A291237 p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - S)(1 - 2 S)(1 - 4 S).

7, 35, 162, 721, 3139, 13504, 57707, 245671, 1043634, 4428053, 18774815, 79573152, 337178159, 1428553243, 6052037010, 25638260873, 108608846171, 460082737472, 1948961747155, 8255982722783, 34973020586946, 148148373971341, 627567262233463, 2658419223345984

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 (7, -11, -6, 11, 7, 1).

Formula

G.f.: (-7 + 14 x + 6 x^2 - 14 x^3 - 7 x^4)/(-1 + 7 x - 11 x^2 - 6 x^3 + 11 x^4 + 7 x^5 + x^6).

a(n) = 7*a(n-1) - 11*a(n-2) - 6*a(n-3) + 11*a(n-4) + 7*a(n-5) + a(n-6) for n >= 7.

Mathematica

z = 60; s = x/(1 - x^2); p = (1 - s)(1 - 2s)(1 - 4 s);
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291237 *)

Crossrefs

Cf. A000035, A291219.

Sequence in context: A243382 A242577 A163348 * A037099 A055421 A373928

Adjacent sequences: A291234 A291235 A291236 * A291238 A291239 A291240

Keyword

nonn,easy

Author

Clark Kimberling, Aug 28 2017

Status

approved