cp's OEIS Frontend

A145189 Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) and a(1)=15.

%I A145189 #8 Jun 02 2025 00:38:10
%S A145189 15,3420,40001698260,64008151994095341241755497070780,
%T A145189 262244184463346778261182615794616508638576477409715732397097802610370956164308073990185129764340
%N A145189 Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) and a(1)=15.
%C A145189 General formula for continued cotangent recurrences type:
%C A145189 a(n+1)=a(n)3+3*a(n) and a(1)=k is following:
%C A145189 a(n)=Floor[((k+Sqrt[k^2+4])/2)^(3^(n-1))]
%C A145189 k=1 see A006267
%C A145189 k=2 see A006266
%C A145189 k=3 see A006268
%C A145189 k=4 see A006267(n+1)
%C A145189 k=5 see A006269
%C A145189 k=6 see A145180
%C A145189 k=7 see A145181
%C A145189 k=8 see A145182
%C A145189 k=9 see A145183
%C A145189 k=10 see A145184
%C A145189 k=11 see A145185
%C A145189 k=12 see A145186
%C A145189 k=13 see A145187
%C A145189 k=14 see A145188
%C A145189 k=15 see A145189
%F A145189 a(n+1)=a(n)3+3*a(n) and a(1)=14
%F A145189 a(n)=Floor[((14+Sqrt[14^2+4])/2)^(3^(n-1))]
%t A145189 a = {}; k = 15; Do[AppendTo[a, k]; k = k^3 + 3 k, {n, 1, 6}]; a
%t A145189 or
%t A145189 Table[Floor[((15 + Sqrt[229])/2)^(3^(n - 1))], {n, 1, 5}] (*Artur Jasinski*)
%t A145189 NestList[#^3+3#&,15,5] (* _Harvey P. Dale_, Aug 20 2017 *)
%Y A145189 A006267, A006266, A006268, A006269, A145180, A145181, A145182, A145183, A145184, A145185, A145186, A145187, A145188, A145189
%K A145189 nonn
%O A145189 1,1
%A A145189 _Artur Jasinski_, Oct 03 2008