A391361 Number of irreducible elements of Z[sqrt(-6)] of norm n up to association.

0, 0, 0, 1, 0, 1, 2, 0, 1, 2, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 2, 0, 0, 3, 0, 0, 0, 0, 0, 2, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0

Offset

1,7

Comments

Note that Q(sqrt(-6)) has class number 2.

Note that a is irreducible if and only if (a) is either a prime ideal of the product of two (not necessarily distinct) non-principal prime ideals; the same applies in every quadratic field whose class group has exponent 2. We have:

- {N(P) : P principal prime ideal} = {p : p prime == 1, 7 (mod 24)} U {p^2 : p prime == 13, 17, 19, 23 (mod 24)}. There are two prime ideals with norm p for each prime p == 1, 7 (mod 24), and only one with norm p^2;

- {N(P) : P non-principal prime ideal} = {2, 3} U {p : p prime == 5, 11 (mod 24)}. There are two prime ideals with norm p for each prime p == 5, 11 (mod 24), and only one with norm 2 or 3.

As a result:

- a(4) = a(6) = a(9) = a(p^2) = 1 for primes p == 13, 17, 19, 23 (mod 24);

- a(p) = 2 for primes p == 1, 7 (mod 24);

- a(2*p) = a(3*p) = 2 for primes p == 5, 11 (mod 24);

- a(p^2) = 3 for primes p == 5, 11 (mod 24);

- a(p*q) = 4 for distinct primes p == 5, 11 (mod 24);

- a(n) = 0 for every other n.

Links

Jianing Song, Table of n, a(n) for n = 1..10000

Example

Write Pp = (p, x+y*sqrt(-6)) and Pp' = (p, x-y*sqrt(-6)) for p == 5, 11 (mod 24), where (x,y) is the unique positive solution to x^2 + 6*y^2 = p^2. Write P2 = (2, sqrt(-6)) and P3 = (3, sqrt(-6)).
There are 2 irreducible elements with norm 10, corresponding to P2*P5' = (2-sqrt(-6)) and P2*P5' = (2+sqrt(-6)).
There are 2 irreducible elements with norm 15, corresponding to P3*P5' = (3+sqrt(-6)) and P3*P5' = (3-sqrt(-6)).
There are 2 irreducible elements with norm 22, corresponding to P2*P11' = (4+sqrt(-6)) and P2*P11' = (4-sqrt(-6)).
There are 3 irreducible elements with norm 25, corresponding to P5^2 = (1+2*sqrt(-6)), P5*P5' = (5), and P5^2' = (1-2*sqrt(-6)).
There are 2 irreducible elements with norm 33, corresponding to P3*P11' = (3-2*sqrt(-6)) and P3*P11' = (3+2*sqrt(-6)).
There are 4 irreducible elements with norm 55, corresponding to P5*P11 = (7-sqrt(-6)), P5*P11' = (1-3*sqrt(-6)), P5'*P11 = (1+3*sqrt(-6)), and P5'*P11' = (7+sqrt(-6)).
There are 2 irreducible elements with norm 58, corresponding to P2*P29' = (2-3*sqrt(-6)) and P2*P29' = (2+3*sqrt(-6)).
There are 2 irreducible elements with norm 87, corresponding to P3*P29' = (9+sqrt(-6)) and P3*P29' = (9-sqrt(-6)).

Prog

(PARI) A391361(n) = {
if(bigomega(n) >= 3, return(0));
my(f = factor(n));
if(n==4 || n==6 || n==9, return(1));
if(#f~ == 1, if((f[1, 1] % 24 == 1 || f[1, 1] % 24 == 7) && f[1, 2] == 1, return(2)); if(kronecker(-24, f[1, 1]) == -1 && f[1, 2] == 2, return(1)); if((f[1, 1] % 24 == 5 || f[1, 1] % 24 == 11) && f[1, 2] == 2, return(3)));
if(#f~ == 2 && (f[2, 1] % 24 == 5 || f[2, 1] % 24 == 11), if(f[1, 1] == 2 || f[1, 1] == 3, return(2)); if(f[1, 1] % 24 == 5 || f[1, 1] % 24 == 11, return(4)));
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).

Other sequences related to factorizations in Z[sqrt(-6)]: A391366, A390498, A391364.

Analogs in other quadratic fields with class number 2: A391360 (discriminant -15), A091731 (discriminant -20), A391362 (discriminant 40).

Sequence in context: A380705 A295859 A180160 * A393823 A101661 A322989

Adjacent sequences: A391358 A391359 A391360 * A391362 A391363 A391364

Keyword

nonn,easy

Author

Jianing Song, Dec 07 2025

Status

approved