A032247
"DHK[ 6 ]" (bracelet, identity, unlabeled, 6 parts) transform of 1,1,1,1,...
Original entry on oeis.org
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
A032250
"DHK[ n ](2n)" (bracelet, identity, unlabeled, n parts, evaluated at 2n) transform of 1,1,1,1,...
Original entry on oeis.org
1, 1, 1, 2, 10, 29, 113, 368, 1316, 4490, 15907, 55866, 199550, 714601, 2583575, 9385280, 34311304, 126018592, 465044951, 1722987050, 6407739430, 23909854891, 89493459541, 335911158480, 1264104712300
Offset: 1
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# This is a crude program that assumes the above conjecture is true (which was later proved in Hadjicostas (2019)). It is only valid for n >= 3 because of Bower's special definition of DHK[k] for the cases k=1 and k=2.
with(NumberTheory);
ff := proc (k, x) (1/2)*x^k*(add(phi(n)/(1-x^n)^(k/n), n in Divisors(k))/k-(x+1)/(1-x^2)^floor((1/2)*k+1)); end proc;
gg := proc (k, x) add(Moebius(d)*ff(k/d, x^d), d in Divisors(k)); end proc;
vv := proc (n) simplify(subs(x = 0, diff(gg(n, x), x$(2*n)))/factorial(2*n)); end proc;
for i from 3 to 100 do print(i, vv(i)); end do; # Petros Hadjicostas, Feb 24 2019
A308401
Number of bracelets (turnover necklaces) of length n that have no reflection symmetry and consist of 6 white beads and n-6 black beads.
Original entry on oeis.org
3, 6, 16, 30, 56, 91, 150, 224, 336, 477, 672, 912, 1233, 1617, 2112, 2700, 3432, 4290, 5340, 6552, 8008, 9678, 11648, 13888, 16503, 19448, 22848, 26658, 31008, 35853, 41346, 47424, 54264, 61803, 70224, 79464, 89733, 100947, 113344, 126840, 141680, 157780, 175416, 194480, 215280, 237708
Offset: 9
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 (and thus, a(9) = 3).
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 (and thus, a(9) = 3).
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 (and thus, a(9) = 3). For example, dihedral composition 1 is symmetric because we can draw an axis of symmetry through one of the 1s and 4. In addition, dihedral composition 5 is symmetric because we may draw an axis of symmetry through the numbers 2 and 3.
- Colin Barker, Table of n, a(n) for n = 9..1000
- Hansraj Gupta, Enumeration of incongruent cyclic k-gons, Indian J. Pure and Appl. Math., 10 (1979), no. 8, 964-999.
- Petros Hadjicostas, The aperiodic version of Herbert Kociemba's formula for bracelets with no reflection symmetry, 2019.
- Arnold Knopfmacher and Neville Robbins, Some properties of dihedral compositions, Util. Math. 92 (2013), 207-220.
- Richard H. Reis, A formula for C(T) in Gupta's paper, Indian J. Pure and Appl. Math., 10 (1979), no. 8, 1000-1001.
- Frank Ruskey, Necklaces, Lyndon words, De Bruijn sequences, etc.
- Vladimir S. Shevelev, Necklaces and convex k-gons, Indian J. Pure and Appl. Math., 35 (2004), no. 5, 629-638.
- Vladimir S. Shevelev, Necklaces and convex k-gons, Indian J. Pure and Appl. Math., 35 (2004), no. 5, 629-638.
- Duncan M. Y. Sommerville, On certain periodic properties of cyclic compositions of numbers, Proc. London Math. Soc. S2-7(1) (1909), 263-313.
- Index entries for linear recurrences with constant coefficients, signature (2,1,-3,-1,1,4,-3,-3,4,1,-1,-3,1,2,-1).
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a(n) = (1/12)* (sumdiv(gcd(n, 6), d, eulerphi(d)*binomial((n/d) - 1, (6/d) - 1))) - (1/2)*binomial(floor(n/2), 3); \\ Michel Marcus, May 28 2019
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Vec(x^9*(3 + x^2 + x^3 + x^4) / ((1 - x)^6*(1 + x)^3*(1 - x + x^2)*(1 + x + x^2)^2) + O(x^50)) \\ Colin Barker, Jun 02 2019
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.
Original entry on oeis.org
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, 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, 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
Offset: 1
The triangle begins (with rows for n >= 1 and columns for k >= 1) as follows:
0;
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, 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, 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;
...
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.
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).
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.
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.
Showing 1-4 of 4 results.
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