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A139035
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Primes of the form 4*k+3 with primitive root -2.
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7
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7, 23, 47, 71, 79, 103, 167, 191, 199, 239, 263, 271, 311, 359, 367, 383, 463, 479, 487, 503, 599, 607, 647, 719, 743, 751, 823, 839, 863, 887, 967, 983, 991, 1031, 1039, 1063, 1087, 1151, 1223, 1231, 1279, 1303, 1319, 1367, 1439, 1447, 1487, 1511, 1543, 1559
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OFFSET
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1,1
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COMMENTS
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Original name: Primes with semiprimitive root 2.
If p is a prime, then we call r a semiprimitive root if it has order (p-1)/2 and there is no x for which a^x is congruent to -1 (mod p). So +/- r^k, 0 <= k <= (p-3)/2 is a complete set of nonzero residues (mod p).
If r=2, then (-1/p)=-1 and, consequently, a(n)==-1(mod 4).
Besides, (2/a(n))=1. Indeed, since 2^((p-1)/2)=1 (mod p), then 2==2^((p+1)/2)=(2^((p+1)/4))^2. Therefore, (a(n))^2==1(mod 16) and thus a(n)==-1(mod 8). This yields that residues 1,2,...,2^((p-3)/2) are quadratic residues modulo a(n), while -1,-2,...,-2^((p-3)/2) are quadratic nonresidues modulo a(n). Primitive root of a(n) is greater than or equal to 3. All terms are in A115591.
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LINKS
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FORMULA
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Prime p is in the sequence iff p==-1(mod 8) and A002326((p-1)/2)=(p-1)/2. A sufficient condition: if p==-1 (mod 8) and (p-1)/2 is prime, then p is in the sequence (the converse statement, generally speaking, is not true).
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MATHEMATICA
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Reap[For[p = 3, p <= 10^4, p = NextPrime[p], rp = MultiplicativeOrder[2, p]; rm = MultiplicativeOrder[-2, p]; If[rp != p-1 && rm == p-1, Sow[p]]] ][[2, 1]] (* Jean-François Alcover, Sep 03 2016, after Joerg Arndt *)
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PROG
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(PARI)
{ forprime (p=3, 10^4,
rp = znorder(Mod(+2, p));
rm = znorder(Mod(-2, p));
if ( (rp!=p-1) && (rm==p-1), print1(p, ", ") );
); }
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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