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

0,8

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.

Links

Table of n, a(n) for n=0..65.

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.

Example

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.

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 *)

Crossrefs

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

Sequence in context: A105422 A166291 A162986 * A321198 A128584 A327853

Adjacent sequences: A319200 A319201 A319202 * A319204 A319205 A319206

Keyword

sign,tabl

Author

Wolfdieter Lang, Oct 29 2018

Status

approved