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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)

Such primes are the exceptional p for which x^64 == -2 (mod p) has a solution, as x^64 == -2 (mod p) is soluble for *every* p with ord_p(-2) odd.

But if ord_p(-2) is even and p - 1 = 2^r.j with j odd, then x^64 == -2 (mod p) is soluble if and only if ord_p(-2) is not divisible by 2^(r-5). See comment at A163185 for explanation.

Most primes p for which x^64 == -2 (mod p) has a solution (A051101) have ord_p(-2) odd (so belong to A163183). Thus 25601 (first element of current sequence, and 827th element of A051101) is the first element where A051101 and A163183 differ.

Complement of A051101 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.