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)=14
11

%I #6 Mar 31 2012 10:22:10

%S 14,2786,21624372014,10111847525912679844192131854786,

%T 1033930953043290626825587838528711318150300040875029341893199068078185510802565166824630504014

%N Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) and a(1)=14

%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 Essentially the same as A006266. [From _R. J. Mathar_, Mar 18 2009]

%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 = 14; Do[AppendTo[a, k]; k = k^3 + 3 k, {n, 1, 6}]; a

%t or

%t Table[Floor[((14 + Sqrt[200])/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