OFFSET
1,1
COMMENTS
Given the nine quadratic integer rings with complex numbers which are also unique factorization domains, it stands to reason that most primes in Z remain inert in some of these quadratic rings and split in others.
So if we add up Legendre(H_i, p), where H_i iterates over the Heegner numbers (A003173 multiplied by -1) and p is an odd prime, we will generally find this sum to be odd and greater than -9 but less than 9.
Indeed the first occurrence of a(n) = -9 corresponds to the prime 3167, and the first occurrence of a(n) = 9 corresponds to the prime 15073. The first occurrence of a(n) = -7 or 7 corresponds to the prime 709.
The only even values in this sequence correspond the primes of A003173 (all nine except for 1). The only 0 corresponds to the prime 11; the three instances of -4 correspond to the primes 2, 3, 7; one instance of -2 for 19; two instances of 2 for 43 and 67; and one instance of 4 for 163.
Although (1 + i) is a ramifying ideal in Z[i], Kronecker(-1, 2) = 1, so here 2 is counted as splitting (+1) rather than ramifying (0) in Z[i].
FORMULA
a(n) = Sum_{i = 1..9} Kronecker(H_i, prime(n)), where Kronecker(a, p) is the Kronecker symbol (the Legendre symbol for all odd primes) and H_i is the i-th Heegner number (A003173 multiplied by -1).
EXAMPLE
We see that 3 = (1 - sqrt(-2))(1 + sqrt(-2)) = (1/2 - sqrt(-11)/2)(1/2 + sqrt(-11)/2) but is prime in Z[i], O_(Q(sqrt(-7))), O_(Q(sqrt(-19))), O_(Q(sqrt(-43))), O_(Q(sqrt(-67))) and O_(Q(sqrt(-163))).
The fact that 3 = -sqrt(-3)^2 does not matter for our purpose here. So 3 splits in two of these domains but is prime in six of them.
As 3 is the second prime, a(2) is therefore -6 + 2 = -4.
MATHEMATICA
Table[Plus@@KroneckerSymbol[{-1, -2, -3, -7, -11, -19, -43, -67, -163}, Prime[n]], {n, 100}]
CROSSREFS
KEYWORD
sign
AUTHOR
Alonso del Arte, Jul 07 2019
STATUS
approved