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A040076
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Smallest m >= 0 such that n*2^m + 1 is prime, or -1 if no such m exists.
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21
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0, 0, 1, 0, 1, 0, 2, 1, 1, 0, 1, 0, 2, 1, 1, 0, 3, 0, 6, 1, 1, 0, 1, 2, 2, 1, 2, 0, 1, 0, 8, 3, 1, 2, 1, 0, 2, 5, 1, 0, 1, 0, 2, 1, 2, 0, 583, 1, 2, 1, 1, 0, 1, 1, 4, 1, 2, 0, 5, 0, 4, 7, 1, 2, 1, 0, 2, 1, 1, 0, 3, 0, 2, 1, 1, 4, 3, 0, 2, 3, 1, 0, 1, 2, 4, 1, 2, 0, 1, 1, 8, 7, 2, 582, 1, 0, 2, 1, 1, 0, 3, 0
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OFFSET
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1,7
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COMMENTS
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Sierpiński showed that a(n) = -1 infinitely often. John Selfridge showed that a(78557) = -1 and it is conjectured that a(n) >= 0 for all n < 78557.
Determining a(131072) = a(2^17) is equivalent to finding the next Fermat prime after F_4 = 2^16 + 1. - Jeppe Stig Nielsen, Jul 27 2019
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LINKS
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EXAMPLE
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1*(2^0)+1=2 is prime, so a(1)=0;
3*(2^1)+1=5 is prime, so a(3)=1;
For n=7, 7+1 and 7*2+1 are composite, but 7*2^2+1=29 is prime, so a(7)=2.
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MATHEMATICA
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Do[m = 0; While[ !PrimeQ[n*2^m + 1], m++ ]; Print[m], {n, 1, 110} ]
sm[n_]:=Module[{k=0}, While[!PrimeQ[n 2^k+1], k++]; k]; Array[sm, 120] (* Harvey P. Dale, Feb 05 2020 *)
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CROSSREFS
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For the corresponding primes see A050921.
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KEYWORD
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easy,nice,sign
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AUTHOR
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STATUS
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approved
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