A289976 p-INVERT of (0,0,1,2,3,5,8,...), the Fibonacci numbers preceded by two zeros, where p(S) = 1 - S - S^2.

0, 0, 1, 1, 2, 5, 9, 18, 36, 70, 137, 268, 522, 1017, 1980, 3852, 7492, 14568, 28321, 55051, 106999, 207952, 404134, 785366, 1526186, 2965752, 5763103, 11198858, 21761463, 42286357, 82169547, 159668921, 310262351, 602888757, 1171506956, 2276419286

Offset: 0

Author

Clark Kimberling, Aug 21 2017

Keywords

• nonn
• easy

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, -1, -2, -1, 1)

Crossrefs

Cf. A000045, A289975, A289780.

Programs

Mathematica

z = 60; s = x^3/(1 - x - x^2); p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* 0,0,1,2,3,5,... *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A289976 *)

Formula

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

a(n) = 2*a(n-1) + a(n-2) - a(n-3) - 2*a(n-4) - a(n-5) + a(n-6).