A291038 p-INVERT of (1,0,0,1,0,0,1,0,0,...), where p(S) = (1 - 2 S)^2.

4, 12, 32, 84, 216, 544, 1348, 3300, 8000, 19236, 45936, 109056, 257604, 605820, 1419232, 3313396, 7711944, 17900320, 41445764, 95746260, 220735616, 507934276, 1166792160, 2676017408, 6128381316, 14015556588, 32012831648, 73033858964, 166434905016

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 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 (4, -4, 2, -4, 0, -1)

Formula

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

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

Mathematica

z = 60; s = x/(x - x^3); p = (1 - 2 s)^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A079978  *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291038 *)
u/4 (* A291039 *)

Crossrefs

Cf. A079978, A290616, A291039.

Sequence in context: A048776 A135248 A205976 * A271898 A276178 A120369

Adjacent sequences: A291035 A291036 A291037 * A291039 A291040 A291041

Keyword

nonn,easy

Author

Clark Kimberling, Sep 14 2017

Status

approved