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

0,2

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).

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).

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 *)

Crossrefs

Cf. A001906, A289780, A298804.

Sequence in context: A356340 A120902 A054441 * A102285 A218985 A129162

Adjacent sequences: A289800 A289801 A289802 * A289804 A289805 A289806

Keyword

nonn,easy

Author

Clark Kimberling, Aug 12 2017

Status

approved