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

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

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The Brauer-Siegel theorem gives some information on the asymptotics of the class number in this situation ( see en.wikipedia.org/wiki/Brauer-Siegel_theorem ). Note that to apply it in your situation, you would have to get some hold on the regulators of your sequence of quadratics fields. –  Damian Rössler Oct 29 '11 at 13:57
I'm no expert, but I suspect that the reason that the class numbers seem, vaguely, to be going "big small big small..." alternately in the Mersenne list is that, when $n=2m$ is even, you get a unit $u=2^m+x$ of about as small a height as you could ever hope for, which makes the regulator small so the class group has to be big by Brauer-Siegel. –  Kevin Buzzard Oct 29 '11 at 15:16