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A226366
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Numbers k such that 5*2^k + 1 is a prime factor of a Fermat number 2^(2^m) + 1 for some m.
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15
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OFFSET
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1,1
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COMMENTS
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No other terms below 5330000.
The reason all terms are odd is that if k is even, then 5*2^k + 1 == (-1)*(-1)^k + 1 = (-1)*1 + 1 = 0 (mod 3). So if k is even, then 3 divides 5*2^k + 1, and since 3 divides no other Fermat number than F_0=3 itself, we do not have a Fermat factor. - Jeppe Stig Nielsen, Jul 21 2019
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LINKS
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MATHEMATICA
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lst = {}; Do[p = 5*2^n + 1; If[PrimeQ[p] && IntegerQ@Log[2, MultiplicativeOrder[2, p]], AppendTo[lst, n]], {n, 7, 3313, 2}]; lst
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PROG
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(PARI) isok(n) = my(p = 5*2^n + 1, z = znorder(Mod(2, p))); isprime(p) && ((z >> valuation(z, 2)) == 1); \\ Michel Marcus, Nov 10 2018
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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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