<pre>
6210001000

For other numbers of digits:

 n=1:	no sequence possible
 n=2:	no sequence possible
 n=3:	no sequence possible
 n=4:	1210, 2020
 n=5:	21200
 n=6:	no sequence possible
 n=7:	3211000
 n=8:	42101000
 n=9:	521001000
 n=10:	6210001000
 n>10:	(n-4), 2, 1, 0 * (n-7), 1, 0, 0, 0

 No 1, 2, or 3 digit numbers are possible.  Letting x_i be the ith
 digit, starting with 0, we see that (1) x''0 + ... + x''n = n+1 and
 (2) 0'''x''0 + ... + n'''x''n = n+1, where n+1 is the number of digits.

I'll first prove that x_0 > n-3 if n>4.  Assume not, then this
implies that at least four of the x_i with i>0 are non-zero.  But
then we would have sum''i i*x''i ><code> 10 by (2), impossible unless n</code>9,
but it isn't possible in this case (51111100000 isn't valid).

 Now I'll prove that x''0 < n-1.  x''0 clearly can't equal n; assume
 x''0 <code> n-1 </code><code>> x''{n-1} </code> 1 by (2) if n>3.  Now only one of the
 remaining x''i may be non-zero, and we must have that x''0 + ... + x_n
 <code> n+1, but since x''0 + x''{n-1} </code> n <code></code>> the remaining x_i <code> 1 </code>=> by
 (2) that x''2 <code> 1.  But this can't be, since x''{n-1} </code> 1 <code></code>> x_1>0.
 Now assuming x''0 <code> n-2 we conclude that x''{n-2} </code> 1 by (2) if n>5
 <code></code>> x''1 + ... + x''{n-3} + x''{n-1} + x''n = 2 and 1*x_1 + ... +
 (n-3)'''x''{n-3} + (n-1)'''x''{n-1} + n*x''n <code> 3 </code><code>> x''1</code>1 and x_2=1,
 contradiction.

Case n>5:

We have that x''0 <code> n-3 and if n></code>7 <code></code>> x''{n-3}<code>1 </code><code>> x_1</code>2 and
x''2<code>1 by (1) and (2).  For the case n</code>6 we see that x''{n-3}=2
leads to an easy contradiction, and we get the same result.  The
cases n=4,5 are easy enough to handle, and lead to the two solutions
above.
--
    -- clong@cnj.digex.com (Chris Long)
</pre>
