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%I #27 Sep 30 2022 14:37:57
%S 1,2,4,9,24,72,225,702,2160,6561,19764,59292,177633,532170,1595052,
%T 4782969,14346720,43040160,129127041,387400806,1162241784,3486784401,
%U 10460412252,31381236756,94143533121,282430067922,847289140884,2541865828329,7625595890664
%N a(n) = (3^n + 2*3^(n/2)*cos(n*Pi/6))/3.
%C From an Internat. Math. Olympiad problem.
%C Comment from _Neil Fernandez_, Apr 29 2014: The question was proposed at the IMO in 1992 and got on to the long list, as question 19, but not the shortlist.
%D E. Lozansky and C. Rousseau, Winning Solutions, Springer, 1996; see p. 154.
%H The Art of Problem Solving, <a href="https://artofproblemsolving.com/community/c4022_1992_imo_longlists">IMO Longlists 1992</a> (From _Neil Fernandez_, Jul 29 2014)
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (6,-12,9).
%F G.f.: (1 - 4*x + 4*x^2)/((1 - 3*x)*(1 - 3*x + 3*x^2)).
%F a(n) = Sum_{k=0..n} binomial(n, k) * Sum_{j=0..floor(k/3)} binomial(k, 3*j). - _Joseph M. Shunia_, Jul 25 2022
%K nonn,easy
%O 0,2
%A _N. J. A. Sloane_, Feb 18 2004
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