A081858 Numbers k such that 2*k+1 divides 2^k-1.

0, 3, 8, 11, 15, 20, 23, 35, 36, 39, 44, 48, 51, 56, 63, 68, 75, 83, 95, 96, 99, 111, 116, 119, 120, 128, 131, 135, 140, 155, 156, 168, 170, 176, 179, 183, 191, 200, 204, 215, 216, 219, 224, 228, 231, 239, 243, 251, 260, 280, 284, 288, 296, 299, 300, 303, 308

Offset: 1

Author

Benoit Cloitre, Apr 11 2003

Keywords

• nonn

Comments

From Chris Boyd, Mar 16 2014: (Start)

n is a term if and only if n=0, 2n+1 is a prime of the form 8k+-1, or 2n+1 is an Euler pseudoprime satisfying 2^n == 1 mod 2n+1.

Case 1: 0 is a term. Case 2, 2n+1 prime: by Euler's criterion and the quadratic character of 2, 2^n == 1 mod 2n+1 only if 2n+1 is of the form 8k+-1. Case 3, 2n+1 composite: 2^n == 1 mod 2n+1 only if 2n+1 is one of the subset of Euler pseudoprimes satisfying 2^n == 1 mod 2n+1.

The first term for which 2n+1 is a qualifying Euler pseudoprime is n=170.

The first Euler pseudoprime that does not correspond to a term is 3277, because 2^((3277-1)/2) == -1 mod 3277. (End)

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A002326, A014664.

Programs

Mathematica

Join[{0}, Select[Range[300], PowerMod[2, #, 2*# + 1] === 1 &]] (* Amiram Eldar, Jun 02 2022 *)

PARI

isok(n) = !((2^n-1) % (2*n+1)); \\ Michel Marcus, Dec 04 2013

PARI

for(n=0,400,if(n%znorder(Mod(2,2*n+1))==0,print1(n","))) \\ Chris Boyd, Mar 16 2014, after Michael Somos in A002326

Formula

k such that A002326(k)|k: since 2^k == 1 mod 2*k+1, k must be a multiple of the order of 2 mod 2*k+1.

Extensions

Formula corrected by Chris Boyd, Mar 16 2014