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Let k be any quadratic field such that all prime factors of n are inert in k, O_k be the corresponding ring of integers and G(n) = (O_k/(nO_k))* be the multiplicative group of integers in O_k modulo n; then a(n) is the exponent of G(n).
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%I #13 Feb 03 2020 05:06:55

%S 1,3,8,6,24,24,48,12,24,24,120,24,168,48,24,24,288,24,360,24,48,120,

%T 528,24,120,168,72,48,840,24,960,48,120,288,48,24,1368,360,168,24,

%U 1680,48,1848,120,24,528,2208,24,336,120,288,168,2808,72,120,48,360,840,3480

%N Let k be any quadratic field such that all prime factors of n are inert in k, O_k be the corresponding ring of integers and G(n) = (O_k/(nO_k))* be the multiplicative group of integers in O_k modulo n; then a(n) is the exponent of G(n).

%C The exponent of a finite group G is the least positive integer k such that x^k = e for every x in G, where e is the group identity.

%C Let k be any quadratic field; if a prime p is inert in k, then O_k/(pO_k) is a finite field with order p^2, so the exponent is p^2 - 1. By induction we have: G(p^e) is isomorphic to C_(p^(e-1)) X C_((p^2-1)*p^(e-1)) is p is an odd prime; G(2^e) is isomorphic to C_3 if e = 1 and C_2 X C_(2^(e-2)) X C(3*2^(e-1)) if e >= 2. By the Chinese Remainder Theorem, if n = Product_{i=1..m} (p_i)^(e_i), then G(n) = G((p_1)^(e_1)) X G((p_2)^(e_2)) X ... G((p_m)^(e_m)). The order of G(n) is A007434(n).

%C a(n) is divisible by 24 unless n = 1, 2, 3, 4, 8.

%H Jianing Song, <a href="/A306933/b306933.txt">Table of n, a(n) for n = 1..10000</a>

%F a(p^e) = (p^2-1)*p^(e-1); if n = Product_{i=1..m} (p_i)^(e_i), then a(n) = lcm(a((p_1)^(e_1)), a((p_2)^(e_2)), ..., a((p_m)^(e_m))). [Simplified by _Jianing Song_, Feb 02 2020]

%e Let n = 10 = 2 * 5 and k = Q[sqrt(-3)]; then both 2 and 5 are inert in k. G(10) = (O_k/(10O_k))* = (O_k/(2O_k))* X (O_k/(5O_k))* = C_3 X C_24, the exponent of which is 24, so a(10) = 24.

%e Let n = 45 = 3^2 * 5 and k = Q[sqrt(2)]; then both 3 and 5 are inert in k. G(45) = (O_k/(45O_k))* = (O_k/(3^2*O_k))* X (O_k/(5O_k))* = (C_3 X C_24) X C_24, the exponent of which is 24, so a(45) = 24.

%o (PARI) a(n) = my(r=1, f=factor(n)); for(j=1, #f[, 1], my(p=f[j, 1], e=f[j, 2]); r=lcm(r, (p^2-1)*p^(e-1))); r \\ Corrected by _Jianing Song_, Feb 02 2020

%Y Cf. A007434.

%K nonn

%O 1,2

%A _Jianing Song_, Mar 16 2019