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A250205 Riesel problem in base 6: Least k > 0 such that n*6^k-1 is prime, or 0 if no such k exists. 2
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,13

COMMENTS

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.

LINKS

Eric Chen, Table of n, a(n) for n = 1..1000

Gary Barnes, Riesel conjectures and proofs

FORMULA

a(A024898(n)) = 1. - Michel Marcus, Mar 16 2015

MAPLE

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

MATHEMATICA

(* 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}]

PROG

(PARI) a(n) = if(n%5==1 && n>1, 0, for(k = 1, 10000, if(ispseudoprime(n*6^k-1), return(k))))

CROSSREFS

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

KEYWORD

nonn

AUTHOR

Eric Chen, Mar 13 2015

STATUS

approved

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Last modified March 8 20:52 EST 2021. Contains 341953 sequences. (Running on oeis4.)