A211662 Number of iterations log_3(log_3(log_3(...(n)...))) such that the result is < 2.

0, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2

Offset

1,9

Links

Table of n, a(n) for n=1..107.

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} 3) = a(E_{i=1..n-1} 3)+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)=3 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^9+x^19683+…)/(1-x).

Example

a(n)=0, 1, 2, 3, 4, for n=1, 2, 3^2, 3^3^2, 3^3^3^2 =1, 2, 9, 19683, 3^19683

Crossrefs

Cf. A001069, A010096, A211661, A211664, A211666, A211668, A211669.

Sequence in context: A329683 A130130 A046698 * A007395 A036453 A040000

Adjacent sequences: A211659 A211660 A211661 * A211663 A211664 A211665

Keyword

nonn

Author

Hieronymus Fischer, Apr 30 2012

Status

approved