%I #19 Jun 12 2014 13:22:47
%S 1,108,243,108,675,972,1323,243,300,3267,972,4563,5292,
%T 6075,867,972,1083,2700,11907,13068,14283,675,2028,588,
%U 22707,24300,25947,29403,31212,3675,972,4107,38988,41067,45387,47628,49923,1452,6075,6348,59643
%N Discriminants of cubic domains for cubefree n.
%C The table in Alaca & Williams (2004) skips over n = 4, 9, 16 but includes 12, 18 and 20; then there is a footnote to the table explaining that Q(4^(1/3)) and Q(16^(1/3)) work out to be subdomains of Q(2^(1/3)), and similarly for Q(9^(1/3)) and Q(3^(1/3)) and for Q(18^(1/3)) and Q(12^(1/3)).
%D Şaban Alaca & Kenneth S. Williams, Introductory Algebraic Number Theory. Cambridge: Cambridge University Press (2004): 176177, Theorem 7.3.2 on the former page, Table 1 on the latter page.
%F Set m = A004709(n), then express it as m = h * k^2, where k = A000188(m), the square root of the largest square dividing m, and h = m/k^2 = A007913(m). Then:
%F a(n) = 3h^2 * k^2 if m == +1 (mod 9), otherwise a(n) = 27h^2 * k^2.
%F This formula is from Theorem 7.3.2 in Alaca & Williams (2004).
%e a(7) = 1323 because the seventh cubefree number is 7 and Q(7^(1/3)) has 1323 for a discriminant.
%e a(8) = 243 because the eighth cubefree number is 9 and Q(9^(1/3)) is a subdomain of Q(3^(1/3)), which has a discriminant of 243.
%t DeleteCases[Table[Boole[FreeQ[FactorInteger[n], {_, k_ /; k > 2}]] * NumberFieldDiscriminant[n^(1/3)], {n, 100}], 0]
%Y Cf. A004709 (cubefree numbers).
%K sign
%O 1,2
%A _Alonso del Arte_, May 24 2014
