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%I A261096 #23 Sep 24 2015 01:39:50
%S A261096 0,1,1,2,0,2,3,4,3,3,4,5,0,2,4,5,2,1,5,5,5,6,3,5,4,1,4,6,7,7,4,0,0,3,
%T A261096 7,7,8,6,12,1,3,2,8,6,8,9,10,13,13,2,1,9,10,9,9,10,11,14,12,18,0,10,
%U A261096 11,6,8,10,11,8,15,16,19,19,11,8,7,11,11,11,12,9,16,17,20,18,0,9,11,10,7,10,12,13,18,17,14,21,22,1,1,10,6,6,9,13,13,14,19,6,15,22,23,2,0,14,7,9,8,14,12,14
%N A261096 A(i,j) = rank (in A055089) of the composition of the i-th and the j-th permutation in table A055089, which lists all finite permutations in reversed colexicographic ordering.
%C A261096 The square array A(row>=0, col>=0) is read by downwards antidiagonals as: A(0,0), A(0,1), A(1,0), A(0,2), A(1,1), A(2,0), A(0,3), A(1,2), A(2,1), A(3,0), ...
%C A261096 A(i,j) gives the rank (in ordering used by table A055089) of the permutation which is obtained by composing permutations p and q listed as the i-th and the j-th permutation in irregular table A055089 (note that the identity permutation is the 0th). Here the convention is that "permutations act of the left", thus, if p1 and p2 are permutations, then the product of p1 and p2 (p1 * p2) is defined such that (p1 * p2)(i) = p1(p2(i)) for i=1...
%C A261096 Each row and column is a permutation of A001477, because this is the Cayley table ("multiplication table") of an infinite enumerable group, namely, that subgroup of the infinite symmetric group (S_inf) which consists of permutations moving only finite number of elements.
%H A261096 Antti Karttunen, <a href="/A261096/b261096.txt">Table of n, a(n) for n = 0..10439; the first 144 antidiagonals of array</a>
%H A261096 Wikipedia, <a href="https://en.wikipedia.org/wiki/Cayley_table">Cayley table</a>
%F A261096 By conjugating with related permutations and arrays:
%F A261096 A(i,j) = A056019(A261097(A056019(i),A056019(j))).
%F A261096 A(i,j) = A060119(A261216(A060126(i),A060126(j))).
%F A261096 A(i,j) = A060120(A261217(A060127(i),A060127(j))).
%e A261096 The top left corner of the array:
%e A261096 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, ...
%e A261096 1, 0, 4, 5, 2, 3, 7, 6, 10, 11, 8, 9, 18, ...
%e A261096 2, 3, 0, 1, 5, 4, 12, 13, 14, 15, 16, 17, 6, ...
%e A261096 3, 2, 5, 4, 0, 1, 13, 12, 16, 17, 14, 15, 19, ...
%e A261096 4, 5, 1, 0, 3, 2, 18, 19, 20, 21, 22, 23, 7, ...
%e A261096 5, 4, 3, 2, 1, 0, 19, 18, 22, 23, 20, 21, 13, ...
%e A261096 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 14, ...
%e A261096 7, 6, 10, 11, 8, 9, 1, 0, 4, 5, 2, 3, 20, ...
%e A261096 8, 9, 6, 7, 11, 10, 14, 15, 12, 13, 17, 16, 0, ...
%e A261096 9, 8, 11, 10, 6, 7, 15, 14, 17, 16, 12, 13, 21, ...
%e A261096 10, 11, 7, 6, 9, 8, 20, 21, 18, 19, 23, 22, 1, ...
%e A261096 11, 10, 9, 8, 7, 6, 21, 20, 23, 22, 18, 19, 15, ...
%e A261096 12, 13, 14, 15, 16, 17, 2, 3, 0, 1, 5, 4, 8, ...
%e A261096 ...
%e A261096 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.
%e A261096 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.
%Y A261096 Transpose: A261097.
%Y A261096 Row 0 & Column 0: A001477 (identity permutation).
%Y A261096 Row 1: A261098.
%Y A261096 Column 1: A004442.
%Y A261096 Main diagonal: A261099.
%Y A261096 Cf. tables A055089, A195663.
%Y A261096 Cf. also A261216, A261217 (similar arrays, but using different orderings of permutations).
%Y A261096 Permutations used in conjugation-formulas: A056019, A060119, A060120, A060126, A060127.
%K A261096 nonn,tabl
%O A261096 0,4
%A A261096 _Antti Karttunen_, Aug 26 2015