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%I #7 Nov 22 2012 12:24:19
%S 7,364,48229636,112186849649044142700364,
%T 1411971263214164889494039458947084336929208169473485667118006013929636
%N Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) and a(1)=7
%C General formula for continued cotangent recurrences type:
%C a(n+1)=a(n)3+3*a(n) and a(1)=k is following:
%C a(n)=Floor[((k+Sqrt[k^2+4])/2)^(3^(n-1))]
%C k=1 see A006267
%C k=2 see A006266
%C k=3 see A006268
%C k=4 see A006267(n+1)
%C k=5 see A006269
%C k=6 see A145180
%C k=7 see A145181
%C k=8 see A145182
%C k=9 see A145183
%C k=10 see A145184
%C k=11 see A145185
%C k=12 see A145186
%C k=13 see A145187
%C k=14 see A145188
%C k=15 see A145189
%F a(n+1)=a(n)^3 + 3*a(n) and a(1)=7
%F a(n)=Floor[((7+Sqrt[7^2+4])/2)^(3^(n-1))]
%t a = {}; k = 7; Do[AppendTo[a, k]; k = k^3 + 3 k, {n, 1, 6}]; a
%t or
%t Table[Floor[((7 + Sqrt[53])/2)^(3^(n - 1))], {n, 1, 5}] (*Artur Jasinski*)
%Y A006267, A006266, A006268, A006269, A145180, A145181, A145182, A145183, A145184, A145185, A145186, A145187, A145188, A145189
%K nonn
%O 1,1
%A _Artur Jasinski_, Oct 03 2008