%I
%S 0,1,0,3,1,2,0,7,3,5,1,6,2,4,0,15,7,12,3,13,5,9,1,14,6,10,11,2,4,8,0,
%T 31,15,26,7,28,12,20,3,29,13,22,23,5,9,17,1,30,14,24,25,6,27,10,11,18,
%U 19,2,21,4,8,16,0
%N Triangle of permutations that assign sonabecs (A227960) to their complements.
%C Subgroups of nimber addition (sona, A190939) have complements (defined using their Walsh spectrum). All sona in the same sonabec (A227960) have complements in a unique sonabec, which thus can be called its complement.
%C The permutation in row n of this triangle assigns complementary sonabecs of size 2^n to each other. (It is thus self inverse.)
%C Even rows contain fixed points, because some sonabecs with weight 2^(n/2) are their own complements. E.g. in row 4 the fixed points are 3, 5, 10 and 11.
%C Each row contains the row before as a subsequence.
%C 0 is always complement with A076766(n)1, so each row ends with 0, and the left column is A0767661 (not A000225).
%C Triangle begins:
%C k = 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
%C n
%C 0 0
%C 1 1 0
%C 2 3 1 2 0
%C 3 7 3 5 1 6 2 4 0
%C 4 15 7 12 3 13 5 9 1 14 6 10 11 2 4 8 0
%H Tilman Piesk, <a href="/A227962/b227962.txt">Rows 0...7, flattened</a>
%H Tilman Piesk, <a href="/A227962/a227962_1.txt">Rows 0...7</a> (<a href="/A227962/a227962_3.txt">the same with emphasis on subsequences</a>)
%H Tilman Piesk, <a href="http://pastebin.com/raw.php?i=RFbBYVfx">Complement pairs for n=0...7</a>
%H Tilman Piesk, <a href="https://commons.wikimedia.org/wiki/File:Z2%5E4;_Lattice_of_subgroups_Hasse_diagram;_equivalence_classes.svg">Graphic for n=4</a>, complements are symmetric to each other
%H Tilman Piesk, <a href="http://en.wikiversity.org/wiki/Subgroups_of_nimber_addition">Subgroups of nimber addition</a> (Wikiversity)
%e a(4;1)=7 and a(4;7)=1, so 1 and 7 are complements for n=4.
%e a(4;3)=3, so 3 is its own complement for n=4.
%K nonn,tabf
%O 0,4
%A _Tilman Piesk_, Aug 04 2013
