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%I #30 Dec 15 2017 14:58:02
%S 1,6,54,857,15058,269394,4831929,86699846,1555750918,27916779057,
%T 500946173586,8989114087586,161303106727729,2894466805243782,
%U 51939099383032278,932009322077220809,16724228697975221074
%N a(n) = F(2n+1)^3 - F(3n)^2 - F(6n-2), where the F(i) are Fibonacci numbers.
%C The ratio of two consecutive terms of this sequence, as n goes to infinity, is phi^6 = 8*phi+5 = 9+4*sqrt(5) where phi is the golden ratio=1.618...
%H G. C. Greubel, <a href="/A171681/b171681.txt">Table of n, a(n) for n = 1..500</a>
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (20,-35,-35,20,-1).
%F a(n) = 20*a(n-1) - 35*a(n-2) - 35*a(n-3) + 20*a(n-4) - a(n-5). - _R. J. Mathar_, Nov 23 2010
%F G.f.: x*(1-14*x-31*x^2+22*x^3-2*x^4) / ((1+x)*(x^2-3*x+1)*(x^2-18*x+1)).
%F a(n+1) = (-2*(-1)^n + A134493(n+1) + 3*A001519(n+2))/5. - _R. J. Mathar_, Nov 23 2010
%e d(3) = 54 since F(7)^3 = F(9)^2 + F(16) + 54.
%t Table[(1/5)*(3*Fibonacci[2*n + 1] + Fibonacci[6*n - 5] + 2*(-1)^n), {n, 1, 10}] (* _G. C. Greubel_, Apr 18 2016 *)
%t LinearRecurrence[{20,-35,-35,20,-1},{1,6,54,857,15058},20] (* _Harvey P. Dale_, Dec 15 2017 *)
%K nonn,easy
%O 1,2
%A _Carmine Suriano_, Dec 15 2009
%E Simplified the definition. - _N. J. A. Sloane_, Nov 24 2010