A330341
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.
Original entry on oeis.org
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, 360, 0, 1, 27, 384, 2235, 5670, 6300, 2520, 0, 0, 29, 980, 9380, 36960, 68880, 60480, 20160, 0, 1, 75, 2904, 38484, 217152, 604800, 876960, 635040, 181440
Offset: 1
Triangle begins:
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, 360;
0, 1, 27, 384, 2235, 5670, 6300, 2520;
0, 0, 29, 980, 9380, 36960, 68880, 60480, 20160;
...
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\\ here U(n, k) is A208544(n, k) for n > 1.
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;
T(n, k)={sum(j=1, k, (-1)^(k-j)*binomial(k, j)*U(n, j))}
A056346
Number of bracelets of length n using exactly six different colored beads.
Original entry on oeis.org
0, 0, 0, 0, 0, 60, 1080, 11970, 105840, 821952, 5874480, 39713550, 258136200, 1631273220, 10096734312, 61536377700, 370710950400, 2213749658880, 13132080672480, 77509456944318, 455754569692680
Offset: 1
For a(6)=60, pair up the 120 permutations of BCDEF, each with its reverse, such as BCDEF-FEDCB. Precede the first of each pair with an A, such as ABCDEF. These are the 60 arrangements, all chiral. If we precede the second of each pair with an A, such as AFEDCB, we get the chiral partner of each. - _Robert A. Russell_, Sep 27 2018
- M. R. Nester (1999). Mathematical investigations of some plant interaction designs. PhD Thesis. University of Queensland, Brisbane, Australia. [See A056391 for pdf file of Chap. 2]
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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}];
a[n_] := T[n, 6];
Array[a, 21] (* Jean-François Alcover, Nov 05 2017, after Andrew Howroyd *)
k=6; Table[k! DivisorSum[n, EulerPhi[#] StirlingS2[n/#,k]&]/(2n) + k!(StirlingS2[Floor[(n+1)/2], k] + StirlingS2[Ceiling[(n+1)/2], k])/4, {n,1,30}] (* Robert A. Russell, Sep 27 2018 *)
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a(n) = my(k=6); (k!/4) * (stirling(floor((n+1)/2),k,2) + stirling(ceil((n+1)/2),k,2)) + (k!/(2*n))*sumdiv(n, d, eulerphi(d)*stirling(n/d,k,2)); \\ Michel Marcus, Sep 29 2018
A019537
Number of special orbits for dihedral group of degree n.
Original entry on oeis.org
1, 2, 4, 14, 61, 414, 3416, 34274, 394009, 5113712, 73758368, 1170495180, 20263806277, 380048113202, 7676106638884, 166114210737254, 3834434327929981, 94042629562443206, 2442147034770292496, 66942194906543381336, 1931543452346146410965, 58519191359170883258606
Offset: 1
Manfred Goebel (goebel(AT)informatik.uni-tuebingen.de)
For a(3) = 4, the necklaces are AAA, AAB, ABB, and ABC. Last one is chiral. For a(4) = 14, the necklacess are AAAA, AAAB, AABB, ABAB, ABBB, ABAC, ABCB, ACBC, AABC, ABBC, ABCC, ABCD, ABDC, and ACBD. Last six are chiral. - _Robert A. Russell_, May 31 2018
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Table[Sum[(k!/(2n)) DivisorSum[n, EulerPhi[#] StirlingS2[n/#,k] &] + (k!/4) (StirlingS2[Floor[(n+1)/2],k] + StirlingS2[Ceiling[(n+1)/2],k]), {k, 1, n}], {n, 1, 40}] (* Robert A. Russell, May 31 2018 *)
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a(n) = sum(k=1, n, (k!/4)*(stirling(floor((n+1)/2),k,2) + stirling(ceil((n+1)/2),k,2)) + (k!/(2*n))*sumdiv(n, d, eulerphi(d)*stirling(n/d,k,2))); \\ Michel Marcus, Jun 06 2018
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