%I #28 Jan 11 2020 15:57:47
%S 3,23,25165823
%N a(n)= ackb(n,3) where ackb is the AckermannBurnell function.
%C ackb(x,z) =
%C { x+2 for z=0
%C {
%C { z for x=0, z>0
%C {
%C { ackb(ackb(x1,z), z1) for x,z > 0
%C This version of the Ackermann function was created with the goal of creating the fastest growth with the least total number of operators, recursive calls, and conditional tests. Check the link for more details.
%C The reason we take ackb(n,3) is that it is the only sequence that can have its own entry in the OEIS.
%C a(3) has 7575669 decimal digits and is too big to be included in the data section.
%H Robert Munafo, <a href="http://mrob.com/pub/math/ln2deep.html">Versions of Ackermann's functions</a>
%F a(n) = ackb(x,3) = (3/2) ie3(sqrt(8), x, 8/3)  1 where ie3(a, b, c) = a^(a^( ... a^c))) (with b copies of a).
%F For proof, check the link above.
%e a(1) = ackb(1,3) = (3/2) ie3(sqrt(8), 1, 8/3)  1 = (3/2)sqrt(8)^(8/3)  1 = (3/2)2^((3/2)(8/3))  1 = (3/2)16  1 = 23.
%t ie3[a_, b_, c_] := Nest[a^# &, c, b]; Table[(3/2) ie3[Sqrt@8, x, 8/3]  1, {x, 0, 2}] (* _Michael De Vlieger_, Dec 01 2015 *)
%Y Cf. A083329.
%K nonn
%O 0,1
%A _Natan Arie Consigli_, Nov 28 2015
