%I #26 Aug 23 2021 12:05:00
%S 1,11,79,545,3739,25631,175681,1204139,8253295,56568929,387729211,
%T 2657535551,18215019649,124847601995,855718194319,5865179758241,
%U 40200540113371,275538601035359,1888569667134145,12944449068903659,88722573815191471,608113567637436641
%N Numbers w such that (F(2*n-1)^2, -F(2*n)^2, w) are primitive solutions of the Diophantine equation 2*x^3 + 2*y^3 + z^3 = 1, where F(n) is the n-th Fibonacci number (A000045).
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (8,-8,1).
%F a(n) = (2*F(2*n)^6 - 2*F(2*n-1)^6 + 1)^(1/3).
%F From _Colin Barker_, Oct 01 2020: (Start)
%F G.f.: x*(1 + 3*x - x^2) / ((1 - x)*(1 - 7*x + x^2)).
%F a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3) for n>3.
%F (End)
%F a(n) = 2*A003482(n) + 1. - _Hugo Pfoertner_, Oct 01 2020
%F a(n) = A033888(n) - A064170(n+2). - _Flávio V. Fernandes_, Jan 10 2021
%e 2*(F(3)^2)^3 + 2*(-F(4)^2)^3 + 11^3 = 2*4^3 + 2*(-9)^3 + 11^3 = 1, 11 is a term.
%t Table[(2*Fibonacci[2n]^6 - 2*Fibonacci[2n-1]^6 + 1)^(1/3), {n, 22}]
%t LinearRecurrence[{8,-8,1},{1,11,79},30] (* _Harvey P. Dale_, Aug 23 2021 *)
%Y Cf. A000045, A000578, A003482, A056573, A337928.
%K nonn,easy
%O 1,2
%A _XU Pingya_, Sep 30 2020