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A291011 p-INVERT of (1,1,1,1,1,...), where p(S) = (1 - S)^2 (1 - 2 S). 2
4, 15, 52, 172, 552, 1736, 5384, 16536, 50440, 153112, 463176, 1397720, 4210568, 12668568, 38083528, 114414424, 343587336, 1031482904, 3095956040, 9291013848, 27879595144, 83652416920, 250985562312, 753015407192, 2259167856392, 6777755227416, 20333785775944 (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 A291000 for a guide to related sequences.

LINKS

Clark Kimberling, Table of n, a(n) for n = 0..999

Index entries for linear recurrences with constant coefficients, signature (7, -16, 12)

FORMULA

G.f.: (-4 + 13 x - 11 x^2)/((-1 + 2 x)^2 (-1 + 3 x)).

a(n) = 7*a(n-1) - 16*a(n-2) + 12*a(n-3) for n >= 4.

a(n) = (8*3^n - 2^(n-1)*(8+n)). - Colin Barker, Aug 23 2017

MATHEMATICA

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

Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)

Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291011 *)

LinearRecurrence[{7, -16, 12}, {4, 15, 52}, 30] (* Harvey P. Dale, Sep 23 2017 *)

PROG

(PARI) Vec((4 - 13*x + 11*x^2) / ((1 - 2*x)^2*(1 - 3*x)) + O(x^30)) \\ Colin Barker, Aug 23 2017

CROSSREFS

Cf. A000012, A289780, A291000.

Sequence in context: A005492 A003013 A117202 * A137213 A027853 A132894

Adjacent sequences:  A291008 A291009 A291010 * A291012 A291013 A291014

KEYWORD

nonn,easy

AUTHOR

Clark Kimberling, Aug 23 2017

STATUS

approved

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Last modified September 17 16:32 EDT 2021. Contains 347487 sequences. (Running on oeis4.)