

A227962


Triangle of permutations that assign sonabecs (A227960) to their complements.


1



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, 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, 19, 2, 21, 4, 8, 16, 0
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OFFSET

0,4


COMMENTS

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.
The permutation in row n of this triangle assigns complementary sonabecs of size 2^n to each other. (It is thus self inverse.)
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.
Each row contains the row before as a subsequence.
0 is always complement with A076766(n)1, so each row ends with 0, and the left column is A0767661 (not A000225).
Triangle begins:
k = 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
n
0 0
1 1 0
2 3 1 2 0
3 7 3 5 1 6 2 4 0
4 15 7 12 3 13 5 9 1 14 6 10 11 2 4 8 0


LINKS

Tilman Piesk, Rows 0...7, flattened
Tilman Piesk, Rows 0...7 (the same with emphasis on subsequences)
Tilman Piesk, Complement pairs for n=0...7
Tilman Piesk, Graphic for n=4, complements are symmetric to each other
Tilman Piesk, Subgroups of nimber addition (Wikiversity)


EXAMPLE

a(4;1)=7 and a(4;7)=1, so 1 and 7 are complements for n=4.
a(4;3)=3, so 3 is its own complement for n=4.


CROSSREFS

Sequence in context: A201671 A226590 A261349 * A255615 A056931 A139569
Adjacent sequences: A227959 A227960 A227961 * A227963 A227964 A227965


KEYWORD

nonn,tabf


AUTHOR

Tilman Piesk, Aug 04 2013


STATUS

approved



