A245441 a(1)=3, then a(n) = smallest odd k > Ceiling(a(n-1)/2) such that k*2^n-1 is prime.

3, 3, 3, 3, 7, 17, 13, 27, 25, 15, 25, 23, 21, 15, 9, 17, 15, 21, 51, 35, 19, 33, 25, 39, 57, 57, 81, 45, 45, 213, 111, 57, 31, 131, 99, 83, 45, 27, 25, 107, 55, 33, 33, 35, 67, 141, 91, 89, 69, 41, 129, 89, 147, 101, 195, 129, 79, 77, 45, 77, 69, 53, 61

Offset: 1

Author

Pierre CAMI, Jul 22 2014

Keywords

• nonn

Comments

A126715(n) = smallest odd k such that k*2^n-1 is prime, the primes are not always in increasing order.

Here the primes k*2^n-1 are always in increasing order.

The ratio sum_{1..N}a(n)/sum_{1..N}n is near 2*log(2) as N increases.

The ratio a(n)/n is always < 8 for n from 1 to 6000.

Examples

			3*2^1-1 = 5 is prime, a(1)=3 by definition.
3*2^2-1 = 11 is prime, 3 > 3/2 so a(2) = 3.
3*2^3-1 = 23 is prime, so a(3) = 3.
3*2^4-1 = 47 is prime, so a(4) = 3.
3*2^5-1 = 95 is composite.
5*2^5-1 = 159 is composite.
7*2^5-1 = 223 is prime so a(5) = 7.
		

Links

• Pierre CAMI, Table of n, a(n) for n = 1..6000

Crossrefs

Cf. A126715.

Programs

PARI

a=[3]; for(n=2, 100, k=floor(a[n-1]/2)+2; if(k%2==0, k++); t=2^n; while(!isprime(k*t-1), k+=2); a=concat(a, k)); a \\ Colin Barker, Jul 22 2014