A292326 p-INVERT of (1,1,1,0,0,0,0,0,0,0,0,...), where p(S) = (1 - S)^3.

3, 9, 25, 63, 153, 359, 819, 1830, 4018, 8694, 18582, 39298, 82350, 171186, 353338, 724719, 1478061, 2999175, 6057687, 12183945, 24411935, 48740193, 96998325, 192459996, 380812692, 751557756, 1479686972, 2906717460, 5698014924, 11147786740, 21769549380

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).

Links

Clark Kimberling, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (3, 0, -2, -6, 0, 4, 6, 3, 1)

Formula

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

a(n) = 3*a(n-1) - 2*a(n-3) - 6*a(n-4) + 4*a(n-6) + 6*a(n-7) + 3*a(n-8) + a(n-9) for n >= 10.

Mathematica

z = 60; s = x + x^2 + x^3; p = (1 - s)^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292326 *)
LinearRecurrence[{3, 0, -2, -6, 0, 4, 6, 3, 1}, {3, 9, 25, 63, 153, 359, 819, 1830, 4018}, 40] (* Harvey P. Dale, Nov 01 2019 *)

Crossrefs

Cf. A292324.

Sequence in context: A065971 A145127 A096260 * A195417 A295142 A002064

Adjacent sequences: A292323 A292324 A292325 * A292327 A292328 A292329

Keyword

nonn,easy

Author

Clark Kimberling, Sep 15 2017

Status

approved