A291240 p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - S^3)^2.

0, 0, 2, 0, 6, 3, 12, 18, 24, 63, 66, 173, 222, 438, 722, 1146, 2142, 3213, 5958, 9327, 16210, 26898, 44400, 75875, 123252, 210240, 344160, 578052, 958200, 1588404, 2650008, 4370292, 7285684, 12022704, 19960488, 33008505, 54594504, 90368550, 149168350

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 (0,6,2,-15,-6,19,6,-15,-2,6,0,-1)

Crossrefs

Cf. A000035, A291219.

Programs

Mathematica

z = 60; s = x/(1 - x^2); p = (1 - s^3)^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291240 *)
LinearRecurrence[{0, 6, 2, -15, -6, 19, 6, -15, -2, 6, 0, -1}, {0, 0, 2, 0, 6, 3, 12, 18, 24, 63, 66, 173}, 40] (* Vincenzo Librandi, Aug 29 2017 *)

Formula

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

a(n) = 6*a(n-2) + 2*a(n-3) - 15*a(n-4) - 6*a(n-5) + 19*a(n-6) + 6*a(n-7) - 15*a(n-8) - 2*a(n-9) + 6*a(n-10) - a(n-12) for n >= 13.