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%I #9 Jun 13 2015 00:52:37
%S 1,40,495,8616,124011,1905804,28383615,427423824,6403870611,
%T 96118424820,1441505073735,21624751859256,324361491427611,
%U 4865500724823324,72982158337539855,1094735196058294944,16421015247083935011,246315330264968309700,3694729496968781349975
%N Sequence generated from Z/6Z addition table considered as a matrix.
%C A140044 = the analogous sequence for Z/5Z; A095897 for Z/4Z and A007070 for Z/3Z.
%H <a href="/index/Rec">Index entries for linear recurrences with constant coefficients</a>, signature (12,93,-576,-2592,5184,19440).
%F Let X = the Z/6Z addition table considered as a matrix: [0,1,2,3,4,5; 1,2,3,4,5,0; 2,3,4,5,0,1; 3,4,5,0,1,2; 4,5,0,1,2,3; 5,0,1,2,3,4]. a(n) = term (1,2) of X^n.
%F G.f.: -x*(216*x^4-468*x^3-78*x^2+28*x+1) / ((3*x+1)*(6*x-1)*(6*x+1)*(15*x-1)*(12*x^2-1)). - _Colin Barker_, May 25 2013
%e a(3) = 495 since term (1,2) of X^3 = 495.
%p a:= n-> (Matrix(6, (i, j)-> irem(i+j-2, 6))^n)[1, 2]:
%p seq(a(n), n=0..20); # _Alois P. Heinz_, May 25 2013
%Y Cf. A140044, A095897, A007070.
%K nonn,easy
%O 1,2
%A _Gary W. Adamson_ and _Roger L. Bagula_, May 02 2008
%E More terms from _Colin Barker_, May 25 2013
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