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, 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

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.

Offset is 2, since exactly two colors are required, ergo at least two beads.

T[2n,n] equals A045611. Row sums equal A103442.

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.

Links

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

Example

Table starts as
  1;
  1,1;
  1,2,1;
  1,2,2,1;
  ...

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[n-i]], Dihedral]]], {n, 2, 16}, {i, 1, n-1}]

Crossrefs

Cf. A052307, A045611, A077078, A077079, A103442.

Sequence in context: A111007 A176353 A103691 * A081206 A333161 A156044

Adjacent sequences: A103438 A103439 A103440 * A103442 A103443 A103444

Keyword

nonn,tabl

Author

Wouter Meeussen, Feb 06 2005

Status

approved