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.

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, -21, -7, 21, 14, -5, -6, 0, 1, -14, -56, -16, 32, 20, -6, -7, 0, 1, 72, -30, -108, -30, 45, 27, -7, -8, 0, 1, 138, 210, -45, -180, -50, 60, 35, -8, -9, 0, 1

Offset: 0

Author

Wolfdieter Lang, Oct 29 2018

Keywords

• sign
• tabl

Comments

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.

This gives the inverse Matrix (with upper diagonals filled with 0's) of the Riordan matrix from A104578 for any finite dimension.

The row sums give A321204, and the alternating row sums give A321205.

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.

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.

Examples

			The triangle T(n, k) 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:      2  -2   -3    0   1
5;      5   5   -3   -4   0  1
6:     -2  12    9   -4  -5  0  1
7:    -21  -7   21   14  -5 -6  0  1
8:    -14 -56  -16   32  20 -6 -7  0  1
9:     72 -30 -108  -30  45 27 -7 -8  0  1
10:   138 210  -45 -180 -50 60 35 -8 -9  0  1
...
Recurrence from A- and Z-sequence: 5 =  T(5, 0) = -(-2 + (-3)); 9 = T(6, 2) = 5 - (- 4 + 0).
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.
		

Crossrefs

Cf. A104578, A319201, A319204, A321204, A321205.

Programs

Mathematica

(* The function RiordanArray is defined in A256893. *)
nmax = 10;
R = RiordanArray[1/(1 - #^2 - #^3)&, #/(1 - #^2 - #^3)&, nmax+1];
M = Inverse[PadRight[#, nmax+1]& /@ R];
T[n_, k_] := M[[n+1, k+1]];
Table[T[n, k], {n, 0, nmax}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 19 2019 *)

Formula

Recurrence from the Z- and A-sequence: T(n, k) = 0 if n < k; T(0, 0) = 1;

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.

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.

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.

G.f. of column sequences Gcol(k, x) = x^k*f(x)^{k+1}, for k >= 0.