A291229 p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - S)(1 - 2 S).

3, 7, 18, 45, 111, 272, 663, 1611, 3906, 9457, 22875, 55296, 133611, 322751, 779490, 1882341, 4545159, 10974256, 26496255, 63970947, 154444914, 372871721, 900206067, 2173312512, 5246877459, 12667142455, 30581283762, 73829906397, 178241414367, 430313249360

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 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 (3, 0, -3, -1)

Formula

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

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

Mathematica

z = 60; s = x/(1 - x^2); p = (1 - s)(1 - 2 s);
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291229 *)

Crossrefs

Cf. A000035, A291219.

Sequence in context: A178035 A000226 A291734 * A036883 A247296 A191652

Adjacent sequences: A291226 A291227 A291228 * A291230 A291231 A291232

Keyword

nonn,easy

Author

Clark Kimberling, Aug 25 2017

Status

approved