

A001069


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


16



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
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OFFSET

1,4


COMMENTS

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)


LINKS



FORMULA

With the exponentiation definition E_{i=1..n} c(i) := c(1)^(c(2)^(c(3)^(...(c(n1)^(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..n1} 2) +1, for n>=1.
G.f.: g(x) = 1/(1x)*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+...)/(1x). (End)


EXAMPLE

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, ...


MATHEMATICA

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


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



STATUS

approved



