A382057 Z-sequence for the Riordan triangle A125166.

8, -37, 181, -865, 4105, -19441, 92017, -435457, 2060641, -9751105, 46142785, -218350081, 1033243777, -4889362177, 23136710401, -109484089345, 518084273665, -2451601105921, 11601100993537, -54896999325697, 259775389992961, -1229270344003585, 5816969724063745, -27526196280360961

Offset

0,1

Comments

For the Z-sequence of a Riordan trangle R(G(x), F(x)=x*Fhat(x)) see the first W. Lang link in A006232, where also references are given,

The Z-sequence implies a recurrence formula for R(n, 0) using the previous row entries of R.

R(n, 0) = Sum_{j=0..n-1} Z(j)*R(n-1, j), for n >= 1, and R(0, 0) = G(0), usually 1.

The o.g.f. of the Z-sequence of R is GZ(y) = (1/F^{[-1]}(y))*(1 - 1/G(F^{[-1]}(y))), with the composition inverse F^{[-1]} of F.

Links

Table of n, a(n) for n=0..23.

Index entries for linear recurrences with constant coefficients, signature (-7,-12,-6).

Formula

O.g.f.: (8 + 19*x + 18*x^2 + 6*x^3)/((1 + x)*(1 + 6*x+ 6*x^2)).

Example

The Riordan triangle A125166 has row n = 3 [64, 36, 10, 1], hence R(0, 4) = 8*64 - 37*36 + 10*181 - 1*865 = 125 = 5^3.

Mathematica

LinearRecurrence[{-7, -12, -6}, {8, -37, 181, -865}, 24] (* Stefano Spezia, Mar 26 2025 *)

Crossrefs

Cf. A006232, A125166.

Sequence in context: A221891 A221063 A221633 * A201452 A128246 A257215

Adjacent sequences: A382054 A382055 A382056 * A382058 A382059 A382060

Keyword

sign,easy

Author

Wolfdieter Lang, Mar 25 2025

Status

approved