A080873 a(n)*a(n+3) - a(n+1)*a(n+2) = 5, given a(0)=a(1)=1, a(2)=2.

1, 1, 2, 7, 19, 69, 188, 683, 1861, 6761, 18422, 66927, 182359, 662509, 1805168, 6558163, 17869321, 64919121, 176888042, 642633047, 1751011099, 6361411349, 17333222948, 62971480443, 171581218381, 623353393081, 1698478960862

Offset

0,3

Links

Seiichi Manyama, Table of n, a(n) for n = 0..2000

Index entries for linear recurrences with constant coefficients, signature (0,10,0,-1).

Formula

For n > 1: a(2*n-1) = 3*a(2*n-2) + 2*a(2*n-3) - a(2*n-4); a(2n) = 3*a(2*n-1) - a(2*n-2).

G.f.: (1 + x - 8*x^2 - 3*x^3) / (1 - 10*x^2 + x^4). - N. J. A. Sloane, Jul 19 2005

a(n+4) = 10*a(n+2) - a(n). [Richard Choulet, Dec 04 2008]

a(n) = (1/24*(3 + 3*3^(1/2)*2^(1/2) + 6*2^(1/2) + 2*3^(1/2))/(3^(1/2)*2^(1/2) + 2))*(sqrt(3) + sqrt(2))^n + (1/16*3^(1/2)*2^(1/2) + 1/8*2^(1/2) + 1/4 + 1/6*3^(1/2))*(sqrt(3) - sqrt(2))^n + (1/24*(3*3^(1/2)*2^(1/2) - 6*2^(1/2) + 3 - 2*3^(1/2))/(3^(1/2)*2^(1/2) + 2))*( - sqrt(3) - sqrt(2))^n + (1/16*3^(1/2)*2^(1/2) - 1/8*2^(1/2) + 1/4 - 1/6*3^(1/2))*( - sqrt(3) + sqrt(2))^n. [Richard Choulet, Dec 04 2008]

Mathematica

CoefficientList[Series[(-3x^3-8x^2+x+1)/(x^4-10x^2+1), {x, 0, 30}], x] (* or *) LinearRecurrence[{0, 10, 0, -1}, {1, 1, 2, 7}, 30] (* Harvey P. Dale, Feb 27 2023 *)

Crossrefs

Cf. A080871, A080872, A080874, A080875.

Sequence in context: A164979 A243279 A362097 * A126162 A054423 A137990

Adjacent sequences: A080870 A080871 A080872 * A080874 A080875 A080876

Keyword

nonn

Author

Paul D. Hanna, Feb 22 2003

Status

approved