OFFSET
1,1
COMMENTS
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.
Also contains some primes == 1 (mod 8), given in A163184. So sequence is a union of A007520 and A163184.
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.
EXAMPLE
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.
MAPLE
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;
PROG
(PARI) lista(nn) = forprime(p=3, nn, if(znorder(Mod(-2, p))%2, print1(p, ", "))); \\ Jinyuan Wang, Mar 23 2020
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Christopher J. Smyth, Jul 22 2009
STATUS
approved