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%I #15 May 18 2021 04:07:46
%S 1,-4,-3,256,125,144,-16807,16777216,-19683,4000000,-2357947691,
%T 5308416,1792160394037,1157018619904,1265625,18446744073709551616,
%U 2862423051509815793,1586874322944,-5480386857784802185939,1048576000000000000,205924456521,5829995856912430117421056
%N Discriminant of the (2n)-th cyclotomic field Q(zeta_(2n)). Equivalently, discriminant of the (2n)-th cyclotomic polynomial.
%C Note that Q(zeta_n) = Q(zeta_(2n)) for odd n, so this sequence is A004124 with redundant values removed.
%C a(n) is negative <=> phi(2n) == 2 (mod 4) <=> n = 2 or n is of the form p^e, where p is a prime congruent to 3 modulo 4.
%H Jianing Song, <a href="/A344407/b344407.txt">Table of n, a(n) for n = 1..200</a>
%F a(n) = A004124(2n). See A004124 for its formula.
%F For n >= 2, a(p) = (-1)^((p-1)/2) * A130614(n), where p = prime(n).
%e n = 2: Q(zeta_4) = Q(i) has discriminant -4;
%e n = 3: Q(zeta_6) = Q(sqrt(-3)) has discriminant -3;
%e n = 4: Q(zeta_8) = Q(sqrt(2), i) has discriminant 256;
%e n = 5: Q(zeta_10) = Q(exp(2*Pi*i/5)) has discriminant 125;
%e n = 6: Q(zeta_12) = Q(sqrt(3), i) has discriminant 144.
%o (PARI) vector(25,n,poldisc(polcyclo(2*n)))
%Y Cf. A004124, A062570 (degree of Q(zeta_(2n))).
%K sign,easy
%O 1,2
%A _Jianing Song_, May 17 2021
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