A250204 Sierpiński 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, 0, 1, 1, 1, 4, 0, 1, 1, 1, 1, 0, 2, 1, 1, 1, 0, 5, 1, 4, 1, 0, 1, 1, 1, 2, 0, 1, 2, 1, 1, 0, 1, 2, 1, 1, 0, 1, 5, 5, 2, 0, 1, 1, 1, 3, 0, 2, 1, 1, 7, 0, 1, 1, 2, 1, 0, 2, 1, 1, 1, 0, 2, 1, 8, 1, 0, 1, 2, 1, 1, 0, 7, 1, 1, 4, 0, 4, 1, 2, 1, 0, 2, 5, 1, 2, 0, 1, 1, 2, 3, 0, 1, 1, 9, 2, 0, 1, 1, 1, 1, 0, 1, 6, 1, 2, 0, 1, 3, 1, 4, 0, 1, 2, 23, 1, 0, 4

Offset

1,8

Comments

a(5k+4) = 0, since (5k+4)*6^n+1 is always divisible by 5, but there are infinitely many numbers not in the form 5k+4 such that a(n) = 0. For example, a(174308) = 0 since 174308*6^n+1 is always divisible by 7, 13, 31, 37, or 97 (See A123159). Conjecture: if n is not in the form 5k+4 and n < 174308, then a(n) > 0.

However, according to the Barnes link no primes n*6^k+1 are known for n = 1296, 7776 and 46656, so these may be counterexamples. - Robert Israel, Mar 17 2015

Links

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

Gary Barnes, Sierpinski conjectures and proofs

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 = 4 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:
L:= [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==4, 0, for(k = 1, 10000, if(ispseudoprime(n*6^k+1), return(k))))

Crossrefs

Cf. A040076, A046067, A078680, A033809, A123159.

Cf. A250205 (Least k > 0 such that n*6^k-1 is prime).

Sequence in context: A321316 A185690 A298248 * A096459 A378833 A293301

Adjacent sequences: A250201 A250202 A250203 * A250205 A250206 A250207

Keyword

nonn

Author

Eric Chen, Mar 11 2015

Status

approved