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%I A321198 #17 Dec 10 2023 23:37:34
%S A321198 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,
%T A321198 21,-21,14,-5,6,0,1,42,-56,40,-32,20,-6,7,0,1,-96,114,-108,66,-45,27,
%U A321198 -7,8,0,1,222,-270,225,-180,100,-60,35,-8,9,0,1
%N 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.
%C A321198 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.
%C A321198 This gives the inverse matrix (with upper diagonals filled with zeros) of the Riordan matrix from A321196 for any finite dimension.
%C A321198 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.
%C A321198 The row sums give A321199. The alternating row sums give A321200.
%C A321198 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.
%C A321198 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.
%F A321198 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.
%F A321198 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.
%F A321198 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).
%F A321198 G.f of column k: Gcol(k, x) = x^k*f(x)^{k+1}.
%e A321198 The triangle begins:
%e A321198 n\k 0 1 2 3 4 5 6 7 8 9 10 ...
%e A321198 -----------------------------------------------------------
%e A321198 0: 1
%e A321198 1: 0 1
%e A321198 2: 1 0 1
%e A321198 3: -1 2 0 1
%e A321198 4: 2 -2 3 0 1
%e A321198 5: -5 5 -3 4 0 1
%e A321198 6: 8 -12 9 -4 5 0 1
%e A321198 7: -21 21 -21 14 -5 6 0 1
%e A321198 8: 42 -56 40 -32 20 -6 7 0 1
%e A321198 9: -96 114 -108 66 -45 27 -7 8 0 1
%e A321198 10: 222 -270 225 -180 100 -60 35 -8 9 0 1
%e A321198 ...
%e A321198 ------------------------------------------------------------
%e A321198 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.
%e A321198 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.
%t A321198 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 *)
%Y A321198 Cf. A001005, A176806, A321196, A321199, A321200, A321201, A321203.
%K A321198 sign,tabl
%O A321198 0,8
%A A321198 _Wolfdieter Lang_, Nov 12 2018