A292404 p-INVERT of (1,0,0,1,0,0,0,0,0,0,...), where p(S) = 1 - S^4.

0, 0, 0, 1, 0, 0, 4, 1, 0, 6, 8, 1, 4, 28, 12, 2, 56, 66, 16, 71, 220, 120, 76, 496, 560, 218, 816, 1821, 1148, 1200, 4396, 4847, 2816, 8386, 15536, 11122, 14716, 39256, 42760, 33346, 82480, 135292, 109760, 161931, 353256, 385528, 369380, 794378, 1198288

Offset: 0

Author

Clark Kimberling, Sep 30 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).

Links

• Clark Kimberling, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 1, 0, 0, 4, 0, 0, 6, 0, 0, 4, 0, 0, 1)

Crossrefs

Cf. A292402.

Programs

Mathematica

z = 60; s = x + x^4; p = 1 - s^4;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292404 *)

Formula

G.f.: -((x^3 (1 + x)^4 (1 - x + x^2)^4)/((-1 + x + x^4) (1 + x + x^4) (1 + x^2 + 2 x^5 + x^8))).

a(n) = a(n-4) + 4*a(n-7) + 6*a(n-10) + 4*a(n-13) + a(n-16) for n >= 17.