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

%I #5 Apr 02 2018 16:35:14

%S 1,1,2,1,1,1,1,4,1,5,1,34,7,1,1,1,2,2,1,1,1,1,4,24,8,1,10,7,1,1,2,1,3,

%T 2,1,1,1,2,1,1,1,1,28,5,2,2484,1,2,1,1

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

%e 2*((6*1)^1) - 1 = 11 prime, so a(1)=1

%e 2*((6*2)^1) - 1 = 23 prime, so a(2)=1

%e 2*((6*3)^1) - 1 = 35 = 5*7

%e 2*((6*3)^2) - 1 = 647 prime, so a(3)=2

%t lk[n_]:=Module[{k=1},While[!PrimeQ[2((6n)^k)-1],k++];k]; Array[lk,50] (* _Harvey P. Dale_, Apr 02 2018 *)

%K nonn

%O 1,3

%A _Pierre CAMI_, Oct 13 2004

%E Corrected by _Harvey P. Dale_, Apr 02 2018