A145182 Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) and a(1)=8.

8, 536, 153992264, 3651713626720249047672536, 48695646535829720063008633136610768101443687873746944465180200686293744264

Offset: 1

Author

Artur Jasinski, Oct 03 2008

Keywords

• nonn

Comments

General formula for continued cotangent recurrences 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

The next term has 222 digits. - Harvey P. Dale, Mar 02 2018

Crossrefs

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

Programs

Mathematica

a = {}; k = 7; Do[AppendTo[a, k]; k = k^3 + 3 k, {n, 1, 6}]; a
or
Table[Floor[((8 + Sqrt[68])/2)^(3^(n - 1))], {n, 1, 5}] (*Artur Jasinski*)
RecurrenceTable[{a[1]==8,a[n]==a[n-1]^3+3a[n-1]},a,{n,5}] (* Harvey P. Dale, Mar 02 2018 *)

Formula

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

a(n)=Floor[((8+Sqrt[8^2+4])/2)^(3^(n-1))]