%I #40 Dec 16 2023 17:45:34
%S 5,10,50,325,2210,15130,103685,710650,4870850,33385285,228826130,
%T 1568397610,10749957125,73681302250,505019158610,3461452808005,
%U 23725150497410,162614600673850,1114577054219525,7639424778862810
%N a(n) = Lucas(4n) + 3, or 5*Fibonacci(2n-1)*Fibonacci(2n+1).
%D Hugh C. Williams, Edouard Lucas and Primality Testing, John Wiley and Sons, 1998, p. 75.
%H G. C. Greubel, <a href="/A081076/b081076.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (8,-8,1).
%F a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3).
%F G.f.: 5*(1 -6*x +2*x^2)/((1-x)*(1-7*x+x^2)). - _Colin Barker_, Jun 22 2012
%F Sum_{n>=0} 1/a(n) = phi/5, where phi = A001622 = (1 + sqrt(5))/2. - _Diego Rattaggi_, Apr 22 2020
%F From _G. C. Greubel_, May 26 2020: (Start)
%F a(n) = 5*A064170(n+1).
%F a(n) = Lucas(n)^4 - 4*(-1)^n*Lucas(n)^2 + 5.
%F E.g.f.: 3*exp(x) + 2*exp(7*x/2)*cosh(3*sqrt(5)*x/2). (End)
%p luc := proc(n) option remember: if n=0 then RETURN(2) fi: if n=1 then RETURN(1) fi: luc(n-1)+luc(n-2): end: for n from 0 to 30 do printf(`%d,`,luc(4*n)+3) od: # _James A. Sellers_, Mar 05 2003
%t Table[LucasL[4 n] + 3, {n,0,30}] (* _Wesley Ivan Hurt_, Nov 20 2014 *)
%o (PARI) Vec(-5*(2*x^2-6*x+1)/((x-1)*(x^2-7*x+1)) + O(x^30)) \\ _Michel Marcus_, Dec 23 2014
%o (Magma) [Lucas(4*n) +3: n in [0..30]]; // _G. C. Greubel_, May 26 2020
%o (Sage) [lucas_number2(4*n,1,-1) + 3 for n in (0..30)] # _G. C. Greubel_, May 26 2020
%o (GAP) List([0..30], n-> Lucas(1, -1, 4*n)[2] +3 ); # _G. C. Greubel_, May 26 2020
%Y Cf. A000032 (Lucas numbers), A000045 (Fibonacci numbers), A056854.
%K nonn,easy
%O 0,1
%A _R. K. Guy_, Mar 04 2003
%E More terms from _James A. Sellers_, Mar 05 2003