cp's OEIS Frontend

A145185 Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) and a(1)=11.

%I A145185 #8 Jun 02 2025 00:37:56
%S A145185 11,1364,2537720636,16342986943522226847837781364,
%T A145185 4365101043708483494615466932242949707161871659736799144058331102381689400753867700636
%N A145185 Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) and a(1)=11.
%C A145185 General formula for continued cotangent recurrences type:
%C A145185 a(n+1)=a(n)3+3*a(n) and a(1)=k is following:
%C A145185 a(n)=Floor[((k+Sqrt[k^2+4])/2)^(3^(n-1))]
%C A145185 k=1 see A006267
%C A145185 k=2 see A006266
%C A145185 k=3 see A006268
%C A145185 k=4 see A006267(n+1)
%C A145185 k=5 see A006269
%C A145185 k=6 see A145180
%C A145185 k=7 see A145181
%C A145185 k=8 see A145182
%C A145185 k=9 see A145183
%C A145185 k=10 see A145184
%C A145185 k=11 see A145185
%C A145185 k=12 see A145186
%C A145185 k=13 see A145187
%C A145185 k=14 see A145188
%C A145185 k=15 see A145189
%F A145185 a(n+1)=a(n)^3+3*a(n) and a(1)=11
%F A145185 a(n)=Floor[((11+Sqrt[11^2+4])/2)^(3^(n-1))]
%t A145185 a = {}; k = 11; Do[AppendTo[a, k]; k = k^3 + 3 k, {n, 1, 6}]; a
%t A145185 or
%t A145185 Table[Floor[((11 + Sqrt[125])/2)^(3^(n - 1))], {n, 1, 5}] (*Artur Jasinski*)
%t A145185 RecurrenceTable[{a[1]==11,a[n]==a[n-1]^3+3*a[n-1]},a,{n,5}] (* _Harvey P. Dale_, Jul 23 2012 *)
%Y A145185 A006267, A006266, A006268, A006269, A145180, A145181, A145182, A145183, A145184, A145185, A145186, A145187, A145188, A145189
%K A145185 nonn
%O A145185 1,1
%A A145185 _Artur Jasinski_, Oct 03 2008