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A215540
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Least k such that (2*n-1)*2^k + 1 is a prime factor of a Fermat number 2^(2^m) + 1 for some m, or 0 if no such value exists.
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0
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1, 41, 7, 14, 67, 18759, 20, 229, 147, 6838, 41
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OFFSET
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1,2
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COMMENTS
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a(n) >= 7 for n > 1.
a(39279) = 0. No n < 39279 with a(n)=0 is known.
a(12)>2500000, a(13)>2500000, a(14)=455, a(15)=57 (see Ballinger and Keller link).
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LINKS
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MATHEMATICA
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lst = {}; Do[k = 1; While[True, p = n*2^k + 1; If[PrimeQ[p] && IntegerQ@Log[2, MultiplicativeOrder[2, p]], AppendTo[lst, k]; Break[]]; k++], {n, 1, 9, 2}]; lst
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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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