A291394 p-INVERT of (1,1,0,0,0,0,...), where p(S) = (1 - S)(1 - 3 S).

4, 17, 66, 254, 968, 3679, 13962, 52957, 200812, 761396, 2886768, 10944725, 41494856, 157319353, 596443614, 2261290498, 8573204920, 32503490435, 123230092830, 467200760741, 1771292578424, 6715480046152, 25460317920096, 96527393973769, 365963135802988

Offset

0,1

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 (4, 1, -6, -3)

Formula

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

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

Mathematica

z = 60; s = x + x^2; p = (1 - s)(1 - 3s);
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291394 *)

Crossrefs

Cf. A019590, A291382.

Sequence in context: A045992 A217539 A046723 * A244616 A030529 A266862

Adjacent sequences: A291391 A291392 A291393 * A291395 A291396 A291397

Keyword

nonn,easy

Author

Clark Kimberling, Sep 06 2017

Status

approved