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%I A032249 #32 May 28 2019 01:46:22
%S A032249 5,14,42,90,197,368,680,1152,1926,3044,4740,7100,10494,15072,21384,
%T A032249 29680,40755,54994,73502,96854,126555,163424,209456,265792,335036,
%U A032249 418728,520200,641496,786828,958848,1162800,1402080
%N A032249 "DHK[ 8 ]" (bracelet, identity, unlabeled, 8 parts) transform of 1,1,1,1,...
%C A032249 Here, a(n) is the number of aperiodic bracelets with k = 8 black beads and n-k = n-8 white beads that have no reflection symmetry. We conjecture that we can use Herbert Kociemba's formula from the documentation of sequences A008804 and A032246 to derive the g.f. of (a(n): n >= 1). See below for more details. - _Petros Hadjicostas_, Feb 24 2019
%H A032249 C. G. Bower, <a href="/transforms2.html">Transforms (2)</a>
%H A032249 Petros Hadjicostas, <a href="/A032249/a032249.pdf">The aperiodic version of Herbert Kociemba's formula for bracelets with no reflection symmetry</a>, 2019.
%F A032249 From _Petros Hadjicostas_, Feb 24 2019, proven in Hadjicostas (2019): (Start)
%F A032249 Let gf(k, x) = x^k/2 * ( (1/k)*Sum_{n|k} phi(n)/(1 - x^n)^(k/n) - (1 + x)/(1 -x^2)^floor(k/2 + 1) ) be Herbert Kociemba's formula for the g.f. of the number of all bracelets with k black beads and n-k white beads that have no reflection symmetry.
%F A032249 We conjecture that g.f. = Sum_{n>=1} a(n)*x^n = gf(8,x) - gf(4, x^2).
%F A032249 (End)
%F A032249 G.f.: (x^k/(2*k)) * Sum_{d|k} mu(d) * (1/(1 - x^d)^(k/d) - k*(1 + x^d)/(1 - x^(2*d))^floor(k/(2*d) + 1)) with k = 8. - _Petros Hadjicostas_, May 24 2019
%F A032249 a(n) = (1/16)* Sum_{d|gcd(n, 8)} mu(d) * (binomial((n/d) - 1, (8/d) - 1) - 8 * binomial(floor(b(n,d)/2), floor(4/d))) for n >= 11, where b(n,d) = n/d + ((-1)^(8/d) - 1)/2. (Thus, b(n,d) = n/d for d = 1, 2, 4, and b(n, d) = n/d - 1 for d = 8.) - _Petros Hadjicostas_, May 27 2019
%Y A032249 Cf. A008804, A032246, A032247, A032248.
%K A032249 nonn
%O A032249 11,1
%A A032249 _Christian G. Bower_

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