A056344 Number of bracelets of length n using exactly four different colored beads.

0, 0, 0, 3, 24, 136, 612, 2619, 10480, 41388, 159780, 614058, 2341920, 8919816, 33905188, 128907279, 490213680, 1866127840, 7111777860, 27140369148, 103721218000, 396974781456, 1521577377012, 5840547488954

Offset: 1

Author

Marks R. Nester

Keywords

• nonn

Comments

Turning over will not create a new bracelet.

Examples

			For a(4)=3, the arrangements are ABCD, ABDC, and ACBD, all chiral, their reverses being ADCB, ACDB, and ADBC respectively.
		

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]

Crossrefs

Column 4 of A273891.

Cf. A056284 (oriented), A056490 (achiral), A305543 (chiral).

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, 4];
Array[a, 24] (* Jean-François Alcover, Nov 05 2017, after Andrew Howroyd *)
k=4; 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 27 2018 *)

Formula

a(n) = A032275(n) - 4*A027671(n) + 6*A000029(n) - 4.

From Robert A. Russell, Sep 27 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=4 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=4 is the number of colors.

a(n) = (A056284(n) + A056490(n)) / 2 = A056284(n) - A305543(n) = A305543(n) + A056490(n). (End)