cp's OEIS Frontend

A316506 a(n) is the rank of the multiplicative group of Gaussian integers modulo n.

%I A316506 #44 Jun 21 2025 20:00:18
%S A316506 0,1,1,2,2,2,1,3,2,3,1,3,2,2,3,3,2,2,1,4,2,2,1,4,2,3,2,3,2,4,1,3,2,3,
%T A316506 3,3,2,2,3,5,2,3,1,3,3,2,1,4,2,3,3,4,2,2,3,4,2,3,1,5,2,2,3,3,4,3,1,4,
%U A316506 2,4,1,4,2,3,3,3,2,4,1,5,2,3,1,4,4,2,3
%N A316506 a(n) is the rank of the multiplicative group of Gaussian integers modulo n.
%C A316506 The rank of a finitely generated group rank(G) is defined to be the size of the minimal generating sets of G.
%C A316506 Let p be an odd prime and (Z[i]/nZ[i])* be the multiplicative group of Gaussian integers modulo n, then: (Z[i]/p^e*Z[i])* = (C_((p-1)*p^(e-1)))^2 if p == 1 (mod 4); C_(p^(e-1)) X C_(p^(e-1)*(p^2-1)) if p == 3 (mod 4). (Z[i]/2Z[i])* = C_2, (Z[i]/2^e*Z[i])* = C_4 X C_(2^(e-2)) X C_(2^(e-1)) for e >= 2. If n = Product_{i=1..k} (p_i)^(e_i), then (Z[i]/nZ[i])* = (Z[i]/(p_1)^(e_1)*Z[i])* X (Z[i]/(p_2)^(e_2)*Z[i])* X ... X (Z[i]/(p_k)^(e_k)*Z[i])*.
%C A316506 The order of (Z[i]/nZ[i])* is A079458(n) and the exponent of it is A227334(n).
%C A316506 {a(n)} is not additive: (Z[i]/2Z[i])* = C_2, (Z[i]/9Z[i])* = C_3 X C_24, so (Z[i]/18Z[i])* = C_6 X C_24, a(18) < a(2) + a(9). The same problem occurs for a(36), a(54) and a(72) and so on. But note that (Z[i]/63Z[i])* = C_3 X C_24 X C_48 and a(63) = a(7) + a(9).
%C A316506 A079458(n)/A227334(n) is always an integer, and is 1 if and only if (Z[i]/nZ[i])* is cyclic, that is, rank((Z[i]/nZ[i])*) = a(n) = 0 or 1, and n has a primitive root in (Z[i]/nZ[i])*. a(n) = 1 if and only if n = 2 or a prime congruent to 3 mod 4. - _Jianing Song_, Jan 08 2019
%C A316506 From _Jianing Song_, Oct 03 2022: (Start)
%C A316506 More generally, let pi be a prime element of Z[i] of norm p or p^2 for prime p, then:
%C A316506 - for p == 1 (mod 4), (Z[i]/(pi^e)Z[i])* = C_((p-1)*p^(e-1));
%C A316506 - for p == 3 (mod 4), (Z[i]/(pi^e)Z[i])* = C_(p^(e-1)) X C_(p^(e-1)*(p^2-1));
%C A316506 - for p = 2, (Z[i]/(pi^e)Z[i])* = C_1 for e = 1, C_2 for e = 2, C_4 X C_(2^floor((e-3)/2)) X C_(2^ceiling((e-3)/2)) for e >= 3.
%C A316506 For a more general result see my link below. (End)
%H A316506 Jianing Song, <a href="/A316506/b316506.txt">Table of n, a(n) for n = 1..10000</a>
%H A316506 Jianing Song, <a href="/A306933/a306933.pdf">Structure of (R/P^e)* for R the ring of integers of a quadratic field and P a prime ideal of R</a>
%F A316506 Let p be an odd prime, then: a(p^e) = 2 if p == 1 (mod 4) or p == 3 (mod 4), e >= 2; a(p) = 1 if p == 3 (mod 4). a(2) = 1, a(4) = 2, a(2^e) = 3 for e >= 3.
%e A316506 (Z[i]/1Z[i])* = C_1 (has rank 0);
%e A316506 (Z[i]/2Z[i])* = C_2 (has rank 1);
%e A316506 (Z[i]/3Z[i])* = C_8 (has rank 1);
%e A316506 (Z[i]/4Z[i])* = C_2 X C_4 (has rank 2);
%e A316506 (Z[i]/5Z[i])* = C_4 X C_4 (has rank 2);
%e A316506 (Z[i]/6Z[i])* = C_2 X C_8 (has rank 2);
%e A316506 (Z[i]/7Z[i])* = C_48 (has rank 1);
%e A316506 (Z[i]/8Z[i])* = C_2 X C_4 X C_4 (has rank 3);
%e A316506 (Z[i]/9Z[i])* = C_3 X C_24 (has rank 2);
%e A316506 (Z[i]/10Z[i])* = C_2 X C_4 X C_4 (has rank 3).
%o A316506 (PARI) rad(n) = factorback(factorint(n)[, 1]);
%o A316506 grad(n)=
%o A316506 {
%o A316506     my(r=1, f=factor(n));
%o A316506     for(j=1, #f[, 1], my(p=f[j, 1], e=f[j, 2]);
%o A316506         if(p==2&e==1, r*=2);
%o A316506         if(p==2&e==2, r*=4);
%o A316506         if(p==2&e>=3, r*=8);
%o A316506         if(p%4==1, r*=(rad(p-1))^2);
%o A316506         if(p%4==3&e==1, r*=rad(p^2-1));
%o A316506         if(p%4==3&e>=2, r*=p^2*rad(p^2-1));
%o A316506     );
%o A316506     return(r);
%o A316506 }
%o A316506 a(n)=if(n>1, vecmax(factor(grad(n))[, 2]), 0);
%Y A316506 Cf. A046072, A079458, A227334, A302254.
%Y A316506 Equivalent in the ring of Eisenstein integers: A319447.
%K A316506 nonn
%O A316506 1,4
%A A316506 _Jianing Song_, Jul 05 2018