OFFSET
1,1
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) = 2.
Numbers of the form k^2 where k is in A002476 U {9}. That is to say, numbers of the form k^2 where k = 9 or is a prime 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.
LINKS
FORMULA
a(n) = A002476(n-1)^2 for n >= 3.
EXAMPLE
169 is a term since the one (and only one) cyclic cubic field with that discriminant is Q[x]/(x^3 - x^2 - 4x - 1).
PROG
(PARI) isA343022(n) = if(issquare(n), my(k=sqrtint(n)); k==9 || (isprime(k) && k%3==1), 0)
CROSSREFS
Discriminants and their square roots of cyclic cubic fields:
Exactly 1 associated cyclic cubic field: this sequence, A002476 U {9}.
KEYWORD
nonn,easy
AUTHOR
Jianing Song, Apr 02 2021
STATUS
approved