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a(n) = -(sin(2*n*arccos(sqrt(3))))^2.
20

%I #14 Mar 02 2022 13:24:48

%S 0,24,2400,235224,23049600,2258625624,221322261600,21687323011224,

%T 2125136332838400,208241673295152024,20405558846592060000,

%U 1999536525292726728024,195934173919840627286400

%N a(n) = -(sin(2*n*arccos(sqrt(3))))^2.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (99,-99,1)

%F a(n) = 99*a(n-1) - 99*a(n-2) + a(n-3), n > 2.

%F Binet formula: a(n) = -1/2 + (1/4)(49 + 20*sqrt(6))^n + (1/4)(49 - 20*sqrt(6))^n.

%F a(n) = 24*A108741(n).

%F From _R. J. Mathar_, Aug 23 2012: (Start)

%F G.f.: -24*x*(1+x) / ( (x-1)*(x^2-98*x+1) ).

%F a(n) = A132596(2n). (End)

%t Table[ -Round[N[Sin[2 n ArcCos[Sqrt[3]]]^2, 100]], {n, 0, 20}]

%t OR

%t Table[Round[N[ -1/2 + (1/4) (49 + 20 Sqrt[6])^n + (1/4) (49 - 20 Sqrt[6])^n]], {n, 0, 6}]

%t OR

%t Clear[a]; a[n_] := a[n] = 99 a[n - 1] - 99 a[n - 2] + a[n - 3]; a[0] = 0; a[1] = 24; a[2] = 2400; Table[a[n], {n, 0, 10}]

%o (PARI) a(n)=([0,1,0;0,0,1;1,-99,99]^n*[0;24;2400])[1,1] \\ _Charles R Greathouse IV_, Jun 11 2015

%Y Cf. A001079, A108741, A132592, A146311, A146312, A146313.

%K nonn,easy

%O 0,2

%A _Artur Jasinski_, Feb 10 2010