A291405 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - 2 S^2 - 2 S^4.

0, 2, 4, 8, 24, 52, 120, 290, 672, 1592, 3760, 8824, 20800, 48976, 115296, 271588, 639488, 1505816, 3546032, 8350064, 19662944, 46302800, 109033952, 256754760, 604609280, 1423740736, 3352643712, 7894846528, 18590881280, 43778039424, 103089066752

Offset

0,2

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 A291382 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, 2, 4, 4, 8, 12, 8, 2)

Formula

G.f.: -((2 x (1 + x)^2 (1 + x^2 + 2 x^3 + x^4))/(-1 + 2 x^2 + 4 x^3 + 4 x^4 + 8 x^5 + 12 x^6 + 8 x^7 + 2 x^8)).

a(n) = 2*a(n-2) + 4*a(n-3) + 4*a(n-4) + 5*a(n-5) + 12*a(n-6) + 8*a(n-7) + 2*a(n-8) for n >= 9.

Mathematica

z = 60; s = x + x^2; p = 1 - 2 s^2 - 2 s^4;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291405 *)
u / 2 (* A291406 *)

Crossrefs

Cf. A019590, A291382, A291406.

Sequence in context: A359039 A078222 A078223 * A306420 A361726 A333971

Adjacent sequences: A291402 A291403 A291404 * A291406 A291407 A291408

Keyword

nonn,easy

Author

Clark Kimberling, Sep 07 2017

Status

approved