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Search: id:A103441
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| A103441 |
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Triangle read by rows: T(n,k) = number of bracelets of n beads (necklaces that can be flipped over) with exactly two colors and k white beads for which the set of distances among the white beads are different. |
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+0
2
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| 1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 3, 3, 3, 1, 1, 3, 4, 4, 3, 1, 1, 4, 5, 7, 5, 4, 1, 1, 4, 7, 10, 10, 7, 4, 1, 1, 5, 8, 16, 13, 16, 8, 5, 1, 1, 5, 10, 20, 26, 26, 20, 10, 5, 1, 1, 6, 12, 28, 35, 35, 35, 28, 12, 6, 1, 1, 6, 14, 34, 57, 74, 74, 57, 34, 14, 6, 1, 1, 7, 16, 47, 73, 120, 85, 120, 73
(list; table; graph; listen)
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OFFSET
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2,5
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COMMENT
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If two bracelets can be made to coincide by rotation or flipping over they nescessarily have the same set of distances, but the reverse is obviously not true. Table starts as 1; 1,1; 1,2,1; 1,2,2,1; Offset is 2, since exactly two colors are required, ergo at least two beads. T[2n,n] equals A045611. Row sums equal A103442.
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EXAMPLE
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Same as A052307, except for bracelets such as {0,0,0,1,1,0,1,1} and{0,0,1,0,0,1,1,1}, that both have the same set of distances between the "1" beads: 4 d[0]+ 4 d[1]+ 2 d[2]+ 4 d[3]+ 2 d[4], where d[k] represents the unidirectional distance between two beads k places apart.
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MATHEMATICA
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Needs[DiscreteMath`NewCombinatorica`]; f[bi_]:=DeleteCases[bi Range[Length[bi]], 0]; dist[li_, l_]:=Plus@@Flatten[Outer[d[Min[ #, l-# ]&@Mod[Abs[ #1-#2], l, 0]]&, li, li]]; Table[Length[Union[(dist[f[ #1], n]&)/@ListNecklaces[n, Join[1+0*Range[i], 0*Range[n-i]], Dihedral]]], {n, 2, 16}, {i, 1, n-1}]
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CROSSREFS
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Cf. A052307, A045611, A077078, A077079, A103442.
Sequence in context: A117147 A111007 A103691 this_sequence A081206 A156044 A048570
Adjacent sequences: A103438 A103439 A103440 this_sequence A103442 A103443 A103444
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KEYWORD
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nonn,tabl
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AUTHOR
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Wouter Meeussen (wouter.meeussen(AT)pandora.be), Feb 06 2005
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