A263768 Number of necklaces with n beads colored white or red, where the number of white beads is odd and at least three and turning over is allowed.

1, 1, 3, 4, 8, 11, 22, 33, 62, 101, 189, 324, 611, 1087, 2055, 3770, 7154, 13363, 25481, 48174, 92204, 175791, 337593, 647325, 1246862, 2400841, 4636389, 8956059, 17334800, 33570815, 65108061, 126355335, 245492243, 477284181, 928772649, 1808538354, 3524337979, 6872209823

Offset: 3

Author

David Eppstein, Oct 25 2015

Keywords

• nonn

Comments

a(n) is also the number of non-isomorphic n-vertex undirected graphs forming an odd cycle with any number of degree-1 vertices attached to each cycle vertex. To transform a necklace into a graph of this type, create a cycle vertex for each white bead and a pendant vertex for each red bead, with each pendant vertex attached to the next clockwise cycle vertex. Since these are exactly the graphs of the n-vertex and n-edge linear thrackles, a(n) is also the number of non-isomorphic linear thrackles.

For any n there is a unique n-bead necklace where the number of white beads is 1. Hence this sequence is one less than the number of n-bead (0,1) bracelets with an odd number of 0's. - Andrew Howroyd, Feb 28 2017

Examples

			For n=5 the a(5)=3 solutions are: five white beads (a 5-cycle), three white beads and two red beads with the two red beads adjacent (a triangle with two pendant vertices attached at one triangle vertex), and three white beads and two red beads with the two red beads separated (a triangle with two of its vertices having a single pendant vertex attached).
		

Links

• Andrew Howroyd, Table of n, a(n) for n = 3..100
• Bernd Mulansky and Andreas Potschka, A zonogon approach for computing small polygons of maximum perimeter, arXiv:2404.01841 [math.OC], 2024. See p. 9. See also Math. Program., 2025.

Crossrefs

Cf. A227910, A007147, A000016.

Programs

Mathematica

Table[1/2*(2^Quotient[n - 1, 2] + Total@ Map[(Mod[#, 2]*EulerPhi[#]*2^(n/#) &), Divisors[n]]/(2 n)) - 1, {n, 3, 40}] (* Michael De Vlieger, Apr 10 2024, after Jean-François Alcover at A007147 *)

Formula

a(n) = (A000016(n) + A016116(n-1)) / 2 - 1. - Andrew Howroyd, Feb 28 2017

a(n) = A007147(n) - 1. - Bernd Mulansky, Mar 08 2023

Extensions

a(21)-a(40) from Andrew Howroyd, Feb 28 2017