A292401 p-INVERT of (1,0,2,0,2,0,2,0,2,0,...), where p(S) = (1 - S)^2.

2, 3, 8, 17, 34, 71, 144, 289, 578, 1147, 2264, 4449, 8706, 16975, 32992, 63937, 123586, 238323, 458600, 880753, 1688482, 3231639, 6175728, 11785313, 22460802, 42754283, 81290424, 154396097, 292953858, 555334047, 1051781312, 1990373249, 3763583618

Offset

0,1

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 (2, 1, 0, -3, -2, -1)

Formula

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

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

Mathematica

z = 60; s = x (x^2 + 1)/(1 - x^2); p = (1 - s)^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* abs. values of A176742 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292401 *)

Crossrefs

Cf. A176742, A292400.

Sequence in context: A158921 A064954 A267223 * A132333 A182889 A256169

Adjacent sequences: A292398 A292399 A292400 * A292402 A292403 A292404

Keyword

nonn,easy

Author

Clark Kimberling, Sep 30 2017

Status

approved