A215492 a(n) = 21*a(n-2) + 7*a(n-3), with a(0)=0, a(1)=3, and a(2)=6.

0, 3, 6, 63, 147, 1365, 3528, 29694, 83643, 648270, 1964361, 14199171, 45789471, 311933118, 1060973088, 6871121775, 24463966674, 151720368891, 561841152579, 3357375513429, 12860706786396, 74437773850062, 293576471108319, 1653218198356074, 6686170310225133

Offset

0,2

Comments

We have a(n)=B(n;3), where B(n;d), n=1,2,..., d \in C, denote one of the quasi-Fibonacci numbers defined in the comments to A121449 and in the Witula-Slota-Warzynski paper. Its conjugate sequences A(n;3) and C(n;3) are discussed in A121458 and A215484 respectively. Similarly as in A121458 we deduce that each of the following elements a(3*n), a(3*n+1), a(3*n+2) is divided by 3*7^n for every n=0,1,... .

Links

G. C. Greubel, Table of n, a(n) for n = 0..1000

Roman Witula, Damian Slota and Adam Warzynski, Quasi-Fibonacci Numbers of the Seventh Order, J. Integer Seq., 9 (2006), Article 06.4.3.

Index entries for linear recurrences with constant coefficients, signature (0, 21, 7).

Formula

a(n) = (1/7)*((c(1)-c(4))*(1+3*c(1))^n + (c(2)-c(1))*(1+3*c(2))^n + (c(4)-c(2))*(1+3*c(4))^n), where c(j):=2*cos(2*Pi*j/7) (for the proof see Witula-Slota-Warzynski paper).

G.f.: (3*x+6*x^2)/(1-21*x^2-7*x^3).

Mathematica

LinearRecurrence[{0, 21, 7}, {0, 3, 6}, 50]
CoefficientList[Series[(3 x + 6 x^2)/(1 - 21 x^2 - 7 x^3), {x, 0, 33}], x] (* Vincenzo Librandi, Sep 18 2015 *)

Prog

(PARI) concat(0, Vec((3+6*x)/(1-21*x^2-7*x^3)+O(x^99))) \\ Charles R Greathouse IV, Oct 01 2012
(Magma) I:=[0, 3, 6]; [n le 3 select I[n] else 21*Self(n-2)+7*Self(n-3): n in [1..30]]; // Vincenzo Librandi, Sep 18 2015

Crossrefs

Cf. A121458, A215484, A121449, A085810, A215404, A077998, A006054, A033304, A052975, A094789, A005021, A121442, A121458.

Sequence in context: A137134 A137090 A137138 * A125817 A053948 A270745

Adjacent sequences: A215489 A215490 A215491 * A215493 A215494 A215495

Keyword

nonn,easy

Author

Roman Witula, Aug 13 2012

Status

approved