A056342 Number of bracelets of length n using exactly two different colored beads.
0, 1, 2, 4, 6, 11, 16, 28, 44, 76, 124, 222, 378, 685, 1222, 2248, 4110, 7683, 14308, 27010, 50962, 96907, 184408, 352696, 675186, 1296856, 2493724, 4806076, 9272778, 17920858, 34669600, 67159048, 130216122, 252745366, 490984486, 954637556, 1857545298, 3617214679, 7048675958, 13744694926, 26818405350
Offset: 1
Keywords
Examples
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
References
- 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]
Links
- G. C. Greubel, Table of n, a(n) for n = 1..3000
Programs
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Mathematica
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 *) 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 *)
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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
Formula
a(n) = A000029(n) - 2.
From Robert A. Russell, Sep 26 2018: (Start)
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.
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)
Extensions
More terms from Joerg Arndt, Jun 10 2016
Comments