A291723 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = 1 - S^3.

0, 0, 1, 0, 3, 1, 3, 6, 2, 15, 9, 21, 36, 27, 85, 72, 141, 222, 231, 513, 540, 945, 1422, 1741, 3222, 3876, 6337, 9339, 12447, 20809, 27135, 42546, 62195, 86709, 136866, 187278, 286113, 417303, 595852, 910431, 1281810, 1926984, 2810883, 4064571, 6097464

Offset: 0

Author

Clark Kimberling, Sep 08 2017

Keywords

• nonn
• easy

Comments

Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the "INVERT" transform of s, so that p-INVERT is a generalization of the "INVERT" transform (e.g., A033453).

See A291728 for a guide to related sequences.

Links

• Clark Kimberling, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (0, 0, 1, 0, 3, 0, 3, 0, 1)

Crossrefs

Cf. A154272, A291728.

Programs

Mathematica

z = 60; s = x + x^3; p = 1 - s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A154272 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291723 *)
LinearRecurrence[{0,0,1,0,3,0,3,0,1},{0,0,1,0,3,1,3,6,2},60] (* Harvey P. Dale, Jun 07 2022 *)

Formula

G.f.: -((x^2 (1 + x^2)^3)/((1 - x + x^2) (-1 + x + x^3) (1 + 2 x + 2 x^2 + x^3 + x^4))).

a(n) = a(n-3) + 3*a(n-5) + 3*a(n-7) + a(n-9) for n >= 10.