%I #12 Mar 13 2016 10:53:51
%S 1,-1,16,353,6535,117764,2114521,37946999,680940352,12219002585,
%T 219261167071,3934482164084,70601418203761,1266891046596143,
%U 22733437423387120,407934982581860369,7320096249069704311,131353797500724143204,2357048258764099246537
%N a(n) = F(2*n)^3 - F(3*n-1)^2 - F(6*n-8).
%C Previous name was: If a(n) is a term of this sequence, it represents the remainder of the expression of the cube of a Fibonacci number in terms of a square of a Fibonacci number and another Fibonacci number; if F(n) is the n-th Fibonacci number, then F(2*n)^3 = F(3*n-1)^2 + F(6*n-8) + a(n).
%C The limit of the ratio of two consecutive members of the sequence as n goes to infinity, is Phi^8 = 8*Phi+5 = 9+4*sqrt(5) where Phi is the golden ratio = 1.618...
%H Colin Barker, <a href="/A171680/b171680.txt">Table of n, a(n) for n = 1..700</a>
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (20,-35,-35,20,-1).
%F From _Colin Barker_, Mar 13 2016: (Start)
%F a(n) = fibonacci(2*n)^3-fibonacci(3*n-1)^2-fibonacci(6*n-8).
%F a(n) = 20*a(n-1)-35*a(n-2)-35*a(n-3)+20*a(n-4)-a(n-5) for n>5.
%F G.f.: x*(1-21*x+71*x^2+33*x^3-20*x^4) / ((1+x)*(1-18*x+x^2)*(1-3*x+x^2)).
%F (End)
%e a(4) = 353 since F(8)^3 = F(11)^2 + F(16) + 353.
%o (PARI) vector(26, n, fibonacci(2*n)^3-fibonacci(3*n-1)^2-fibonacci(6*n-8)) \\ _Colin Barker_, Mar 13 2016
%o (PARI) Vec(x*(1-21*x+71*x^2+33*x^3-20*x^4)/((1+x)*(1-18*x+x^2)*(1-3*x+x^2)) + O(x^25)) \\ _Colin Barker_, Mar 13 2016
%Y Cf. A000045.
%K sign,easy
%O 1,3
%A _Carmine Suriano_, Dec 15 2009
%E New name base on _Colin Barker_'s formula from _Joerg Arndt_, Mar 13 2016