A289846 p-INVERT of (1,0,1,0,1,0,1,0,1,...) (A059841), where p(S) = 1 - S - S^2.

1, 2, 4, 9, 18, 39, 80, 170, 353, 744, 1553, 3262, 6824, 14313, 29970, 62823, 131596, 275782, 577777, 1210704, 2536657, 5315210, 11136700, 23334969, 48893202, 102446199, 214654136, 449764562, 942387569, 1974580920, 4137324929, 8668915558, 18163921856

Offset: 0

Author

Clark Kimberling, Aug 14 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 (1, 3, -1, -1)

Crossrefs

Cf. A059841, A289780.

Programs

Mathematica

z = 60; s = x/(1 - x^2); p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A059841 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289846 *)

Formula

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

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