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A195665
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Consecutive bit-permutations of nonnegative integers.
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1
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0, 1, 0, 2, 1, 3, 0, 1, 4, 5, 2, 3, 6, 7, 0, 2, 4, 6, 1, 3, 5, 7, 0, 4, 1, 5, 2, 6, 3, 7, 0, 4, 2, 6, 1, 5, 3, 7, 0, 1, 2, 3, 8, 9, 10, 11, 4, 5, 6, 7, 12, 13, 14, 15, 0, 2, 1, 3, 8, 10, 9, 11, 4, 6, 5, 7, 12, 14, 13, 15, 0, 1, 4, 5, 8, 9, 12
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OFFSET
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0,4
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COMMENTS
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All rows of this array are infinite permutations of the nonnegative integers. Row m (counted from 0) is always generated by modifying the sequence of nonnegative integers in the following way: The sequence of integers is written in reverse binary. Than the finite permutation p_m (row m of array A055089) is applied on the digits of all entries.
The rows of the top left n! X 2^n submatrix describe the rotations and reflections of the n-hypercube that preserve the binary digit sums of the vertex numbers. With permutation composition these permutations form the symmetric group S_n.
Applying such a permutation on the binary string of a Boolean function gives the string of a function in the same big equivalence class (compare A227723).
Triangle row m has length 2^n for m in the interval [(n-1)!,n![. The rest of the array row repeats the same pattern. The first digit of the rest is the digit before plus one.
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LINKS
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EXAMPLE
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Top left corner of array:
0 1 2 3 4 5 6 7
0 2 1 3 4 6 5 7
0 1 4 5 2 3 6 7
0 2 4 6 1 3 5 7
0 4 1 5 2 6 3 7
0 4 2 6 1 5 3 7
The entry in row 2, column 5 (both counted from 0) is 3: 5 in reverse binary is 101, permutation p_2 applied on 101 gives 110, 110 from reverse binary to decimal is 3.
Corresponding rows of the triangle:
0 1
0 2 1 3
0 1 4 5 2 3 6 7
0 2 4 6 1 3 5 7
0 4 1 5 2 6 3 7
0 4 2 6 1 5 3 7
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CROSSREFS
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The finite permutations in A055089 are applied on the reverse binary digits.
Row n!-1 of the triangle is the n-bit bit-reversal permutation. Compare A030109.
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KEYWORD
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nonn,tabf
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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