1,1

General formula for continued cotangent recurences type:

a(n+1)=a(n)3+3*a(n) and a(1)=k is following:

a(n)=Floor[((k+Sqrt[k^2+4])/2)^(3^(n-1))]

k=1 see A006267

k=2 see A006266

k=3 see A006268

k=4 see A006267(n+1)

k=5 see A006269

k=6 see A145180

k=7 see A145181

k=8 see A145182

k=9 see A145183

k=10 see A145184

k=11 see A145185

k=12 see A145186

k=13 see A145187

k=14 see A145188

k=15 see A145189

Essentially the same as A006266. [From R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Mar 18 2009]

a(n+1)=a(n)3+3*a(n) and a(1)=14

a(n)=Floor[((14+Sqrt[14^2+4])/2)^(3^(n-1))]

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

or

Table[Floor[((14 + Sqrt[200])/2)^(3^(n - 1))], {n, 1, 5}] (*Artur Jasinski*)

A006267, A006266, A006268, A006269, A145180, A145181, A145182, A145183, A145184, A145185, A145186, A145187, A145188, A145189

Sequence in context: A080690 A079918 A013719 this_sequence A064075 A157824 A159372

Adjacent sequences: A145185 A145186 A145187 this_sequence A145189 A145190 A145191

nonn

Artur Jasinski (grafix(AT)csl.pl), Oct 03 2008