A291738 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = 1 - S - S^4.

1, 1, 2, 4, 6, 13, 23, 43, 76, 138, 244, 444, 795, 1444, 2600, 4705, 8474, 15307, 27583, 49797, 89800, 162088, 292388, 527663, 951922, 1717692, 3098937, 5591589, 10088361, 18202665, 32841990, 59256835, 106914493, 192904396, 348050363, 627980316, 1133045001

Offset: 0

Author

Clark Kimberling, Sep 11 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 A291728 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 (1, 0, 1, 1, 0, 4, 0, 6, 0, 4, 0, 1)

Crossrefs

Cf. A154272, A291728.

Programs

Mathematica

z = 60; s = x + x^3; p = 1 - s - s^4;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A154272 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291738 *)

Formula

G.f.: -(((1 + x^2) (1 + x + x^2) (1 + x + x^3) (1 - 2 x + 2 x^2 - x^3 + x^4))/(-1 + x + x^3 + x^4 + 4 x^6 + 6 x^8 + 4 x^10 + x^12)).

a(n) = a(n-1) + a(n-3) + a(n-4) + 4* a(n-6) + 6*a(n-8) + 4*a(n-10) + a(n-12) for n >= 13.