OFFSET
1,4
COMMENTS
Note that Q(sqrt(-15)) 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, 4 (mod 15)} U {p^2 : p prime == 7, 11, 13, 14 (mod 15)}. There are two prime ideals with norm p for each prime p == 1, 4 (mod 15), and only one with norm p^2;
- {N(P) : P non-principal prime ideal} = {3, 5} U {p : p prime == 2, 8 (mod 15)}. There are two prime ideals with norm p for each prime p == 2, 8 (mod 15), and only one with norm 3 or 5.
As a result:
- a(9) = a(15) = a(25) = a(p^2) = 1 for primes p == 7, 11, 13, 14 (mod 15);
- a(p) = 2 for primes p == 1, 4 (mod 15);
- a(3*p) = a(5*p) = 2 for primes p == 2, 8 (mod 15);
- a(p^2) = 3 for primes p == 2, 8 (mod 15);
- a(p*q) = 4 for distinct primes p == 2, 8 (mod 15);
- 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(-15))/2) and Pp' = (p, (x-y*sqrt(-15))/2) for p == 2, 8 (mod 15), where (x,y) is the unique positive solution to x^2 + 15*y^2 = 4*p^2. (Note that x == 2 (mod 4) and 8|y for p > 2). Write P3 = (3, sqrt(-15)) and P5 = (5, sqrt(-15)).
There are 3 irreducible elements with norm 4, corresponding to P2^2 = ((1+sqrt(-15))/2), P2*P2' = (2), and P2'^2 = ((1-sqrt(-15))/2).
There are 2 irreducible elements with norm 6, corresponding to P2*P3 = ((3+sqrt(-15))/2) and P2'*P3 = ((3-sqrt(-15))/2).
There are 2 irreducible elements with norm 10, corresponding to P2*P5 = ((5+sqrt(-15))/2) and P2'*P5 = ((5-sqrt(-15))/2).
There are 4 irreducible elements with norm 34, corresponding to P2*P17 = (7+2*sqrt(-5)), P2*P17' = (8+sqrt(-5)), P2'*P17 = (8-sqrt(-5)), and P2'*P17' = (7-2*sqrt(-5)).
There are 4 irreducible elements with norm 46, corresponding to P2*P23 = ((7+3*sqrt(-15))/2), P2*P17' = ((13+sqrt(-15))/2), P2'*P17 = ((13-sqrt(-15))/2), and P2'*P17' = ((7-3*sqrt(-15))/2).
There are 2 irreducible elements with norm 51, corresponding to P3*P17 = (6+sqrt(-15)) and P3*P17' = (6-sqrt(-15)).
There are 2 irreducible elements with norm 69, corresponding to P3*P23 = (3+2*sqrt(-15)) and P3*P23' = (3-2*sqrt(-15)).
There are 2 irreducible elements with norm 85, corresponding to P5*P17 = (5+2*sqrt(-15)) and P5*P17' = (5-2*sqrt(-15)).
There are 4 irreducible elements with norm 94, corresponding to P2*P47 = ((19-sqrt(-15))/2), P2*P47' = ((1+5*sqrt(-15))/2), P2'*P47 = ((1-5*sqrt(-15))/2), and P2'*P47' = ((19+sqrt(-15))/2).
PROG
(PARI) A391360(n) = {
if(bigomega(n) >= 3, return(0));
my(f = factor(n));
if(n==9 || n==15 || n==25, return(1));
if(#f~ == 1, if((f[1, 1] % 15 == 1 || f[1, 1] % 15 == 4) && f[1, 2] == 1, return(2)); if(kronecker(-15, f[1, 1]) == -1 && f[1, 2] == 2, return(1)); if((f[1, 1] % 15 == 2 || f[1, 1] % 15 == 8) && f[1, 2] == 2, return(3)));
if(n==6 || n==10, return(2));
if(#f~ == 2 && (f[2, 1] % 15 == 2 || f[2, 1] % 15 == 8), if(f[1, 1] == 3 || f[1, 1] == 5, return(2)); if(f[1, 1] % 15 == 2 || f[1, 1] % 15 == 8, return(4)));
return(0);
}
CROSSREFS
Cf. A191018 (primes decomposing in Q(sqrt(-15))), A191062 (primes remaining inert in Q(sqrt(-15))), A033212 ((p) is the product of two principal ideals), A106859 ((p) is the product of two non-principal ideals).
KEYWORD
nonn,easy
AUTHOR
Jianing Song, Dec 07 2025
STATUS
approved