A290999 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - 6*S^2.

0, 6, 12, 54, 168, 606, 2052, 7134, 24528, 84726, 292092, 1007814, 3476088, 11991246, 41362932, 142682094, 492178848, 1697768166, 5856430572, 20201701974, 69685556808, 240379623486, 829187031012, 2860272179454, 9866479513968, 34034319925206, 117401037420252

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 A291000 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 (2,5).

Formula

G.f.: 6*x/(1 - 2*x - 5*x^2).

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

a(n) = 6*A002532(n).

a(n) = sqrt(3/2)*((1+sqrt(6))^n - (1-sqrt(6))^n). - Colin Barker, Aug 23 2017

Mathematica

z = 60; s = x/(1 - x); p = 1 - s^6;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A290999 *)
LinearRecurrence[{2, 5}, {0, 6}, 30] (* Harvey P. Dale, Mar 25 2018 *)

Prog

(Magma) [n le 2 select 6*(n-1) else 2*Self(n-1) +5*Self(n-2): n in [1..41]]; // G. C. Greubel, Apr 25 2023
(SageMath)
A290999=BinaryRecurrenceSequence(2, 5, 0, 6)
[A290999(n) for n in range(41)] # G. C. Greubel, Apr 25 2023

Crossrefs

Cf. A000012, A002532, A033453, A289780, A290997, A290998, A291000.

Sequence in context: A256584 A117866 A365691 * A094060 A341893 A324529

Adjacent sequences: A290996 A290997 A290998 * A291000 A291001 A291002

Keyword

nonn,easy

Author

Clark Kimberling, Aug 22 2017

Status

approved