A291252 p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - S^3)^3.

0, 0, 3, 0, 9, 6, 18, 36, 40, 126, 135, 351, 513, 936, 1755, 2682, 5373, 8260, 15525, 25731, 44511, 78030, 129564, 229617, 381438, 664038, 1121144, 1910790, 3263796, 5500110, 9404820, 15824790, 26910426, 45388638, 76700664, 129564945, 218084256, 368095230

Offset: 0

Author

Clark Kimberling, Aug 29 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, 9, 3, -36, -18, 81, 45, -117, -59, 117, 45, -81, -18, 36, 3, -9, 0, 1)

Crossrefs

Cf. A000035, A291219.

Programs

Mathematica

z = 60; s = x/(1 - x^2); p = (1 - s^3)^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291252 *)

Formula

G.f.: -((x^2 (3 - 18 x^2 - 3 x^3 + 45 x^4 + 9 x^5 - 59 x^6 - 9 x^7 + 45 x^8 + 3 x^9 - 18 x^10 + 3 x^12))/((-1 + x + x^2)^3 (1 + x - x^2 - x^3 + x^4)^3)).

a(n) = 9*a(n-2) + 3*a(n-3) - 36*a(n-4) - 18*a(n-5) + 81*a(n-6) + 45*a(n-7) - 117*a(n-8) - 59*a(n-9) + 117*a(n-10) + 45*a(n-11) - 81*a(n-12) - 18*a(n-13) + 36*a(n-14) + 3*a(n-15) - 9*a(n-16) + a(n-18) for n >= 19.