A056346 - OEIS

%I #29 Oct 13 2018 09:29:57

%S 0,0,0,0,0,60,1080,11970,105840,821952,5874480,39713550,258136200,

%T 1631273220,10096734312,61536377700,370710950400,2213749658880,

%U 13132080672480,77509456944318,455754569692680

%N Number of bracelets of length n using exactly six different colored beads.

%C Turning over will not create a new bracelet.

%D 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]

%F a(n) = A056341(n) - 6*A032276(n) + 15*A032275(n) - 20*A027671(n) + 15*A000029(n) - 6.

%F From _Robert A. Russell_, Sep 27 2018: (Start)

%F a(n) = (k!/4) * (S2(floor((n+1)/2),k) + S2(ceiling((n+1)/2),k)) + (k!/2n) * Sum_{d|n} phi(d) * S2(n/d,k), where k=6 is the number of colors and S2 is the Stirling subset number A008277.

%F G.f.: (k!/4) * x^(2k-2) * (1+x)^2 / Product_{i=1..k} (1-i x^2) - Sum_{d>0} (phi(d)/2d) * Sum_{j} (-1)^(k-j) * C(k,j) * log(1-j x^d), where k=6 is the number of colors. (End)

%e 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

%t 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 T[n_, k_] := Sum[(-1)^i*Binomial[k, i]*t[n, k - i], {i, 0, k - 1}];

%t a[n_] := T[n, 6];

%t Array[a, 21] (* _Jean-François Alcover_, Nov 05 2017, after _Andrew Howroyd_ *)

%t 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 *)

%o (PARI) 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

%Y Column 6 of A273891.

%Y Equals (A056286 + A056492) / 2 = A056286 - A305545 = A305545 + A056492.

%Y Cf. A008277.

%K nonn

%O 1,6

%A _Marks R. Nester_