A289918 p-INVERT of (1,0,0,1,0,0,1,0,0,...), where p(S) = (1 - S^2).

0, 1, 0, 1, 2, 1, 4, 4, 6, 11, 12, 22, 30, 42, 68, 91, 140, 205, 292, 443, 634, 936, 1380, 1999, 2960, 4316, 6324, 9300, 13576, 19949, 29216, 42785, 62790, 91917, 134784, 197548, 289402, 424331, 621708, 911218, 1335586, 1957086, 2868620, 4203927, 6161084

Offset

0,5

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

Formula

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

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

Mathematica

z = 60; s = x/(x - x^3); p = (1 - s^2);
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A079978  *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A289918 *)

Crossrefs

Cf. A079978, A289919.

Sequence in context: A081243 A261297 A308034 * A127480 A141446 A339407

Adjacent sequences: A289915 A289916 A289917 * A289919 A289920 A289921

Keyword

nonn,easy

Author

Clark Kimberling, Sep 14 2017

Extensions

Recurrence corrected by Colin Barker, Sep 14 2017

Status

approved