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%I #48 Mar 18 2023 08:49:14
%S 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,
%T 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,
%U 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
%N A Padovan convolution triangle.
%C A Padovan convolution triangle. See A000931 for the Padovan sequence.
%C Row sums are tribonacci numbers A000073(n+2). Antidiagonal sums are A008346. The first columns are A000931(n+3), A228577.
%C From _Wolfdieter Lang_, Oct 30 2018: (Start)
%C The alternating row sums give A001057(n+1), for n >= 0.
%C The inverse of this Riordan triangle is given in A319203.
%C 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.
%C For the Cayley-Hamilton formula for 3 X 3 matrices with Det M = +1, sigma(3,2) = +1 and Tr(M) = x see A321196.
%C (End)
%H Tomislav Došlic and Luka Podrug, <a href="https://arxiv.org/abs/2203.11761">Tilings of a Honeycomb Strip and Higher Order Fibonacci Numbers</a>, arXiv:2203.11761 [math.CO], 2022.
%H Milan Janjić, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL21/Janjic/janjic93.html">Words and Linear Recurrences</a>, J. Int. Seq. 21 (2018), #18.1.4.
%F Riordan array (1/(1 - x^2 - x^3), x/(1 - x^2 - x^3)).
%F 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
%F From _Wolfdieter Lang_, Oct 30 2018: (Start)
%F The Riordan property T = (G(x), x*G(x)) with G(x)= 1/(1-x^2-x^3) implies the following.
%F G.f. of row polynomials R(n, x) is G(x,z) = 1/(1- x*z - z^2 - z^3).
%F G.f. of column sequence k: x^k/(1 - x^2 - x^3)^(k+1), k >= 0.
%F Boas-Buck recurrence (see the Aug 10 2017 remark in A046521, also for the reference):
%F 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.
%F 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).
%F T(0, 0) = 1, T(n, k) = 0 for n < k, and
%F T(n, 0) = Sum_{j=0..n-1} Z(j)*T(n-1, j), for n >= 1, and
%F T(n, k) = Sum_{j=0..n-k} A(j)*T(n-1, k-1+j), for n >= m >= 1.
%F (End)
%e From _Wolfdieter Lang_, Oct 30 2018: (Start)
%e The triangle T begins:
%e n\k 0 1 2 3 4 5 6 7 8 9 10 ...
%e --------------------------------------
%e 0: 1
%e 1: 0 1
%e 2: 1 0 1
%e 3: 1 2 0 1
%e 4: 1 2 3 0 1
%e 5: 2 3 3 4 0 1
%e 6: 2 6 6 4 5 0 1
%e 7: 3 7 12 10 5 6 0 1
%e 8: 4 12 16 20 15 6 7 0 1
%e 9: 5 17 30 30 30 21 7 8 0 1
%e 10: 7 24 45 60 50 42 28 8 9 0 1
%e ...
%e 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).
%e Recurrence: T(6, 2) = T(5, 1) + T(4, 2) + T(3, 2) = 3 + 3 + 0 = 6.
%e Z- and A- recurrence with A319202 = {1, 0, 1, 1, -1, -3, 0, ...}:
%e 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.
%e 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.
%e (End)
%t 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 *)
%o (Sage) # uses[riordan_array from A256893]
%o riordan_array( 1/(1 - x^2 - x^3), x/(1 - x^2 - x^3), 8) # _Peter Luschny_, Nov 09 2018
%Y Cf. A000073, A000931, A001057, A001608, A008346, A058265, A228577, A319202, A319203 (inverse), A321196.
%K easy,nonn,tabl
%O 0,8
%A _Paul Barry_, Mar 16 2005
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