A291251 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - 3 S + 2 S^3.

0, 3, -2, 15, -18, 76, -126, 405, -802, 2241, -4884, 12696, -29100, 72903, -171490, 421683, -1005030, 2448356, -5873706, 14243001, -34280258, 82936965, -199930344, 483172656, -1165648152, 2815517835, -6794932418, 16408304343, -39606671610, 95629756540

Offset: 0

Author

Clark Kimberling, Aug 29 2017

Keywords

• easy
• sign

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).

Putting s = (0, -1, 0, -1, 0, -1, ...) gives (|a(n)|). For given s, it would be of interest to know conditions on p that imply that t(s) has terms that are all positive (or all nonnegative, or strictly increasing, or alternating, as in the present case.)

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 (0, 6, -2, -6, 0, 1)

Crossrefs

Cf. A000035, A291219.

Programs

Mathematica

z = 60; s = x/(1 - x^2); p = 1 - 3 s^2 + 2 s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291251 *)

Formula

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

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