A391441 2-rank of the form class group of the order of real quadratic fields with discriminant A079896(n).

0, 0, 1, 0, 0, 0, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 1, 0, 0, 1, 1, 2, 0, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 2, 0, 0, 1, 2, 1, 0, 1, 0, 0, 1, 2, 1, 0, 1, 1, 1, 1, 1, 0, 2, 1, 1, 0, 0, 1, 1, 2, 0, 2, 1, 0, 2, 2, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 2, 0, 0, 1, 1, 2, 1, 1, 1, 0, 1, 1, 1, 2, 1, 2

Offset

1,23

Comments

Let Cl+(D) be the form class group and the class group of the quadratic order of discriminant D. Claim: Cl+(D) has 2-rank omega(|D|) - t, where omega = A001221, and t = 0 if 32|D, t = 2 if D == 4 (mod 16), t = 1 otherwise.

Proof. Each class in Cl+(D) whose square is the identity contains two of the following quadratic forms:

(i) a*x^2 + c*y^2, where a > 0, gcd(a,c) = 1;

(ii) a*x^2 + a*y^2 + c*y^2, where a > 0, gcd(a,c) = 1.

See Section 3.3 of the Rick L. Shepherd link below. The proof assumes that D is fundamental but this is not really needed.

As a result, we only need to count the solutions to

(i) -4*a*c = D, where a > 0, gcd(a,c) = 1;

(ii) a^2 - 4*a*c = D, where a > 0, gcd(a,c) = 1.

Indeed, if r is the 2-rank of Cl+(D) and N is the number of solutions, then N = 2^r * 2 = 2^(r+1).

In (i), a can be any unitary divisor of |D|/4, so the number of solutions to (i) is 2^(omega(|D|/4)) if 4|D, and 0 otherwise.

To find the number of solutions to (ii), write D = 2^e * s, s odd. Note that e = v2(a) + v2(a - 4*c), where v2 is the 2-adic valuation.

- If e = 0, then a can be any unitary divisor of |D|, so the number of solutions is 2^(omega(|D|)).

- If e = 2, then a must be even and c must be odd, and so a == 2 (mod 4). We have {a,a-4*c} == {2,6} (mod 8), so D = a * (a-4*c) == 12 (mod 16), in which case a can be twice any unitary divisor of s. The number of solutions is 2^(omega(s)) = 2^(omega(|D|)) - 1 if D == 12 (mod 16), and 0 if D == 4 (mod 16).

- Obviously there is no solution if e = 3, since v2(a) + v2(a - 4*c) = 2 if v2(a) = 1 and >= v2(a) + 2 if v2(a) >= 2.

- There is also no solution if e = 4: we must have v2(a) = 2, but then a - 4*c is divisible by 8.

- If e >= 5, then v2(a) = 2 or v2(a) = e-2, and so a can be 4 times any unitary divisor of s or 2^(e-2) times any unitary divisor of s. The number of solutions is 2^(omega(s)) * 2 = 2^(omega(|D|)).

We conclude that the number of solutions to (i), (ii) is

- 2^(omega(|D|)-1) if D == 4 (mod 16);

- 2^(omega(|D|)+1) if 32|D;

- 2^(omega(|D|)) otherwise.

Links

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

Rick L. Shepherd, Binary quadratic forms and genus theory, Master of Arts Thesis, University of North Carolina at Greensboro, 2013.

Formula

For D = A079896(n), a(n) = omega(|D|) - t, where omega = A001221, and t = 0 if 32|D, t = 2 if D == 4 (mod 16), t = 1 otherwise.

Example

Let D = A079896(23) = 60. It is not divisible by 32 nor congruent to 4 modulo 16, so a(23) = omega(60) - 1 = 2. Indeed, Cl+(60) = C_2 X C_2.
Let D = A079896(219) = 480. It is divisible by 32, so a(219) = omega(480) = 3. Indeed, Cl+(480) = C_2 X C_2 X C_2.
Let D = A079896(238) = 520. It is not divisible by 32 nor congruent to 4 modulo 16, so a(238) = omega(520) - 1 = 2. Indeed, Cl+(520) = C_2 X C_2.

Prog

(PARI) r(D) = omega(D) - if(D%32==0, 0, if(D%16==4, 2, 1)) \\ gives 2-rank of Cl+(D)
for(D=1, 1000, if(D%4<=1 && !issquare(D), print1(r(D), ", ")))

Crossrefs

Cf. A319659 (for imaginary quadratic fields).

For a list of sequences related to the class groups of real quadratic fields, see A390079.

Sequence in context: A287179 A236511 A235924 * A393337 A097304 A136745

Adjacent sequences: A391438 A391439 A391440 * A391442 A391443 A391444

Keyword

nonn

Author

Jianing Song, Dec 09 2025

Status

approved