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
KEYWORD
nonn
AUTHOR
Eric Chen, Mar 13 2015
STATUS
approved