A321196 -id:A321196 - cp's OEIS frontend

A321198 Triangular Riordan matrix T = R^(-1) for triangular Riordan matrix R = (1/(1 + x^2 - x^3), x/(1 + x^2 - x^3)) given in A321196.

1, 0, 1, 1, 0, 1, -1, 2, 0, 1, 2, -2, 3, 0, 1, -5, 5, -3, 4, 0, 1, 8, -12, 9, -4, 5, 0, 1, -21, 21, -21, 14, -5, 6, 0, 1, 42, -56, 40, -32, 20, -6, 7, 0, 1, -96, 114, -108, 66, -45, 27, -7, 8, 0, 1, 222, -270, 225, -180, 100, -60, 35, -8, 9, 0, 1

Offset: 0

Wolfdieter Lang, Nov 12 2018

Riordan triangle T = (f(t), t*f(t)), with f(t) = F^{[-1]}(t)/t, where F^{[-1]}(t) is the compositional inverse of t = F(x) = x/(1 + x^2 - x^3). The expansion of f(t) is given by {(-1)^n*A001005(n)}_{n >= 0}, the sequence of column k = 0.
This gives the inverse matrix (with upper diagonals filled with zeros) of the Riordan matrix from A321196 for any finite dimension.
The inverse of the Riordan matrix (1/(1 + x^2 + x^3), x/(1 + x^2 + x^3)) is obtained from the triangle t(n, k) = (-1)^(n-k)*T(n, k), with vanishing upper diagonals.
The row sums give A321199. The alternating row sums give A321200.
The finite A- and Z-sequences of this inverse Riordan triangle of (F(x)/x, F(x)) are A = [1, 0, +1, -1] generated by 1/(F(x)/x), and Z = [0, +1, -1] generated by 1/F(x) - 1/x. See the W. Lang link for A- and Z- sequences in A006232 with references.
For the Boas-Buck column recurrences of Riordan triangles see the Aug 10 2017 remark in A046521, also for two references. For this Bell-type Riordan triangle the Boas-Buck sequence b is generated by B(t) = (log(f(t)))' = (1/(1/f(t) - t^2*f(t) + 2*t^3*f(t)^2) - 1)/t, and b(n) = (-1)^(n)*A176806(n+1), for n >= 0, because the parity of e_3 in the rows n of A321201 coincides with the one of n. See A321203 for the multinomials with negative signs for odd row numbers.

			The triangle begins:
n\k     0     1     2     3    4    5   6   7  8  9  10 ...
-----------------------------------------------------------
0:      1
1:      0     1
2:      1     0     1
3:     -1     2     0     1
4:      2    -2     3     0    1
5:     -5     5    -3     4    0    1
6:      8   -12     9    -4    5    0   1
7:    -21    21   -21    14   -5    6   0   1
8:     42   -56    40   -32   20   -6   7   0  1
9:    -96   114  -108    66  -45   27  -7   8  0  1
10:   222  -270   225  -180  100  -60  35  -8  9  0   1
...
------------------------------------------------------------
Recurrence (from A- and Z-sequence): Z: T(5, 0) = T(4, 1) - T(4, 2) = -2 - 3 = -5. A: T(5, 2) = T(4, 1) + T(4, 3) - T(4, 4) = - 2 + 0 - 1 = -3.
Recurrence column k = 2 (Boas-Buck type sequence b = (-1)^(n+1)* = {0, 2, -3, 6, ...}): T(5, 2) = (3/3)*(b(2)*T(2, 2) + b(1)*T(3, 2) + b(0)*T(4, 2)) = -3*1 + 2*0 + 0*4 = -3.

Cf. A001005, A176806, A321196, A321199, A321200, A321201, A321203.

T[n_, k_] := T[n - 1, k - 1] + T[n - 1, k + 1] - T[n - 1, k + 2] /; n >= k >= 0;Table[T[n, k], {n, 0, 10}, {k, 0, n}] (* James C. McMahon, Dec 10 2023 *)

Recurrence (from A-and Z-sequences): T(n, k) = 0 for n < k, T(0, 0) = 1. Z: T(n, 0) = T(n-1, 1) - T(n-1, 2), n >= 1; A: T(n, k) = T(n-1, k-1) + T(n-1, k+1) - T(n-1, k+2), n >= k >= 0.
Recurrence for column k (Boas-Buck type): T(n, n) = 1; T(n, k) = ((k+1)/(n-k))*Sum_{j=k..n-1} b(n-1-j)*T(j, k), n >= m+1 >= 1. For b see the Boas-Buck comment above.
G.f of row polynomials R(n, x) := Sum_{k=0..n} T(n, k)*x^k: G(x, z) = f(z)/(1 - x*z*f(z)), with f(z) = F^{[-1]}(z)/z, where F^{[-1]}(z) is the compositional inverse of z = F(y) = y/(1 + y^2 - y^3).
G.f of column k: Gcol(k, x) = x^k*f(x)^{k+1}.

A104578 A Padovan convolution triangle.

Original entry on oeis.org

1, 0, 1, 1, 0, 1, 1, 2, 0, 1, 1, 2, 3, 0, 1, 2, 3, 3, 4, 0, 1, 2, 6, 6, 4, 5, 0, 1, 3, 7, 12, 10, 5, 6, 0, 1, 4, 12, 16, 20, 15, 6, 7, 0, 1, 5, 17, 30, 30, 30, 21, 7, 8, 0, 1, 7, 24, 45, 60, 50, 42, 28, 8, 9, 0, 1, 9, 36, 70, 95, 105, 77, 56, 36, 9, 10, 0, 1, 12, 50, 111, 160, 175, 168, 112, 72

Offset: 0

Paul Barry, Mar 16 2005

A Padovan convolution triangle. See A000931 for the Padovan sequence.
Row sums are tribonacci numbers A000073(n+2). Antidiagonal sums are A008346. The first columns are A000931(n+3), A228577.
From Wolfdieter Lang, Oct 30 2018: (Start)
The alternating row sums give A001057(n+1), for n >= 0.
The inverse of this Riordan triangle is given in A319203.
The row polynomials R(n, x) := Sum_{k=0..n} T(n, k)*x^k, with R(-1, x) = 0, appear in the Cayley-Hamilton formula for nonnegative powers of a 3 X 3 matrix with Det M = sigma(3;3) = x1*x2*x3 = +1, sigma(3; 2) := x1*x2 + x1*x*3 + x2*x^3  = -1 and Tr M = sigma(3; 1) = x1 + x2 = x, where x1, x2, and  x3, are the eigenvalues of M, and sigma the elementary symmetric functions, as M^n = R(n-2, x)*M^2  + (R(n-3, x) + R(n-4, x))*M + R(n-3, x)*1_3, for n >= 3, where M^0 = 1_3 is the 3 X 3 unit matrix.
For the Cayley-Hamilton formula for 3 X 3 matrices with Det M = +1, sigma(3,2) = +1 and Tr(M) = x see A321196.
(End)

			From _Wolfdieter Lang_, Oct 30 2018: (Start)
The triangle T begins:
    n\k   0  1  2  3  4  5  6  7  8  9 10 ...
    --------------------------------------
    0:    1
    1:    0  1
    2:    1  0  1
    3:    1  2  0  1
    4:    1  2  3  0  1
    5:    2  3  3  4  0  1
    6:    2  6  6  4  5  0  1
    7:    3  7 12 10  5  6  0  1
    8:    4 12 16 20 15  6  7  0  1
    9:    5 17 30 30 30 21  7  8  0  1
   10:    7 24 45 60 50 42 28  8  9  0  1
   ...
Cayley-Hamilton formula for the tribonacci Q-matrix TQ(x) =[[x,1,1], [1,0,0], [0,1,0]] with Det(TQ) = +1, sigma(3, 2) = -1, and Tr(TQ) = x. For n = 3: TQ(x)^3 = R(1, x)*TQ(x)^2  + (R(0 x) + R(-1, x))*TQ(x) + R(0, x)*1_3 = x*TQ(x)^2 + TQ(x) + 1_3. For x = 1 see also A058265 (powers of the tribonacci constant).
Recurrence: T(6, 2) = T(5, 1) + T(4, 2) + T(3, 2) = 3 + 3 + 0 = 6.
Z- and A- recurrence with A319202 = {1, 0, 1, 1, -1, -3, 0, ...}:
  T(5, 0) = 0*1 + 1*2 + 1*3 + (-1)*0 + (-3)*1 = 2; T(5,2) = 1*2 + 0*3 + 1*0 + 1*1 = 3.
Boas-Buck type recurrence with b = {0, 2, 3, ...}: T(5, 2) = ((1+2)/(5-2)) * (3*1 + 2*0 + 0*3) = 1*3 = 3.
(End)

Tomislav Došlic and Luka Podrug, Tilings of a Honeycomb Strip and Higher Order Fibonacci Numbers, arXiv:2203.11761 [math.CO], 2022.
Milan Janjić, Words and Linear Recurrences, J. Int. Seq. 21 (2018), #18.1.4.

Cf. A000073, A000931, A001057, A001608, A008346, A058265, A228577, A319202, A319203 (inverse), A321196.

T[n_, k_] /; 0 <= k <= n := T[n, k] = T[n-1, k-1] + T[n-2, k] + T[n-3, k]; T[0, 0] = 1; T[, ] = 0; Table[T[n, k], {n, 0, 12}, {k, 0, n}] (* Jean-François Alcover, Jun 11 2019 *)

# uses[riordan_array from A256893]
riordan_array( 1/(1 - x^2 - x^3), x/(1 - x^2 - x^3), 8) # Peter Luschny, Nov 09 2018

Riordan array (1/(1 - x^2 - x^3), x/(1 - x^2 - x^3)).
T(n,k) = T(n-1,k-1) + T(n-2,k) + T(n-3,k), T(0,0)=1, T(n,k)=0 if k > n or if k < n. - Philippe Deléham, Jan 08 2014
From Wolfdieter Lang, Oct 30 2018: (Start)
The Riordan property  T = (G(x), x*G(x)) with G(x)= 1/(1-x^2-x^3) implies the following.
G.f. of row polynomials R(n, x) is G(x,z) = 1/(1- x*z - z^2 - z^3).
G.f. of column sequence k:  x^k/(1 - x^2 - x^3)^(k+1), k >= 0.
Boas-Buck recurrence (see the Aug 10 2017 remark in A046521, also for the reference):
T(n, k) = ((k+1)/(n-k))*Sum_{j=k..n-1} b(n-1-j)*T(j, k), for n >= 1, k = 0,1, ..., n-1, and input T(n,n) = 1, for n >= 0. Here b(n) = [x^n]*(d/dx)log(G(x)) = A001608(n+1), for n >= 0.
Recurrences from the A- and Z- sequences (see the W. Lang link under A006232 with references), which are A(n) = A319202(n) and Z(n) = A(n+1).
  T(0, 0) = 1, T(n, k) = 0 for n < k, and
  T(n, 0) = Sum_{j=0..n-1} Z(j)*T(n-1, j), for n >= 1, and
  T(n, k) = Sum_{j=0..n-k} A(j)*T(n-1, k-1+j), for n >= m >= 1.
(End)

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A321198 Triangular Riordan matrix T = R^(-1) for triangular Riordan matrix R = (1/(1 + x^2 - x^3), x/(1 + x^2 - x^3)) given in A321196.

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A321197 a(n) gives the A-sequence for the Riordan matrix (1/(1 + x^2 - x^3), x/(1 + x^2 - x^3)) from A321196.

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A104578 A Padovan convolution triangle.

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