A250205 Riesel problem in base 6: Least k > 0 such that n*6^k-1 is prime, or 0 if no such k exists.

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

Offset: 1

Author

Eric Chen, Mar 13 2015

Keywords

• nonn

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

Crossrefs

Cf. A040081, A046069, A050412, A108129.

Cf. A250204 (Least k > 0 such that n*6^k+1 is prime).

Programs

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}]

PARI

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

Formula

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