A059053 Number of chiral pairs of necklaces with n beads and two colors (color complements being equivalent); i.e., turning the necklace over neither leaves it unchanged nor simply swaps the colors.

0, 0, 0, 0, 0, 0, 0, 1, 2, 7, 12, 31, 58, 126, 234, 484, 906, 1800, 3402, 6643, 12624, 24458, 46686, 90157, 172810, 333498, 641340, 1238671, 2388852, 4620006, 8932032, 17302033, 33522698, 65042526, 126258960, 245361172, 477091232

Offset: 0

Author

Henry Bottomley, Dec 21 2000

Keywords

• nonn

Comments

Number of chiral pairs of set partitions of a cycle of n elements using exactly two different elements. - Robert A. Russell, Oct 02 2018

Examples

			For a(7) = 1, the chiral pair is AAABABB-AAABBAB.
For a(8) = 2, the chiral pairs are AAAABABB-AAAABBAB and AAABAABB-AAABBAAB.
		

Links

• Andrew Howroyd, Table of n, a(n) for n = 0..1000
• Illustration of initial terms
• Index entries for sequences related to necklaces

Crossrefs

Column 2 of A320647 and A320742.

Cf. A056295 (oriented), A056357 (unoriented), A052551(n-2) (achiral).

Programs

Mathematica

Prepend[Table[DivisorSum[n, EulerPhi[#] StirlingS2[n/# + If[Divisible[#,2],1,0], 2] &] / (2n) - StirlingS2[1+Floor[n/2],2] / 2, {n, 1, 40}],0] (* Robert A. Russell, Oct 02 2018 *)

PARI

a(n) = {if(n<1, 0, (sumdiv(n, k, eulerphi(2*k) * 2^(n/k)) / (2*n) - 2^(n\2))/2)}; \\ Andrew Howroyd, Nov 03 2019

Formula

a(n) = A000013(n) - A000011(n) = A000011(n) - A016116(n) = (A000013(n) - A016116(n))/2.

From Robert A. Russell, Oct 02 2018: (Start)

a(n) = (A056295(n)-A052551(n-2)) / 2 = A056295(n) - A056357(n) = A056357(n) - A052551(n-2).

a(n) = -S2(1+floor(n/2),2) + (1/2n) * Sum_{d|n} phi(d) * S2(n/d+[2|d],2), where S2 is a Stirling subset number A008277.

G.f.: -x(1+2x)/(2-4x^2) - Sum_{d>0} phi(d) * log(1-2x^d) / (2d*(2-[2|d])).

(End)

Extensions

Name clarified by Robert A. Russell, Oct 02 2018