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A032247
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"DHK[ 6 ]" (bracelet, identity, unlabeled, 6 parts) transform of 1,1,1,1,...
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6
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3, 6, 16, 29, 56, 90, 150, 222, 336, 474, 672, 908, 1233, 1612, 2112, 2693, 3432, 4282, 5340, 6542, 8008, 9666, 11648, 13874, 16503, 19432, 22848, 26639, 31008, 35832, 41346, 47400, 54264, 61776, 70224, 79434, 89733, 100914
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OFFSET
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9,1
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COMMENTS
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Here, a(n) is the number of aperiodic bracelets with k = 6 black beads and n-k = n-6 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
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LINKS
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FORMULA
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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.
We prove in the note that g.f. = Sum_{n>=1} a(n)*x^n = gf(6, x) - gf(3, x^2).
(End)
G.f.: x^9*(3 + 4*x^2 + 4*x^4 + 2*x^6 - 2*x^7 + x^8) / ((1 - x)^6*(1 + x)^3*(1 - x + x^2)*(1 + x^2)*(1 + x + x^2)^2).
a(n) = 2*a(n-1) - a(n-3) - 2*a(n-5) + 3*a(n-6) - 2*a(n-7) + a(n-8) + a(n-9) - 2*a(n-10) + 3*a(n-11) - 2*a(n-12) - a(n-14) + 2*a(n-16) - a(n-17) for n > 25.
(End)
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 = 6.
a(n) = (1/12)* Sum_{d|gcd(n, 6)} mu(d) * (binomial((n/d) - 1, (6/d) - 1) - 6*binomial(floor(b(n,d)/2), floor(3/d))) for n >= 6, where b(n,d) = n/d + ((-1)^(6/d) - 1)/2. (Thus, b(n,d) = n/d for d = 1, 3, and b(n, d) = n/d - 1 for d = 2, 6.) (End)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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