A291741 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = (1 - S)(1 + S^2).

1, 0, 1, 1, 1, 4, 5, 7, 11, 12, 19, 30, 42, 68, 98, 137, 205, 292, 429, 644, 936, 1380, 2024, 2936, 4316, 6324, 9260, 13625, 19949, 29216, 42841, 62701, 91917, 134784, 197485, 289547, 424331, 621708, 911255, 1335378, 1957086, 2868620, 4203998, 6161329

Offset: 0

Author

Clark Kimberling, Sep 12 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, -1, 2, -2, 3, -1, 3, 0, 1)

Crossrefs

Cf. A154272, A291728.

Programs

Mathematica

z = 60; s = x + x^3; p = (1 - s) (1 + s^2);
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A154272 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291741 *)

Formula

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

a(n) = a(n-1) - a(n-2) + 2*a(n-3) - 2*a(n-4) + 3*a(n-5) - a(n-6) + 3*a(n-7) + a(n-9) for n >= 10.