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A292536 p-INVERT of the squares (A000290), where p(S) = 1 + S - 3 S^2. 1
1, 8, 48, 255, 1310, 6773, 35260, 183740, 956765, 4980320, 25924725, 134956612, 702554244, 3657326875, 19039098206, 99112598721, 515954630808, 2685927132776, 13982245762937, 72787973059648, 378915453775913, 1972536332660240, 10268516498713448 (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 A292479 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, -14, 26, -10, 5, -1)

FORMULA

G.f.: -(((1 + x) (-1 - 6 x^2 + x^3))/(1 - 7 x + 14 x^2 - 26 x^3 + 10 x^4 - 5 x^5 + x^6)).

a(n) = 7*a(n-1) - 14*a(n-2) + 26*a(n-3) - 10*a(n-4) + 5*a(n-5) - a(n-6) for n >= 7.

MATHEMATICA

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

Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)

Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292536 *)

CROSSREFS

Cf. A000290, A292479.

Sequence in context: A079785 A225977 A305782 * A242668 A002697 A285063

Adjacent sequences:  A292533 A292534 A292535 * A292537 A292538 A292539

KEYWORD

nonn,easy

AUTHOR

Clark Kimberling, Oct 04 2017

STATUS

approved

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Last modified April 9 10:52 EDT 2020. Contains 333348 sequences. (Running on oeis4.)