A291143 p-INVERT of (1,1,1,1,1,...), where p(S) = (1 - S^3)^3.

0, 0, 3, 9, 18, 36, 81, 189, 430, 954, 2097, 4602, 10080, 21996, 47796, 103473, 223308, 480584, 1031571, 2208807, 4718610, 10058580, 21398715, 45438270, 96313626, 203812110, 430615240, 908455203, 1913845374, 4026531804, 8460687861, 17756508321, 37223049942

Offset

0,3

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..1000

Index entries for linear recurrences with constant coefficients, signature (9,-36,87,-144,171,-147,90,-36,8)

Formula

a(n) = 9*a(n-1) - 36 a(n-2) + 87*a(n-3) - 144*a(n-4) + 171*a(n-5) - 147*a(n-6) + 90*a(n-7) - 36*a(n-8) + 8*a(n-9) for n >= 10.

G.f.: x^2*(3 - 18*x + 45*x^2 - 63*x^3 + 54*x^4 - 27*x^5 + 7*x^6) / ((1 - 2*x)^3*(1 - x + x^2)^3). - Colin Barker, Aug 24 2017

Mathematica

z = 60; s = x/(1 - x); p = (1 - s^3)^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291143 *)
LinearRecurrence[{9, -36, 87, -144, 171, -147, 90, -36, 8}, {0, 0, 3, 9, 18, 36, 81, 189, 430}, 40] (* Vincenzo Librandi, Aug 29 2017 *)

Prog

(PARI) concat(vector(2), Vec(x^2*(3 - 18*x + 45*x^2 - 63*x^3 + 54*x^4 - 27*x^5 + 7*x^6) / ((1 - 2*x)^3*(1 - x + x^2)^3) + O(x^30))) \\ Colin Barker, Aug 24 2017

Crossrefs

Cf. A000012, A289780, A291000.

Sequence in context: A028723 A213291 A264365 * A057578 A271370 A015635

Adjacent sequences: A291140 A291141 A291142 * A291144 A291145 A291146

Keyword

nonn,easy

Author

Clark Kimberling, Aug 24 2017

Status

approved