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A228846
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Largest m such that (2k+1)*2^m + 1 is a prime factor of the Fermat number 2^(2^n) + 1.
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4
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1, 2, 4, 8, 16, 7, 8, 9, 11, 16, 14, 14
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OFFSET
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0,2
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COMMENTS
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a(n) >= n + 2 for n >= 2.
a(n) = A228845(n) if 2^(2^n) + 1 is prime or semiprime.
a(n) = max (A007814(p_i-1)), where p_i are the prime factors of 2^(2^n)+1. - Ralf Stephan, Sep 09 2013
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LINKS
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EXAMPLE
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F(5) = 641*6700417 and max(A007814(640),A007814(6700416))=7, so a(5)=7.
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CROSSREFS
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KEYWORD
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nonn,hard,more
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AUTHOR
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STATUS
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approved
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