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A139035
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Primes with semiprimitive root 2.
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3
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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, 1567, 1583, 1607, 1663, 1759, 1783, 1823, 1847, 1871, 1879
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,1
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COMMENTS
| 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
| V. Shevelev, On the Newman sum over multiples of a prime with a primitive or semiprimitive root 2, arXiv:0710.1354
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FORMULA
| 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).
A006694((a(n)-1)/2)=2 and A064287((a(n)-1)/2)=1
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CROSSREFS
| Cf. A006694, A002326, A064287, A001917, A001122, A133954, A115591
Sequence in context: A176557 A000353 A097149 * A002146 A184882 A073577
Adjacent sequences: A139032 A139033 A139034 * A139036 A139037 A139038
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KEYWORD
| nonn
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AUTHOR
| Vladimir Shevelev (shevelev(AT)bgu.ac.il), May 31 2008, Jun 06 2008
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