A032249 "DHK[ 8 ]" (bracelet, identity, unlabeled, 8 parts) transform of 1,1,1,1,...

5, 14, 42, 90, 197, 368, 680, 1152, 1926, 3044, 4740, 7100, 10494, 15072, 21384, 29680, 40755, 54994, 73502, 96854, 126555, 163424, 209456, 265792, 335036, 418728, 520200, 641496, 786828, 958848, 1162800, 1402080

Offset: 11

Author

Christian G. Bower

Keywords

• nonn

Comments

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

Links

• C. G. Bower, Transforms (2)
• Petros Hadjicostas, The aperiodic version of Herbert Kociemba's formula for bracelets with no reflection symmetry, 2019.

Crossrefs

Cf. A008804, A032246, A032247, A032248.

Formula

From Petros Hadjicostas, Feb 24 2019, proven in Hadjicostas (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 conjecture that g.f. = Sum_{n>=1} a(n)*x^n = gf(8,x) - gf(4, x^2).

(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 = 8. - Petros Hadjicostas, May 24 2019

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