

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



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, 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, 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
(list;
table;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

2,5


COMMENTS

Offset is 2, since exactly two colors are required, ergo at least two beads.
T[2n,n] equals A077078. Row sums equal A103692.


LINKS

Table of n, a(n) for n=2..89.


EXAMPLE

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.
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]; 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[ni]], Dihedral], SameTest>(Abs[ #1#2]<10^18&)]], {n, 2, 16}, {i, 1, n1}]


CROSSREFS

Cf. A077078; A005648; A052307, A103692.
Sequence in context: A117147 A111007 A176353 * A103441 A081206 A333161
Adjacent sequences: A103688 A103689 A103690 * A103692 A103693 A103694


KEYWORD

nonn,tabl


AUTHOR

Wouter Meeussen, Feb 12 2005


STATUS

approved



