Generate the minimal remainder sequence Every number can be represented using an infinitely long remainder sequence.  For example, if we take the number 7, and perform 7mod2, then 7mod3, then 7mod4, and so on, we get 1,1,3,2,1,0,7,7,7,7,.....
However, we need the shortest possible remainder subsequence that can still be used to distinguish it from all lower numbers.  Using 7 again, [1,1,3] is the shortest subsequence, because all of the previous subsequences don't start with [1,1,3]:
0: 0,0,0,0...
1: 1,1,1,1...
2: 0,2,2,2...
3: 1,0,3,3...
4: 0,1,0,4...
5: 1,2,1,0...
6: 0,0,2,1...

Note that [1,1] doesn't work to represent 7, because it can also be used to represent 1.  However, you should output [1] with an input of 1.
Input/Output
Your input is a non-negative integer.  You must output a sequence or list of the minimal-length sequence of remainders as defined above.
Test cases:
0: 0
1: 1
2: 0,2
3: 1,0
4: 0,1
5: 1,2
6: 0,0,2
7: 1,1,3
8: 0,2,0
9: 1,0,1
10: 0,1,2
11: 1,2,3
12: 0,0,0,2
30: 0,0,2,0
42: 0,0,2,2
59: 1,2,3,4
60: 0,0,0,0,0,4
257: 1,2,1,2,5,5
566: 0,2,2,1,2,6,6
1000: 0,1,0,0,4,6,0,1
9998: 0,2,2,3,2,2,6,8,8,10
9999: 1,0,3,4,3,3,7,0,9,0

Here are the first 10,000 sequences, in case you are interested (the line numbers are off by 1).
This is a code-golf, so make it as short as you can in your favorite language.  Fake bonus points for any answers that are fast!
Jelly, 14 bytes

‘Ræl\&gt;iṠ2»2r⁸%


This uses the fact a the solution (if any) of a system of linear congruences is unique modulo the LCM of the moduli. Try it online! or verify all test cases.

How it works

‘Ræl\&gt;iṠ2»2r⁸%  Main link. Argument: n

‘               Increment; yield n+1.
 R              Range; yield [1, ..., n+1].
  æl\           Cumulatively reduce by LCM.
                This yields [LCM(1), ..., LCM(1, ..., n+1)].
     &gt;          Compare all LCMs with n.
      iṠ        Find the first index of sign(n).
                This yields the first m such that LCM(2, ..., m) &gt; n if n &gt; 0, and
                0 if n == 0.
        2»      Take the maximum of the previous result and 2, mapping 0 to 2.
          2r    Yield the range from 2 up to and including the maximum.
            ⁸%  Compute n modulo each integer in that range.