

A103441


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.


1



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
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OFFSET

2,5


COMMENTS

If two bracelets can be made to coincide by rotation or flipping over they necessarily 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.


LINKS

Table of n, a(n) for n=2..88.


EXAMPLE

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.


MATHEMATICA

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[ni]], Dihedral]]], {n, 2, 16}, {i, 1, n1}]


CROSSREFS

Cf. A052307, A045611, A077078, A077079, A103442.
Sequence in context: A111007 A176353 A103691 * A081206 A156044 A180980
Adjacent sequences: A103438 A103439 A103440 * A103442 A103443 A103444


KEYWORD

nonn,tabl


AUTHOR

Wouter Meeussen, Feb 06 2005


STATUS

approved



