cp's OEIS Frontend

A145188 Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) and a(1)=14.

%I A145188 #8 Jun 02 2025 00:38:03
%S A145188 14,2786,21624372014,10111847525912679844192131854786,
%T A145188 1033930953043290626825587838528711318150300040875029341893199068078185510802565166824630504014
%N A145188 Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) and a(1)=14.
%C A145188 General formula for continued cotangent recurrences type:
%C A145188 a(n+1)=a(n)3+3*a(n) and a(1)=k is following:
%C A145188 a(n)=Floor[((k+Sqrt[k^2+4])/2)^(3^(n-1))]
%C A145188 k=1 see A006267
%C A145188 k=2 see A006266
%C A145188 k=3 see A006268
%C A145188 k=4 see A006267(n+1)
%C A145188 k=5 see A006269
%C A145188 k=6 see A145180
%C A145188 k=7 see A145181
%C A145188 k=8 see A145182
%C A145188 k=9 see A145183
%C A145188 k=10 see A145184
%C A145188 k=11 see A145185
%C A145188 k=12 see A145186
%C A145188 k=13 see A145187
%C A145188 k=14 see A145188
%C A145188 k=15 see A145189
%C A145188 Essentially the same as A006266. [From _R. J. Mathar_, Mar 18 2009]
%F A145188 a(n+1)=a(n)3+3*a(n) and a(1)=14
%F A145188 a(n)=Floor[((14+Sqrt[14^2+4])/2)^(3^(n-1))]
%t A145188 a = {}; k = 14; Do[AppendTo[a, k]; k = k^3 + 3 k, {n, 1, 6}]; a
%t A145188 or
%t A145188 Table[Floor[((14 + Sqrt[200])/2)^(3^(n - 1))], {n, 1, 5}] (*Artur Jasinski*)
%Y A145188 A006267, A006266, A006268, A006269, A145180, A145181, A145182, A145183, A145184, A145185, A145186, A145187, A145188, A145189
%K A145188 nonn
%O A145188 1,1
%A A145188 _Artur Jasinski_, Oct 03 2008