login
A145185
Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) and a(1)=11
11
11, 1364, 2537720636, 16342986943522226847837781364, 4365101043708483494615466932242949707161871659736799144058331102381689400753867700636
OFFSET
1,1
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
FORMULA
a(n+1)=a(n)^3+3*a(n) and a(1)=11
a(n)=Floor[((11+Sqrt[11^2+4])/2)^(3^(n-1))]
MATHEMATICA
a = {}; k = 11; Do[AppendTo[a, k]; k = k^3 + 3 k, {n, 1, 6}]; a
or
Table[Floor[((11 + Sqrt[125])/2)^(3^(n - 1))], {n, 1, 5}] (*Artur Jasinski*)
RecurrenceTable[{a[1]==11, a[n]==a[n-1]^3+3*a[n-1]}, a, {n, 5}] (* Harvey P. Dale, Jul 23 2012 *)
KEYWORD
nonn
AUTHOR
Artur Jasinski, Oct 03 2008
STATUS
approved