cp's OEIS Frontend

A145182 Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) and a(1)=8.

%I A145182 #8 Jun 02 2025 00:37:36
%S A145182 8,536,153992264,3651713626720249047672536,
%T A145182 48695646535829720063008633136610768101443687873746944465180200686293744264
%N A145182 Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) and a(1)=8.
%C A145182 General formula for continued cotangent recurrences type:
%C A145182 a(n+1)=a(n)3+3*a(n) and a(1)=k is following:
%C A145182 a(n)=Floor[((k+Sqrt[k^2+4])/2)^(3^(n-1))]
%C A145182 k=1 see A006267
%C A145182 k=2 see A006266
%C A145182 k=3 see A006268
%C A145182 k=4 see A006267(n+1)
%C A145182 k=5 see A006269
%C A145182 k=6 see A145180
%C A145182 k=7 see A145181
%C A145182 k=8 see A145182
%C A145182 k=9 see A145183
%C A145182 k=10 see A145184
%C A145182 k=11 see A145185
%C A145182 k=12 see A145186
%C A145182 k=13 see A145187
%C A145182 k=14 see A145188
%C A145182 k=15 see A145189
%C A145182 The next term has 222 digits. - _Harvey P. Dale_, Mar 02 2018
%F A145182 a(n+1)=a(n)3+3*a(n) and a(1)=8
%F A145182 a(n)=Floor[((8+Sqrt[8^2+4])/2)^(3^(n-1))]
%t A145182 a = {}; k = 7; Do[AppendTo[a, k]; k = k^3 + 3 k, {n, 1, 6}]; a
%t A145182 or
%t A145182 Table[Floor[((8 + Sqrt[68])/2)^(3^(n - 1))], {n, 1, 5}] (*Artur Jasinski*)
%t A145182 RecurrenceTable[{a[1]==8,a[n]==a[n-1]^3+3a[n-1]},a,{n,5}] (* _Harvey P. Dale_, Mar 02 2018 *)
%Y A145182 A006267, A006266, A006268, A006269, A145180, A145181, A145182, A145183, A145184, A145185, A145186, A145187, A145188, A145189
%K A145182 nonn
%O A145182 1,1
%A A145182 _Artur Jasinski_, Oct 03 2008