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Array read by antidiagonals: T(n,k) = number of chiral pairs of necklaces with n beads using a maximum of k colors.
11

%I #26 Sep 28 2018 23:34:51

%S 0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,4,3,0,0,0,0,10,15,12,1,0,0,0,20,45,

%T 72,38,2,0,0,0,35,105,252,270,117,6,0,0,0,56,210,672,1130,1044,336,14,

%U 0,0,0,84,378,1512,3535,5270,3795,976,30,0

%N Array read by antidiagonals: T(n,k) = number of chiral pairs of necklaces with n beads using a maximum of k colors.

%C An orientable necklace when turned over does not leave it unchanged. Only one necklace in each pair is included in the count.

%C 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

%H Andrew Howroyd, <a href="/A293496/b293496.txt">Table of n, a(n) for n = 1..1275</a>

%F T(n,k) = (A075195(n,k) - A284855(n,k)) / 2.

%F From _Robert A. Russell_, Sep 28 2018: (Start)

%F T(n, k) = -(k^floor((n+1)/2) + k^ceiling((n+1)/2)) / 4 + (1/2n) * Sum_{d|n} phi(d) * k^(n/d)

%F G.f. for column k: -(kx/4)*(kx+x+2)/(1-kx^2) - Sum_{d>0} phi(d)*log(1-kx^d)/2d. (End)

%e Array begins:

%e ==========================================================

%e n\k | 1 2 3 4 5 6 7 8

%e ----+-----------------------------------------------------

%e 1 | 0 0 0 0 0 0 0 0 ...

%e 2 | 0 0 0 0 0 0 0 0 ...

%e 3 | 0 0 1 4 10 20 35 56 ...

%e 4 | 0 0 3 15 45 105 210 378 ...

%e 5 | 0 0 12 72 252 672 1512 3024 ...

%e 6 | 0 1 38 270 1130 3535 9156 20748 ...

%e 7 | 0 2 117 1044 5270 19350 57627 147752 ...

%e 8 | 0 6 336 3795 23520 102795 355656 1039626 ...

%e 9 | 0 14 976 14060 106960 556010 2233504 7440216 ...

%e 10 | 0 30 2724 51204 483756 3010098 14091000 53615016 ...

%e ...

%e For T(3,4)=4, the chiral pairs are ABC-ACB, ABD-ADB, ACD-ADC, and BCD-BDC.

%e For T(4,3)=3, the chiral pairs are AABC-AACB, ABBC-ACBB, and ABCC-ACCB. - _Robert A. Russell_, Sep 28 2018

%t b[n_, k_] := (1/n)*DivisorSum[n, EulerPhi[#]*k^(n/#) &];

%t c[n_, k_] := If[EvenQ[n], (k^(n/2) + k^(n/2 + 1))/2, k^((n + 1)/2)];

%t T[_, 1] = T[1, _] = 0; T[n_, k_] := (b[n, k] - c[n, k])/2;

%t Table[T[n - k + 1, k], {n, 1, 11}, {k, n, 1, -1}] // Flatten (* _Jean-François Alcover_, Oct 11 2017, translated from PARI *)

%o (PARI) \\ here b(n,k) is A075195 and c(n,k) is A284855

%o b(n, k) = (1/n) * sumdiv(n, d, eulerphi(d)*k^(n/d));

%o c(n, k) = if(n % 2 == 0, (k^(n/2) + k^(n/2+1))/2, k^((n+1)/2));

%o T(n, k) = (b(n, k) - c(n, k)) / 2;

%Y Columns 2..6 are A059076, A278639, A278640, A278641, A278642.

%Y Cf. A075195, A081720, A284855.

%K nonn,tabl

%O 1,18

%A _Andrew Howroyd_, Oct 10 2017