OFFSET
1,18
COMMENTS
An orientable necklace when turned over does not leave it unchanged. Only one necklace in each pair is included in the count.
The number of chiral bracelets. An achiral bracelet is the same as its reverse, while a chiral bracelet is equivalent to its reverse. - Robert A. Russell, Sep 28 2018
LINKS
Andrew Howroyd, Table of n, a(n) for n = 1..1275
FORMULA
From Robert A. Russell, Sep 28 2018: (Start)
T(n, k) = -(k^floor((n+1)/2) + k^ceiling((n+1)/2)) / 4 + (1/2n) * Sum_{d|n} phi(d) * k^(n/d)
G.f. for column k: -(kx/4)*(kx+x+2)/(1-kx^2) - Sum_{d>0} phi(d)*log(1-kx^d)/2d. (End)
EXAMPLE
Array begins:
==========================================================
n\k | 1 2 3 4 5 6 7 8
----+-----------------------------------------------------
1 | 0 0 0 0 0 0 0 0 ...
2 | 0 0 0 0 0 0 0 0 ...
3 | 0 0 1 4 10 20 35 56 ...
4 | 0 0 3 15 45 105 210 378 ...
5 | 0 0 12 72 252 672 1512 3024 ...
6 | 0 1 38 270 1130 3535 9156 20748 ...
7 | 0 2 117 1044 5270 19350 57627 147752 ...
8 | 0 6 336 3795 23520 102795 355656 1039626 ...
9 | 0 14 976 14060 106960 556010 2233504 7440216 ...
10 | 0 30 2724 51204 483756 3010098 14091000 53615016 ...
...
For T(3,4)=4, the chiral pairs are ABC-ACB, ABD-ADB, ACD-ADC, and BCD-BDC.
For T(4,3)=3, the chiral pairs are AABC-AACB, ABBC-ACBB, and ABCC-ACCB. - Robert A. Russell, Sep 28 2018
MATHEMATICA
b[n_, k_] := (1/n)*DivisorSum[n, EulerPhi[#]*k^(n/#) &];
c[n_, k_] := If[EvenQ[n], (k^(n/2) + k^(n/2 + 1))/2, k^((n + 1)/2)];
T[_, 1] = T[1, _] = 0; T[n_, k_] := (b[n, k] - c[n, k])/2;
Table[T[n - k + 1, k], {n, 1, 11}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Oct 11 2017, translated from PARI *)
PROG
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Andrew Howroyd, Oct 10 2017
STATUS
approved