A291383 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - 2 S - 2 S^2.

2, 8, 28, 98, 344, 1208, 4240, 14884, 52248, 183408, 643824, 2260040, 7933504, 27849280, 97760384, 343171984, 1204649632, 4228727296, 14844261824, 52108375328, 182918006400, 642104016000, 2254002082560, 7912309005376, 27774878417792, 97499209219328

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).

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 (2, 4, 4, 2)

Formula

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

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

Mathematica

z = 60; s = x + x^2; p = 1 - 2 s - 2 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291383 *)
u / 2  (* A291384 *)

Crossrefs

Cf. A019590, A291382, A291384.

Sequence in context: A060995 A106731 A318010 * A277653 A066796 A104934

Adjacent sequences: A291380 A291381 A291382 * A291384 A291385 A291386

Keyword

nonn,easy

Author

Clark Kimberling, Sep 04 2017

Status

approved