A291257 a(n) = (1/2)*A291228(n).

1, 3, 9, 28, 85, 261, 797, 2440, 7461, 22827, 69821, 213588, 653345, 1998573, 6113529, 18701072, 57205769, 174990195, 535287793, 1637423756, 5008812525, 15321754293, 46868623381, 143369215128, 438560602669, 1341539064795, 4103713486629, 12553092811972

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 A291219 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 (2, 4, -2, -1)

Formula

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

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

a(n) = (1/2)*A291228(n) for n >= 0.

Mathematica

z = 60; s = x/(1 - x^2); p = 1 - 2 s - 2 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291228 *)
u/2         (* A291257 *)

Crossrefs

Cf. A000035, A291219, A291228.

Sequence in context: A027090 A033139 A291731 * A005354 A084084 A091140

Adjacent sequences: A291254 A291255 A291256 * A291258 A291259 A291260

Keyword

nonn,easy

Author

Clark Kimberling, Aug 25 2017

Status

approved