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a(n) = smallest k such that (6*k-3)*2^n-1 is prime.
4

%I #12 May 15 2017 23:52:59

%S 1,1,1,1,3,1,1,5,8,3,1,15,2,3,2,8,3,1,9,10,2,6,12,7,10,10,4,5,8,36,3,

%T 10,25,1,6,8,4,1,11,20,3,6,1,10,28,6,36,20,12,15,4,31,25,8,1,6,9,19,8,

%U 16,12,10,2,1,17,11,19,11,9,5,21,22,3,4,13,1

%N a(n) = smallest k such that (6*k-3)*2^n-1 is prime.

%C As N increases, (Sum_{n=1..N} a(n)) / (Sum_{n=1..N} n) tends to log(2)/3 as seen by plotting data; this is consistent with the prime number theorem as the probability that (6*x-3)*2^n - 1 is prime is ~ 3/(n*log(2)) if n is great enough, so after n*log(2)/3 try n*log(2)/3*(3/n*log(2))=1.

%C For n=1 to 14000, a(n)/n is always < 3.

%H Pierre CAMI, <a href="/A285808/b285808.txt">Table of n, a(n) for n = 1..14000</a>

%t Table[k = 1; While[! PrimeQ[(6 k - 3)*2^n - 1], k++]; k, {n, 76}] (* _Michael De Vlieger_, Apr 27 2017 *)

%Y Cf. A126717, A284325.

%K nonn

%O 1,5

%A _Pierre CAMI_, Apr 27 2017