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%I #6 Aug 20 2017 19:47:27
%S 15,3420,40001698260,64008151994095341241755497070780,
%T 262244184463346778261182615794616508638576477409715732397097802610370956164308073990185129764340
%N Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) and a(1)=15
%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)=14
%F a(n)=Floor[((14+Sqrt[14^2+4])/2)^(3^(n-1))]
%t a = {}; k = 15; Do[AppendTo[a, k]; k = k^3 + 3 k, {n, 1, 6}]; a
%t or
%t Table[Floor[((15 + Sqrt[229])/2)^(3^(n - 1))], {n, 1, 5}] (*Artur Jasinski*)
%t NestList[#^3+3#&,15,5] (* _Harvey P. Dale_, Aug 20 2017 *)
%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
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