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Number of iterations log(log(log(...(n)...))) such that the result is < 1.
2

%I #9 Apr 23 2023 12:17:50

%S 1,1,2,2,2,2,2,2,2,2,2,2,2,2,2,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,

%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

%N Number of iterations log(log(log(...(n)...))) such that the result is < 1.

%C Same as A211661 for n < 16.

%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(ceiling(E_{i=1..n} e)) = a(ceiling(E_{i=1..n-1} e))+1, for n>=1.

%F G.f.: g(x) = (1/(1-x))*Sum_{k>=0} x^(ceiling(E_{i=1..k} e)). The explicit first terms of the g.f. are g(x) = (x + x^3 + x^16 + x^3814280 + ...)/(1-x).

%e a(n)=1, 2, 3, 4, for n=1, ceiling(e), ceiling(e^e), ceiling(e^e^e), = 1, 3, 16, 3814280, respectively.

%Y Cf. A001069, A010096, A211662, A211664, A211666, A211668, A211669.

%K base,nonn

%O 1,3

%A _Hieronymus Fischer_, Apr 30 2012