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A308583 Triangle read by rows: T(n,k) = number of aperiodic chiral bracelets (turnover necklaces with no reflection symmetry and period n) with n beads, k of which are white and n - k are black, for n >= 1 and 1 <= k <= n.

%I A308583 #46 May 28 2025 04:30:37
%S A308583 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,1,0,0,0,0,0,2,2,2,0,0,
%T A308583 0,0,0,3,4,4,3,0,0,0,0,0,4,6,10,6,4,0,0,0,0,0,5,10,16,16,10,5,0,0,0,0,
%U A308583 0,7,14,28,29,28,14,7,0,0,0,0,0,8,20,42,56,56,42,20,8,0,0,0,0,0,10,26,64,90,113,90,64,26,10,0,0,0,0,0,12,35,90,150,197,197,150,90,35,12,0,0,0,0,0,14,44,126,222,340,368,340,222,126,44,14,0,0,0
%N A308583 Triangle read by rows: T(n,k) = number of aperiodic chiral bracelets (turnover necklaces with no reflection symmetry and period n) with n beads, k of which are white and n - k are black, for n >= 1 and 1 <= k <= n.
%C A308583 For k = 1, 4, or a prime, the columns of this triangular array are exactly the same as the corresponding columns for the triangular array A180472. In other words, all chiral bracelets with n beads, k of which are white and n - k are black, are aperiodic if k = 1, 4, or a prime.
%C A308583 Note that, T(n, k) is also the number of aperiodic dihedral compositions of n with k parts and no reflection symmetry. Since T(n, k) = T(n, n - k), T(n, k) is also the number of aperiodic dihedral compositions of n with n - k parts and no reflection symmetry.
%H A308583 Petros Hadjicostas, <a href="/A308583/a308583.pdf">Formulas for chiral bracelets</a>, 2019.
%H A308583 Arnold Knopfmacher and Neville Robbins, <a href="https://www.researchgate.net/publication/260006088_Some_Properties_of_Dihedral_Compositions">Some properties of dihedral compositions</a>, Util. Math. 92 (2013), 207-220.
%H A308583 Frank Ruskey, <a href="http://combos.org/necklace">Necklaces, Lyndon words, De Bruijn sequences, etc.</a>
%F A308583 T(n, k) = Sum_{d|gcd(n,k)} mu(d) * A180472(n/d, k/d) for 1 <= k <= n.
%F A308583 T(n, k) = T(n, n - k) for 1 <= k <= n - 1.
%F A308583 T(n, k) = (1/(2*k)) * Sum_{d|gcd(n, k)} mu(d) * (binomial(n/d - 1, k/d - 1) - k * binomial(floor(b(n, k, d)/2), floor(k/(2*d)))) for 1 <= k <= n, where b(n, k, d) = (n/d) + ((-1)^(k/d) - 1)/2.
%F A308583 T(n, k) = (1/(2*n)) * Sum_{d|gcd(n, k)} mu(d) * (binomial(n/d, k/d) - n * binomial(floor(b(n, k, d)/2), floor(k/(2*d)))) for 1 <= k <= n, where b(n, k, d) = (n/d) + ((-1)^(k/d) - 1)/2.
%F A308583 G.f. for column k >= 1: (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)).
%F A308583 Bivariate g.f.: Sum_{n,k >= 1} T(n, k)*x^n*y^k = (1/2) * Sum_{d >= 1} mu(d) * (1 - (1 + x^d) * (1 + x^d*y^d) / (1 - x^(2*d) * (1 + y^(2*d)))) - (1/2) * Sum_{d >= 1} (mu(d)/d) * log(1 - x^d * (1 + y^d)).
%e A308583 The triangle begins (with rows for n >= 1 and columns for k >= 1) as follows:
%e A308583   0;
%e A308583   0, 0;
%e A308583   0, 0,  0;
%e A308583   0, 0,  0,  0;
%e A308583   0, 0,  0,  0,  0;
%e A308583   0, 0,  1,  0,  0,  0;
%e A308583   0, 0,  1,  1,  0,  0,   0;
%e A308583   0, 0,  2,  2,  2,  0,   0,  0;
%e A308583   0, 0,  3,  4,  4,  3,   0,  0,  0;
%e A308583   0, 0,  4,  6, 10,  6,   4,  0,  0,  0;
%e A308583   0, 0,  5, 10, 16, 16,  10,  5,  0,  0,  0;
%e A308583   0, 0,  7, 14, 28, 29,  28, 14,  7,  0,  0, 0;
%e A308583   0, 0,  8, 20, 42, 56,  56, 42, 20,  8,  0, 0, 0;
%e A308583   0, 0, 10, 26, 64, 90, 113, 90, 64, 26, 10, 0, 0, 0;
%e A308583   ...
%e A308583 Notice, for example, that T(14, 6) = 90 <> 91 = A180472(14, 6). Out of the 91 chiral bracelets with 6 W and 8 B beads, only WWBWBBBWWBWBBB is periodic.
%e A308583 Using Frank Ruskey's website (listed above) to generate bracelets of fixed content (6, 3) with string length n = 9 and alphabet size 2, we get the following A005513(n = 9) = 7 bracelets: (1) WWWWWWBBB, (2) WWWWWBWBB, (3) WWWWBWWBB, (4) WWWWBWBWB, (5) WWWBWWWBB, (6) WWWBWWBWB, and (7) WWBWWBWWB. From these, bracelets 1, 4, 5, and 7 have reflection symmetry, while bracelets 2, 3 and 6 have no reflection symmetry. Because chiral bracelets 2, 3, and 6 are aperiodic as well, we have T(9, 3) = 3 = T(9, 6).
%e A308583 Starting with a black bead, we count that bead and how many white beads follow (in one direction), and continue this process until we count all beads around the circle. We thus use MacMahon's correspondence to get the following dihedral compositions of n = 9 into 3 parts: (1) 1 + 7 + 1, (2) 1 + 2 + 6, (3) 1 + 3 + 5, (4) 2 + 5 + 2, (5) 4 + 1 + 4, (6) 2 + 3 + 4, and (7) 3 + 3 + 3. Again, dihedral compositions 1, 4, 5, and 7 are symmetric (have reflection symmetry), while dihedral compositions 2, 3, and 6 are not symmetric. In addition, chiral dihedral compositions 2, 3, and 6 are aperiodic as well, and so (again) T(9, 3) = 3.
%e A308583 We may also start with a white bead and count that bead and how many black beads follow (in one direction), and continue this process until we count all beads around the circle. We thus use MacMahon's correspondence again to get the following (conjugate) dihedral compositions of n = 9 into 6 parts: (1) 1 + 1 + 1 + 1 + 1 + 4, (2) 1 + 1 + 1 + 1 + 2 + 3, (3) 1 + 1 + 1 + 2 + 1 + 3, (4) 1 + 1 + 1 + 2 + 2 + 2, (5) 1 + 1 + 2 + 1 + 1 + 3, (6) 1 + 1 + 2 + 1 + 2 + 2, and (7) 1 + 2 + 1 + 2 + 1 + 2. Again, dihedral compositions 1, 4, 5, and 7 have reflection symmetries, while dihedral compositions 2, 3, and 6 do not have reflection symmetries. Chiral dihedral compositions 2, 3, and 6 are aperiodic as well, and hence T(9, 6) = 3.
%Y A308583 Cf. A032239 (row sums for n >= 3), A180472.
%Y A308583 Cf. A001399 (column k = 3 with a different offset), A008804 (column k = 4 with a different offset), A032246 (column k = 5), A032247 (column k = 6), A032248 (column k = 7), A032249 (column k = 8).
%K A308583 nonn,tabl
%O A308583 1,30
%A A308583 _Petros Hadjicostas_, Jun 08 2019