0, 1, 1, 2, 0, 2, 3, 5, 3, 3, 4, 4, 0, 2, 4, 5, 3, 1, 4, 5, 5, 6, 2, 5, 5, 3, 4, 6, 7, 7, 4, 1, 2, 1, 7, 7, 8, 6, 14, 0, 0, 0, 8, 6, 8, 9, 11, 15, 15, 1, 2, 9, 11, 9, 9, 10, 10, 12, 14, 22, 3, 10, 10, 6, 8, 10, 11, 9, 13, 16, 23, 23, 11, 9, 7, 10, 11, 11, 12, 8, 17, 17, 21, 22, 0, 8, 11, 11, 9, 10, 12, 13, 19, 16, 13, 20, 19, 1, 1, 10, 7, 8, 7, 13, 13, 14, 18, 8, 12, 18, 18, 2, 0, 12, 6, 6, 6, 14, 12, 14
Offset: 0
The top left corner of the array:
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, ...
1, 0, 5, 4, 3, 2, 7, 6, 11, 10, 9, 8, 19, ...
2, 3, 0, 1, 5, 4, 14, 15, 12, 13, 17, 16, 8, ...
3, 2, 4, 5, 1, 0, 15, 14, 16, 17, 13, 12, 21, ...
4, 5, 3, 2, 0, 1, 22, 23, 21, 20, 18, 19, 16, ...
5, 4, 1, 0, 2, 3, 23, 22, 19, 18, 20, 21, 11, ...
6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 14, ...
7, 6, 11, 10, 9, 8, 1, 0, 5, 4, 3, 2, 23, ...
8, 9, 6, 7, 11, 10, 12, 13, 14, 15, 16, 17, 2, ...
9, 8, 10, 11, 7, 6, 13, 12, 17, 16, 15, 14, 20, ...
10, 11, 9, 8, 6, 7, 18, 19, 20, 21, 22, 23, 17, ...
11, 10, 7, 6, 8, 9, 19, 18, 23, 22, 21, 20, 5, ...
12, 13, 14, 15, 16, 17, 8, 9, 6, 7, 11, 10, 0, ...
...
For A(1,2) (row=1, column=2, both starting from zero), we take as permutation p the permutation which has rank=1 in the ordering used by A060117, which is a simple transposition (1 2), which we can extend with fixed terms as far as we wish (e.g., like {2,1,3,4,5,...}), and as permutation q we take the permutation which has rank=2 (in the same list), which is {1,3,2}. We compose these from the left, so that the latter one, q, acts first, thus c(i) = p(q(i)), and the result is permutation {2,3,1}, which is listed as the 5th one in A060117, thus A(1,2) = 5.
For A(2,1) we compose those two permutations in opposite order, as d(i) = q(p(i)), which gives permutation {3,1,2} which is listed as the 3rd one in A060117, thus A(2,1) = 3.
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