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

%I #19 Nov 14 2018 13:04:27

%S 1,1,3,1,1,2,3,9,12,8,3,4,3,1,36,25,8,12,19,21,3,12,19,40,9,14,1,14,2,

%T 18,81,56,49,38,38,26,3,33,103,12,67,12,11,8,48,79,2,43,136,82,12,46,

%U 78,31,117,126,34,4,27,49,83,3,57,234,12,10,116,128,53,13

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

%C For n from 1 to 2000, a(n)/prime(n) is always < 1.8.

%C As N increases, (Sum_{n=1..N} a(n)) / (Sum_{n=1..N} prime(n)) tends to log(2)/3; this is consistent with the prime number theorem as the probability that x*2^n-1 is prime with odd x divisible by 3 is ~ 3/(n*log(2)) and after n*log(2)/3 try (n*log(2)/3)*(3/(n*log(2)) = 1.

%H Pierre CAMI, <a href="/A287218/b287218.txt">Table of n, a(n) for n = 1..2000</a>

%F a(n) = A285808(A000040(n)).

%t sk[n_]:=Module[{k=1,t=2^Prime[n]},While[!PrimeQ[(6k-3)*t-1],k++];k]; Array[ sk,70] (* _Harvey P. Dale_, Nov 14 2018 *)

%Y Subsequence of A285808.

%Y Cf. A284325, A284631.

%K nonn

%O 1,3

%A _Pierre CAMI_, May 22 2017