A211664 Number of iterations (...f_4(f_3(f_2(n))))...) such that the result is < 1, where f_j(x):=log_j(x).

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

1,2

Links

Table of n, a(n) for n=1..90.

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..infinity} 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).

Example

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.

Sequence in context: A109038 A208246 A320857 * A182434 A185679 A366721

Adjacent sequences: A211661 A211662 A211663 * A211665 A211666 A211667

Keyword

base,nonn

Author

Hieronymus Fischer, Apr 30 2012

Status

approved