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%I A211661 #8 Mar 08 2020 19:46:33
%S A211661 1,1,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,3,3,3,3,3,3,3,3,
%T A211661 3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,
%U A211661 3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3
%N A211661 Number of iterations log_3(log_3(log_3(...(n)...))) such that the result is < 1.
%C A211661 For n<16 same as A211663.
%F A211661 With the exponentiation definition E_{i=1..n} c(i) := c(1)^(c(2)^(c(3)^(...(c(n-1)^(c(n))))...))); E_{i=1..0} := 1; example: E_{i=1..4} 3 = 3^(3^(3^3)) = 3^(3^27), we get:
%F A211661 a(E_{i=1..n} 3) = a(E_{i=1..n-1} 3)+1, for n>=1.
%F A211661 G.f.: g(x)= 1/(1-x)*sum_{k=0..infinity} x^(E_{i=1..k} 3). The explicit first terms of the g.f. are
%F A211661 g(x)=(x+x^3+x^27+x^7625597484987+…)/(1-x).
%e A211661 a(n)=1, 2, 3, 4, 5 for n=1, 3, 3^3, 3^3^3, 3^3^3^3 =1, 3, 27, 7625597484987, 3^7625597484987
%t A211661 Table[Length[NestWhileList[Log[3,#]&,n,#>=1&]],{n,90}]-1 (* _Harvey P. Dale_, Mar 08 2020 *)
%Y A211661 Cf. A001069, A010096, A211664, A211666, A211668, A211669.
%K A211661 base,nonn
%O A211661 1,3
%A A211661 _Hieronymus Fischer_, Apr 30 2012

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