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A103477
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Positive integers k for which 1 + 3*2^(k+2) divides the Fermat number 1 + 2^2^k.
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2
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OFFSET
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1,1
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COMMENTS
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The smallest prime factor any Fermat number 2^(2^k)+1 can have is 3*2^(k+2)+1. - Jeppe Stig Nielsen, Dec 03 2018
On Keller's linked page, to find the terms, you run through the tables and find all rows with k = 3 and with n exactly 2 greater than m, then that m belongs to this sequence. - Jeppe Stig Nielsen, Dec 04 2018
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LINKS
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EXAMPLE
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a(1)=207 because 207 is the smallest positive integer k for which 1 + 3*2^(k+2) divides the Fermat number 1 + 2^2^k.
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MATHEMATICA
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aQ[n_] := PowerMod[2, 2^n, 1 + 3*2^(n+2)] == 3*2^(n+2); Select[Range[100000], aQ] (* Amiram Eldar, Dec 04 2018 *)
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PROG
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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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Serhat Sevki Dincer (mesti_mudam(AT)yahoo.com), Feb 07 2005
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EXTENSIONS
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STATUS
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approved
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