A292322 p-INVERT of (1,0,0,1,0,0,1,0,0,...), where p(S) = 1 - S - S^2 - S^3.

1, 2, 4, 8, 17, 36, 73, 152, 317, 653, 1355, 2812, 5818, 12061, 25001, 51786, 107323, 222409, 460824, 954942, 1978840, 4100398, 8496827, 17606974, 36484494, 75602461, 156661630, 324629762, 672690133, 1393931744, 2888469094, 5985414154, 12402824741

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 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^2 + 4 x^3 - 2 x^4 - x^5 - 3 x^6 + x^7 + x^9)).

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 - s^2 - s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A079978  *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292322 *)

Crossrefs

Cf. A079978, A292320.

Sequence in context: A182900 A202843 A247297 * A008999 A052903 A308745

Adjacent sequences: A292319 A292320 A292321 * A292323 A292324 A292325

Keyword

nonn,easy

Author

Clark Kimberling, Sep 15 2017

Status

approved