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a(n) = 8*a(n-1) - 5*a(n-2), with a(0)=0, a(1)=1.
3

%I #26 Jun 17 2022 03:25:05

%S 0,1,8,59,432,3161,23128,169219,1238112,9058801,66279848,484944779,

%T 3548158992,25960548041,189943589368,1389745974739,10168249851072,

%U 74397268934881,544336902223688,3982708873115099,29139986473802352,213206347424843321,1559950847029734808

%N a(n) = 8*a(n-1) - 5*a(n-2), with a(0)=0, a(1)=1.

%H G. C. Greubel, <a href="/A190977/b190977.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (8,-5).

%F a(n) = ((4 + sqrt(11))^n - (4 - sqrt(11))^n)/(2*sqrt(11)). - _Giorgio Balzarotti_, May 28 2011

%F G.f.: x/(1 - 8*x + 5*x^2). - _Philippe Deléham_, Oct 12 2011

%F From _G. C. Greubel_, Jun 17 2022: (Start)

%F a(n) = 5^((n-1)/2)*ChebyshevU(n-1, 4/sqrt(5)).

%F E.g.f.: (1/sqrt(11))*exp(4*x)*sinh(sqrt(11)*x). (End)

%t LinearRecurrence[{8,-5}, {0,1}, 50]

%o (Magma) [n le 2 select n-1 else 8*Self(n-1) -5*Self(n-2): n in [1..41]]; // _G. C. Greubel_, Jun 17 2022

%o (SageMath) [sum( (-1)^k*binomial(n-k-1, k)*5^k*8^(n-2*k-1) for k in (0..((n-1)//2))) for n in (0..40)] # _G. C. Greubel_, Jun 17 2022

%Y Cf. A190958 (index to generalized Fibonacci sequences).

%K nonn

%O 0,3

%A _Vladimir Joseph Stephan Orlovsky_, May 24 2011