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

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, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2

Offset

1,100

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..105.

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: A114295 A004216 A076489 * A297037 A127322 A340172

Adjacent sequences: A211663 A211664 A211665 * A211667 A211668 A211669

Keyword

base,nonn

Author

Hieronymus Fischer, Apr 30 2012

Status

approved