login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

Number of bracelets of length n using exactly two different colored beads.
7

%I #37 Oct 13 2018 09:30:30

%S 0,1,2,4,6,11,16,28,44,76,124,222,378,685,1222,2248,4110,7683,14308,

%T 27010,50962,96907,184408,352696,675186,1296856,2493724,4806076,

%U 9272778,17920858,34669600,67159048,130216122,252745366,490984486,954637556,1857545298,3617214679,7048675958,13744694926,26818405350

%N Number of bracelets of length n using exactly two 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]

%H G. C. Greubel, <a href="/A056342/b056342.txt">Table of n, a(n) for n = 1..3000</a>

%F a(n) = A000029(n) - 2.

%F From _Robert A. Russell_, Sep 26 2018: (Start)

%F a(n) = (A052823(n) + A027383(n-2)) / 2 = A059076(n) + A027383(n-2).

%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=2 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=2 is the number of colors. (End)

%e For a(6)=11, the arrangements are AAAAAB, AAAABB, AAABAB, AAABBB, AABAAB, AABBBB, ABABAB, ABABBB, ABBABB, ABBBBB, and AABABB, the last being chiral. Its reverse is AABBAB. - _Robert A. Russell_, Sep 26 2018

%t a[n_] := (1/4)*(Mod[n, 2] + 3)*2^Quotient[n, 2] + DivisorSum[n, EulerPhi[#]*2^(n/#)&]/(2*n) - 2; Array[a, 41] (* _Jean-François Alcover_, Nov 05 2017 *)

%t k=2; 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 26 2018 *)

%o (PARI) a(n) = my(k=2); (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 28 2018

%Y Column 2 of A273891.

%Y Equals A052823 - A059076.

%Y Cf. A008277, A027383.

%K nonn

%O 1,3

%A _Marks R. Nester_

%E More terms from _Joerg Arndt_, Jun 10 2016