[chris_overseas]

Sure they can, by setting the rotate bits to 0xC (1100). Have a play with the interactive widget near the bottom of the article.

Interestingly, there is some redundancy with this encoding meaning you can't represent a full 2^12 unique values. Eg 0x1 could be represented by 0x1 and 0x0 in the rotate field, 0x4 with 0x1 in the rotate field, 0x10 with 0x2, or 0x40 with 0x3. The same applies for every other combination - this means that there are only (2^12)/4 = 1024 unique values that are representable in total.

[Dylan16807]

You didn't do that math right. Out of every 256 values at a particular offset, only a quarter of them are under 64 and can be encoded with rotate+1.

Doing a quick brute force test, it appears that there are 3073 unique values. I suppose that makes sense. Each new rotate introduces 192 new values that have at least one of the new bits set, and 192 * 16 = 3072. Then you have the number 0.