This post is split into three parts because of Atmos's character limit. This is part 3.
Part 1: https://projectatmos.space/profile/1p8WCZnqqG6N3ZOsJxBgUTo/p1SJCxFjA6yCA52Hz
Part 2: https://projectatmos.space/profile/1p8WCZnqqG6N3ZOsJxBgUTo/p1zcNRQIrjGHqXHN0
The last two describers can only be
3,3 if there are at least two describer 1s (otherwise the sum wouldn't be equal to the length). Because there is no describer higher than 3, the 1 and 3 automatically need to be described by the two 3s and the only other describer that can occur is 2. It can't occur 3 times or more, because then you would need a describer 4 and
3,3 wouldn't be the last two anymore. It can't occur 2 times, because then you would need another 3 and the 3 would occur 4 times. So it can only occur once or not at all. That limits the length of the number to at most 10 digits. Again, these can be tested and I did that, the list is above.
The last two describers can only be
4,4 if there are at most 3 of any number in the describers, including the 1. But if there are at most three 1s, that means that the sum can't be equal to the length anymore: For every 4 you need two 1s, so for two 4s you would need four 1s already (which would need to be described by a 5). The same logic also excludes
6,6 and any other pair of describers that are equal and higher than 3 from being the last two describers.
If the last two describers were
5,7, … (the second to last describer 4 or more and the last describer higher), then you would need as many describer 1s as the last two describers added minus 4 (because you need to balance the difference from
2,2,2,2,2,…). But that would mean for
4,5 already five 1s, which would mean that the 1 would need to be described by at least a 6, which doesn't occur. Increasing the last describer does nothing to change this and increasing the second to last describer only makes the difference worse.
If the last two describers are
2,6, … (2 and a number higher than 3), that means that there are exactly as many 1s as the last digit minus two (again, to balance the difference from 2 so that the sum fits). So in the case of
2,4 at the end the number would be
1?1?2?(2?(2?(…)))4?, with a so far unknown amount of 2s in the middle. In this case it's two 1s, so the describer for the 1s would need to be 3. But there is no 3 in this number. The same happens for
2,5 at the end, the 1s would need to be described by a 4. And so on.
So if the last two digits of a self-describing number with at least 14 digits can't be
2,x with x>2 or
x,x with x>3 or
x,y with y>x>3, then that leaves
3,x with x>3 as the only option.
Since x is the highest describer, it needs to describe either 1 or 2. The differences from 2 need to be balanced with 1s, so you need x-2+1 1s (x-2 to balance the x and 1 to balance the 3), plus as many 1s as there are additional 3s. That means that the 1 needs to be described by x-2+1+1 plus the number of additional 3s, so at least x. Since x is already the biggest describer that means that you have to have exactly x-1 1s and only the one 3. So only the number of 2s is now unknown.
The highest describer left is 3, because x is already taken by 1. That means that 2 can occur at most two times. But 2 is also needed at least twice, to describe 3 and x.
So in summary, you have to have exactly x-1 1s, two 2s, one 3 and one x as describers. But that's exactly the pattern:
Now it's just a matter of filling in the described digits: 1 needs to be described by x, 2 needs to be described by 3, 3 and x need to be described by the two 2s, that leaves only the 1s to fill. Those are arbitrary, as long as the described digit doesn't appear anywhere else in the number. Because I only select one set of arbitrary digits per set of digits that "matter", that means that this pattern describes all self-describing numbers with 14 or more digits. The ones with less digits were figured out by just letting a computer try all candidates.