cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-5 of 5 results.

A032248 "DHK[ 7 ]" (bracelet, identity, unlabeled, 7 parts) transform of 1,1,1,1,...

Original entry on oeis.org

4, 10, 28, 56, 113, 197, 340, 544, 856, 1284, 1896, 2709, 3816, 5247, 7128, 9504, 12540, 16302, 21001, 26728, 33748, 42185, 52364, 64448, 78832, 95725, 115600, 138720, 165648, 196707, 232560, 273600, 320601, 374034, 434796, 503448, 581020, 668173, 766084
Offset: 10

Views

Author

Keywords

Comments

From Petros Hadjicostas, Feb 24 2019: (Start)
When k is odd >= 3, the DHK[k] transform of sequence c = (c(n): n >= 1), whose g.f. is C(x) = Sum_{n>=1} c(n)*x^n, has g.f. Sum_{n>=1} (DHK[k] c)n*x^n = (1/2)*Sum{d|k} mu(d)*((1/k)*C(x^d)^(k/d) - C(x^d)*C(x^(2*d))^((k/d) - 1)/2)).
For the current sequence we have k = 7 and c(n) = 1 for all n >= 1. Hence, C(x) = x/(1-x) and A(x) = Sum_{n>=1} a(n)*x^n = (x^k/2)*Sum_{d|k} mu(d)*((1/k)*(1-x^d)^(-k/d) - (1-x^d)^(-1)*(1-x^(2*d))^(-((k/d) - 1)/2)).
The latter g.f. agrees with Herbert Kociemba's formula found in the documentations of sequences and A008804 and A032246 only when k is an odd prime. The reason is that (DHK[k] c)_n, with c=(1,1,1,...), is the number of aperiodic bracelets without reflection symmetry with k black beads and n-k white beads, while Herbert Kociemba's formula (in the documentations of sequences and A008804 and A032246) counts all (periodic and aperiodic) bracelets without reflection symmetry with k black beads and n-k white beads. Hence, in the case k is an odd prime, the two formulas agree.
When k is even, the g.f. of the DHK[k] transform of sequence c = (c(n): n >= 1) is much more complicated.
Note that Herbert Kociemba's formula for counting all (periodic and aperiodic) bracelets with no reflection symmetry is still valid even when k is even; e.g., see sequence A008804 for the case k=4. For k = 4, all bracelets with 4 black beads and n-k = n-4 white beads that have no reflection symmetry are aperiodic, but this is not true anymore for k even >= 6.
(End)

Crossrefs

Cf. A001399, A008804, A032246, A032247, A032250. Column k = 7 of A180472.

Programs

  • Mathematica
    LinearRecurrence[{3,0,-8,6,6,-8,1,0,-1,8,-6,-6,8,0,-3,1},{4,10,28,56,113,197,340,544,856,1284,1896,2709,3816,5247,7128,9504},40] (* Harvey P. Dale, Jul 08 2024 *)
  • PARI
    Vec(x^10*(4 - 2*x - 2*x^2 + 4*x^3 + x^4 - 2*x^5 + x^6) / ((1 - x)^7*(1 + x)^3*(1 + x + x^2 + x^3 + x^4 + x^5 + x^6)) + O(x^40)) \\ Colin Barker, Feb 25 2019

Formula

G.f.: x^7*(1/(14*(1 - x)^7) - 1/((2*(1 - x))*(1 - x^2)^3) + 3/(7*(1 - x^7))). - Petros Hadjicostas, Feb 24 2019
a(n) = 3*a(n-1) - 8*a(n-3) + 6*a(n-4) + 6*a(n-5) - 8*a(n-6) + a(n-7) - a(n-9) + 8*a(n-10) - 6*a(n-11) - 6*a(n-12) + 8*a(n-13) - 3*a(n-15) + a(n-16) for n>25. - Colin Barker, Feb 25 2019

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

Original entry on oeis.org

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

Views

Author

Keywords

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

Crossrefs

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

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

Views

Author

Keywords

Comments

From Petros Hadjicostas, Feb 24 2019: (Start)
Let ff(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.
Let gg(k, x) be the generating function of the number of all aperiodic bracelets with k black beads and n-k white beads that have no reflection symmetry.
We conjecture that gg(k, x)= Sum_{d|k} mu(d)*ff(k/d, x^d).
For n >= 3, a(n) is the coefficient of x^(2*n) of the Taylor expansion of gg(n, x) around x=0. [Bower has special definitions for DHK[1] and DHK[2].]
(End)

Crossrefs

Programs

  • Maple
    # 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

Views

Author

Petros Hadjicostas, May 24 2019

Keywords

Comments

Bracelets that have no reflection symmetry are also known as chiral bracelets.
Here, for n >= 6, a(n) is also the number of dihedral compositions of n with 6 parts that have no reflection symmetry. Taking the MacMahon conjugates of these dihedral compositions, we see that a(n) is also the number of dihedral compositions of n into n-6 parts that have no reflection symmetry.
A cyclic composition b_1 + b_2 + ... + b_k of n into k parts is an equivalent class of (linear) compositions of n into k parts (placed on a circle) such that two such (linear) compositions are equivalent iff one can be obtained from the other by a rotation. Such compositions were first studied extensively by Sommerville (1909).
A dihedral composition b_1 + b_2 + ... + b_k of n into k parts is an equivalent class of (linear) compositions of n into k parts (placed on a circle) such that two such (linear) compositions are equivalent iff one can be obtained from the other by a rotation or a reversal of order. Such compositions were studied, for example, by Knopfmacher and Robbins (2013).
Given a bracelet of length n with k white beads and n-k black beads, we may get the corresponding dihedral composition using MacMahon's correspondence: start with a white bead and count that bead and the black beads that follow (in one direction), and call that b_1; then start with the next white bead and count that one and the black beads that follow, and call that b_2; repeat this process until you reach the k-th white bead and count that one and the black beads that follow, and call that b_k. The corresponding dihedral composition is b_1 + b_2 + ... + b_k.
If in the previous paragraph (given a bracelet of length n with k white beads and n-k black beads), we replace the white beads with black beads and the black beads with white beads, we get a dihedral composition of n into n-k parts: c_1 + c_2 + ... + c_{n-k}. These two dihedral compositions (which correspond to the same bracelet) are called "conjugate" compositions. See p. 273 in Sommerville (1909) for an explanation of "conjugate" compositions in the context of cyclic compositions.
Symmetric cyclic compositions of a positive integer n were first studied by Sommerville (1909, pp. 301-304). It can be proved that the study of necklaces with reflection symmetry using beads of two colors is equivalent to the study of symmetric cyclic compositions of a positive integer. Clearly all the necklaces with reflection symmetry are all the bracelets (turnover necklaces) with reflection symmetry. See also the comments for sequences A119963, A292200, and A295925.

Examples

			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.
		

Crossrefs

Programs

  • PARI
    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
    
  • PARI
    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

Formula

G.f.: (x^k/2) * (-(1 + x)/(1 - x^2)^floor((k/2) + 1) + (1/k) * Sum_{m|k} phi(m)/(1 - x^m)^(k/m)) with k = 6. (This formula is due to Herbert Kociemba.)
a(n) = A005513(n) - A058187(n-6) = A005513(n) - binomial(floor(n/2), 3) for n >= 6.
a(n) = -(1/2)*binomial(floor(n/2), 3) + (1/12)* Sum_{d|gcd(n, 6)} phi(d)*binomial((n/d) - 1, (6/d) - 1) for n >= 6. (This is a modification of formulas found in Gupta (1979) and Shevelev (2004).)
From Colin Barker, May 26 2019: (Start)
G.f.: x^9*(3 + x^2 + x^3 + x^4) / ((1 - x)^6*(1 + x)^3*(1 - x + x^2)*(1 + x + x^2)^2).
a(n) = 2*a(n-1) + a(n-2) - 3*a(n-3) - a(n-4) + a(n-5) + 4*a(n-6) - 3*a(n-7) - 3*a(n-8) + 4*a(n-9) + a(n-10) - a(n-11) - 3*a(n-12) + a(n-13) + 2*a(n-14) - a(n-15) for n > 23. (End)

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

Views

Author

Petros Hadjicostas, Jun 08 2019

Keywords

Comments

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.
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.

Examples

			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.
		

Crossrefs

Cf. A032239 (row sums for n >= 3), A180472.
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).

Formula

T(n, k) = Sum_{d|gcd(n,k)} mu(d) * A180472(n/d, k/d) for 1 <= k <= n.
T(n, k) = T(n, n - k) for 1 <= k <= n - 1.
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.
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.
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)).
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)).
Showing 1-5 of 5 results.