A257379 Smallest odd number k such that k*n*2^n - 1 is a prime number.

3, 1, 1, 3, 3, 1, 3, 3, 5, 5, 9, 5, 7, 7, 3, 17, 11, 11, 7, 9, 11, 15, 3, 7, 9, 67, 3, 45, 3, 1, 33, 21, 15, 23, 17, 3, 7, 9, 19, 15, 17, 63, 51, 3, 9, 33, 53, 61, 13, 45, 75, 39, 83, 43, 7, 19, 13, 41, 5, 19, 31, 165, 13, 27, 3, 13, 135, 33, 31, 15, 33, 87

Offset: 1

Author

Pierre CAMI, Apr 21 2015

Keywords

• nonn

Comments

Conjecture: a(n) exists for every n.

The conjecture follows from Dirichlet's theorem on primes in arithmetic progressions. - Robert Israel, Jan 05 2016

As N increases, (Sum_{n=1..N} k) / (Sum_{n=1..N} n) approaches 0.833.

If k=1 then n*2^n-1 is a Woodall prime (see A002234).

Generalized Woodall primes have the form n*b^n-1, I propose to name the primes k*n*2^n-1 generalized Woodall primes of the second type.

Examples

			1*1*2^1 - 1 = unity, 3*1*2^1 - 1 = 5, which is prime, so a(1) = 3.
1*2*2^2 - 1 = 7, which is prime, so a(2) = 1.
1*3*2^3 - 1 = 23, which is prime, so a(3) = 1.
		

Links

• Pierre CAMI, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A002234, A256787, A257378, A266909.

Programs

Maple

Q:= proc(m) local k;
  for k from 1 by 2 do if isprime(k*m-1) then return k fi od
end proc:
seq(Q(n*2^n),n=1..100); # Robert Israel, Jan 05 2016

Mathematica

Table[k = 1; While[!PrimeQ[k*n*2^n - 1], k += 2]; k, {n, 72}] (* Michael De Vlieger, Apr 21 2015 *)

PARI

a(n) = k=1; while(!isprime(k*n*2^n-1), k+=2); k \\ Colin Barker, Apr 21 2015