A032247 "DHK[ 6 ]" (bracelet, identity, unlabeled, 6 parts) transform of 1,1,1,1,...
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
Offset: 9
Keywords
Links
- C. G. Bower, Transforms (2)
- Petros Hadjicostas, The aperiodic version of Herbert Kociemba's formula for bracelets with no reflection symmetry, 2019.
Formula
From Petros Hadjicostas, Feb 24 2019: (Start)
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)
From Colin Barker, Feb 25 2019: (Start)
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)
From Petros Hadjicostas, May 25 2019: (Start)
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)
Comments