = Tribonacci numbers. XXXX Matrix (Riordan array) Sum of the elements on the rising diagonal = k-bonacci numbers. What family of steps are required to find a Riordan array, such that the sum of the elements on its main diagonal (or on its row) is the k-bonacci sequence F (k) n?

28 Summarizing Pascal Matrix Sum of the elements on the rising diagonal = Fibonacci numbers. Delannoy Matrix Sum of the elements on the rising diagonal = Tribonacci numbers. XXXX Matrix (Riordan array) Sum of the elements on the rising diagonal = k-bonacci numbers. What family of steps are required to find a Riordan array, such that the sum of the elements on its main diagonal (or on its row) is the k-bonacci sequence F (k) n?

29 Combinatorial Interpretation? k-bonacci numbers The k-bonacci numbers are defined by the recurrence F n (k) = F (k) n 1 +F(k) n 2 + +F(k) n k, n 1, with initial values F (k) 1 = F(k) 2 = = F(k) (k 1) = 0 and F (k) 0 = 1. k = 3, tribonacci: 1, 1, 2, 4, 7, 13, 24,... k-bonacci polynomials. F n (k) (x) = x k 1 F (k) n 1 (x)+xk 2 F (k) n 2 + +F(k) n k, n 1,

30 Combinatorial Interpretation? k-bonacci numbers The k-bonacci numbers are defined by the recurrence F n (k) = F (k) n 1 +F(k) n 2 + +F(k) n k, n 1, with initial values F (k) 1 = F(k) 2 = = F(k) (k 1) = 0 and F (k) 0 = 1. k = 3, tribonacci: 1, 1, 2, 4, 7, 13, 24,... k-bonacci polynomials. F n (k) (x) = x k 1 F (k) n 1 (x)+xk 2 F (k) n 2 + +F(k) n k, n 1,

31 Combinatorial Interpretation? k-bonacci numbers The k-bonacci numbers are defined by the recurrence F n (k) = F (k) n 1 +F(k) n 2 + +F(k) n k, n 1, with initial values F (k) 1 = F(k) 2 = = F(k) (k 1) = 0 and F (k) 0 = 1. k = 3, tribonacci: 1, 1, 2, 4, 7, 13, 24,... k-bonacci polynomials. F n (k) (x) = x k 1 F (k) n 1 (x)+xk 2 F (k) n 2 + +F(k) n k, n 1,

32 Outline 1. Lattice Path - Pascal Matrix - Delannoy Matrix 2. (a, b, c, d) weighted paths 3. The General Case 4. Schröder Matrix 5. Some additional comments

33 (a,b,c,d)-weighted paths 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 = (1,0),V = (0,1),D 1 = (1,1),D 2 = (1,2)}, where each step is labelled with weights a,b,c and d, respectively. Example a a b a d b d c c

34 (a,b,c,d)-weighted paths D 3 (n,k) = ad 3 (n 1,k)+bD 3 (n,k 1) +cd 3 (n 1,k 1)+dD 3 (n 1,k 2), with k 2,n 1 and initial conditions D 3 (0,k) = b k and D 3 (n,0) = a n. a c d b

35 (a,b,c,d)-weighted paths Theorem The number of (a,b,c,d)-lattice paths is given by D 3 (n,k) = = n j j=0 l=0 n n j j=0 l=0 ( n j ( n j )( j l )( ) n+k 2j +l a n j c l d j l b k 2j+l. k 2j +l )( )( n j k n+2j +l l n ) a j c l d n j l b k 2n+2j+l. Proof. Then For n 1, let W (3) n (z) := D 3 (n,i)z i. i=0 W (3) n (z) = aw (3) n 1 (z)+bzw(3) n (z)+czw (3) n 1 (z)+dz2 W (3) n 1 (z).

36 (a,b,c,d)-weighted paths Proof. W (3) n (z) = ( a+cz +dz 2 1 bz ( a+cz +dz 2 = 1 bz ) ( W (3) a+cz +dz 2 n 1 (z) = 1 bz ) n 1 1 bz = (a+cz +dz2 ) n (1 bz) n+1. Therefore, by the binomial theorem [ z k] W n (3) = [z k]( (a+cz +dz 2 ) n (1 bz) (n+1)) = n j j=0 l=0 ( n j )( j l ) n W (3) 0 (z) )( ) n+k 2j +l a n j c l d j l b k 2j+l. k 2j +l

37 (a,b,c,d)-weighted paths Definition [ Let H 3 := H 3 (a,b,c,d) := d (3) n,k = d (3) n,k ], where n,k N { D 3 (n k,k), if n k; 0, if n < k. Theorem The infinite triangular array H 3 (a,b,c,d) has a Riordan array expression given by ( ) 1 +dz2 H 3 = H 3 (a,b,c,d) =,za+cz. 1 bz 1 bz

38 (a,b,c,d)-weighted paths Definition [ Let H 3 := H 3 (a,b,c,d) := d (3) n,k = d (3) n,k ], where n,k N { D 3 (n k,k), if n k; 0, if n < k. Theorem The infinite triangular array H 3 (a,b,c,d) has a Riordan array expression given by ( ) 1 +dz2 H 3 = H 3 (a,b,c,d) =,za+cz. 1 bz 1 bz

39 (a,b,c,d)-weighted paths Proposition Let A 3 (z) be the generating function for the rows sums of the Riordan array H 3. Then A 3 (z) = 1 1 (a+b)z cz 2 dz 3. Let a+b = p(x),c = q(x),d = r(x). Let F (3) n (x) be the n-th row sum of H 3. A 3 (z) = i=0 F (3) i (x)z i = 1 1 p(x)z q(x)z 2 r(x)z 3. If p = x 2,q = x and r = 1, we get the tribonacci polynomials.

40 (a,b,c,d)-weighted paths Proposition Let A 3 (z) be the generating function for the rows sums of the Riordan array H 3. Then A 3 (z) = 1 1 (a+b)z cz 2 dz 3. Let a+b = p(x),c = q(x),d = r(x). Let F (3) n (x) be the n-th row sum of H 3. A 3 (z) = i=0 F (3) i (x)z i = 1 1 p(x)z q(x)z 2 r(x)z 3. If p = x 2,q = x and r = 1, we get the tribonacci polynomials.

41 (a,b,c,d)-weighted paths Theorem For all integer n 3, the polynomials F n (3) (x) satisfy the following recurrence: where F (3) 0 F (3) n (x) = p(x)f (3) n 1 (x)+q(x)f(3) n 2 (x)+r(x)f(3) n 3 (x), (x) = 1,F(3) 1 (x) = p(x),f(3) 2 (x) = p2 (x)+q(x).

42 (a,b,c,d)-weighted paths Theorem For all integer n 0, and for any polynomials p(x),q(x),r(x) F n (3) n 2 (x) = i=0 j=0 F (3) n (x) = 0 j i n i ( )( n i j i i j ( )( n i i j i j j ) q i j (x)r j (x)p n 2i j (x). ) q i 2j (x)r j (x)p n 2i+j (x).

43 (a, b, c, d)-weighted paths - Combinatorial Interpretation For any polynomial F (3) n (x), we have F (3) n (x) = n k=0 ω (3) n,k, n 0, where ω (3) n,k (x) := 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 = (0,1),D 1 = (1,1),D 2 = (1,2)}, such that a(x)+b(x) = p(x),c(x) = q(x) and d(x) = r(x).

44 Tribonacci paths The tribonacci polynomialst n (x): T n (x) = x 2 T n 1 (x)+xt n 2 (x)+t n 3 (x), for n 2. n=0 T n (x)z n = 1 1 x 2 z xz 2 z 3. a(x) = x 2,b(x) = 0,c(x) = x and d(x) = 1 H 3 (x 2,0,x,1): x x x x 3 x x 2 3x 5 x x 6x 4 4x 7 x x 3 10x 6 5x 9 x = T 0 (x) T 1 (x) T 2 (x) T 3 (x) T 4 (x) T 5 (x) T 6 (x)..

45 Tribonacci paths Example T 4 (x) = x 8 +3x 5 +3x 2 = 4 i=0 ω(3) 4,i. Note that, ω (3) 0,4 = ω(3) 1,3 = 0,ω(3) 2,2 = 3x2,ω (3) 3,1 = 3x5 and ω (3) 4,0 = x8.

46 Outline 1. Lattice Path - Pascal Matrix - Delannoy Matrix 2. (a, b, c, d) weighted paths 3. The General Case 4. Schröder Matrix 5. Some additional comments

47 The General Case Let A s = (a 1,a 2,...,a s ) be a vector of weights. We denote by M m (n,k) the set of A m+1 -weighted paths from the point (0,0) to the point (k,n), with step set S m = {H = (1,0),V = (0,1),D 1 = (1,1),...,D m 1 = (1,m 1)}, where each step is labelled with weights a 1,a 2,...,a m+1, respectively. D m (n,k) = M m (n,k).

48 The General Case Lemma The numbers D m (n,k) satisfy the following (m+1)-term recurrence relation m 1 D m (n,k) = a 1 D m (n 1,k)+a 2 D m (n,k 1)+ a j+2 D m (n 1,k j) with k m 1,n 1 and initial conditions D m (0,k) = a k 2 and D m (n,0) = a n 1. j=1 S m = {H = (1,0),V = (0,1),D 1 = (1,1),...,D m 1 = (1,m 1)},

49 Riordan Arrays and Generalized k-bonacci numbers Theorem The number of A m+1 -lattice paths is given by D m (n,k) = n n j 1 j 1 =0j 2 =0 n m 2 j=1 j i j m 1 =0 ( n+k u n ( )( n n j1 j 1 j 2 ) a j 1 1 a j 2 3 a j m 1 ) ( n m 2 j=1 j i j m 1 m a n m 1 i=1 j i m+1 a2 k u, ) where m 1 u = (m 1)(n j 1 )+ (i m)j i. i=2

50 Riordan Arrays and Generalized k-bonacci numbers Definition [ Let H m := H m (a 1,a 2,...,a m+1 ) := d (m) n,k = d (m) n,k ] n,k N, where { D m (n k,k), if n k; 0, if n < k. Theorem The infinite triangular array H m has a Riordan array expression given by ( 1 H m = 1 a 2 z,za 1 +a 3 z +a 4 z 2 + +a m+1 z m 1 ). 1 a 2 z

51 Riordan Arrays and Generalized k-bonacci numbers Proposition The k-bonacci triangle T k defined by the following Riordan array ( ) 1 + +zk 1 T k =,z1+z, 1 z 1 z satisfies that the sum on the rising diagonal is the k-bonacci sequence F n (k), where F n (k) = F (k) n 1 +F(k) n 2 + +F(k) n k.

52 Riordan Arrays and Generalized k-bonacci numbers Proposition Let A m (z) be the generating function for the rows sums of the Riordan array H m. Then A m (z) = 1 1 (a 1 +a 2 )z a 3 z 2 a m+1 z m. Let F (k) n (x) be the n-th row sum of H k (generalized k-bonacci polynomials): A k (z) = i=0 F (k) i (x)z i = 1 1 p 1 (x)z p 2 (x)z 2 p k (x)z k, where p 1 (x) = a 1 +a 2, and p j (x) = a j+1,j = 2,...,k.

53 Riordan Arrays and Generalized k-bonacci numbers Proposition Let A m (z) be the generating function for the rows sums of the Riordan array H m. Then A m (z) = 1 1 (a 1 +a 2 )z a 3 z 2 a m+1 z m. Let F (k) n (x) be the n-th row sum of H k (generalized k-bonacci polynomials): A k (z) = i=0 F (k) i (x)z i = 1 1 p 1 (x)z p 2 (x)z 2 p k (x)z k, where p 1 (x) = a 1 +a 2, and p j (x) = a j+1,j = 2,...,k.

54 Riordan Arrays and Generalized k-bonacci numbers Theorem The polynomials F (k) n (x) satisfy the following recurrence for n k F (k) n (x) = p 1 (x)f (k) n 1 (x)+p 2(x)F (k) n 2 (x)+ +p k(x)f (k) n k (x), where F (k) 0 (x) = 1,F (k) i (x) = 0 for i = 1, 2,..., (k 1).

55 Combinatorial Interpretation and Generalized k-bonacci numbers For any polynomial F (m) n (x), we have F (m) n (x) = n k=0 ω (m) n,k, n 0, where ω (m) n,k (x) := h(m) n,k = D m(n k,k) is the sum of weights of A m+1 -weighted paths from (0,0) to (k,n k) with step set S m = {H = (0,0),V = (0,1),D 1 = (1,1),D 2 = (1,2),..., D m 1 = (1,m 1)}, such that a 1 (x)+a 2 (x) = p 1 (x), and a i (x) = p i 1 (x) for i = 1,...,m+1.

56 Riordan Arrays and Generalized k-bonacci numbers Example The tetrabonacci numbers: l 0 = 1, l 1 = 1, l 2 = 2, l 3 = 4 l n+4 = l n+3 +l n+2 +l n+1 +l n, for n 0. The tetrabonacci polynomials R n (x) are defined by the recurrence relation R 0 (x) =1, R 1 (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+1 (x)+r n (x), for n 0. If a 1 (x) = x 3,a 2 (x) = 0,a 3 (x) = x 2,a 4 (x) = x and a 5 (x) = 1, we obtain the tetrabonacci polynomials from the row sums of the Riordan array H 4 (x 3,0,x 2,x,1).

57 Riordan Arrays and Generalized k-bonacci numbers Example The tetrabonacci numbers: l 0 = 1, l 1 = 1, l 2 = 2, l 3 = 4 l n+4 = l n+3 +l n+2 +l n+1 +l n, for n 0. The tetrabonacci polynomials R n (x) are defined by the recurrence relation R 0 (x) =1, R 1 (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+1 (x)+r n (x), for n 0. If a 1 (x) = x 3,a 2 (x) = 0,a 3 (x) = x 2,a 4 (x) = x and a 5 (x) = 1, we obtain the tetrabonacci polynomials from the row sums of the Riordan array H 4 (x 3,0,x 2,x,1).

58 Riordan Arrays and Generalized k-bonacci numbers Example R 3 = x 9 +2x 5 +x x x 3 x 3 x 2 x 2 } {{ } 2x 5 x 3 x 3 x 3 x 9 Figure: (x 3,x 2,0,x,1)-weighted paths and tretranacci polynomials. J. L. R., V. Sirvent. A generalization of the k-bonacci sequence from Riordan arrays. The Electronic Journal of Combinatorics, 22(1), P1.38, 2015.

59 Outline 1. Lattice Path - Pascal Matrix - Delannoy Matrix 2. (a, b, c, d) weighted paths 3. The General Case 4. Schröder Matrix 5. Some additional comments

60 Delannoy Matrix Inverse Matrix Combinatorial Interpretation - Lattice Path

61 Schröder matrix as inverse of Delannoy matrix H 2 = Delannoy matrix. H 1 2 = Inverse of Delannoy matrix..

62 Schröder matrix as inverse of Delannoy matrix H 2 = Delannoy matrix R = Schröder Matrix. Combinatorial Interpretation?

63 Schröder matrix as inverse of Delannoy matrix b b 2 2b+c H 2 = b 3 3b 2 +2cb 3b+2c 1 0 b 4 4b 3 +3cb 2 6b 2 +6cb+c 2 4b+3c 1. Generalized Delannoy matrix b R = b 2 +cb 2b +c b 3 +3cb 2 +2c 2 b 3b 2 +5cb +2c 2 3b+2c 1 0. Generalized Schröder Matrix. Combinatorial Interpretation?

64 Schröder matrix J 3 (n,k):= denote the set of weighted paths from the point (0,0) to the point (n k,n), whose step set is R 3 = {V = (0,1),D = (1,1),H 1 = (1,0)}, such that each step is labelled with weights c,d,e 1, respectively; and that its last step is not a horizontal step and that never falling below line y = x. L 3 (n,k):= total sum of the weights of all weight paths in J 3 (n,k). S.-L. Yang, S.-N. Zheng, S.-P. Yuan and T.-X. He. Schröder matrix as inverse of Delannoy matrix. Linear Algebra Appl. 439(11): , 2013.

65 Schröder matrix

66 Schröder matrix Generalized Schröder Matrix S 3 (c,d,e 1 ) = [L 3 (n,k)] n,k N, The unsigned matrix H 1 3 (1,1,1) is equal to S 3(1,1,1).

67 Problem b b 2 2b+c H 3 (a,b,c,d) = b 3 3b 2 +2cb +d 3b +2c 1 0 b 4 4b 3 +3cb 2 +2db 6b 2 +6cb+c 2 +2d 4b+3c 1... Generalized Lattice Matrix (4 steps) b S 3 = b 2 +cb 2b+c b 3 +3cb 2 +2c 2 b db 3b 2 +5cb+2c 2 d 3b+2c Generalized Inverse Lattice Matrix (4 Steps). Combinatorial Interpretation?

68 Generalized Schröder matrix J 4 (n,k):= the family of weighted lattice paths from the point (0,0) to the point (n k,n), whose step set is R 4 = {V = (0,1),D = (1,1),H 1 = (1,0),H 2 = (2,0)}, such that each step is labelled with weights c,d,e 1,e 2, respectively, the path never falls below the line y = x, and its last step is not a horizontal step. L 4 (n,k):= total sum of the weights of all weight paths in J 4 (n,k). S 4 (c,d,e 1,e 2 ) := (L 4 (n,k)) n,k N.

69 Schröder matrix Example (0,0) (2,3) y c (2,3) y = x e 2 c c x

70 Schröder matrix L 4 (3,1) = 2e 2 1c 3 +e 2 c 3 +5de 1 c 2 +3d 2 c. y (2,3) y (2,3) y (2,3) y (2,3) c c d c e1 d c c d e1 d d d c x x x x y (2,3) y (2,3) y (2,3) y (2,3) c d c d e1 e1 c c d d e1 e1 c c c c x x x x y (2,3) y (2,3) y (2,3) y = x c c d e1 e1 e2 e1 c c c c c c x x x

71 Schröder matrix L 4 (n+1,k+1) = a 0 L 4 (n,k)+a 1 L 4 (n,k+1)+a 2 L 4 (n+1,k+2) + +a n L 4 (n,n), n,k 0; where a 0 = c, a 1 = ce 1 +d, a 2 = e 1 (ce 1 +d)+e 2 c,... y (0,n) e 2 d e 1 (n k,n+1) c (n k,n) x

72 Schröder matrix In general, for every integer n 2. a n = e 1 a n 1 +e 2 a n 2 A (4) (x) = a i x i = i=0 c +dx 1 e 1 x e 2 x 2. If the point is on line y = x, we have L 4 (n +1,0) = b 0 L 4 (n,0)+b 1 L 4 (n,1)+b 2 L 4 (n +1,2) + +b n L 4 (n,n), n,k 0. Note that b 0 = d, b 1 = de 1, b 2 = d(e 2 1 +e 2),... In general, b n = e 1 b n 1 +e 2 b n 2 for every integer n 2. Therefore, Z (4) (x) = b i x i = i=0 d 1 e 1 x e 2 x 2.

73 Schröder matrix In general, for every integer n 2. a n = e 1 a n 1 +e 2 a n 2 A (4) (x) = a i x i = i=0 c +dx 1 e 1 x e 2 x 2. If the point is on line y = x, we have L 4 (n +1,0) = b 0 L 4 (n,0)+b 1 L 4 (n,1)+b 2 L 4 (n +1,2) + +b n L 4 (n,n), n,k 0. Note that b 0 = d, b 1 = de 1, b 2 = d(e 2 1 +e 2),... In general, b n = e 1 b n 1 +e 2 b n 2 for every integer n 2. Therefore, Z (4) (x) = b i x i = i=0 d 1 e 1 x e 2 x 2.

74 Schröder matrix In general, for every integer n 2. a n = e 1 a n 1 +e 2 a n 2 A (4) (x) = a i x i = i=0 c +dx 1 e 1 x e 2 x 2. If the point is on line y = x, we have L 4 (n +1,0) = b 0 L 4 (n,0)+b 1 L 4 (n,1)+b 2 L 4 (n +1,2) + +b n L 4 (n,n), n,k 0. Note that b 0 = d, b 1 = de 1, b 2 = d(e 2 1 +e 2),... In general, b n = e 1 b n 1 +e 2 b n 2 for every integer n 2. Therefore, Z (4) (x) = b i x i = i=0 d 1 e 1 x e 2 x 2.

75 Schröder matrix A-sequence and Z-sequence Theorem If c 0 then the matrix S 4 (c,d,e 1,e 2 ) is a Riordan array given by S 4 (c,d,e 1,e 2 ) = ( c c +dx,x1 e 1x e 2 x 2 ) 1. c +dx Theorem The inverse of the Riordan matrix H 4 and the Riordan matrix S 4 are related by H 4 (1,b,c 1,c 2 ) 1 = S 4 (1, b, c 1, c 2 ).

76 General Case J m+2 (n,k):= the set of weighted paths from the point (0,0) to the point (n k,n), with step set R m+2 = {V = (0,1),D = (1,1),H 1 = (1,0),H 2 = (2,0),...,H m = (m,0)}, where each step is labelled with weights c,d,e 1,...,e m, respectively, and the path never falls below y = x and its last step is not horizontal. L m+2 (n,k):= total sum of the weights of all weight paths in J m+2 (n,k).

77 Schröder matrix Lemma The numbers L m+2 (n,k) satisfy the following recurrence relation L m+2 (n+1,k +1) = a 0 L m+2 (n,k)+a 1 L m+2 (n,k +1) + +a n L m+2 (n,n), n,k 0. where a n = e 1 a n 1 +e 2 a n 2 + +e m a n m, n m, i 1 a 0 = c, a 1 = ce 1 +d, a i = e i j a j, 2 i m 1. L m+2 (n +1,0) = b 0 L m+2 (n,0)+b 1 L m+2 (n,1)+ +b n L m+2 (n,n), where b n = e 1 b n 1 +e 2 b n 2 + +e m b n m, for every n m and j=0 i 1 b 0 = d, b i = e i j b j, 1 i m 1. j=0

78 Schröder matrix Lemma The generating functions A (m+2) (x) and Z (m+2) (x) of the sequences (a i ) i N and (b i ) i N are and A (m+2) (x) := Z (m+2) (x) := a i x i = i=0 b i x i = i=0 c +dx 1 e 1 x e 2 x 2 e m x m, d 1 e 1 x e 2 x 2 e m x m.

79 Schröder matrix Theorem The infinite matrix S m+2 is a Riordan matrix where A (m+2) (x) and Z (m+2) (x) are the generating functions of the A-sequence and Z-sequence, respectively. Moreover, if c 0 then the matrix S m+2 (c,d,e 1,e 2,...,e m ) is a Riordan array given by S m+2 (c,d,e 1,e 2,...,e m ) ( c = c +dx,x1 e 1x e 2 x 2 e m x m ) 1. c +dx

80 Schröder matrix Theorem The inverse of the Riordan matrix H m+2 and the Riordan matrix S m+2 are related by H m+2 (1,b,c 1,c 2,...,c m ) 1 = S m+2 (1, b, c 1, c 2,..., c m ). J. L. R., V. Sirvent. Generalized Schröder matrix and its combinatorial interpretation. Linear and Multilinear Algebra, To Appear, 2017.

81 Outline 1. Lattice Path - Pascal Matrix - Delannoy Matrix 2. (a, b, c, d) weighted paths 3. The General Case 4. Schröder Matrix 5. Some additional comments

82 Some additional comments Given a positive real sequence A = {a n } n 0, we say that A is unimodal if there exists an integer 0 j such that a 0 a j 1 a j a j+1. The sequence A is called log-concave if it satisfies, for all 1 j, a 2 j a j 1 a j+1.

83 Some additional comments Example {( n k)} 0 k n ( 0 0) ( 1 ) ( 1 0 1) ( 2 ) ( 2 ) ( ) ( 3 ) ( 3 ) ( ) ( 3 3). ( n ) ( n ) ( n ) ( n n)

84 Some additional comments Example { (n k ) } k 0 k n/2 ( 0 0) ( 1 ) ( 1 0 1) ( 2 ) ( 2 ) ( ) ( 3 ) ( 3 ) ( ) ( 3 3). ( n ) ( n ) ( n ) ( n n) S. Tanny, M. Zuker. On a unimodality sequence of binomial coefficients. Discrete Math. 9(1974),

85 Some additional comments Example Belbachir and Szalay (2008): any sequence of binomial coefficients lying along a ray in Pascal s triangle is log-concave, thus unimodal { (u0 +αk v 0 +βk) } k ( 0 0) ( 1 ) ( 1 0 1) ( 2 ) ( 2 ) ( ) ( 3 ) ( 3 ) ( 3 ) ( ). ( n ) ( n ) ( n ) ( n n) H. Belbachir, L. Szalay. Unimodal rays in the ordinary and generalized Pascal triangles. J. Integer Seq. 11(2008), Article X.-T. Su, Y. Wang. On unimodality problems in Pascal s triangle. Electron. J. Combin. 15(2008), # R113. Y. Yu. Confirming two conjectures of Su and Wang on binomial coefficients. Adv. in Appl. Math. 43(2009),

86 Some additional comments Example Stirling Matrix { n m} }n,k 0 Log-concave: { { n 0 ai k 0 +bi} }i B.-X. Zhu. Some positivities in certain triangular arrays. Proceedings of the American Mathematical Society. 142(9)(2014) Bonus Example! Example Adiprasito, Huh and Katz solved the Rota conjecture on the log-concavity of the characteristic polynomial of matroids.

87 Some additional comments Example Stirling Matrix { n m} }n,k 0 Log-concave: { { n 0 ai k 0 +bi} }i B.-X. Zhu. Some positivities in certain triangular arrays. Proceedings of the American Mathematical Society. 142(9)(2014) Bonus Example! Example Adiprasito, Huh and Katz solved the Rota conjecture on the log-concavity of the characteristic polynomial of matroids.

88 Some additional comments Delannoy Matrix? H 2 = Conjecture Let n,p,r be positive integers and q Z, with n p, 0 p < r and q+r > 0 then the sequence {d(n qk,p +rk)} 0 k satisfies: If q > 0 then the sequence is log-concave and therefore unimodal. If q < 0 and q < r then the sequence is log-concave and therefore unimodal. If q < 0 and q = r then the sequence is log-concave and increasing. If q < 0 and r < q then the sequence is increasing and asymptotically log-convex..

89 Some additional comments Theorem Let q Z, r N, p Z + with 0 p < r and q +r > 0. Any sequence {d(n qk,p +rk)} 0 k (n p)/(q+r) lying along all finite rays in the generalized Delannoy matrix is log-concave, thus unimodal. (a,b,c,d)-matrix? General case?

90 Some additional comments Theorem Let q Z, r N, p Z + with 0 p < r and q +r > 0. Any sequence {d(n qk,p +rk)} 0 k (n p)/(q+r) lying along all finite rays in the generalized Delannoy matrix is log-concave, thus unimodal. (a,b,c,d)-matrix? General case?

91 Some additional comments Delannoy Matrix H 2 = Central Delannoy Numbers: {D n } n = {1,3,13,63,321,1683,8989,48639,...} Liu and Wang, 2007: proved that the central Delannoy sequence is log-convex. Given a positive real sequence A = {a n } n 0, we say that A is log-convex if it satisfies, for all 1 j, a 2 j a j 1 a j+1.

92 Some additional comments Inverse of Delannoy matrix (Schröder matrix) H 1 = Central sequence: {F n } n = {1,3,22,194,1838,18082,182054, ,...} F n counts the number of lattice paths from the point (0,0) to the point (n,2n), whose step set is {V = (0,1),D = (1,1),H = (1,0)} such that its last step is not a horizontal step and that never falling below line y = x.

93 Some additional comments Inverse of Delannoy matrix (Schröder matrix) H 1 = Central sequence: {F n } n = {1,3,22,194,1838,18082,182054, ,...} F n counts the number of lattice paths from the point (0,0) to the point (n,2n), whose step set is {V = (0,1),D = (1,1),H = (1,0)} such that its last step is not a horizontal step and that never falling below line y = x.

94 Some additional comments F 2 = 22 (2,4) (2,4) (2,4) (2,4) (2,4) (2,4) (2,4) (2,4) (2,4) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (2,4) (2,4) (2,4) (2,4) (2,4) (2,4) (2,4) (2,4) (2,4) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (2,4) (2,4) (2,4) (2,4) (0,0) (0,0) (0,0) (0,0) Figure: Combinatorial interpretation for F n.

95 Some additional comments Theorem The sequence F n is log-convex. Proof. Computational Method... F n = n j=0 ( n 2n j ( )( ) 2n 1 3n j 1 j 2n 1 + n+1 ( 2n 1 2n j j )( 3n j 2 2n 1 )), n 0.

96 Some additional comments Theorem The sequence F n is log-convex. Proof. Computational Method... F n = n j=0 ( n 2n j ( )( ) 2n 1 3n j 1 j 2n 1 + n+1 ( 2n 1 2n j j )( 3n j 2 2n 1 )), n 0.

97 Some additional comments From the Kauers s multivariate guessing library we obtain that p n F n = q n F n 1 +r n F n 2, n 2 with the initial values F 0 = 1 and F 1 = 3, and p n = 20n 4 72n 3 +85n 2 39n +6, q n = 220n 4 902n n 2 711n +129, r n = 20n 4 92n 3 +89n 2 +36n 45.

98 Some additional comments Let {z n } n be a sequence such that a(n)z n+1 = b(n)z n +c(n)z n 1, (1) where a(n),b(n) and c(n) are positive for n 1. Lemma (Liu and Wang, 2007) Let {z n } n be defined by (1) and λ n = b(n)+ b 2 (n)+4a(n)c(n). 2a(n) Suppose that z 0,z 1,z 2,z 3 is log-convex and that the inequality a(n)λ n 1 λ n+1 b(n)λ n 1 c(n) 0 is true for n 2. Then the sequence {z n } n is log-convex. Wolfram Mathematica

99 (1, 1, 1, 1)-weighted paths d a a b b a d c c Central sequence: H 3 (1,1,1,1) = { d (3) 2n,n } n N = {1,3,15,81,459,2673,...}

100 (1, 1, 1, 1)-weighted paths Theorem (Barry, 2013) Let (d(t),h(t)) = (d(t),tf(t)) be an element of the Riordan group of matrices R (with f(0) 0). Let d n,k denote the (n,k)-th element of this matrix, and let v(t) denote the power series (with v(0) = 0) ( ) t v(t) =. f(t) Then the generating function of the central term sequence d 2n,n is given by d(v(t)) d f(v(t)) dt v(t). The { generating } function of the central sequence d (3) 2n,n = {1,3,15,81,459,2673,...} is n N n=0 d (3) 2n,n zn = D 3 (n,n)z n = n= z 3z 2.

101 (1, 1, 1, 1)-weighted paths Theorem (Barry, 2013) Let (d(t),h(t)) = (d(t),tf(t)) be an element of the Riordan group of matrices R (with f(0) 0). Let d n,k denote the (n,k)-th element of this matrix, and let v(t) denote the power series (with v(0) = 0) ( ) t v(t) =. f(t) Then the generating function of the central term sequence d 2n,n is given by d(v(t)) d f(v(t)) dt v(t). The { generating } function of the central sequence d (3) 2n,n = {1,3,15,81,459,2673,...} is n N n=0 d (3) 2n,n zn = D 3 (n,n)z n = n= z 3z 2.

102 (1, 1, 1, 1)-weighted paths Grand Motzkin path: y x

103 (1, 1, 1, 1)-weighted paths := the number of grand Motzkin path such that each horizontal step is colored with one of 3 colors and each up diagonal step is colored with one of 3 specific colors. g (3,3) n Corollary The number of 3 H 3 U -Grand Motzkin path is equal to the number of (1,1,1,1)-weighted paths from the point (0,0) to the point (n,n), i.e., g (3,3) n = D 3 (n,n).

104 (1, 1, 1, 1)-weighted paths Dziemiańczuk, 2013: 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 = {H = (1,0),V = (0,1), D 1 = (1,1),D 2 = ( 1,1)} Therefore S(0,n)z n = n=0 S(0,n) = g (3,3) n 1 1 6z 3z 2. = D 3 (n,n).

105 (1, 1, 1, 1)-weighted paths

106 (1, 1, 1, 1)-weighted paths

107 (1, 1, 1, 1)-weighted paths

108 Some additional comments Is there any Bijection? M. Dziemiańczuk. Counting lattice paths with four types of steps. Graphs Combin. (2013). S = {H = (1,0),V = (0,1), D 1 = (1,1),D 2 = ( 1,1)} R(n,k) = [z k] 2 n+2 (1+z) n 1 6z 3z 2 (1 z + 1 6z 3z 2 ) n+2 (1 z 1 6z 3z 2 ) n+2, where R(n,k) is the family of lattice paths from the origin to (n, k) whose points lie entirely in the integer rectangle of lattice points {(i,j) : 0 i n,0 j k}.

109 Some additional comments Carlitz, L. Some q-analogues of certain combinatorial numbers, SIAM J. Math. Anal. 4(3), (1973) A(n,r) = A(n 1,r 1)+q n A(n,r 1)+q r A(n 1,r) C. Wagner. The Carlitz lattice path polynomials, Discrete Mathematics 222 (2000)

110 Some additional comments The valleys and peaks of a Dyck path P are the local minima and local maxima, respectively. We say that a Dyck path P is non-decreasing if the y-coordinates of valleys of the path P form a non-decreasing sequence.

111 Some additional comments Non-decreasing sequence Dyck path. The number of non-decreasing Dyck paths of length 2n is the Fibonacci number F 2n 1. E. Barcucci, A. Del Lungo, S. Fezzi, and R. Pinzani, Nondecreasing Dyck paths and q-fibonacci numbers. Discrete Math. 170(1 3)(1997),

112 Some additional comments Deutsch and Prodinger, 2003, give a bijection between the non-decreasing Dyck path of length 2n and the direct column-convex polyomino (dccp) of area n (a) (b)

113 Some additional comments The valleys of a Motzkin path P are the local minima. We say that a Motzkin path P is non-decreasing if the y-coordinates of all valleys (including also the left valleys and the right valleys) of the path P form a non-decreasing sequence.

114 Some additional comments l(p) := the length of P, h(p) := the number of horizontal steps of P r(p) := the number of rises (up-diagonal steps) of P p(p) := the number of peaks of P. F(x,y,z,q) := x l(p) y h(p) z r(p) q p(p). P NM Theorem The generating function F(x,y,z,q) is given by the following equation F(x,y,z,q) = x(xy 1)(xzq +y)(x 2 z 1) 1 2xy +x 2 y 2 2x 2 z qx 2 z +2x 3 yz +qx 3 yz x 4 y 2 z +x 4 z 2 x 5 yz 2.

115 Some additional comments l(p) := the length of P, h(p) := the number of horizontal steps of P r(p) := the number of rises (up-diagonal steps) of P p(p) := the number of peaks of P. F(x,y,z,q) := x l(p) y h(p) z r(p) q p(p). P NM Theorem The generating function F(x,y,z,q) is given by the following equation F(x,y,z,q) = x(xy 1)(xzq +y)(x 2 z 1) 1 2xy +x 2 y 2 2x 2 z qx 2 z +2x 3 yz +qx 3 yz x 4 y 2 z +x 4 z 2 x 5 yz 2.

116 Some additional comments l(p) := the length of P, h(p) := the number of horizontal steps of P r(p) := the number of rises (up-diagonal steps) of P p(p) := the number of peaks of P. F(x,y,z,q) := x l(p) y h(p) z r(p) q p(p). P NM Theorem The generating function F(x,y,z,q) is given by the following equation F(x,y,z,q) = x(xy 1)(xzq +y)(x 2 z 1) 1 2xy +x 2 y 2 2x 2 z qx 2 z +2x 3 yz +qx 3 yz x 4 y 2 z +x 4 z 2 x 5 yz 2.

117 Some additional comments Proof. Symbolic Method: Figure: Factorizations of any non-decreasing Motzkin path.

118 Some additional comments If y = 1 = z = q we obtain the generating function for the non-decreasing Motzkin paths respect to the length: x(x 2 1) 2 F(x,1,1,1) = 1 2x 2x 2 +3x 3 x 5 = x +2x 2 +4x 3 +9x 4 +21x 5 +49x x 7 + If y = 0 and q = 1 = z then F(x,0,1,1) = x2 (1 x 2 ) 1 3x 3 +x 4 = F 2n 1 x 2n. n=1

119 Some additional comments If y = 1 = z = q we obtain the generating function for the non-decreasing Motzkin paths respect to the length: x(x 2 1) 2 F(x,1,1,1) = 1 2x 2x 2 +3x 3 x 5 = x +2x 2 +4x 3 +9x 4 +21x 5 +49x x 7 + If y = 0 and q = 1 = z then F(x,0,1,1) = x2 (1 x 2 ) 1 3x 3 +x 4 = F 2n 1 x 2n. n=1

120 Some additional comments A path which is a prefix of a non-decreasing Motzkin path is called non-decreasing Motzkin prefix. The prefixes of the classical Dyck paths are called Ballot paths m n,k := number of non-decreasing Motzkin prefixes of length n and height k. M = [m n,k ] n,k M =

121 Some additional comments A path which is a prefix of a non-decreasing Motzkin path is called non-decreasing Motzkin prefix. The prefixes of the classical Dyck paths are called Ballot paths m n,k := number of non-decreasing Motzkin prefixes of length n and height k. M = [m n,k ] n,k M =

122 Some additional comments A path which is a prefix of a non-decreasing Motzkin path is called non-decreasing Motzkin prefix. The prefixes of the classical Dyck paths are called Ballot paths m n,k := number of non-decreasing Motzkin prefixes of length n and height k. M = [m n,k ] n,k M =

123 Some additional comments A path which is a prefix of a non-decreasing Motzkin path is called non-decreasing Motzkin prefix. The prefixes of the classical Dyck paths are called Ballot paths m n,k := number of non-decreasing Motzkin prefixes of length n and height k. M = [m n,k ] n,k M =

124 Some additional comments A path which is a prefix of a non-decreasing Motzkin path is called non-decreasing Motzkin prefix. The prefixes of the classical Dyck paths are called Ballot paths m n,k := number of non-decreasing Motzkin prefixes of length n and height k. M = [m n,k ] n,k M =

125 Some additional comments A path which is a prefix of a non-decreasing Motzkin path is called non-decreasing Motzkin prefix. The prefixes of the classical Dyck paths are called Ballot paths m n,k := number of non-decreasing Motzkin prefixes of length n and height k. M = [m n,k ] n,k M =

126 Some additional comments Example m 4,2 = 9 paths. Figure: Non-decreasing Motzkin prefixes of length 4 and height 2.

127 Some additional comments Theorem Let T k (x) be the generating function defined be Then T k (x) = T k (x) := d n,k x n. n=0 1 x 2x 2 +x 3 ( 1 2x 2x 2 +3x 3 x 5 x (1 x) 2 ) k (1+x) 1 2x x 2 +2x 3 x 4.

128 Some additional comments Theorem The matrix M is a Riordan array given by ( 1 x 2x 2 +x 3 (1 x) 2 ) (1+x) M = 1 2x 2x 2 +3x 3 x 5,x 1 2x x 2 +2x 3 x 4. Additional Properties of the non-decreasing Motzkin paths...

129 Some additional comments Theorem The matrix M is a Riordan array given by ( 1 x 2x 2 +x 3 (1 x) 2 ) (1+x) M = 1 2x 2x 2 +3x 3 x 5,x 1 2x x 2 +2x 3 x 4. Additional Properties of the non-decreasing Motzkin paths...

130 Thank you!!! Gracias!! Figure: Churros and Chocolate; Chocolatería San Ginés, 17-Jul-2017!