A343023 Number of cyclic cubic fields with discriminant n^2.

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

Offset: 1

Author

Jianing Song, Apr 02 2021

Keywords

• nonn
• easy

Comments

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.

Examples

			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.
		

Links

• Jianing Song, Table of n, a(n) for n = 1..16000
• LMFDB, Cubic fields
• Wikipedia, Cubic field
• Ka Lun Wong, Maximal Unramified Extensions of Cyclic Cubic Fields, (2011), Theses and Dissertations, 2781.

Crossrefs

Cf. A160498, A121940, A343000 (discriminants of cyclic cubic fields), A343001 (indices of positive terms).

Programs

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))

Formula

a(n) = A160498(n)/2 for n > 1.