A289781 p-INVERT of the positive Fibonacci numbers (A000045), where p(S) = 1 - S - S^2.

1, 3, 9, 27, 80, 237, 701, 2073, 6129, 18120, 53569, 158367, 468181, 1384083, 4091760, 12096453, 35760689, 105719157, 312537041, 923951760, 2731474161, 8075043963, 23872213729, 70573310907, 208635540400, 616788246957, 1823408134821, 5390532719313

Offset: 0

Author

Clark Kimberling, Aug 10 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 A289780 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, 1, -3, -1)

Crossrefs

Cf. A000045, A289780, A289975, A289976, A289977.

Programs

Mathematica

z = 60; s = x/(1 - x - x^2); p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000045 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289781 *)

Formula

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

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