This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
The top left corner of the array:
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, ...
1, 0, 4, 5, 2, 3, 7, 6, 10, 11, 8, 9, 18, ...
2, 3, 0, 1, 5, 4, 12, 13, 14, 15, 16, 17, 6, ...
3, 2, 5, 4, 0, 1, 13, 12, 16, 17, 14, 15, 19, ...
4, 5, 1, 0, 3, 2, 18, 19, 20, 21, 22, 23, 7, ...
5, 4, 3, 2, 1, 0, 19, 18, 22, 23, 20, 21, 13, ...
6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 14, ...
7, 6, 10, 11, 8, 9, 1, 0, 4, 5, 2, 3, 20, ...
8, 9, 6, 7, 11, 10, 14, 15, 12, 13, 17, 16, 0, ...
9, 8, 11, 10, 6, 7, 15, 14, 17, 16, 12, 13, 21, ...
10, 11, 7, 6, 9, 8, 20, 21, 18, 19, 23, 22, 1, ...
11, 10, 9, 8, 7, 6, 21, 20, 23, 22, 18, 19, 15, ...
12, 13, 14, 15, 16, 17, 2, 3, 0, 1, 5, 4, 8, ...
...
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 A055089, 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 4th one in A055089, thus A(1,2) = 4.
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 A055089, thus A(2,1) = 3.
The top left corner of the array: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, ... 1, 0, 3, 2, 5, 4, 7, 6, 9, 8, 11, 10, 13, ... 2, 4, 0, 5, 1, 3, 8, 10, 6, 11, 7, 9, 14, ... 3, 5, 1, 4, 0, 2, 9, 11, 7, 10, 6, 8, 15, ... 4, 2, 5, 0, 3, 1, 10, 8, 11, 6, 9, 7, 16, ... 5, 3, 4, 1, 2, 0, 11, 9, 10, 7, 8, 6, 17, ... 6, 7, 12, 13, 18, 19, 0, 1, 14, 15, 20, 21, 2, ... 7, 6, 13, 12, 19, 18, 1, 0, 15, 14, 21, 20, 3, ... 8, 10, 14, 16, 20, 22, 2, 4, 12, 17, 18, 23, 0, ... 9, 11, 15, 17, 21, 23, 3, 5, 13, 16, 19, 22, 1, ... 10, 8, 16, 14, 22, 20, 4, 2, 17, 12, 23, 18, 5, ... 11, 9, 17, 15, 23, 21, 5, 3, 16, 13, 22, 19, 4, ... 12, 18, 6, 19, 7, 13, 14, 20, 0, 21, 1, 15, 8, ... ...
N:= 100: # to get a(0) to a(N)
M:= 0: A[0]:= 0: count:= 0:
for m from 2 while count < N do
P:= remove(t -> t[1]=1, combinat:-permute(m));
P:= map(t -> ListTools:-Reverse(subs([seq(i=m+1-i,i=1..m)],t)),P);
R:= select(t -> max(map(nops,convert(P[t],disjcyc))) = 2, [$1..nops(P)]);
for r in R do
count:= count+1;
A[count]:= r+M;
if count = N then break fi;
od:
M:= M + nops(P);
od:
seq(A[i],i=0..count); # Robert Israel, Oct 28 2015
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