A292320 p-INVERT of (1,0,0,1,0,0,1,0,0,...), where p(S) = 1 - S - S^3.

1, 1, 2, 4, 6, 12, 22, 36, 67, 122, 209, 377, 681, 1193, 2130, 3823, 6764, 12043, 21531, 38252, 68076, 121456, 216126, 384691, 685636, 1220767, 2173346, 3871747, 6894873, 12276852, 21866387, 38941846, 69344928, 123500513, 219943018, 391676701, 697538335

Offset

0,3

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, 0, 4, -2, 0, -3, 1, 0, 1)

Formula

G.f.: -((1 + x^2 - 2 x^3 + x^6)/(-1 + x + 4 x^3 - 2 x^4 - 3 x^6 + x^7 + x^9)).

a(n) = a(n-1) + 4*a(n-3) - 2*a(n-4) - 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^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A079978  *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292320 *)

Crossrefs

Cf. A079978, A290616.

Sequence in context: A073660 A102588 A005303 * A057575 A354580 A196700

Adjacent sequences: A292317 A292318 A292319 * A292321 A292322 A292323

Keyword

nonn,easy

Author

Clark Kimberling, Sep 14 2017

Status

approved