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Least k such that 2*((6*n)^k) + 1 is prime.
0

%I #10 Sep 16 2019 06:06:44

%S 1,3,1,2,1,1,2,1,1,3,2,2,1,4,1,1,2,5,1,1,4,2,1,5,3,1,2,1,1,8,1,3,1,1,

%T 1,1,66,1,6,2,3,3,5,2,1,4,7,1,9,1,1,3,6,2,1,1,96,4,1,2,1,62,1,1,9,5,

%U 159,4,1,7,1,4,1,8,2,2,2,1,2,8,2,2,1,1,1,1,54,8,1,5,1,16,1,1,2,1,7,2,4,1,1

%N Least k such that 2*((6*n)^k) + 1 is prime.

%e 2*((6*1)^1) + 1 = 13 prime, so a(1)=1

%e 2*((6*2)^1) + 1 = 25 = 5^2

%e 2*((6*2)^2) + 1 = 289 = 17^2

%e 2*((6*2)^3) + 1 = 3457 prime, so a(2)=3

%t lk[n_]:=Module[{k=1},While[!PrimeQ[2(6n)^k+1],k++];k]; Array[lk,110] (* _Harvey P. Dale_, May 16 2012 *)

%K nonn

%O 1,2

%A _Pierre CAMI_, Oct 13 2004

%E Extended by _R. J. Mathar_, Nov 13 2009