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

%I #13 May 06 2017 20:41:30

%S 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,

%T 14,65,8,58,114,98,63,45,124,36,72,516,28,45,43,183,2,25,142,68,249,

%U 30,324,155,188,200,334,56,87,178,98,110,22,25,24,70,2,271,17,498,412,750,877

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

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

%H Pierre CAMI, <a href="/A285933/b285933.txt">Table of n, a(n) for n = 1..1000</a>

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

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

%t 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 *)

%o (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

%Y Cf. A285808.

%K nonn

%O 1,3

%A _Pierre CAMI_, Apr 29 2017