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%I #24 Feb 17 2024 12:35:27
%S 1,-1,9,-11,59,-267,433,-2041,9753,-15443,73547,-349875,555073,
%T -2641297,12569673,-19938491,94883195,-451526331,716237425,
%U -3408408073,16219834713,-25728821507,122437560587,-582652240611,924236100865,-4398227463841,20930155058697,-33200601349355
%N a(n) = 3^(-1-floor(n/3))*A215829(n).
%C The Berndt-type sequence number 8a for the argument 2Pi/9 (see A215829 for details).
%C From the recurrence relation for A215829 it can be proved that a(3*n+2) is divisible by 3, a(3*n) is congruent to 1 modulo 3, and a(3*n+1) is congruent to 2 modulo 3, which implies that a(3*n)+a(3*n+1) is divisible by 3.
%H <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,-33,0,0,105,0,0,1).
%F G.f.: (3*x^8-11*x^7+35*x^6-30*x^5-26*x^4-22*x^3-9*x^2+x-1)/(x^9+105*x^6-33*x^3-1). [_Colin Barker_, Oct 28 2012]
%e We have a(6)+3*a(3)=400, while a(30)+3*a(3) is divisible by 1000.
%t LinearRecurrence[{0,0,-33,0,0,105,0,0,1},{1,-1,9,-11,59,-267,433,-2041,9753},30] (* _Harvey P. Dale_, Nov 29 2013 *)
%o (Magma) i:=28; I:=[3,-3,27]; A215829:=[m le 3 select I[m] else -3*Self(m-1)+9*Self(m-2)+3*Self(m-3): m in [1..i]]; [3^(-1-Floor((n-1)/3))*A215829[n]: n in [1..i]]; // _Bruno Berselli_, Oct 02 2012
%K sign,easy
%O 0,3
%A _Roman Witula_, Aug 24 2012