This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A319203 #19 Dec 10 2023 23:37:42
%S A319203 1,0,1,-1,0,1,-1,-2,0,1,2,-2,-3,0,1,5,5,-3,-4,0,1,-2,12,9,-4,-5,0,1,
%T A319203 -21,-7,21,14,-5,-6,0,1,-14,-56,-16,32,20,-6,-7,0,1,72,-30,-108,-30,
%U A319203 45,27,-7,-8,0,1,138,210,-45,-180,-50,60,35,-8,-9,0,1
%N A319203 Triangular Riordan matrix T = R^(-1) for triangular Riordan matrix R = (1/(1 - x^2 - x^3), x/(1 - x^2 - x^3)) = A104578.
%C A319203 This is the lower triangular Riordan matrix (f(t), t*f(t)), with f(t) = F^{[-1]}(t)/t, where F(x) = x/(1 - x^2 - x^3). The expansion of f(t) is given in A319201, the sequence of column k = 0.
%C A319203 This gives the inverse Matrix (with upper diagonals filled with 0's) of the Riordan matrix from A104578 for any finite dimension.
%C A319203 The row sums give A321204, and the alternating row sums give A321205.
%C A319203 The 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 from 1/F(x) - 1/x. See the link W. Lang link for A- and Z- sequences in A006232 with references.
%C A319203 For the Boas-Buck recurrence of Riordan triangles see the Aug 10 2017 remark in A046521, also for the reference. For this Bell-type triangle the 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 is given in A319204.
%F A319203 Recurrence from the Z- and A-sequence: T(n, k) = 0 if n < k; T(0, 0) = 1;
%F A319203 T(n, 0) = -(T(n-1, 1) + T(n-1, 2)), for n >= 1; and T(n, m) = T(n-1, k-1) - T(n-1, k+1) - T(n-1, k+2), for n>=1 and k >= 1.
%F A319203 Boas-Buck recurrence with B(n) = A319204(n): 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.
%F A319203 G.f. of row polynomials R(n,x) = Sum_{k=0..n} T(n, k)*x^k is G(x,z) = f(z)/(1-x*z*f(z)) with the expansion of f given in A319201.
%F A319203 G.f. of column sequences Gcol(k, x) = x^k*f(x)^{k+1}, for k >= 0.
%e A319203 The triangle T(n, k) begins:
%e A319203 n\k 0 1 2 3 4 5 6 7 8 9 10 ...
%e A319203 -------------------------------------------------
%e A319203 0: 1
%e A319203 1: 0 1
%e A319203 2: -1 0 1
%e A319203 3: -1 -2 0 1
%e A319203 4: 2 -2 -3 0 1
%e A319203 5; 5 5 -3 -4 0 1
%e A319203 6: -2 12 9 -4 -5 0 1
%e A319203 7: -21 -7 21 14 -5 -6 0 1
%e A319203 8: -14 -56 -16 32 20 -6 -7 0 1
%e A319203 9: 72 -30 -108 -30 45 27 -7 -8 0 1
%e A319203 10: 138 210 -45 -180 -50 60 35 -8 -9 0 1
%e A319203 ...
%e A319203 Recurrence from A- and Z-sequence: 5 = T(5, 0) = -(-2 + (-3)); 9 = T(6, 2) = 5 - (- 4 + 0).
%e A319203 Recurrence of Boas-Buck type, with B = [0,-2,-3, 6, ...] = A319204: 9 = T(6, 2) = ((2+1)/(6-2))*(6*1 + (-3)*0 + (-2)*(-3) + 0*(-3)) = (3/4)*12 = 9.
%t A319203 (* The function RiordanArray is defined in A256893. *)
%t A319203 nmax = 10;
%t A319203 R = RiordanArray[1/(1 - #^2 - #^3)&, #/(1 - #^2 - #^3)&, nmax+1];
%t A319203 M = Inverse[PadRight[#, nmax+1]& /@ R];
%t A319203 T[n_, k_] := M[[n+1, k+1]];
%t A319203 Table[T[n, k], {n, 0, nmax}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Jul 19 2019 *)
%Y A319203 Cf. A104578, A319201, A319204, A321204, A321205.
%K A319203 sign,tabl
%O A319203 0,8
%A A319203 _Wolfdieter Lang_, Oct 29 2018