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A264929 a(n)= ackb(n,3) where ackb is the Ackermann-Burnell function.

%I A264929 #28 Jan 11 2020 15:57:47
%S A264929 3,23,25165823
%N A264929 a(n)= ackb(n,3) where ackb is the Ackermann-Burnell function.
%C A264929 ackb(x,z) =
%C A264929 { x+2 for z=0
%C A264929 {
%C A264929 { z for x=0, z>0
%C A264929 {
%C A264929 { ackb(ackb(x-1,z), z-1) for x,z > 0
%C A264929 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 A264929 The reason we take ackb(n,3) is that it is the only sequence that can have its own entry in the OEIS.
%C A264929 a(3) has 7575669 decimal digits and is too big to be included in the data section.
%H A264929 Robert Munafo, <a href="http://mrob.com/pub/math/ln-2deep.html">Versions of Ackermann's functions</a>
%F A264929 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 A264929 For proof, check the link above.
%e A264929 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 A264929 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 A264929 Cf. A083329.
%K A264929 nonn
%O A264929 0,1
%A A264929 _Natan Arie Consigli_, Nov 28 2015