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Triangle of permutations that assign sona-becs (A227960) to their complements.
1

%I #27 Aug 24 2021 06:29:10

%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 sona-becs (A227960) to their complements.

%C Subgroups of nimber addition (sona, A190939) have complements (defined using their Walsh spectrum). All sona in the same sona-bec (A227960) have complements in a unique sona-bec, which thus can be called its complement.

%C The permutation in row n of this triangle assigns complementary sona-becs of size 2^n to each other. (It is thus self-inverse.)

%C Even rows contain fixed points, because some sona-becs 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 A076766-1 (not A000225).

%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 T(4,1)=7 and T(4,7)=1, so 1 and 7 are complements for n=4.

%e T(4,3)=3, so 3 is its own complement for n=4.

%e Triangle begins:

%e k = 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

%e n

%e 0 0

%e 1 1 0

%e 2 3 1 2 0

%e 3 7 3 5 1 6 2 4 0

%e 4 15 7 12 3 13 5 9 1 14 6 10 11 2 4 8 0

%K nonn,tabf

%O 0,4

%A _Tilman Piesk_, Aug 04 2013