cp's OEIS Frontend

A145187 Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) with a(1)=13.

%I A145187 #7 Mar 31 2012 10:22:10
%S A145187 13,2236,11179326964,1397162674037779847605429310236,
%T A145187 2727350312258670490076364505418491429134385511825631286349491134548728023756939667650354964
%N A145187 Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) with a(1)=13.
%C A145187 General formula for continued cotangent recurrences of the form a(n+1)=a(n)^3+3*a(n) with a(1)=k is a(n)=floor(((k+sqrt(k^2+4))/2)^(3^(n-1))).
%C A145187 For k=1 see A006267
%C A145187 k=2 see A006266
%C A145187 k=3 see A006268
%C A145187 k=4 see A006267(n+1)
%C A145187 k=5 see A006269
%C A145187 k=6 see A145180
%C A145187 k=7 see A145181
%C A145187 k=8 see A145182
%C A145187 k=9 see A145183
%C A145187 k=10 see A145184
%C A145187 k=11 see A145185
%C A145187 k=12 see A145186
%C A145187 k=13 see A145187
%C A145187 k=14 see A145188
%C A145187 k=15 see A145189
%F A145187 a(n+1)=a(n)^3+3*a(n) and a(1)=13
%F A145187 a(n)=Floor[((13+Sqrt[13^2+4])/2)^(3^(n-1))]
%t A145187 a = {}; k = 13; Do[AppendTo[a, k]; k = k^3 + 3 k, {n, 1, 6}]; a
%t A145187 or
%t A145187 Table[Floor[((13 + Sqrt[173])/2)^(3^(n - 1))], {n, 1, 5}] (*Artur Jasinski*)
%Y A145187 Cf. A006267, A006266, A006268, A006269, A145180, A145181, A145182, A145183, A145184, A145185, A145186, A145187, A145188, A145189
%K A145187 nonn
%O A145187 1,1
%A A145187 _Artur Jasinski_, Oct 03 2008