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A250205
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Riesel problem in base 6: Least k > 0 such that n*6^k-1 is prime, or 0 if no such k exists.
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2
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1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 2, 1, 1, 0, 1, 1, 1, 2, 0, 1, 1, 2, 1, 0, 2, 1, 1, 1, 0, 1, 1, 2, 2, 0, 4, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 4, 1, 3, 0, 1, 1, 6, 2, 0, 5, 1, 1, 1, 0, 6, 2, 1, 1, 0, 1, 2, 10, 1, 0, 1, 3, 1, 1, 0, 1, 1, 2, 1, 0, 1, 8, 1, 1, 0, 1, 2, 2, 4, 0, 49, 1, 1, 1, 0, 2, 1, 1, 1, 0, 2, 1, 6, 2, 0, 1, 1, 1, 1, 0, 5, 1, 1, 2, 0, 1, 10, 2, 1
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OFFSET
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1,13
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COMMENTS
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a(5j+1) = 0 except for a(1), since (5j+1)*6^k-1 is always divisible by 5, but there are infinitely many numbers not in the form 5j+1 such that a(n) = 0.
a(n) = 0 for n == 84687 mod 10124569, because then n*6^k-1 is always divisible by at least one of 7, 13, 31, 37, 97. - Robert Israel, Mar 17 2015
Conjecture: if n is not in the form 5j+1 and n < 84687, then a(n) > 0.
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LINKS
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Eric Chen, Table of n, a(n) for n = 1..1000
Gary Barnes, Riesel conjectures and proofs
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FORMULA
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a(A024898(n)) = 1. - Michel Marcus, Mar 16 2015
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MAPLE
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N:= 1000: # to get a(1) to a(N), using k up to 10000
a[1]:= 1:
for n from 2 to N do
if n mod 5 = 1 then a[n]:= 0
else
for k from 1 to 10000 do
if isprime(n*6^k-1) then
a[n]:= k;
break
fi
od
fi
od:
seq(a[n], n=1..N); # Robert Israel, Mar 17 2015
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MATHEMATICA
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(* m <= 10000 is sufficient up to n = 1000 *)
a[n_] := For[k = 1, k <= 10000, k++, If[PrimeQ[n*6^k - 1], Return[k]]] /. Null -> 0; Table[a[n], {n, 1, 120}]
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PROG
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(PARI) a(n) = if(n%5==1 && n>1, 0, for(k = 1, 10000, if(ispseudoprime(n*6^k-1), return(k))))
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CROSSREFS
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Cf. A040081, A046069, A050412, A108129.
Cf. A250204 (Least k > 0 such that n*6^k+1 is prime).
Sequence in context: A104234 A321926 A037870 * A326017 A290307 A206588
Adjacent sequences: A250202 A250203 A250204 * A250206 A250207 A250208
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KEYWORD
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nonn
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AUTHOR
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Eric Chen, Mar 13 2015
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STATUS
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approved
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