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A343023
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Number of cyclic cubic fields with discriminant n^2.
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9
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0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 2, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 1, 0, 0, 0
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OFFSET
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1,63
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COMMENTS
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Equivalently, number of cubic fields with discriminant n^2. That is to say, it makes no difference if the word "cyclic" is omitted from the title.
Let D be a discriminant of a cubic field F, then F is a cyclic cubic field if and only if D is a square. For D = k^2, k must be 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), in which case there are exactly 2^(t-1) = 2^(omega(k)-1) (cyclic) cubic fields with discriminant D. See Page 17, Theorem 2.7 of the Ka Lun Wong link.
Each term is 0 or a power of 2.
The first occurrence of 2^t is 9*A121940(t) for t >= 1.
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LINKS
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FORMULA
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EXAMPLE
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a(7) = 1 since there is only 1 (cyclic) cubic field with discriminant 7^2 = 49 is Q[x]/(x^3 - x^2 + x + 1).
a(63) = 2 since there are 2 (cyclic) cubic fields with discriminant 63^2 = 3969: Q[x]/(x^3 - 21x - 28) and Q[x]/(x^3 - 21x - 35).
a(819) = 4 since there are 4 (cyclic) cubic fields with discriminant 819^2 = 670761: Q[x]/(x^3 - 273x - 91), Q[x]/(x^3 - 273x - 728), Q[x]/(x^3 - 273x - 1547) and Q[x]/(x^3 - 273x - 1729).
a(35) = 0 since it is not of form (p_1)*(p_2)*...*(p_t) or 9*(p_1)*(p_2)*...*(p_{t-1}) with distinct primes p_i == 1 (mod 3). Indeed, there are no (cyclic) cubic fields with discriminant 35^2 = 1225.
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PROG
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(PARI) a(n) = if(n<=1, 0, my(L=factor(n), w=omega(n)); for(i=1, w, if(!((L[i, 1]%3==1 && L[i, 2]==1) || L[i, 1]^L[i, 2] == 9), return(0))); 2^(w-1))
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CROSSREFS
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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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