A285933 a(n) = smallest k such that (6*k-3)*2^n-1 and (6*k-3)*2^n+1 are twin primes.

1, 1, 2, 3, 14, 1, 2, 10, 8, 3, 17, 28, 62, 8, 58, 20, 64, 1, 12, 75, 14, 6, 197, 41, 128, 63, 14, 65, 8, 58, 114, 98, 63, 45, 124, 36, 72, 516, 28, 45, 43, 183, 2, 25, 142, 68, 249, 30, 324, 155, 188, 200, 334, 56, 87, 178, 98, 110, 22, 25, 24, 70, 2, 271, 17, 498, 412, 750, 877

Offset: 1

Author

Pierre CAMI, Apr 29 2017

Keywords

• nonn

Comments

Conjecture: a(n) is ~ (n*log(2))^2/9 as n increases.

Examples

			(6*1-3)*2^1-1 = 5, (6*1-3)*2^1+1 = 7; 5 and 7 are twin primes so a(1) = 1.
(6*1-3)*2^2-1 = 11, (6*1-3)*2^2+1 = 13; 11 and 13 are twin primes so a(2) = 1.
		

Links

• Pierre CAMI, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A285808.

Programs

Mathematica

Table[k = 1; While[Times @@ Boole@ PrimeQ[(6 k - 3) 2^n + {-1, 1}] < 1, k++]; k, {n, 69}] (* Michael De Vlieger, May 04 2017 *)

PARI

a(n) = {my(k=1); while (!isprime((6*k-3)*2^n-1) || !isprime((6*k-3)*2^n+1), k++); k;} \\ Michel Marcus, May 01 2017