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Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) and a(1)=8
11

%I #6 Mar 02 2018 13:14:49

%S 8,536,153992264,3651713626720249047672536,

%T 48695646535829720063008633136610768101443687873746944465180200686293744264

%N Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) and a(1)=8

%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

%C The next term has 222 digits. - _Harvey P. Dale_, Mar 02 2018

%F a(n+1)=a(n)3+3*a(n) and a(1)=8

%F a(n)=Floor[((8+Sqrt[8^2+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[((8 + Sqrt[68])/2)^(3^(n - 1))], {n, 1, 5}] (*Artur Jasinski*)

%t RecurrenceTable[{a[1]==8,a[n]==a[n-1]^3+3a[n-1]},a,{n,5}] (* _Harvey P. Dale_, Mar 02 2018 *)

%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