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a(n) = 5*a(n-1) + 3*a(n-2) where a(0) = a(1) = 1.
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%I #18 Jan 01 2024 11:38:29

%S 1,1,8,43,239,1324,7337,40657,225296,1248451,6918143,38336068,

%T 212434769,1177182049,6523214552,36147618907,200307738191,

%U 1109981547676,6150830952953,34084099407793,188872989897824,1046617247712499

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

%D Thomas Koshy, "Fibonacci and Lucas Numbers with Applications", Wiley, 2001

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (5, 3).

%F a(n)=A015536(n+1)-4*A015536(n). G.f.: (1-4x)/(1-5x-3x^2). [From _R. J. Mathar_, Jul 08 2009]

%e a(2) = 5*a(1) + 3*a(0) = 5*1 + 3*1 = 8 which is the third term in the sequence.

%t Transpose[NestList[Flatten[{Rest[#],ListCorrelate[{3,5},#]}]&, {1,1},40]][[1]] (* _Harvey P. Dale_, Mar 23 2011 *)

%Y Cf. A000045, A072264.

%K nonn

%O 0,3

%A _Parthasarathy Nambi_, Nov 07 2005

%E More terms from _Robert G. Wilson v_, Nov 10 2005