A152414 a(n) = least positive k such that k*2^n*(2^n-1) - 1 or k*2^n*(2^n-1) + 1 is prime.

1, 1, 2, 1, 1, 3, 3, 6, 1, 1, 4, 2, 5, 3, 9, 8, 4, 1, 3, 4, 36, 5, 2, 4, 10, 4, 18, 3, 21, 9, 6, 1, 6, 8, 12, 2, 51, 1, 2, 2, 21, 6, 6, 12, 1, 5, 5, 3, 10, 1, 11, 53, 9, 4, 3, 2, 1, 5, 12, 10, 9, 8, 5, 9, 7, 6, 62, 29, 16, 51, 12, 3, 30, 56, 2, 23, 70, 3, 23

Offset: 1

Author

Pierre CAMI, Dec 03 2008

Keywords

• nonn

Comments

All values in the b-file are verified and primes certified using PFGW from Primeform group.

Examples

			For n = 1, 1*2^1*(2^1-1)+1 = 3 is prime, so a(1) = 1.
For n = 2, 1*2^2*(2^2-1)-1 = 11 is prime, as well as 1*2^2*(2^2-1)+1 = 13, so a(2) = 1.
For n = 3, k = 2 is the least k satisfying the condition: 2*2^3*(2^3-1)+1 = 113 is prime, so a(3) = 2.
		

Links

• Pierre CAMI, Table of n, a(n) for n=1..4000

Crossrefs

Cf. A153091.

Programs

Mathematica

A152414[n_] := Module[{k = 0}, While[NoneTrue[++k*# + {-1, 1}, PrimeQ]] & [4^n-2^n]; k];
Array[A152414, 100] (* Paolo Xausa, Jun 30 2025 *)

PARI

a(n) = {k = 1; while (! (isprime(k*2^n*(2^n-1)+1) || isprime(k*2^n*(2^n-1)-1)), k++); return (k);} \\ Michel Marcus, Mar 07 2013

Formula

From Pierre CAMI, Dec 04 2008: (Start)

Lim_{n->oo} ( (Sum_{i=1..n} a(i)) / (n*(n+1)/2) ) = 1/4.

Lim_{n->oo} ( (Sum_{i=1..n} a(2*i)) / (n*(n+1)) ) = log(2)/4.

Lim_{n->oo} ( (Sum_{i=0..n} a(2*i+1)) / (n*(n+2)) ) = 1/2 - log(2)/4. (End)