A291238 p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - S)^2 (1 - 2 S).

4, 11, 30, 79, 202, 508, 1262, 3109, 7614, 18569, 45152, 109560, 265448, 642463, 1553790, 3755843, 9075302, 21923060, 52949458, 127869209, 308767326, 745537309, 1800065788, 4346043888, 10492781068, 25332654899, 61159842270, 147655261111, 356475233986

Offset: 0

Author

Clark Kimberling, Aug 28 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 A291219 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 (4, -2, -6, 2, 4, 1)

Crossrefs

Cf. A000035, A291219.

Programs

Mathematica

z = 60; s = x/(1 - x^2); p = (1 - s)^2(1 - 2s);
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291238 *)

Formula

G.f.: (-4 + 5 x + 6 x^2 - 5 x^3 - 4 x^4)/((-1 + x + x^2)^2 (-1 + 2 x + x^2)).

a(n) = 4*a(n-1) - 2*a(n-2) - 6*a(n-3) + 2*a(n-4) + 4*a(n-5) + a(n-6) for n >= 7.