A255707 Least number k > 0 such that (2*n-1)^k - 2 is prime, or 0 if no such number exists.

0, 2, 1, 1, 1, 4, 1, 1, 6, 1, 1, 24, 1, 2, 2, 1, 1, 2, 2, 1, 4, 1, 1, 2, 1, 8, 4, 1, 12, 4, 1, 1, 8, 3, 1, 2, 1, 1, 2, 38, 1, 4, 1, 4, 2, 1, 2, 4, 747, 1, 4, 1, 1, 2, 1, 1, 10, 1, 2, 2, 2, 6, 42, 2, 1, 2, 1, 2, 10, 1, 1, 4, 2, 16, 50, 1, 1, 2, 22, 1, 2, 38

Offset: 1

Author

Robert Price, Mar 02 2015

Keywords

• nonn

Links

• Michel Marcus, Table of n, a(n) for n = 1..152 (terms 1..143 from Robert Price)
• Carlos Rivera, Puzzle 887. p(n)^c-2 is prime, The Prime Puzzles and Problems Connection.

Crossrefs

Cf. A079706, A084712, A084713, A084714, A098090, A138066.

Programs

Mathematica

lst = {0}; For[n = 2, n ≤ 143, n++, For[k = 1, k >= 1, k++, If[PrimeQ[(2*n - 1)^k - 2], AppendTo[lst, k]; Break[]]]]; lst

PARI

a(n)=if(n==1,return(0));k=1;while(k,if(ispseudoprime((2*n-1)^k-2),return(k));k++)
vector(50,n,a(n)) \\ Derek Orr, Mar 03 2015

Formula

a(A098090(n)) = 1. - Michel Marcus, Mar 03 2015