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%I #5 May 08 2012 10:41:43
%S 1,2,2,2,2,2,2,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,
%T 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 3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3
%N Number of iterations (...f_4(f_3(f_2(n))))...) such that the result is < 1, where f_j(x):=log_j(x).
%F 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 a(E_{i=1..n} (i+1)) = a(E_{i=1..n-1} (i+1))+1, for n>=1.
%F G.f.: g(x)= 1/(1-x)*sum_{k=0..infinity} x^(E_{i=1..k} (i+1)).
%F The explicit first terms of the g.f. are
%F g(x)=(x+x^2+x^(2^3)+x^(2^3^4)+(x^2^3^4^5)+...)/(1-x) =(x+x^2+x^8+x^2417851639229258349412352+...)/(1-x).
%e a(n)=1, 2, 3, 4, 5 for n=1, 2, 2^3, 2^3^4, 2^3^4^5 =1, 2, 8, 2417851639229258349412352, 2^3^1024.
%Y Cf. A001069, A010096, A084558, A211661, A211666, A211668, A211670.
%K base,nonn
%O 1,2
%A _Hieronymus Fischer_, Apr 30 2012
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