Is the class number of the quadratic field $x^2=3\cdot2^n+1$ $O(n)$?

2011-10-29T13:14:06Z

Numerical evidence suggest that the class number of the quadratic field $x^2=3\cdot2^n+1$ is $O(n)$ while the discriminant is $O(2^n)$.

Here are the class numbers for $n=6 \dots 110$ computed with pari/gp:

1, 2, 1, 2, 1, 4, 1, 3, 2, 4, 1, 1, 1, 2, 3, 1, 2, 1, 2, 2, 2, 4, 48, 1, 3, 2, 1, 4, 2, 2, 1, 8, 4, 4, 4, 5, 1, 2, 6, 35, 3, 80, 25, 2, 4, 24, 4, 4, 12, 8, 2, 1, 24, 12, 2, 8, 8, 1, 8, 4, 13, 104, 4, 2, 1, 8, 4, 8, 400, 2, 4, 1, 1, 4, 2, 2, 2, 4, 4, 2, 10, 80, 2, 16, 2, 16, 2, 372, 4, 32, 4, 46, 8, 6, 8, 12, 6, 1, 4, 4, 4, 4, 8, 4, 12, 4, 8, 8, 4, 72

A Monte Carlo factoring algorithm with linear storage possibly might be used to verify for larger $n$, though I don't have working implementation yet.

The motivation is that small class number and the above algorithm might give a divisor of numbers of these form.

I get similar results for $x^2=4\left(3\cdot2^n-1\right)$

For Mersenne numbers $x^2=2^n-1$ there are large class numbers.

 1, 1, 4, 2, 8, 2, 4, 1, 16, 4, 48, 9, 16, 1, 16, 2, 256, 2, 272, 4, 520, 4, 1920, 32, 576, 5, 4352, 16, 12168, 16, 6656, 4, 24480, 16, 20416, 4, 8560, 72, 241920, 140352, 498720, 4, 206592, 2, 1494528, 384, 4771328, 32, 758016, 128, 11758848, 8, 19031040, 6, 7610368, 1, 70529424, 16, 246776832, 80, 51142144, 2, 615757824, 16, 1378201600, 8, 900177920, 8, 4452143904, 128, 9139057152, 8, 8201582592, 24, 4530836992, 32, 83129328224, 4, 6856657920, 16, 182429183488, 128, 617893922304, 15, 260395347968, 32, 1944611808768, 4, 3750374341632, 32

Is the class number of the quadratic field $x^2=3\cdot2^n+1$ $O(n)$? - MathOverflow

Is the class number of the quadratic field $x^2=3\cdot2^n+1$ $O(n)$?