A385167 Primes p == 3 (mod 4) such that (p+1) * ord(5,p) / ord(2+-i,p) is even. Here ord(a,m) is the multiplicative order of a modulo m.

11, 79, 131, 199, 211, 239, 251, 331, 359, 439, 479, 491, 571, 599, 691, 719, 811, 839, 919, 971, 1039, 1051, 1091, 1171, 1279, 1291, 1319, 1399, 1439, 1451, 1531, 1559, 1571, 1759, 1811, 1879, 1931, 1999, 2011, 2039, 2131, 2239, 2251, 2371, 2399, 2411, 2531, 2719, 2731, 2851, 2879, 2971, 2999

Offset: 1

Author

Jianing Song, Jun 20 2025

Keywords

• nonn
• easy

Comments

Of course if a and m are integers, it doesn't matter if the base ring is Z or Z[i] for ord(a,m).

List of p = A002145(k) such that A385166(k) is even.

Since in this case d(p) divides (p^2-1)/2, 5 must be a quadratic residue modulo p (see A385165).

Examples

			359 is a term since the multiplicative order of 2+-i modulo 359 is 6444, and (360*ord(5,359))/6444 = 10 is even.
		

Links

• Jianing Song, Table of n, a(n) for n = 1..10001

Crossrefs

Cf. A002145, A385165 (list of ord(2+-i,p)), A385166 (list of (p+1) * ord(5,p) / ord(2+-i,p)).

Subsequence of the intersection of A122869 and A385168. Contains A385180 as a subsequence.

Programs

PARI

quot(p) = my(z = znorder(Mod(5,p)), d = divisors((p+1)*z)); for(i=1, #d, if(Mod([2,-1;1,2],p)^d[i] == 1, return((p+1)*z/d[i]))) \\ for a prime p == 3 (mod 4), returns (p+1) * ord(5,p) / ord(2+-i, p)
isA385167(p) = isprime(p) && p%4==3 && quot(p)%2==0