A291228 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - 2 S - 2 S^2.

2, 6, 18, 56, 170, 522, 1594, 4880, 14922, 45654, 139642, 427176, 1306690, 3997146, 12227058, 37402144, 114411538, 349980390, 1070575586, 3274847512, 10017625050, 30643508586, 93737246762, 286738430256, 877121205338, 2683078129590, 8207426973258

Offset: 0

Author

Clark Kimberling, Aug 25 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 A291219 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, 4, -2, -1)

Crossrefs

Cf. A000035, A291219, A291257.

Programs

Mathematica

z = 60; s = x/(1 - x^2); p = 1 - 2 s - 2 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291228 *)
u/2         (* A291257 *)
LinearRecurrence[{2,4,-2,-1},{2,6,18,56},30] (* Harvey P. Dale, Aug 08 2019 *)

Formula

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

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

a(n) = 2*A291257(n) for n >= 0.