A056343 Number of bracelets of length n using exactly three different colored beads.

0, 0, 1, 6, 18, 56, 147, 411, 1084, 2979, 8043, 22244, 61278, 171030, 477929, 1345236, 3795750, 10758902, 30572427, 87149124, 248991822, 713096352, 2046303339, 5883433409, 16944543810, 48879769575

Offset: 1

Author

Marks R. Nester

Keywords

• nonn

Comments

Turning over will not create a new bracelet.

Examples

			For a(4)=6, the arrangements are ABAC, ABCB, ACBC, AABC, ABBC, and ABCC. Only the last three are chiral, their reverses being AACB, ACBB, and ACCB respectively. - _Robert A. Russell_, Sep 26 2018
		

References

• M. R. Nester (1999). Mathematical investigations of some plant interaction designs. PhD Thesis. University of Queensland, Brisbane, Australia. [See A056391 for pdf file of Chap. 2]

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000

Crossrefs

Column 3 of A273891.

Equals (A056283 + A056489) / 2 = A056283 - A305542 = A305542 + A056489.

Programs

Mathematica

t[n_, k_] := (For[t1 = 0; d = 1, d <= n, d++, If[Mod[n, d] == 0, t1 = t1 + EulerPhi[d]*k^(n/d)]]; If[EvenQ[n], (t1 + (n/2)*(1 + k)*k^(n/2))/(2*n), (t1 + n*k^((n + 1)/2))/(2*n)]);
T[n_, k_] := Sum[(-1)^i*Binomial[k, i]*t[n, k - i], {i, 0, k - 1}];
a[n_] := T[n, 3];
Array[a, 26] (* Jean-François Alcover, Nov 05 2017, after Andrew Howroyd *)
k=3; Table[k! DivisorSum[n, EulerPhi[#] StirlingS2[n/#,k]&]/(2n) + k!(StirlingS2[Floor[(n+1)/2], k] + StirlingS2[Ceiling[(n+1)/2], k])/4, {n,1,30}] (* Robert A. Russell, Sep 26 2018 *)

Formula

a(n) = A027671(n) - 3*A000029(n) + 3.

From Robert A. Russell, Sep 26 2018: (Start)

a(n) = (k!/4) * (S2(floor((n+1)/2),k) + S2(ceiling((n+1)/2),k)) + (k!/2n) * Sum_{d|n} phi(d) * S2(n/d,k), where k=3 is the number of colors and S2 is the Stirling subset number A008277.

G.f.: (k!/4) * x^(2k-2) * (1+x)^2 / Product_{i=1..k} (1-i x^2) - Sum_{d>0} (phi(d)/2d) * Sum_{j} (-1)^(k-j) * C(k,j) * log(1-j x^d), where k=3 is the number of colors. (End)