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A081858 Numbers n such that 2n+1 divides 2^n-1. 5
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 (list; graph; refs; listen; history; text; internal format)
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

Table of n, a(n) for n=1..57.

FORMULA

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

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

Cf. A014664.

Sequence in context: A257336 A234431 A310282 * A145837 A111132 A188473

Adjacent sequences:  A081855 A081856 A081857 * A081859 A081860 A081861

KEYWORD

nonn

AUTHOR

Benoit Cloitre, Apr 11 2003

EXTENSIONS

Formula corrected by Chris Boyd, Mar 16 2014

STATUS

approved

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Last modified March 8 20:14 EST 2021. Contains 341953 sequences. (Running on oeis4.)