cp's OEIS Frontend

A145186 Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) and a(1)=12.

%I A145186 #10 May 21 2018 11:14:47
%S A145186 12,1764,5489037036,165382092777963331246695013764,
%T A145186 4523404750894779548516344022127873154658656755028228436816797201835023951822441803129036
%N A145186 Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) and a(1)=12.
%C A145186 General formula for continued cotangent recurrences type:
%C A145186 a(n+1)=a(n)^3+3*a(n) and a(1)=k is following:
%C A145186 a(n)=Floor[((k+Sqrt[k^2+4])/2)^(3^(n-1))]
%C A145186 k=1 see A006267
%C A145186 k=2 see A006266
%C A145186 k=3 see A006268
%C A145186 k=4 see A006267(n+1)
%C A145186 k=5 see A006269
%C A145186 k=6 see A145180
%C A145186 k=7 see A145181
%C A145186 k=8 see A145182
%C A145186 k=9 see A145183
%C A145186 k=10 see A145184
%C A145186 k=11 see A145185
%C A145186 k=12 see A145186
%C A145186 k=13 see A145187
%C A145186 k=14 see A145188
%C A145186 k=15 see A145189
%C A145186 The next term (a(6)) has 263 digits. - _Harvey P. Dale_, May 21 2018
%F A145186 a(n+1)=a(n)3+3*a(n) and a(1)=12.
%F A145186 a(n)=Floor[((12+Sqrt[12^2+4])/2)^(3^(n-1))].
%t A145186 a = {}; k = 12; Do[AppendTo[a, k]; k = k^3 + 3 k, {n, 1, 6}]; a
%t A145186 or
%t A145186 Table[Floor[((12 + Sqrt[148])/2)^(3^(n - 1))], {n, 1, 5}] (*Artur Jasinski*)
%t A145186 NestList[#^3+3#&,12,5] (* _Harvey P. Dale_, May 21 2018 *)
%Y A145186 Cf. A006267, A006266, A006268, A006269, A145180, A145181, A145182, A145183, A145184, A145185, A145186, A145187, A145188, A145189.
%K A145186 nonn
%O A145186 1,1
%A A145186 _Artur Jasinski_, Oct 03 2008