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A083564 a(n) = L(n)*L(2n), where L(n) are the Lucas numbers (A000204).

%I A083564 #43 Jan 01 2024 11:18:01
%S A083564 3,21,72,329,1353,5796,24447,103729,439128,1860621,7880997,33385604,
%T A083564 141421803,599075421,2537719272,10749959329,45537545553,192900159396,
%U A083564 817138154247,3461452823129,14662949371128,62113250430021
%N A083564 a(n) = L(n)*L(2n), where L(n) are the Lucas numbers (A000204).
%C A083564 a(n+1)/a(n) -> (phi)^3 = ((1 + sqrt(5))/2)^3 = 4.236067...
%H A083564 Vincenzo Librandi, <a href="/A083564/b083564.txt">Table of n, a(n) for n = 1..160</a>
%H A083564 C. Pita, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL14/Pita/pita12.html">On s-Fibonomials</a>, J. Int. Seq. 14 (2011) # 11.3.7
%H A083564 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (3, 6, -3, -1).
%F A083564 From _Benoit Cloitre_, Aug 30 2003: (Start)
%F A083564 a(n) = 3*a(n-1) + 6*a(n-2) - 3*a(n-3) - a(n-4);
%F A083564 a(n) = Fibonacci(4*n)/Fibonacci(n) = A000045(4*n)/A000045(n). (End)
%F A083564 a(n) = Lucas(3*n) + (-1)^n*Lucas(n).
%F A083564 From _R. J. Mathar_, Oct 27 2008: (Start)
%F A083564 G.f.: x*(3+12*x-9*x^2-4*x^3)/((1+x-x^2)*(1-4*x-x^2)).
%F A083564 a(n) = A061084(n+1) + 2*A001077(n). (End)
%F A083564 a(n) = (1+phi)^n + (-phi)^n + (2*phi+1)^n + (3-2*phi)^n, phi = (1+sqrt(5))/2. - _Gary Detlefs_, Dec 09 2012
%e A083564 a(4) = Lucas(4)*Lucas(8) = 7*47 = 329.
%t A083564 Table[Fibonacci[n*4]/Fibonacci[n],{n,50}] (* _Vladimir Joseph Stephan Orlovsky_, May 02 2011 *)
%o A083564 (Magma) [Lucas(n)*Lucas(2*n): n in [1..25]]; // _Vincenzo Librandi_, May 03 2011
%Y A083564 Third row of array A028412.
%K A083564 nonn,easy
%O A083564 1,1
%A A083564 _Gary W. Adamson_, Jun 12 2003