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A211666 Number of iterations log_10(log_10(log_10(...(n)...))) such that the result is < 2. 10
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, 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, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 (list; graph; refs; listen; history; text; internal format)
OFFSET

1

COMMENTS

Different from A004216, A057427 and A185114.

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

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

g(x) = (x^q+x^(p^q)+x^(p^p^q)+x^(p^p^p^q)+…)/(1-x).

LINKS

Table of n, a(n) for n=1..85.

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} c := 1; example: E_{i=1..3} 10 = 10^(10^10) = 10^10000000000, we get:

a(E_{i=1..n} 10) = a(E_{i=1..n-1} 10)+1, for n>=1.

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.

The explicit first terms of the g.f. are

g(x)=(x^2+x^100+x^(10^100)+…)/(1-x).

EXAMPLE

a(n)=0, 1, 2, 3 for n=1, 2, 10^2, 10^10^2 =1, 2, 100, 10^100.

CROSSREFS

Cf. A001069, A010096, A211662, A211664, A211668, A211670.

Sequence in context: A204447 A188642 A267084 * A226523 A240351 A249832

Adjacent sequences:  A211663 A211664 A211665 * A211667 A211668 A211669

KEYWORD

base,nonn

AUTHOR

Hieronymus Fischer, Apr 30 2012

STATUS

approved

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Last modified October 19 05:08 EDT 2018. Contains 316336 sequences. (Running on oeis4.)