A291230 p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - S)(1 - 2 S)(1 - 3 S).

6, 25, 96, 351, 1242, 4304, 14706, 49761, 167232, 559303, 1864110, 6197472, 20567262, 68166713, 225713280, 746866143, 2470077378, 8166190192, 26990599050, 89190984033, 294691499808, 973574384231, 3216160413654, 10623856065984, 35092075282998, 115910575744921

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 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 (6, -8, -6, 8, 6, 1)

Formula

G.f.: (-6 + 11 x + 6 x^2 - 11 x^3 - 6 x^4)/(-1 + 6 x - 8 x^2 - 6 x^3 + 8 x^4 + 6 x^5 + x^6).

a(n) = 6*a(n-1) - 8*a(n-2) - 6*a(n-3) + 8*a(n-4) + 6*a(n-5) + a(n-6) for n >= 7.

Mathematica

z = 60; s = x/(1 - x^2); p = (1 - s)(1 - 2 s)(1 - 3 s);
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291230 *)

Crossrefs

Cf. A000035, A291219.

Sequence in context: A092491 A112308 A034336 * A092184 A214955 A286433

Adjacent sequences: A291227 A291228 A291229 * A291231 A291232 A291233

Keyword

nonn,easy

Author

Clark Kimberling, Aug 25 2017

Status

approved