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A056344
Number of bracelets of length n using exactly four different colored beads.
5
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
1,4
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) = 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)
EXAMPLE
For a(4)=3, the arrangements are ABCD, ABDC, and ACBD, all chiral, their reverses being ADCB, ACDB, and ADBC respectively.
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 *)
CROSSREFS
Column 4 of A273891.
Cf. A056284 (oriented), A056490 (achiral), A305543 (chiral).
Sequence in context: A206949 A215636 A056350 * A201231 A212698 A226511
KEYWORD
nonn
STATUS
approved