A056346 Number of bracelets of length n using exactly six different colored beads.

0, 0, 0, 0, 0, 60, 1080, 11970, 105840, 821952, 5874480, 39713550, 258136200, 1631273220, 10096734312, 61536377700, 370710950400, 2213749658880, 13132080672480, 77509456944318, 455754569692680

Offset: 1

Author

Marks R. Nester

Keywords

• nonn

Comments

Turning over will not create a new bracelet.

Examples

			For a(6)=60, pair up the 120 permutations of BCDEF, each with its reverse, such as BCDEF-FEDCB.  Precede the first of each pair with an A, such as ABCDEF.  These are the 60 arrangements, all chiral.  If we precede the second of each pair with an A, such as AFEDCB, we get the chiral partner of each. - _Robert A. Russell_, Sep 27 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]

Crossrefs

Column 6 of A273891.

Equals (A056286 + A056492) / 2 = A056286 - A305545 = A305545 + A056492.

Cf. A008277.

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, 6];
Array[a, 21] (* Jean-François Alcover, Nov 05 2017, after Andrew Howroyd *)
k=6; 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 *)

PARI

a(n) = my(k=6); (k!/4) * (stirling(floor((n+1)/2),k,2) + stirling(ceil((n+1)/2),k,2)) + (k!/(2*n))*sumdiv(n, d, eulerphi(d)*stirling(n/d,k,2)); \\ Michel Marcus, Sep 29 2018

Formula

a(n) = A056341(n) - 6*A032276(n) + 15*A032275(n) - 20*A027671(n) + 15*A000029(n) - 6.

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=6 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=6 is the number of colors. (End)