A291403 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - 2 S^2 - S^4.

0, 2, 4, 7, 20, 42, 92, 214, 472, 1062, 2396, 5361, 12052, 27074, 60764, 136497, 306520, 688292, 1545768, 3471224, 7795184, 17505588, 39311608, 88280985, 198250312, 445204610, 999783508, 2245185343, 5041947516, 11322557726, 25426742788, 57100105470

Offset

0,2

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 A291382 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, 2, 4, 3, 4, 6, 4, 1)

Formula

G.f.: -((x (1 + x)^2 (2 + x^2 + 2 x^3 + x^4))/(-1 + 2 x^2 + 4 x^3 + 3 x^4 + 4 x^5 + 6 x^6 + 4 x^7 + x^8)).

a(n) = 2*a(n-2) + 4*a(n-3) + 3*a(n-4) + 4*a(n-5) + 6*a(n-6) + 4*a(n-7) + a(n-8) for n >= 9.

Mathematica

z = 60; s = x + x^2; p = 1 - 2 s^2 - s^4;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291403 *)

Crossrefs

Cf. A019590, A291382.

Sequence in context: A243049 A247234 A327444 * A101805 A145777 A188497

Adjacent sequences: A291400 A291401 A291402 * A291404 A291405 A291406

Keyword

nonn,easy

Author

Clark Kimberling, Sep 07 2017

Status

approved