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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.

%I A385167 #23 Jun 23 2025 14:41:34
%S A385167 11,79,131,199,211,239,251,331,359,439,479,491,571,599,691,719,811,
%T A385167 839,919,971,1039,1051,1091,1171,1279,1291,1319,1399,1439,1451,1531,
%U A385167 1559,1571,1759,1811,1879,1931,1999,2011,2039,2131,2239,2251,2371,2399,2411,2531,2719,2731,2851,2879,2971,2999
%N 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.
%C A385167 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).
%C A385167 List of p = A002145(k) such that A385166(k) is even.
%C A385167 Since in this case d(p) divides (p^2-1)/2, 5 must be a quadratic residue modulo p (see A385165).
%H A385167 Jianing Song, <a href="/A385167/b385167.txt">Table of n, a(n) for n = 1..10001</a>
%e A385167 359 is a term since the multiplicative order of 2+-i modulo 359 is 6444, and (360*ord(5,359))/6444 = 10 is even.
%o A385167 (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)
%o A385167 isA385167(p) = isprime(p) && p%4==3 && quot(p)%2==0
%Y A385167 Cf. A002145, A385165 (list of ord(2+-i,p)), A385166 (list of (p+1) * ord(5,p) / ord(2+-i,p)).
%Y A385167 Subsequence of the intersection of A122869 and A385168. Contains A385180 as a subsequence.
%K A385167 nonn,easy
%O A385167 1,1
%A A385167 _Jianing Song_, Jun 20 2025