cp's OEIS Frontend

A385166 Let p = A002145(n) be the n-th prime == 3 (mod 4); a(n) = (p+1) * ord(5,p) / ord(2+-i,p) = (p+1) * ord(5,p) / A385165(n). Here ord(a,m) is the multiplicative order of a modulo m.

%I A385166 #17 Jun 21 2025 19:59:56
%S A385166 1,1,2,1,1,1,1,1,1,1,1,2,3,1,1,1,2,5,1,1,1,1,3,2,2,1,1,2,2,1,1,1,11,1,
%T A385166 4,3,10,1,1,1,3,1,2,1,1,1,6,1,6,1,3,1,1,1,4,3,24,1,1,1,1,1,3,1,4,2,1,
%U A385166 1,1,1,1,2,1,1,8,1,27,1,1,1,1,20,3,1,4,1,1
%N A385166 Let p = A002145(n) be the n-th prime == 3 (mod 4); a(n) = (p+1) * ord(5,p) / ord(2+-i,p) = (p+1) * ord(5,p) / A385165(n). Here ord(a,m) is the multiplicative order of a modulo m.
%C A385166 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).
%H A385166 Jianing Song, <a href="/A385166/b385166.txt">Table of n, a(n) for n = 1..10000</a>
%e A385166 The multiplicative order of 2+-i modulo A002145(3) = 11 is A385165(3) = 30, so a(3) = (12*ord(5,11))/30 = 2.
%e A385166 The multiplicative order of 2+-i modulo A002145(13) = 83 is A385165(13) = 2296, so a(13) = (84*ord(5,83))/2296 = 3.
%o A385166 (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 A385166 forprime(p=3, 1e3, if(p%4==3, print1(quot(p), ", ")))
%Y A385166 Cf. A002145, A385165. Primes corresponding to special terms: A385168 (>1), A385167 (even), A385180 (divisible by 4).
%Y A385166 Cf. A211450.
%K A385166 nonn,easy
%O A385166 1,3
%A A385166 _Jianing Song_, Jun 20 2025