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A010096
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log2*(n) (version 1): number of times floor(log_2(x)) is used in floor(log_2(floor(log_2(...(floor(log_2(n)))...)))) = 0.
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15
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1, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4
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OFFSET
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1,2
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COMMENTS
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A possibly simpler definition could be: "Number of iterations log_2(log_2(log_2(...(n)...))) such that the result is < 1".
Changing "< 1" to "<= 1" produces version 3, A230864.
With the only difference in the termination criterion, the definition is essentially the same as version 2, A001069. If we change the definition to "floor(log_2(... = 1" we get A001069. Therefore we get A001069 when subtracting 1 from each term. (End)
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LINKS
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FORMULA
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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>=0} x^(E_{i=1..k} 2).
The explicit first terms of this g.f. are
g(x) = (x + x^2 + x^4 + x^16 + x^65536 + ...)/(1-x). (End)
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EXAMPLE
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Becomes 5 at 65536, 6 at 2^65536, etc.
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MATHEMATICA
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f[n_] := Length@ NestWhileList[ Log[2, #] &, n, # >= 1 &] - 1; Array[f, 105] (* Robert G. Wilson v, Apr 19 2012 *)
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PROG
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(Haskell)
a010096 = length . takeWhile (/= 0) . iterate a000523
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CROSSREFS
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KEYWORD
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nonn,nice
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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