A273891 Triangle read by rows: T(n,k) is the number of n-bead bracelets with exactly k different colored beads.

1, 1, 1, 1, 2, 1, 1, 4, 6, 3, 1, 6, 18, 24, 12, 1, 11, 56, 136, 150, 60, 1, 16, 147, 612, 1200, 1080, 360, 1, 28, 411, 2619, 7905, 11970, 8820, 2520, 1, 44, 1084, 10480, 46400, 105840, 129360, 80640, 20160, 1, 76, 2979, 41388, 255636, 821952, 1481760, 1512000, 816480, 181440

Offset: 1

Author

Marko Riedel, Jun 02 2016

Keywords

• nonn
• tabl

Comments

For bracelets, chiral pairs are counted as one.

Examples

			Triangle begins with T(1,1):
1;
1,  1;
1,  2,    1;
1,  4,    6,     3;
1,  6,   18,    24,     12;
1, 11,   56,   136,    150,     60;
1, 16,  147,   612,   1200,   1080,     360;
1, 28,  411,  2619,   7905,  11970,    8820,    2520;
1, 44, 1084, 10480,  46400, 105840,  129360,   80640,  20160;
1, 76, 2979, 41388, 255636, 821952, 1481760, 1512000, 816480, 181440;
For T(4,2)=4, the arrangements are AAAB, AABB, ABAB, and ABBB, all achiral.
For T(4,4)=3, the arrangements are ABCD, ABDC, and ACBD, whose chiral partners are ADCB, ACDB, and ADBC respectively. - _Robert A. Russell_, Sep 26 2018
		

Links

• Andrew Howroyd, Table of n, a(n) for n = 1..1275
• Marko Riedel, Maple code for A087854 and A273891.

Crossrefs

Columns 1-6: A057427, A056342, A056343, A056344, A056345, A056346.

Row sums give A019537.

Cf. A087854 (oriented), A305540 (achiral), A305541 (chiral).

Programs

Mathematica

(* t = A081720 *) 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}]; Table[T[n, k], {n, 1, 10}, {k, 1, n}] // Flatten (* Jean-François Alcover, Oct 07 2017, after Andrew Howroyd *)
Table[k! DivisorSum[n, EulerPhi[#] StirlingS2[n/#,k]&]/(2n) + k!(StirlingS2[Floor[(n+1)/2], k] + StirlingS2[Ceiling[(n+1)/2], k])/4, {n,1,10}, {k,1,n}] // Flatten (* Robert A. Russell, Sep 26 2018 *)

Formula

T(n,k) = Sum_{i=0..k-1} (-1)^i * binomial(k,i) * A081720(n,k-i). - Andrew Howroyd, Mar 25 2017

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

T(n,k) = (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 S2 is the Stirling subset number A008277.

G.f. for column k>1: (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).

T(n,k) = (A087854(n,k) + A305540(n,k)) / 2 = A087854(n,k) - A305541(n,k) = A305541(n,k) + A305540(n,k).

(End)