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A316506
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a(n) is the rank of the multiplicative group of Gaussian integers modulo n.
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4
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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, 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, 2, 4, 1, 4, 2, 3, 3, 3, 2, 4, 1, 5, 2, 3, 1, 4, 4, 2, 3
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OFFSET
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1,4
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COMMENTS
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The rank of a finitely generated group rank(G) is defined to be the size of the minimal generating sets of G.
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])*.
The order of (Z[i]/nZ[i])* is A079458(n) and the exponent of it is A227334(n).
{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).
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
More generally, let pi be a prime element of Z[i] of norm p or p^2 for prime p, then:
- for p == 1 (mod 4), (Z[i]/(pi^e)Z[i])* = C_((p-1)*p^(e-1));
- for p == 3 (mod 4), (Z[i]/(pi^e)Z[i])* = C_(p^(e-1)) X C_(p^(e-1)*(p^2-1));
- 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.
For a more general result see my link below. (End)
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LINKS
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FORMULA
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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.
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EXAMPLE
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(Z[i]/1Z[i])* = C_1 (has rank 0);
(Z[i]/2Z[i])* = C_2 (has rank 1);
(Z[i]/3Z[i])* = C_8 (has rank 1);
(Z[i]/4Z[i])* = C_2 X C_4 (has rank 2);
(Z[i]/5Z[i])* = C_4 X C_4 (has rank 2);
(Z[i]/6Z[i])* = C_2 X C_8 (has rank 2);
(Z[i]/7Z[i])* = C_48 (has rank 1);
(Z[i]/8Z[i])* = C_2 X C_4 X C_4 (has rank 3);
(Z[i]/9Z[i])* = C_3 X C_24 (has rank 2);
(Z[i]/10Z[i])* = C_2 X C_4 X C_4 (has rank 3).
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PROG
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(PARI) rad(n) = factorback(factorint(n)[, 1]);
grad(n)=
{
my(r=1, f=factor(n));
for(j=1, #f[, 1], my(p=f[j, 1], e=f[j, 2]);
if(p==2&e==1, r*=2);
if(p==2&e==2, r*=4);
if(p==2&e>=3, r*=8);
if(p%4==1, r*=(rad(p-1))^2);
if(p%4==3&e==1, r*=rad(p^2-1));
if(p%4==3&e>=2, r*=p^2*rad(p^2-1));
);
return(r);
}
a(n)=if(n>1, vecmax(factor(grad(n))[, 2]), 0);
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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