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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. 1

%I #8 Oct 28 2021 06:29:59

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

%C T[2n,n] equals A077078. Row sums equal A103692.

%e 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 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]; 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 Cf. A077078; A005648; A052307, A103692.

%K nonn,tabl

%O 2,5

%A _Wouter Meeussen_, Feb 12 2005

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Last modified April 19 16:52 EDT 2024. Contains 371794 sequences. (Running on oeis4.)