cp's OEIS Frontend

A145184 Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) and a(1)=10.

%I A145184 #6 Jun 02 2025 00:37:49
%S A145184 10,1030,1092730090,1304784252725333839617919270,
%T A145184 2221345538213703371536935622204403026741331806706388823688859272519059871168740810
%N A145184 Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) and a(1)=10.
%C A145184 General formula for continued cotangent recurrences type:
%C A145184 a(n+1)=a(n)3+3*a(n) and a(1)=k is following:
%C A145184 a(n)=Floor[((k+Sqrt[k^2+4])/2)^(3^(n-1))]
%C A145184 k=1 see A006267
%C A145184 k=2 see A006266
%C A145184 k=3 see A006268
%C A145184 k=4 see A006267(n+1)
%C A145184 k=5 see A006269
%C A145184 k=6 see A145180
%C A145184 k=7 see A145181
%C A145184 k=8 see A145182
%C A145184 k=9 see A145183
%C A145184 k=10 see A145184
%C A145184 k=11 see A145185
%C A145184 k=12 see A145186
%C A145184 k=13 see A145187
%C A145184 k=14 see A145188
%C A145184 k=15 see A145189
%F A145184 a(n+1)=a(n)3+3*a(n) and a(1)=10
%F A145184 a(n)=Floor[((10+Sqrt[10^2+4])/2)^(3^(n-1))]
%t A145184 a = {}; k = 10; Do[AppendTo[a, k]; k = k^3 + 3 k, {n, 1, 6}]; a
%t A145184 or
%t A145184 Table[Floor[((10 + Sqrt[104])/2)^(3^(n - 1))], {n, 1, 5}] (*Artur Jasinski*)
%Y A145184 A006267, A006266, A006268, A006269, A145180, A145181, A145182, A145183, A145184, A145185, A145186, A145187, A145188, A145189
%K A145184 nonn
%O A145184 1,1
%A A145184 _Artur Jasinski_, Oct 03 2008