A104578 A Padovan convolution triangle.

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

Author

Paul Barry, Mar 16 2005

Keywords

• easy
• nonn
• tabl

Comments

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)

Examples

			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)
		

Links

• 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.

Crossrefs

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

Programs

Mathematica

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 *)

Sage

# 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

Formula

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)