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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.

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.

From Jianing Song, Jun 22 2025: (Start)

The multiplicative order of -2 modulo a(n) is A385228(n).

Contained in primes congruent to 1 or 3 modulo 8 (primes p such that -2 is a quadratic residue modulo p, A033200), and contains primes congruent to 3 modulo 8 (A007520).

Conjecture: this sequence has density 7/24 among the primes (see A014663). (End)

Differs from A051085 first at the 541st entry, at p=15809. - R. J. Mathar, Oct 14 2008

From Christopher J. Smyth, Jul 24 2009: (Start)

Differs from A163183 (primes dividing 2^j+1 for some odd j) at the 827th entry, at p=25601. See comment at A163186 for explanation.

Sequence is union of A163183 and A163186 (primes p such that the equation x^64 = -2 mod p has a solution, and ord_p(-2) is even).

(End)

Complement of A216777 relative to A000040. - Vincenzo Librandi, Sep 17 2012

Complement of A051085 relative to A000040.

Each term p has the form 2^r*j + 1, where r >= 3, j is odd, and ord_p(-2) divides j.