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A306933
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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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2
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1, 3, 8, 6, 24, 24, 48, 12, 24, 24, 120, 24, 168, 48, 24, 24, 288, 24, 360, 24, 48, 120, 528, 24, 120, 168, 72, 48, 840, 24, 960, 48, 120, 288, 48, 24, 1368, 360, 168, 24, 1680, 48, 1848, 120, 24, 528, 2208, 24, 336, 120, 288, 168, 2808, 72, 120, 48, 360, 840, 3480
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OFFSET
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1,2
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COMMENTS
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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.
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).
a(n) is divisible by 24 unless n = 1, 2, 3, 4, 8.
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LINKS
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FORMULA
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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]
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EXAMPLE
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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.
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.
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PROG
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(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
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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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