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%I #47 Nov 23 2023 00:19:53
%S 7,17,23,31,41,43,47,71,73,79,89,97,103,109,113,127,137,151,157,167,
%T 191,193,199,223,229,233,239,241,251,257,263,271,277,281,283,307,311,
%U 313,331,337,353,359,367,383,397,401,409,431,433,439,449,457,463,479
%N Odd primes for which 2 is not a primitive root.
%C Alternately, for these primes p, the polynomial (x^p+1)/(x+1) is reducible over GF(2).
%C The prime p belongs to this sequence if and only if A002326((p-1)/2) != (p-1). If A002326((p-1)/2) = (p-1), then the prime p belongs to the sequence A001122. - _V. Raman_, Dec 01 2012
%C The only primitive root modulo 2 is 1. See A060749. Hence 2 should be added to this sequence in order to obtain the complement of A001122. - _Wolfdieter Lang_, May 19 2014
%H Robert Israel, <a href="/A216838/b216838.txt">Table of n, a(n) for n = 1..10000</a>
%H Anand Bhardwaj, Luisa Degen, Radostin Petkov, and Sidney Stanbury, <a href="https://arxiv.org/abs/2311.13375">A Study of Cunningham Bounds through Rogue Primes</a>, arXiv:2311.13375 [math.NT], 2023.
%p select(t -> isprime(t) and numtheory[order](2,t) <> t-1, [seq](2*i+1,i=1..1000)); # _Robert Israel_, May 20 2014
%t Select[Prime[Range[2, 100]], PrimitiveRoot[#] =!= 2 &] (* _T. D. Noe_, Sep 19 2012 *)
%o (PARI) forprime(p=3, 1000, if(znorder(Mod(2,p))!=p-1, print(p)))
%o (PARI) forprime(p=3, 1000, if(factormod((x^p+1)/(x+1), 2, 1)[1, 1]!=(p-1), print(p)))
%Y Cf. A002326 (multiplicative order of 2 mod 2n+1)
%Y Cf. A001122 (Primes for which 2 is a primitive root)
%Y Cf. A115586 (Primes for which 2 is neither a primitive root nor a quadratic residue).
%K nonn
%O 1,1
%A _V. Raman_, Sep 17 2012
%E Name corrected by _Wolfdieter Lang_, May 19 2014
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