[seqfan] Naturals reordered: a(n)th first digit is not a(n)th digit of S

[Eric Angelini]

Hello Seqfans,
... computed by hand, here is the first (I guess) lexicographically
reordering S of the Natural numbers where every a(n) says:
- "My first digit is _not_ the a(n)th digit of S":
 
S=2,1,4,3,6,5,8,7,10,9,20,22,23,24,11,13,14,15,16,12,17,18,19,21,25,26,27,28,29,30,31,32,33,34,35,37,38,...
 
Example:
10 says that the 10th digit of S is not '1' -- which is true (it's '0')
9 says that the 9th digit of S is not '9' -- which is true (it's '1')
20 says that the 20th digit of S is not '2' -- which is true (it's '1)
...
This sequence is easy to build with a two-line array where:
- the first line is the Natural numbers (one integer per cell)
- the cells of the second line are always filled, from left to right,
and _one digit per cell_, with "the smallest Natural number not
yet in the sequence and not having any digit matching the digit
which starts the integer in the cell right above it";
 
Cells are separated by commas:
N= 1, 2, 3, 4, 5, 6, 7, 8, 9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,...
S= 2, 1, 4, 3, 6, 5, 8, 7, 1,0 , 9, 2,0 , 2,2 , 2,3 , 2,4 , 1,1 , 1,3 , 1,4 , 1,5 , 1,6 , 1,2 , 1,7 , 1,8 ,...
 
... thus, after '9' being put under the cell '11' of N, we cannot
split 11 or 12 or 13 or ... 19 in the next two cells of S because
a '1' would figure in the 12th cell of S -- which is forbidden (as
this '1' would be the same '1' starting '12' above it)... For the
same reason we cannot split '21' after the split of '20' in S
because the '1' of '21' would be written under the cell '15' of N,
sharing thus the said leading '1' with it.
 
If someone could send me a b-file for this, I would be happy.
 
Best,
É.