An Infinite 2-Dimensional Derangement

[Leroy Quet]

Take an infinite matrix where each positive integer occurs exactly once 
in each column and in each row.
(In other words, each row is a permutation of the positive integers, and 
is a derangement of every other row-permutation.)

So, if we let the first row simply be  
1, 2, 3, 4, 5,...,

and then form each additional row by choosing for an element, as we move 
from left to right, the lowest integer not yet in that row NOR in that 
element's column, we have an easy and natural way to generate such a 
matrix.

1, 2, 3, 4, 5, 6, 7, 8...
2, 1, 4, 3, 6, 5, 8, 7...
3, 4, 1, 2, 7, 8, 5, 6...
4, 3, 2, 1, 8, 7, 6, 5...
5, 6, 7, 8, 1, 2, 3, 4...
6, 5, 8, 7, 2, 1, 4, 3...
7, 8, 5, 6, 3, 4, 1, 2...
8, 7, 6, 5, 4, 3, 2, 1...
....

Now, several patterns seem to be in-evidence with the matrix as figured 
so far.

So, the most general question I will ask is, what is a closed-form way to 
calculate each element given only its row and column?


And, oh yeah, the diagonally-read sequence of this table is not in the 
EIS.

1, 2, 2, 3, 1, 3, 4, 4, 4, 4, 5, 3, 1, 3, 5, 6, 6, 2, 2, 6, 6,...

thanks,
Leroy Quet