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A081858
Numbers k such that 2*k+1 divides 2^k-1.
7
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
1,2
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
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.
MATHEMATICA
Join[{0}, Select[Range[300], PowerMod[2, #, 2*# + 1] === 1 &]] (* Amiram Eldar, Jun 02 2022 *)
PROG
(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
CROSSREFS
Sequence in context: A257336 A234431 A310282 * A145837 A111132 A188473
KEYWORD
nonn
AUTHOR
Benoit Cloitre, Apr 11 2003
EXTENSIONS
Formula corrected by Chris Boyd, Mar 16 2014
STATUS
approved