A289977 p-INVERT of (0,0,0,1,2,3,5,8,...), the Fibonacci numbers preceded by three zeros, where p(S) = 1 - S - S^2.

0, 0, 0, 1, 1, 2, 3, 7, 12, 23, 41, 77, 140, 258, 470, 861, 1570, 2867, 5225, 9526, 17352, 31607, 57547, 104766, 190684, 347029, 631476, 1148985, 2090427, 3803044, 6918379, 12585209, 22892932, 41641932, 75744383, 137772396, 250592150, 455792833, 829016539

Offset

0,6

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 A290890 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 (2, 1, -2, 0, -1, -1, 0, 1)

Formula

a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3) - a(n-5) - a(n-6) + a(n-8).

G.f.: x^3*(1 - x)*(1 - x^2 - x^3) / (1 - 2*x - x^2 + 2*x^3 + x^5 + x^6 - x^8). - Colin Barker, Aug 24 2017

Mathematica

z = 60; s = x^4/(1 - x - x^2); p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* 0, 0, 0, 1, 2, 3, 5, ... *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A289977 *)
LinearRecurrence[{2, 1, -2, 0, -1, -1, 0, 1}, {0, 0, 0, 1, 1, 2, 3, 7}, 40] (* Harvey P. Dale, Jul 14 2018 *)

Prog

(PARI) concat(vector(3), Vec(x^3*(1 - x)*(1 - x^2 - x^3) / (1 - 2*x - x^2 + 2*x^3 + x^5 + x^6 - x^8) + O(x^50))) \\ Colin Barker, Aug 24 2017

Crossrefs

Cf. A000045, A289975, A289780.

Sequence in context: A027675 A298353 A054176 * A000630 A036538 A341407

Adjacent sequences: A289974 A289975 A289976 * A289978 A289979 A289980

Keyword

nonn,easy

Author

Clark Kimberling, Aug 21 2017

Status

approved