cp's OEIS Frontend

A103691 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 length (or abs value) of sum of the position vectors of the white beads are different.

%I A103691 #8 Oct 28 2021 06:29:59
%S A103691 1,1,1,1,2,1,1,2,2,1,1,3,3,3,1,1,3,4,4,3,1,1,4,4,6,4,4,1,1,4,7,10,10,
%T A103691 7,4,1,1,5,7,11,11,11,7,5,1,1,5,10,20,26,26,20,10,5,1,1,6,10,16,18,20,
%U A103691 18,16,10,6,1,1,6,14,34,57,74,74,57,34,14,6,1,1,7,14,33,44,53,53,53,44,33
%N A103691 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 length (or abs value) of sum of the position vectors of the white beads are different.
%C A103691 Offset is 2, since exactly two colors are required, ergo at least two beads.
%C A103691 T[2n,n] equals A077078. Row sums equal A103692.
%e A103691 T[8,3]=4 because of the 5 bracelets {1,1,1,0,0,0,0,0}, {0,0,0,0,1,0,1,1}, {0,0,0,1,0,0,1,1},{0,0,0,1,0,1,0,1} and {0,0,1,0,0,1,0,1}, the third and the fourth have equal absolute vector sums, length 1.
%e A103691 Table starts as:
%e A103691   1;
%e A103691   1,1;
%e A103691   1,2,1;
%e A103691   1,2,2,1;
%e A103691   ...
%t A103691 Needs[DiscreteMath`NewCombinatorica`]; f[bi_]:=DeleteCases[bi*Range[Length[bi]], 0]; vec[li_, l_]:= Abs[Plus@@ N[Exp[2*Pi*I*f[li]/l], 24]]; Table[Length[Union[(vec[ #, n]&)/@ ListNecklaces[n, Join[1+0*Range[i], 0*Range[n-i]], Dihedral], SameTest->(Abs[ #1-#2]<10^-18&)]], {n, 2, 16}, {i, 1, n-1}]
%Y A103691 Cf. A077078; A005648; A052307, A103692.
%K A103691 nonn,tabl
%O A103691 2,5
%A A103691 _Wouter Meeussen_, Feb 12 2005