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

%I A001069 #40 Jul 28 2024 18:51:24
%S A001069 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,
%T A001069 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,
%U A001069 3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3
%N A001069 Log2*(n) (version 2): take log_2 of n this many times to get a number < 2.
%C A001069 From _Hieronymus Fischer_, Apr 08 2012: (Start)
%C A001069 In terms of A010096 the definition could read: "Number of iterations log_2(log_2(log_2(...(n)...))) such that the result is < 2".
%C A001069 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)
%H A001069 T. D. Noe, <a href="/A001069/b001069.txt">Table of n, a(n) for n = 1..1000</a>
%F A001069 From _Hieronymus Fischer_, Apr 08 2012: (Start)
%F A001069 a(n) = A010096(n)-1.
%F A001069 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:
%F A001069 a(E_{i=1..n} 2) = a(E_{i=1..n-1} 2) +1, for n>=1.
%F A001069 G.f.: g(x) = 1/(1-x)*Sum_{k >= 1} x^(E_{i=1..k} 2).
%F A001069 The explicit first terms of this g.f. are
%F A001069 g(x) = (x^2+x^4+x^16+x^65536+...)/(1-x).  (End)
%e A001069 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, ...
%t A001069 f[n_] := Length@ NestWhileList[ Log[2, #] &, n, # >= 2 &] - 1; Array[f, 105] (* _Robert G. Wilson v_, Apr 19 2012 *)
%Y A001069 Cf. A010096 (version 1), A230864 (version 3).
%K A001069 nonn,easy
%O A001069 1,4
%A A001069 _N. J. A. Sloane_ and _J. H. Conway_