A005648 -id:A005648 - 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.

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

Offset: 2

Wouter Meeussen, Feb 12 2005

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

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

			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;
  ...

Cf. A077078; A005648; A052307, A103692.

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}]

A115120 Number of imprimitive (periodic) 2n-bead black-white reversible necklaces with n black beads.

Original entry on oeis.org

0, 0, 1, 1, 2, 1, 4, 1, 8, 3, 17, 1, 56, 1, 134, 18, 440, 1, 1434, 1, 4758, 135, 16160, 1, 57254, 16, 200475, 1387, 718152, 1, 2591800, 1, 9398520, 16161, 34324175, 148, 126125330, 1, 465093572, 200476, 1723176740, 1, 6408642359, 1, 23910576236, 2588402, 89494164974, 1, 335923316982, 133, 1264107421202

Offset: 0

Valery A. Liskovets, Jan 17 2006

a(p)=1 for prime p.

A005648[0] = 1; A005648[n_] := (1/2)(Binomial[2 Quotient[n, 2], Quotient[n, 2]] + DivisorSum[n, EulerPhi[#] Binomial[2n/#, n/#] &]/(2n));
A045628[n_] := If[n == 0, 1, Sum[MoebiusMu[n/d] (2n Binomial[2 Quotient[d, 2], Quotient[d, 2]] + Binomial[2d, d]), {d, Divisors[n]}]/(4n)];
a[n_] := A005648[n] - A045628[n];
Table[a[n], {n, 0, 50}] (* Jean-François Alcover, Aug 28 2019 *)

a(n) = A005648(n) - A045628(n).

More terms from Jean-François Alcover, Aug 28 2019

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.

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