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A290897 p-INVERT of the positive integers, where p(S) = 1 - S - S^3. 2
1, 3, 9, 29, 95, 307, 976, 3073, 9645, 30283, 95207, 299625, 943363, 2970320, 9351621, 29439359, 92671625, 291715157, 918275995, 2890621063, 9099375792, 28643956245, 90168412937, 283841284899, 893503898503, 2812659866565, 8853968158791, 27871395427616 (list; graph; refs; listen; history; text; internal format)
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 (7, -19, 27, -19, 7, -1)

FORMULA

a(n) = 7*a(n-1) - 19*a(n-2) + 27*a(n-3) - 19*a(n-4) + 7*a(n-5) - a(n-6).

G.f.: (1 - 4*x + 7*x^2 - 4*x^3 + x^4) / ((1 - 3*x + 4*x^2 - x^3)*(1 - 4*x + 3*x^2 - x^3)). - Colin Barker, Aug 16 2017

MATHEMATICA

z = 60; s = x/(1 - x)^2; p = 1 - s - s^3;

Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)

Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290897 *)

PROG

(PARI) Vec((1 - 4*x + 7*x^2 - 4*x^3 + x^4) / ((1 - 3*x + 4*x^2 - x^3)*(1 - 4*x + 3*x^2 - x^3)) + O(x^30)) \\ Colin Barker, Aug 16 2017

CROSSREFS

Cf. A000027, A290890.

Sequence in context: A300044 A071728 A036550 * A289448 A071732 A289804

Adjacent sequences:  A290894 A290895 A290896 * A290898 A290899 A290900

KEYWORD

nonn,easy

AUTHOR

Clark Kimberling, Aug 15 2017

STATUS

approved

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Last modified May 29 02:45 EDT 2022. Contains 354122 sequences. (Running on oeis4.)