A250204 - OEIS

%I #34 Jun 15 2015 02:29:36

%S 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,

%T 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,

%U 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

%N Sierpiński problem in base 6: Least k > 0 such that n*6^k+1 is prime, or 0 if no such k exists.

%C 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.

%C 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

%H Eric Chen, <a href="/A250204/b250204.txt">Table of n, a(n) for n = 1..1000</a>

%H Gary Barnes, <a href="http://www.noprimeleftbehind.net/crus/Sierp-conjectures.htm">Sierpinski conjectures and proofs</a>

%p N:= 1000: # to get a(1) to a(N), using k up to 10000

%p a[1]:= 1:

%p for n from 2 to N do

%p   if n mod 5 = 4 then a[n]:= 0

%p   else

%p     for k from 1 to 10000 do

%p     if isprime(n*6^k+1) then

%p        a[n]:= k;

%p        break

%p     fi

%p     od

%p   fi

%p od:

%p L:= [seq(a[n],n=1..N)]; # _Robert Israel_, Mar 17 2015

%t (* m <= 10000 is sufficient up to n = 1000 *)

%t a[n_] := For[k = 1, k <= 10000, k++, If[PrimeQ[n*6^k + 1], Return[k]]] /. Null -> 0; Table[a[n], {n, 1, 120}]

%o (PARI) a(n) = if(n%5==4, 0, for(k = 1, 10000, if(ispseudoprime(n*6^k+1), return(k))))

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

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

%K nonn

%O 1,8

%A _Eric Chen_, Mar 11 2015