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A204620
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Numbers k such that 3*2^k + 1 is a prime factor of a Fermat number 2^(2^m) + 1 for some m.
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17
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OFFSET
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1,1
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COMMENTS
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Terms are odd: by Morehead's theorem, 3*2^(2*n) + 1 can never divide a Fermat number.
No other terms below 7516000.
Is this sequence the same as "Numbers k such that 3*2^k + 1 is a factor of a Fermat number 2^(2^m) + 1 for some m"? - Arkadiusz Wesolowski, Nov 13 2018
The last sentence of Morehead's paper is: "It is easy to show that _composite_ numbers of the forms 2^kappa * 3 + 1, 2^kappa * 5 + 1 can not be factors of Fermat's numbers." [a proof is needed]. - Jeppe Stig Nielsen, Jul 23 2019
Any factor of a Fermat number 2^(2^m) + 1 of the form 3*2^k + 1 is prime if k < 2*m + 6. - Arkadiusz Wesolowski, Jun 12 2021
If, for any m >= 0, F(m) = 2^(2^m) + 1 has a prime factor p of the form 3*2^k + 1, then F(m)/p is congruent to 11 mod 30. - Arkadiusz Wesolowski, Jun 13 2021
A number k belongs to this sequence if and only if the order of 2 modulo p is not divisible by 3, where p is a prime of the form 3*2^k + 1 (see Golomb paper). - Arkadiusz Wesolowski, Jun 14 2021
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LINKS
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MATHEMATICA
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lst = {}; Do[p = 3*2^n + 1; If[PrimeQ[p] && IntegerQ@Log[2, MultiplicativeOrder[2, p]], AppendTo[lst, n]], {n, 7, 209, 2}]; lst
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PROG
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(PARI) isok(n) = my(p = 3*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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