A291726 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = (1 - S)^3.

3, 6, 13, 27, 51, 94, 171, 303, 527, 906, 1539, 2586, 4308, 7122, 11692, 19077, 30957, 49986, 80349, 128628, 205146, 326058, 516594, 816076, 1285674, 2020380, 3167464, 4954887, 7734993, 12051616, 18743037, 29099781, 45106223, 69810162, 107887629, 166505313

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 A291728 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 (3, -3, 4, -6, 3, -3, 3, 0, 1)

Formula

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

a(n) = 3*a(n-1) - 3*a(n-2) + 4*a(n-3) - 6*a(n-4) + 3*a(n-5) - 3*a(n-6) + 3*a(n-7) + a(n-9) for n >= 10.

Mathematica

z = 60; s = x + x^3; p = (1 - s)^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A154272 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291726 *)

Crossrefs

Cf. A154272, A291728.

Sequence in context: A215980 A215979 A273226 * A280563 A281283 A281638

Adjacent sequences: A291723 A291724 A291725 * A291727 A291728 A291729

Keyword

nonn,easy

Author

Clark Kimberling, Sep 08 2017

Status

approved