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 A001069 Log2*(n) (version 2): take log_2 of n this many times to get a number < 2. 14

%I

%S 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 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 3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3

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

%C Contribution from _Hieronymus Fischer_, Apr 08 2012: (Start)

%C 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 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 T. D. Noe, <a href="/A001069/b001069.txt">Table of n, a(n) for n = 1..1000</a>

%F Contribution from _Hieronymus Fischer_, Apr 08 2012: (Start)

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

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

%F G.f.: g(x)= 1/(1-x)*sum_{k=1..infinity} x^(E_{i=1..k} 2).

%F The explicit first terms of this g.f. are

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

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

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

%K nonn,easy

%O 1,4

%A _N. J. A. Sloane_ and _J. H. Conway_

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Last modified January 27 09:09 EST 2020. Contains 331293 sequences. (Running on oeis4.)