A264097 Smallest odd number k divisible by 3 such that k*2^n-1 is a prime.

3, 3, 3, 3, 3, 15, 3, 3, 27, 45, 15, 3, 87, 9, 15, 9, 45, 15, 3, 51, 57, 9, 33, 69, 39, 57, 57, 21, 27, 45, 213, 15, 57, 147, 3, 33, 45, 21, 3, 63, 117, 15, 33, 3, 57, 165, 33, 213, 117, 69, 87, 21, 183, 147, 45, 3, 33, 51, 111, 45, 93, 69, 57, 9, 3, 99, 63

Offset: 0

Author

Pierre CAMI, Nov 03 2015

Keywords

• nonn

Comments

As N increases, (Sum_{n=1..N} a(n))/(Sum_{n=1..N} n) appears to tend to 2*log(2), as can be seen by plotting the first 31000 terms.

This observation is consistent with the prime number theorem as the probability that k*2^n-1 is prime where k is a multiple of 3 is 1/(2*(n*log(2)+log(k))) ~ 1/(2*n*log(2)).

Examples

			3*2^0-1=2 prime so a(0)=3.
3*2^1-1=5 prime so a(1)=3.
3*2^2-1=11 prime so a(2)=3.
3*2^3-1=23 prime so a(3)=3.
		

Links

• Pierre CAMI, Table of n, a(n) for n = 0..31000

Crossrefs

Cf. A256787, A085427, A126717.

Programs

Mathematica

Table[k = 3; While[! PrimeQ[k 2^n - 1], k += 6]; k, {n, 0, 68}] (* Michael De Vlieger, Nov 03 2015 *)

PARI

a(n) = {k = 3; while (!isprime(k*2^n-1), k += 6); k;} \\ Michel Marcus, Nov 03 2015