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%I #11 Oct 28 2021 06:30:08
%S 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,
%T 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,
%U 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
%N 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.
%C 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.
%C Offset is 2, since exactly two colors are required, ergo at least two beads.
%C T[2n,n] equals A045611. Row sums equal A103442.
%C 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.
%e Table starts as
%e 1;
%e 1,1;
%e 1,2,1;
%e 1,2,2,1;
%e ...
%t 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}]
%Y Cf. A052307, A045611, A077078, A077079, A103442.
%K nonn,tabl
%O 2,5
%A _Wouter Meeussen_, Feb 06 2005
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