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A343003
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Numbers k such that there are exactly 2 cyclic cubic fields with discriminant k^2.
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7
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63, 91, 117, 133, 171, 217, 247, 259, 279, 301, 333, 387, 403, 427, 469, 481, 511, 549, 553, 559, 589, 603, 657, 679, 703, 711, 721, 763, 793, 817, 871, 873, 889, 927, 949, 973, 981, 1027, 1057, 1099, 1141, 1143, 1147, 1159, 1251, 1261, 1267, 1273, 1333
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OFFSET
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1,1
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COMMENTS
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It makes no difference if the word "cyclic" is omitted from the title because a cubic field is cyclic if and only if its discriminant is a square.
Numbers k such that A160498(k) = 4.
Numbers of the form (i) 9p, where p is a prime congruent to 1 modulo 3; (ii) pq, where p, q are distinct primes congruent to 1 modulo 3.
In general, there are exactly 2^(t-1) (cyclic) cubic fields with discriminant k^2 if and only if k is of the form (p_1)*(p_2)*...*(p_t) or 9*(p_1)*(p_2)*...*(p_{t-1}) with distinct primes p_i == 1 (mod 3), See A343000 for more detailed information.
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LINKS
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FORMULA
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EXAMPLE
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63 is a term since 63^2 = 3969 is the discriminant of the 2 cyclic cubic fields Q[x]/(x^3 - 21x - 28) and Q[x]/(x^3 - 21x - 35).
91 is a term since 91^2 = 8281 is the discriminant of the 2 cyclic cubic fields Q[x]/(x^3 - x^2 - 30x + 64) and Q[x]/(x^3 - x^2 - 30x - 27).
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PROG
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(PARI) isA343003(n) = if(omega(n)==2, if(n==63, 1, my(L=factor(n)); L[2, 1]%3==1 && L[2, 2]==1 && ((L[1, 1]%3==1 && L[1, 2]==1) || L[1, 1]^L[1, 2] == 9)), 0)
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CROSSREFS
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Discriminants and their square roots of cyclic cubic fields:
Exactly 2 associated cyclic cubic fields: A343002, this sequence.
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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