A289803 p-INVERT of the even bisection (A001906) of the Fibonacci numbers, where p(S) = 1 - S - S^2.

1, 5, 23, 103, 456, 2009, 8833, 38803, 170399, 748176, 3284833, 14421533, 63314735, 277968871, 1220356440, 5357681369, 23521603225, 103265890987, 453363808127, 1990383615264, 8738295434881, 38363361811637, 168425013526727, 739429075564711, 3246283590352104

Offset: 0

Author

Clark Kimberling, Aug 12 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
• Rigoberto Flórez, Javier González, Mateo Matijasevick, Cristhian Pardo, José Luis Ramírez, Lina Simbaqueba, and Fabio Velandia, Lattice paths in corridors and cyclic corridors, Contrib. Disc. Math. (2024) Vol. 19. No. 2, 36-55. See p. 11.
• Index entries for linear recurrences with constant coefficients, signature (7,-13,7,-1).

Crossrefs

Cf. A001906, A289780, A298804.

Programs

Mathematica

z = 60; s = x/(1 - 3*x + x^2); p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A001906 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289803 *)

Formula

G.f.: (1 - 2 x + x^2)/(1 - 7 x + 13 x^2 - 7 x^3 + x^4).

a(n) = 7*a(n-1) - 13*a(n-2) + 7*a(n-3) - a(n-4).