So I just wanna know if I got this right
When increasing damage dice (NOT BY SIZE) it just goes by steps
The highest I could find for medium weapons is 2d10, so the next step after that would be 2d12. After that, I looked at when the damage dice change from single to double, specifically from 1d12 to 2d4. The smallest possible dice is 1d2, so I figured that means that the increased dice would have to be two steps above the lowest dice depending on single or double rolled. What this means is that the step after 2d12 would be 3d8, after which is 3d10.
Does this make sense?
Actually there is a more recent thread where one of the revs has explained why it hasn't been FAQ'd. They are trying to decide what to do about the one or two instances that don't fit the existing tables and make a mess of a unified set. They are hoping to figure out the least impacting way to go forward with it if something does needed to be errata'd.
2d10 goes to 2d12, which goes to 6d6
Strictly speaking it's 2(1d10)>2(1d12)>2(3d6) following the 1d10>2d6>3d6 progression but doubling the dice, and allowing that 2d6 is equivalent to 1d12.
This assumes we're simply increasing the dice, not the size, of a weapon (not a natural weapon), and that the weapon is medium sized or smaller. Crazy stuff happens if those aren't true.
If you're choosing a weapon, 2d4 is slightly better than 1d8 and 2d6 is slightly better than a 1d12. While the highest rolls are the same, the lowest roll with two dice is a two, and the average roll is 1/2 a point higher. It's actually more complicated than that, but 1/2 point is the advantage. If people REALLY want the boring proof I can post it.
I'm obviously not grasping something fundamental to the question.
But here's my stab at the answer:
Medium weapons typically fall somewhere on this track:
On this track 2d4 is interchangeable with 1d8 and 1d12 is interchangeable with 2d6. To go higher, just use multiple dice expressions for the last three slots in the progression, always using 1d12 instead of 2d6.