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Is there a simpler characterization of these primes?

Answer from Don Reble, Oct 25 2018: (Start)

Let POT(x) be the largest power of 2 which divides x (A006519).

Apart from the initial 2, this sequence consists of those primes P such that

2 <= POT(the order of 3 modulo P) <= POT(P-1)/8.

The condition "2 <=" ensures that P divides some 3^k+1, and the condition "<= POT(P-1)/8" is so that 3 has an eighth root modulo P. A062117 is the order of 3 modulo prime(n). (End)

Comments from Richard Bumby, Nov 12 2018: (Start)

When considering methods for finding square roots mod p one is led to filtering the nonzero elements by the power of 2 dividing the multiplicative order of the element. The lowest level -- elements of odd order -- have easily computed square roots, and the square roots of other elements can be found if you can discover at least one element at a higher level.

To say that "x^8 = 3 has no solution mod p" is to say that 3 is in one of the top three levels and that there are more than 3 levels (so that 8 divides p-1).

To say that primes "divide numbers of the form 3^k + 1" is to say that -1 is a power of 3 mod p, or that 3 is not at the lowest level. If there are only four levels (9 mod 16), these statements are equivalent. Otherwise, the two statements are different. An interesting case has 3 at the second level, so that (-3) has odd order allowing cube roots of unity to be found quickly.

I was told that Odoni had some results on finding the number of primes with k levels for which a given number (e.g., 3) is at level j, but I never tracked down a reference. If the asymptotic behavior is what one would expect, A045318 and A301916 are really far from being "almost the same", except in the trivial sense of "zero density". (End)

Complement of A045318 relative to A000040. - Vincenzo Librandi, Sep 13 2012

Union of 2, 5, A068231 (primes congruent to 11 modulo 12), prime p == 5 (mod 8) such that 3^((p-1)/4) == 1 (mod p), and primes p == 1 (mod 8) such that 3^((p-1)/8) == 1 (mod p). - Jianing Song, Jun 22 2025

Union of {3} and A301916, because 3^k + 3 = 3*(3^(k-1) + 1). [Comment edited by Jeppe Stig Nielsen, Jul 04 2020.]

Can be used to factor P+1 values where P is a potential prime of the form 3^k+2.

Is this 2 and 3 with A045318? - David A. Corneth, May 04 2018

No, it is not. Primes like 769, 1297, ... are also here but not in A045318. See A320481 for the explanation. - Jeppe Stig Nielsen, Jun 27 2020