A385180 Primes p == 3 (mod 4) such that (p+1) * ord(5,p) / ord(2+-i,p) is divisible by 4. Here ord(a,m) is the multiplicative order of a modulo m.

331, 571, 599, 691, 839, 919, 971, 1039, 1051, 1171, 1279, 1291, 1319, 1399, 1439, 1451, 1571, 1759, 1879, 2131, 2411, 2879, 2971, 3079, 3251, 3331, 3491, 3571, 3691, 3851, 4051, 4079, 4091, 4211, 4519, 4639, 4651, 4679, 4691, 4759, 4919, 4931, 5051, 5119, 5171, 5279, 5479, 5519, 5531

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 divisible by 4.

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

By definition, a term that is in neither A385169 nor A385179 must be congruent to 31 or 79 modulo 80. The smallest such term is p = 1759 (ord(2+-i,p) = ((p+1)/4) * ord(5,p) = 128920); even if 1039 == 79 (mod 80), we have ord(2+-i,p) = ((p+1)/8) * ord(5,p) = 22490 == 2 (mod 4), which means that 1039 is in A385179.

Examples

			571 is a term since the multiplicative order of 2+-i modulo 571 is 40755, and (572*ord(5,571))/40755 = 4 is divisible by 4.
		

Links

• Jianing Song, Table of n, a(n) for n = 1..10000

Crossrefs

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

Subsequence of A385167, which is itself a subsequence of intersection of A122869 and A385168.

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)
isA385180(p) = isprime(p) && p%4==3 && quot(p)%4==0