cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A343002 Discriminants with exactly 2 associated cyclic cubic fields.

Original entry on oeis.org

3969, 8281, 13689, 17689, 29241, 47089, 61009, 67081, 77841, 90601, 110889, 149769, 162409, 182329, 219961, 231361, 261121, 301401, 305809, 312481, 346921, 363609, 431649, 461041, 494209, 505521, 519841, 582169, 628849, 667489, 758641, 762129, 790321, 859329, 900601, 946729, 962361
Offset: 1

Views

Author

Jianing Song, Apr 02 2021

Keywords

Comments

A cubic field is cyclic if and only if its discriminant is a square. Hence all terms are squares.
Numbers of the form k^2 where A160498(k) = 4.
Numbers of the form k^2 where k is 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.
Products of two nonequal terms in A343022.
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.

Examples

			3969 = 63^2 is a term since it is the discriminant of the 2 cyclic cubic fields Q[x]/(x^3 - 21x - 28) and Q[x]/(x^3 - 21x - 35).
8281 = 91^2 is a term since it 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).
		

Crossrefs

Discriminants and their square roots of cyclic cubic fields:
At least 1 associated cyclic cubic field: A343000, A343001.
Exactly 1 associated cyclic cubic field: A343022, A002476 U {9}.
At least 2 associated cyclic cubic fields: A343024, A343025.
Exactly 2 associated cyclic cubic fields: this sequence, A343003.

Programs

  • PARI
    isA343002(n) = if(omega(n)==2, if(n==3969, 1, my(L=factor(n)); L[2,1]%3==1 && L[2,2]==2 && ((L[1,1]%3==1 && L[1,2]==2) || L[1,1]^L[1,2] == 81)), 0)

Formula

a(n) = A343003(n)^2.