OFFSET
1,1
COMMENTS
Note that Q(sqrt(-6)) has class number 2.
The complement of this sequence is the set of numbers m such that the non-principal prime ideal factors (counted with multiplicities) of (m) has a unique way to be grouped in pairs. In other words, m is not a term if and only if:
- either there is only one non-principal prime ideal dividing (m);
- or there are exactly two non-principal prime ideal factors of (m), each with exponent 1.
The same applies in every quadratic field whose class group has exponent 2.
Since (p) is the product of two distinct non-principal prime ideals if and only if p == 5, 11 (mod 24), this sequence consists exactly numbers divisible by 6, 2*p, 3*p or p*q (not necessarily distinct) for primes p, q == 5, 11 (mod 24).
LINKS
Jianing Song, Table of n, a(n) for n = 1..10000
EXAMPLE
The non-principal prime ideal factors of 10 are {P2,P2,P5,P5'}, which can be grouped in two ways {{P2,P2},{P5,P5'}} and {{P2,P5},{P2,P5'}}. Indeed, 10 = 2 * 5 = (2+sqrt(-6)) * (2-sqrt(-6)).
The non-principal prime ideal factors of 33 are {P3,P3,P11,P11'}, which can be grouped in two ways {{P3,P3},{P11,P11'}} and {{P3,P11},{P3,P11'}}. Indeed, 33 = 3 * 11 = (3+2*sqrt(-6)) * (3-2*sqrt(-6)).
The non-principal prime ideal factors of 55 are {P5,P5',P11,P11'}, which can be grouped in three ways {{P5,P5'},{P11,P11'}}, {{P5,P11},{P5',P11'}} and {{P5,P11'},{P5',P11}}. Indeed, 55 = 5 * 11 = (7-sqrt(-6)) * (7+sqrt(-6)) = (1-3*sqrt(-6)) * (1+3*sqrt(-6)).
PROG
(PARI) isA391364(n) = {
if(n % 6 == 0, return(1));
my(f = factor(n), cnt = 0); for(i=1, #f~, if(f[i, 1] % 24 == 5 || f[i, 1] % 24 == 11, cnt+=f[i, 2]); if(cnt >= 2 || ((n % 2 == 0 || n % 3 == 0) && cnt >= 1), return(1)));
return(0);
}
CROSSREFS
Cf. A157437 (primes decomposing in Q(sqrt(-6))), A191059 (prime remaining inert in Q(sqrt(-6))), A033199 ((p) is the product of two principal ideals), A084865 ((p) is the product of two non-principal ideals).
KEYWORD
nonn,easy
AUTHOR
Jianing Song, Dec 07 2025
STATUS
approved