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A056346
Number of bracelets of length n using exactly six different colored beads.
3
0, 0, 0, 0, 0, 60, 1080, 11970, 105840, 821952, 5874480, 39713550, 258136200, 1631273220, 10096734312, 61536377700, 370710950400, 2213749658880, 13132080672480, 77509456944318, 455754569692680
OFFSET
1,6
COMMENTS
Turning over will not create a new bracelet.
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]
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)
EXAMPLE
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
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 *)
PROG
(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
CROSSREFS
Column 6 of A273891.
Equals (A056286 + A056492) / 2 = A056286 - A305545 = A305545 + A056492.
Cf. A008277.
Sequence in context: A223348 A305545 A056352 * A283722 A105252 A269138
KEYWORD
nonn
STATUS
approved