A291223 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S^3 - S^4.

0, 0, 1, 1, 3, 5, 8, 17, 25, 52, 83, 159, 271, 497, 868, 1572, 2762, 4984, 8784, 15799, 27939, 50089, 88831, 158880, 282293, 504179, 896780, 1600335, 2848339, 5080363, 9045953, 16129172, 28726972, 51209648, 91223508, 162594868, 289675121, 516264093

Offset

0,5

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 A291219 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 (0, 4, 1, -5, -1, 4, 0, -1)

Formula

a(n) = 4*a(n-2) + a(n-3) - 5*a(n-4) - a(n-5) + 4*a(n-6) - a(n-8) for n >= 9.

G.f.: x^2*(1 + x - x^2) / (1 - 4*x^2 - x^3 + 5*x^4 + x^5 - 4*x^6 + x^8). - Colin Barker, Aug 25 2017

Mathematica

z = 60; s = x/(1 - x^2); p = 1 - s^3 - s^4;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291223 *)

Prog

(PARI) concat(vector(2), Vec(x^2*(1 + x - x^2) / (1 - 4*x^2 - x^3 + 5*x^4 + x^5 - 4*x^6 + x^8) + O(x^50))) \\ Colin Barker, Aug 25 2017

Crossrefs

Cf. A000035, A291219.

Sequence in context: A056765 A080006 A174011 * A240532 A184434 A196140

Adjacent sequences: A291220 A291221 A291222 * A291224 A291225 A291226

Keyword

nonn,easy

Author

Clark Kimberling, Aug 24 2017

Status

approved