A291217 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S^3.

0, 0, 1, 0, 3, 1, 6, 6, 11, 21, 24, 57, 66, 138, 194, 330, 546, 827, 1452, 2175, 3739, 5826, 9582, 15519, 24807, 40836, 64933, 106584, 170796, 277696, 448980, 724968, 1177181, 1897380, 3080367, 4972113, 8055918, 13029534, 21075947, 34125561, 55169988

Offset: 0

Author

Clark Kimberling, Aug 24 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 (0,3, 1,-3,0,1)

Crossrefs

Cf. A000035, A291000, A291219.

Programs

Magma

I:=[0,0,1,0,3,1]; [n le 6 select I[n] else 3*Self(n-2)+Self(n-3)-3*Self(n-4)+Self(n-6): n in [1..45]]; // Vincenzo Librandi, Aug 25 2017

Mathematica

z = 60; s = x/(1 - x^2); p = 1 - s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291217 *)
LinearRecurrence[{0, 3, 1, -3, 0, 1}, {0, 0, 1, 0, 3, 1}, 50] (* Vincenzo Librandi, Aug 25 2017 *)

Formula

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

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