%I #30 Jan 09 2020 19:23:34
%S 0,0,1,0,0,1,0,1,3,3,0,0,3,12,12,0,1,10,46,90,60,0,0,9,120,480,720,
%T 360,0,1,27,384,2235,5670,6300,2520,0,0,29,980,9380,36960,68880,60480,
%U 20160,0,1,75,2904,38484,217152,604800,876960,635040,181440
%N Triangle read by rows: T(n,k) is the number of n-bead bracelets using exactly k colors with no adjacent beads having the same color.
%C In the case of n = 1, the single bead is considered to be cyclically adjacent to itself giving T(1,1) = 0. If compatibility with A208544 is wanted then T(1,1) should be 1.
%H Andrew Howroyd, <a href="/A330341/b330341.txt">Table of n, a(n) for n = 1..1275</a> (first 50 rows)
%F T(n,k) = Sum_{j=1..k} (-1)^(k-j)*binomial(k,j)*A208544(n,j) for n > 1.
%e Triangle begins:
%e 0;
%e 0, 1;
%e 0, 0, 1;
%e 0, 1, 3, 3;
%e 0, 0, 3, 12, 12;
%e 0, 1, 10, 46, 90, 60;
%e 0, 0, 9, 120, 480, 720, 360;
%e 0, 1, 27, 384, 2235, 5670, 6300, 2520;
%e 0, 0, 29, 980, 9380, 36960, 68880, 60480, 20160;
%e ...
%o (PARI) \\ here U(n, k) is A208544(n, k) for n > 1.
%o U(n, k) = (sumdiv(n, d, eulerphi(n/d)*(k-1)^d)/n + if(n%2, 1-k, k*(k-1)^(n/2)/2))/2;
%o T(n, k)={sum(j=1, k, (-1)^(k-j)*binomial(k, j)*U(n, j))}
%Y Column 3 is A330632.
%Y Row sums are A330621.
%Y Cf. A208544, A273891, A330618.
%K nonn,tabl
%O 1,9
%A _Andrew Howroyd_, Dec 20 2019