OFFSET
1,9
COMMENTS
In other words, the number of n-bead bracelets with beads of exactly k different colors that when turned over are different from themselves. - Andrew Howroyd, Sep 13 2019
LINKS
Andrew Howroyd, Table of n, a(n) for n = 1..1275 (first 50 rows)
FORMULA
T(n,k) = -(k!/4)*(S2(floor((n+1)/2),k) + S2(ceiling((n+1)/2),k)) + (k!/(2 n))*Sum_{d|n} phi(d)*S2(n/d,k), where S2(n,k) is the Stirling subset number A008277.
T(n, k) = Sum_{i=0..k} (-1)^(k-i)*binomial(k,i)*A293496(n, i). - Andrew Howroyd, Sep 13 2019
EXAMPLE
Triangle T(n,k) begins:
0;
0, 0;
0, 0, 1;
0, 0, 3, 3;
0, 0, 12, 24, 12;
0, 1, 35, 124, 150, 60;
0, 2, 111, 588, 1200, 1080, 360;
0, 6, 318, 2487, 7845, 11970, 8820, 2520;
0, 14, 934, 10240, 46280, 105840, 129360, 80640, 20160;
0, 30, 2634, 40488, 254676, 821592, 1481760, 1512000, 816480, 181440;
...
For T(4,3)=3, the chiral pairs are AABC-AACB, ABBC-ACBB, and ABCC-ACCB.
For T(4,4)=3, the chiral pairs are ABCD-ADCB, ABDC-ACDB, and ACBD-ADBC.
MATHEMATICA
Table[(k!/(2n)) DivisorSum[n, EulerPhi[#] StirlingS2[n/#, k] &] - (k!/4) (StirlingS2[Floor[(n+1)/2], k] + StirlingS2[Ceiling[(n+1)/2], k]), {n, 1, 15}, {k, 1, n}] // Flatten
PROG
(PARI) T(n, k) = {-k!*(stirling((n+1)\2, k, 2) + stirling(n\2+1, k, 2))/4 + k!*sumdiv(n, d, eulerphi(d)*stirling(n/d, k, 2))/(2*n)} \\ Andrew Howroyd, Sep 13 2019
CROSSREFS
KEYWORD
AUTHOR
Robert A. Russell, Jun 04 2018
STATUS
approved