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A104578
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A Padovan convolution triangle.
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5
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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
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OFFSET
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0,8
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COMMENTS
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A Padovan convolution triangle. See A000931 for the Padovan sequence.
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)
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LINKS
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FORMULA
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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
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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EXAMPLE
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The triangle T begins:
n\k 0 1 2 3 4 5 6 7 8 9 10 ...
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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)
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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