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%I #40 Feb 15 2024 08:45:44
%S 3,-10,94,-907,8778,-84965,822409,-7960417,77051978,-745816120,
%T 7219044849,-69875948152,676356530853,-6546718419225,63368238651539,
%U -613365874726862,5937007312894778,-57466607266115655,556241684847745354,-5384080019366211797
%N a(n) = -10*a(n-1) - 3*a(n-2) + a(n-3) with a(0) = 3, a(1) = -10, and a(2) = 94.
%C a(n) = (a/b)^n + (b/c)^n + (c/a)^n, where a = cos(Pi/13) + cos(5*Pi/13), b = cos(3*Pi/13) + cos(11*Pi/13), and c = cos(7*Pi/13) + cos(9*Pi/13).
%C The Berndt-type sequence number 11 for the argument 2Pi/13. I am very grateful to Sergey Markelov and LiveJournal for his and its respectively inspiration for creating this sequence.
%H Sergey Markelov, <a href="http://ru-math.livejournal.com/797774.html">Identity for Pi/19 cosines with cube roots</a>, LiveJournal for Mathematics in Russian, 2012 (in Russian).
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (-10,-3,1).
%F a(n) = -10*a(n-1)-3*a(n-2)+a(n-3). G.f.: -(3*x^2+20*x+3) / (x^3-3*x^2-10*x-1). - _Colin Barker_, Jun 01 2013
%t LinearRecurrence[{-10, -3, 1}, {3, -10, 94}, 20] (* _T. D. Noe_, Sep 17 2012 *)
%K sign,easy
%O 0,1
%A _Roman Witula_, Sep 15 2012
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