A287218 a(n) = smallest k such that (6*k-3)*2^prime(n) - 1 is prime.

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, 18, 81, 56, 49, 38, 38, 26, 3, 33, 103, 12, 67, 12, 11, 8, 48, 79, 2, 43, 136, 82, 12, 46, 78, 31, 117, 126, 34, 4, 27, 49, 83, 3, 57, 234, 12, 10, 116, 128, 53, 13

Offset: 1

Author

Pierre CAMI, May 22 2017

Keywords

• nonn

Comments

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

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.

Links

• Pierre CAMI, Table of n, a(n) for n = 1..2000

Crossrefs

Subsequence of A285808.

Cf. A284325, A284631.

Programs

Mathematica

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 *)

Formula

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