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Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) and a(1)=7
11

%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