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A291264 p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - 2 S)^2. 2
4, 12, 36, 104, 292, 804, 2180, 5840, 15492, 40764, 106532, 276792, 715556, 1841748, 4722180, 12066208, 30737924, 78088812, 197892388, 500374024, 1262618148, 3180066180, 7995639940, 20071580784, 50312160388, 125942854044, 314865132324, 786254598872 (list; graph; refs; listen; history; text; internal format)
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 (4, -2, -4, -1)

FORMULA

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

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

a(n) - 4*A054447(n) for n >= 0.

a(n) = ((1-sqrt(2))^(1+n)*(2-3*sqrt(2) + 2*n) + (1+sqrt(2))^(1+n)*(2+3*sqrt(2) + 2*n)) / 4. - Colin Barker, Aug 26 2017

MATHEMATICA

z = 60; s = x/(1 - x^2); p = (1 - 2 s)^2;

Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)

u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291264 *)

u/4  (A054447 *)

PROG

(PARI) Vec(4*(1 - x - x^2) / (1 - 2*x - x^2)^2 + O(x^40)) \\ Colin Barker, Aug 26 2017

CROSSREFS

Cf. A000035, A291219, A291264.

Sequence in context: A162921 A237045 A291732 * A171849 A199937 A290380

Adjacent sequences:  A291261 A291262 A291263 * A291265 A291266 A291267

KEYWORD

nonn,easy

AUTHOR

Clark Kimberling, Aug 26 2017

STATUS

approved

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Last modified July 23 04:27 EDT 2019. Contains 325234 sequences. (Running on oeis4.)