cp's OEIS Frontend

It is conjectured that a(101) = 7392 is the last term. If it would exist, a(102) > 10^6. - Hugo Pfoertner, Dec 01 2019

Intersection of A046005 and A003644. Note that A003644 = A014602 union A014603 union A192322 union {a(n)} union {5460}. - Jianing Song, Jul 12 2018

This sequence lists the negated discriminants of imaginary fields whose class group is isomorphic to (C_2)^r, r > 0.

These are the negated fundamental discriminants in A133288.

Also numbers in A003644 but not in A014602. Equals A014603 union A192322 union A305416 union {5460}.

Fundamental terms of A317987.

k is a term iff the class group of Q[sqrt(-k)], or the form class group of positive binary quadratic forms with discriminant -k is isomorphic to (C_2)^r X C_4.

This is a subsequence of A133676, so it's finite. It seems that this sequence has 161 terms, the largest being 40755.

It seems that this sequence contains 810 terms, the largest being 1154008. In general, it seems that for any t > 0, b(-d) = o(d^t) as -d -> -oo.

For fundamental discriminants -d, we want to determine the size of b(-d), i.e., the size of the smallest prime that decomposes in Q[sqrt(-d)].

Let K = Q[sqrt(-d)], O_K be the ring of integers over K, so O_K is a Dedekind domain. Let E(-d) be the exponent of the ideal class group of O_K (the exponent of a group G is the smallest e > 0 such that x^e = I for all x in G, where I is the group identity).

If Kronecker(-d,p) = 1, it is well-known that p*O_K is the product of two distinct prime ideals of O_K, say, p*O_K = I*I'. By the property of the ideal class group of Q[sqrt(-d)], I^(k*e) must be principal, e = E(-d). Let t*O_K = I^(k*e), then t/p is not an algebraic integer, and the norm of t is p^e. Define f(x,y) = x^2 + (d/4)*y^2 if -d == 0 (mod 4), x^2 + x*y + ((d+1)/4)*y^2 otherwise, it is easy to see f(x,y) = p^(k*e) has integral solutions (x,y) such that gcd(x,y) = 1.

If f(x,y) = p^(k*e) < d, then |y| = 1, so 4*p^(k*e) - d must be a (positive) square. Setting k = 1 gives b(-d) > (d/4)^(1/e) (and furthermore we have: if Kronecker(-d,p) = 1 and p^(k*e) < d, then k = 1, or (p,k,e,d) = (2,2,1,7), (3,2,1,11)).

If E(-d) = 3, then d is in this sequence.

We also have the following observations (not proved):

(a) if e = 2 (i.e., d is in A003644\A014602 = A316743), then b(-d) < d/4;

(b) if e > 2, then b(-d) < sqrt(d/4) (it can be proved by using deeper algebraic number theory that b(-d) < 2*sqrt(d)/Pi).

If these observations are true, this sequence is also the list of d such that b(-d) > (d/4)^(1/3) and d is not in A003644.

Note that 5460 is conjectured to be the largest term in A003644. Therefore, it seems that b(-d) < sqrt(d/4) for all d > 5460; it seems that b(-d) < (d/4)^(1/3) for all d > 1154008.

Among the known terms:

(1) the term d with the largest E(-d) is d = 998328 with E(-d) = 66.

(2) the term d with the largest b(-d) is d = 656755 with b(-d) = 79.

(3) the largest prime is d = 90787 with E(-d) = 23.

2*sqrt(d)/Pi is the so-called "Minkowski's bound" for imaginary quadratic field. For other discriminants -d, there exists a prime p < 2*sqrt(d)/Pi such that Kronecker(-d,p) = 1.

Let K = Q(sqrt(-d)) be an imaginary quadratic field. The ideal class group of O_K (the ring of integers over K) is generated by ideal classes that contain I, where each I divides p*O_K for some p < 2*sqrt(d)/Pi. Note that if Kronecker(-d,p) = -1 (i.e., p is inert in K), then p*O_K is a prime ideal; if p | -d (i.e., p ramifies in K), then p*O_K = I^2, so the order of the ideal class that contains I is <= 2 in the ideal class group. So the ideal class group of Q(sqrt(-d)) necessarily has exponent <= 2 (The exponent of a group G is the smallest e > 0 such that x^e = I for all x in G, where I is the group identity.). So this is a subsequence of A003644.

But there are other d such that the ideal class group of O_K has exponent 2. In fact, the exponent is <= 2 if and only if: for all primes p < 2*sqrt(d)/Pi, either (a) Kronecker(-d,p) = 0 or -1, or (b) Kronecker(-d,p) = 1, and 4*p^2 - d is a square. Here 2*sqrt(d)/Pi can be replaced by sqrt(d); conjecturally, if 2*sqrt(d)/Pi is replaced by sqrt(d/4), we get exactly the sequence A003644.