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A129784
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a(n) = floor(log_10(2^(2^n))).
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0
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0, 1, 2, 4, 9, 19, 38, 77, 154, 308, 616, 1233, 2466, 4932, 9864, 19728, 39456, 78913, 157826, 315652, 631305, 1262611, 2525222, 5050445, 10100890, 20201781, 40403562, 80807124, 161614248, 323228496, 646456993, 1292913986, 2585827972
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OFFSET
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1,3
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COMMENTS
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Starting with 2, n successive squarings yields an (a(n)+1)-digit number.
Dubickas proves that infinitely many terms of this sequence are divisible by 2 or 3 (and hence infinitely many composites). - Charles R Greathouse IV, Feb 04 2016
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LINKS
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EXAMPLE
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a(16) = 19728 because floor(log_10(2^(2^16))) = floor(log_10(2^65536)) = floor(log_10(2.003529930406846*10^19728)) = floor(19728.30179583467) = 19728.
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MATHEMATICA
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PROG
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(PARI) a(n) = floor(log(2^(2^n))/log(10)); \\ Michel Marcus, Dec 30 2015
(PARI) a(n)=logint(2^2^n, 10) \\ impractical except for small n, but avoids rounding; Charles R Greathouse IV, Feb 04 2016
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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