A001069 Log2*(n) (version 2): take log_2 of n this many times to get a number < 2.

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

Offset: 1

Author

N. J. A. Sloane and J. H. Conway

Keywords

• nonn
• easy

Comments

From Hieronymus Fischer, Apr 08 2012: (Start)

In terms of A010096 the definition could read: "Number of iterations log_2(log_2(log_2(...(n)...))) such that the result is < 2".

With the only difference in the termination criterion, the definition is essentially the same as A010096. If we change the definition to " ...number < 1" we get A010096. Therefore we get A010096 when adding 1 to each term. (End)

Examples

			a(n)=1, 2, 3, 4, 5, ... for n=2, 2^2, 2^2^2, 2^2^2^2, 2^2^2^2^2, ... =2, 4, 16, 65536, 2^65536, ...
		

Links

• T. D. Noe, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A010096 (version 1), A230864 (version 3).

Programs

Mathematica

f[n_] := Length@ NestWhileList[ Log[2, #] &, n, # >= 2 &] - 1; Array[f, 105] (* Robert G. Wilson v, Apr 19 2012 *)

Formula

From Hieronymus Fischer, Apr 08 2012: (Start)

a(n) = A010096(n)-1.

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} 2 = 2^(2^(2^2)) = 2^16, we get:

a(E_{i=1..n} 2) = a(E_{i=1..n-1} 2) +1, for n>=1.

G.f.: g(x) = 1/(1-x)*Sum_{k >= 1} x^(E_{i=1..k} 2).

The explicit first terms of this g.f. are

g(x) = (x^2+x^4+x^16+x^65536+...)/(1-x). (End)