A292323 p-INVERT of (1,0,0,1,0,0,1,0,0,...), where p(S) = (1 - S)(1 + S^2).

1, 0, 0, 2, 1, 0, 5, 6, 1, 11, 23, 10, 22, 71, 57, 50, 191, 243, 164, 474, 860, 676, 1175, 2674, 2758, 3225, 7626, 10256, 10313, 20882, 34642, 36384, 57921, 108270, 130025, 170606, 321415, 448093, 540825, 934958, 1468860, 1798559, 2750605, 4605556, 6042649

Offset

0,4

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, 4, -2, 1, -3, 1, 0, 1)

Formula

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

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

Mathematica

z = 60; s = x/(x - x^3); p = (1 - s)(1 + s^2);
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A079978  *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292323 *)

Prog

(PARI) x='x+O('x^99); Vec((1-x+x^2-2*x^3+x^4+x^6)/((1-x-x^3)*(1+x^2-2*x^3+x^6))) \\ Altug Alkan, Oct 05 2017

Crossrefs

Cf. A079978, A292322.

Sequence in context: A298213 A130191 A054651 * A059720 A140589 A331955

Adjacent sequences: A292320 A292321 A292322 * A292324 A292325 A292326

Keyword

nonn,easy

Author

Clark Kimberling, Sep 15 2017

Status

approved