A211362 Inversion sets of finite permutations interpreted as binary numbers.
0, 1, 4, 3, 6, 7, 32, 33, 20, 11, 22, 15, 48, 41, 52, 43, 30, 31, 56, 57, 60, 59, 62, 63, 512, 513, 516, 515, 518, 519, 288, 289, 148, 75, 150, 79, 304, 297, 180, 107, 158, 95, 312, 313, 188, 123, 190, 127, 768, 769, 644, 579, 646, 583, 800
Offset: 0
Keywords
Examples
The 4th finite permutation (2,3,1,4,...) has the inversion set {(1,3),(2,3)}. This set represented by a vector is (0,1,1,zeros...). This vector interpreted as a number is 6. So a(4)=6. The 23rd finite permutation (4,3,2,1,...) has the inversion set {(1,2),(1,3),(2,3),(1,4),(2,4),(3,4)}. This set represented by a vector is (1,1,1,1,1,1,zeros...). This vector interpreted as a number is 63. So a(23)=63. Beginning of corresponding array: n permutation inversion set a(n) 00 1 2 3 4 0 0 0 0 0 0 0 01 2 1 3 4 1 0 0 0 0 0 1 02 1 3 2 4 0 0 1 0 0 0 4 03 3 1 2 4 1 1 0 0 0 0 3 04 2 3 1 4 0 1 1 0 0 0 6 05 3 2 1 4 1 1 1 0 0 0 7 06 1 2 4 3 0 0 0 0 0 1 32 07 2 1 4 3 1 0 0 0 0 1 33 08 1 4 2 3 0 0 1 0 1 0 20 09 4 1 2 3 1 1 0 1 0 0 11 10 2 4 1 3 0 1 1 0 1 0 22 11 4 2 1 3 1 1 1 1 0 0 15 12 1 3 4 2 0 0 0 0 1 1 48 13 3 1 4 2 1 0 0 1 0 1 41 14 1 4 3 2 0 0 1 0 1 1 52 15 4 1 3 2 1 1 0 1 0 1 43 16 3 4 1 2 0 1 1 1 1 0 30 17 4 3 1 2 1 1 1 1 1 0 31 18 2 3 4 1 0 0 0 1 1 1 56 19 3 2 4 1 1 0 0 1 1 1 57 20 2 4 3 1 0 0 1 1 1 1 60 21 4 2 3 1 1 1 0 1 1 1 59 22 3 4 2 1 0 1 1 1 1 1 62 23 4 3 2 1 1 1 1 1 1 1 63
Links
- Tilman Piesk, Table of n, a(n) for n = 0..5039
- Wikipedia, Inversion (discrete mathematics)
Comments