A291255 p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - S - 2 S^2)^2.

2, 7, 18, 55, 144, 404, 1060, 2853, 7442, 19573, 50670, 131368, 337622, 866819, 2213650, 5642899, 14332988, 36335548, 91872760, 231875713, 584030738, 1468631153, 3686943130, 9242753104, 23138167146, 57851432575, 144470316562, 360384852207, 898051760168

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 (2, 7, -10, -16, 10, 7, -2, -1)

Formula

G.f.: (2 + 3 x - 10 x^2 - 10 x^3 + 10 x^4 + 3 x^5 - 2 x^6)/(1 - x - 4 x^2 + x^3 + x^4)^2.

a(n) = 2*a(n-1) + 7*a(n-2) - 10*a(n-3) - 16*a(n-4) + 10*a(n-5) + 7*a(n-6) - 2*a(n-7) - a(n-8) for n >= 9.

Mathematica

z = 60; s = x/(1 - x^2); p = (1 - s - 2 s^2)^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291255 *)

Crossrefs

Cf. A000035, A291219.

Sequence in context: A261331 A046672 A046866 * A294004 A214836 A176813

Adjacent sequences: A291252 A291253 A291254 * A291256 A291257 A291258

Keyword

nonn,easy

Author

Clark Kimberling, Sep 02 2017

Status

approved