A290908 p-INVERT of the positive integers, where p(S) = 1 - 4*S^2.

0, 4, 16, 56, 208, 780, 2912, 10864, 40544, 151316, 564720, 2107560, 7865520, 29354524, 109552576, 408855776, 1525870528, 5694626340, 21252634832, 79315912984, 296011017104, 1104728155436, 4122901604640, 15386878263120, 57424611447840, 214311567528244

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 A290890 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 (4, -2, 4, -1)

Formula

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

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

a(n) = 4*A099486(n) for n >= 0.

Mathematica

z = 60; s = x/(1 - x)^2; p = 1 - 4 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290908 *)
u/4 (* A099486 *)

Crossrefs

Cf. A000027, A099486, A290890.

Sequence in context: A097128 A006079 A218263 * A201619 A197532 A220106

Adjacent sequences: A290905 A290906 A290907 * A290909 A290910 A290911

Keyword

nonn,easy

Author

Clark Kimberling, Aug 17 2017

Status

approved