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

%I #15 Mar 06 2023 10:59:47

%S 1,1,2,7,19,69,188,683,1861,6761,18422,66927,182359,662509,1805168,

%T 6558163,17869321,64919121,176888042,642633047,1751011099,6361411349,

%U 17333222948,62971480443,171581218381,623353393081,1698478960862

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

%H Seiichi Manyama, <a href="/A080873/b080873.txt">Table of n, a(n) for n = 0..2000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,10,0,-1).

%F 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).

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

%F a(n+4) = 10*a(n+2) - a(n). [_Richard Choulet_, Dec 04 2008]

%F 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]

%t 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 *)

%Y Cf. A080871, A080872, A080874, A080875.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Feb 22 2003