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A163183
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Primes dividing 2^j + 1 for some odd j
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5
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3, 11, 19, 43, 59, 67, 83, 107, 131, 139, 163, 179, 211, 227, 251, 281, 283, 307, 331, 347, 379, 419, 443, 467, 491, 499, 523, 547, 563, 571, 587, 617, 619, 643, 659, 683, 691, 739, 787, 811, 827, 859, 883, 907, 947, 971, 1019, 1033, 1049, 1051, 1091, 1097
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OFFSET
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1,1
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COMMENTS
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Also the primes p for which ord_p(-2) is odd, as (-2)^j == 1 (mod p).
All such p are = 1 or 3 mod 8, so sequence is subsequence of A033200, as (-2)^{j+1} == -2 (mod p) implies that (-2/p) = 1, p == 1 or 3 (mod 8).
Claim: Sequence contains all primes = 3 mod 8, so contains A007520 as a subsequence.
Proof: If p = 8r + 3 then 2^{4r+1} == 1 or -1 (mod p). If former, then (2^{2r+1})^2 == 2 (mod p), (2/p) = 1, only true for p == 1 or 7 (mod 8). So p | 2^{4r+1} + 1.
Claim: For every p in sequence and every 2^k, the equation x^{2^k} == -2 (mod p) is soluble. Hence sequence is a subsequence of A033203 (k=1), A051071 (k=2), A051073 (k=3), A051077 (k=4), A051085 (k=5), A051101 (k=6), ....
Proof: Put x == (-2)^u (mod p). Then using (-2)^j == 1 (mod p), we can solve x^{2^k} == -2 (mod p) if can find u and v such that u*2^k + v*j = 1, possible as gcd(2^k, j) = 1.
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LINKS
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EXAMPLE
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11 is in sequence as 11 | 2^5 + 1; 281 (smallest element of the sequence == 1 (mod 8)) is in the sequence as 281 | 2^35 + 1.
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MAPLE
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with(numtheory):A:=3:p:=3: for c to 500 do p:=nextprime(p); if order(-2, p) mod 2=1 then A:=A, p;; fi; od:A;
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PROG
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(PARI) lista(nn) = forprime(p=3, nn, if(znorder(Mod(-2, p))%2, print1(p, ", "))); \\ Jinyuan Wang, Mar 23 2020
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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