cp's OEIS Frontend

A211666 Number of iterations log_10(log_10(log_10(...(n)...))) such that the result is < 2.

%I A211666 #20 May 14 2025 04:24:11
%S A211666 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 A211666 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 A211666 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 A211666 Number of iterations log_10(log_10(log_10(...(n)...))) such that the result is < 2.
%C A211666 Different from A004216, A057427 and A185114.
%C A211666 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 A211666 g(x) = (1/(1-x))*Sum_{k>=1} 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 g(x) = (x^q+x^(p^q)+x^(p^p^q)+x^(p^p^p^q)+...)/(1-x).
%F A211666 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 A211666 a(E_{i=1..n} 10) = a(E_{i=1..n-1} 10)+1, for n>=1.
%F A211666 G.f.: g(x) = (1/(1-x))*Sum_{k>=1} 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 A211666 The explicit first terms of the g.f. are g(x) = (x^2+x^100+x^(10^100)+...)/(1-x).
%e A211666 a(n) = 0, 1, 2, 3 for n = 1, 2, 10^2, 10^10^2 (= 1, 2, 100, 10^100).
%Y A211666 Cf. A001069, A010096, A211662, A211664, A211668, A211670.
%K A211666 base,nonn
%O A211666 1,100
%A A211666 _Hieronymus Fischer_, Apr 30 2012