login

Reminder: The OEIS is hiring a new managing editor, and the application deadline is January 26.

Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) and a(1)=12.
11

%I #10 May 21 2018 11:14:47

%S 12,1764,5489037036,165382092777963331246695013764,

%T 4523404750894779548516344022127873154658656755028228436816797201835023951822441803129036

%N Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) and a(1)=12.

%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 (a(6)) has 263 digits. - _Harvey P. Dale_, May 21 2018

%F a(n+1)=a(n)3+3*a(n) and a(1)=12.

%F a(n)=Floor[((12+Sqrt[12^2+4])/2)^(3^(n-1))].

%t a = {}; k = 12; Do[AppendTo[a, k]; k = k^3 + 3 k, {n, 1, 6}]; a

%t or

%t Table[Floor[((12 + Sqrt[148])/2)^(3^(n - 1))], {n, 1, 5}] (*Artur Jasinski*)

%t NestList[#^3+3#&,12,5] (* _Harvey P. Dale_, May 21 2018 *)

%Y Cf. 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