A211664 Number of iterations (...(log_4(log_3(log_2(n))))...) such that the result is < 1.

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, 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, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3

Offset: 1

Author

Hieronymus Fischer, Apr 30 2012

Keywords

• base
• nonn

Examples

			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).
		

Crossrefs

Cf. A001069, A010096, A084558, A211661, A211666, A211668, A211670.

Formula

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:

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

G.f.: g(x) = (1/(1-x))*Sum_{k>=0} x^(E_{i=1..k} (i+1)).

The explicit first terms of the g.f. are 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).