A291409 p-INVERT of (1,1,0,0,0,0,...), where p(S) = (1 - S^2)(1 - S)^2.

2, 6, 14, 31, 66, 136, 272, 534, 1030, 1958, 3678, 6837, 12594, 23016, 41768, 75325, 135084, 241032, 428112, 757236, 1334292, 2342892, 4100676, 7155937, 12453170, 21616242, 37432010, 64675099, 111512574, 191893120, 329605760, 565166682, 967491754, 1653659282

Offset: 0

Author

Clark Kimberling, Sep 07 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 A291382 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 (2, 2, -2, -5, -2, 4, 4, 1)

Crossrefs

Cf. A019590, A291382.

Programs

Mathematica

z = 60; s = x + x^2; p = (1 - s^2)(1 - s)^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291409 *)
LinearRecurrence[{2,2,-2,-5,-2,4,4,1},{2,6,14,31,66,136,272,534},40] (* Harvey P. Dale, May 12 2024 *)

Formula

G.f.: -(((1 + x) (2 - 2 x^2 - 3 x^3 + x^4 + 3 x^5 + x^6))/((-1 + x + x^2)^3 (1 + x + x^2))).

a(n) = 2*a(n-1) + 2*a(n-2) - 2*a(n-3) - 5*a(n-4) - 2*a(n-5) + 4*a(n-6) + 4*a(n-7) + a(n-8) for n >= 9.