A285808 a(n) = smallest k such that (6*k-3)*2^n-1 is prime.

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

Offset: 1

Author

Pierre CAMI, Apr 27 2017

Keywords

• nonn

Comments

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.

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

Links

• Pierre CAMI, Table of n, a(n) for n = 1..14000

Crossrefs

Cf. A126717, A284325.

Programs

Mathematica

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