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%I #15 May 07 2020 08:32:38
%S 0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,
%T 1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,
%U 1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2
%N Number of iterations log_10(log_10(log_10(...(n)...))) such that the result is < 2.
%C Different from A004216, A057427 and A185114.
%C For a general definition like "Number of iterations log_p(log_p(log_p(...(n)...))) such that the result is < q", where p > 1, q > 0, the resulting g.f. is
%C g(x) = 1/(1-x)*sum_{k=1..infinity} x^(E_{i=1..k} b(i,k)), where b(i,k)=p for i<k and b(i,k)=q for i=k. The explicit first terms of the g.f. are
%C g(x) = (x^q+x^(p^q)+x^(p^p^q)+x^(p^p^p^q)+…)/(1-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} c := 1; example: E_{i=1..3} 10 = 10^(10^10) = 10^10000000000, we get:
%F a(E_{i=1..n} 10) = a(E_{i=1..n-1} 10)+1, for n>=1.
%F G.f.: g(x)= 1/(1-x)*sum_{k=1..infinity} x^(E_{i=1..k} b(i,k)), where b(i,k)=10 for i<k and b(i,k)=2 for i=k.
%F The explicit first terms of the g.f. are
%F g(x) = (x^2+x^100+x^(10^100)+...)/(1-x).
%e a(n) = 0, 1, 2, 3 for n = 1, 2, 10^2, 10^10^2 = 1, 2, 100, 10^100.
%Y Cf. A001069, A010096, A211662, A211664, A211668, A211670.
%K base,nonn
%O 1,100
%A _Hieronymus Fischer_, Apr 30 2012
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