A289975 p-INVERT of the Fibonacci numbers (A000045, including 0), where p(S) = 1 - S - S^2.

0, 1, 1, 4, 7, 18, 37, 85, 183, 407, 888, 1956, 4284, 9409, 20630, 45270, 99289, 217819, 477776, 1048053, 2298912, 5042783, 11061455, 24263687, 53223023, 116746272, 256086074, 561731936, 1232174181, 2702807740, 5928681960, 13004724921, 28526216361

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, 2, -3, -1)

Crossrefs

Cf. A000045, A289781, A289976, A289780.

Programs

Mathematica

z = 60; s = x^2/(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]  (* A289975 *)

Formula

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

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