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 1-8 of 8 results.

A008804 Expansion of 1/((1-x)^2*(1-x^2)*(1-x^4)).

Original entry on oeis.org

1, 2, 4, 6, 10, 14, 20, 26, 35, 44, 56, 68, 84, 100, 120, 140, 165, 190, 220, 250, 286, 322, 364, 406, 455, 504, 560, 616, 680, 744, 816, 888, 969, 1050, 1140, 1230, 1330, 1430, 1540, 1650, 1771, 1892, 2024, 2156, 2300, 2444, 2600, 2756, 2925, 3094, 3276, 3458
Offset: 0

Views

Author

Keywords

Comments

b(n)=a(n-3) is the number of asymmetric nonnegative integer 2 X 2 matrices with sum of elements equal to n, under action of dihedral group D_4(b(0)=b(1)=b(2)=0). G.f. for b(n) is x^3/((1-x)^2*(1-x^2)*(1-x^4)). - Vladeta Jovovic, May 07 2000
If the offset is changed to 5, this is the 2nd Witt transform of A004526 [Moree]. - R. J. Mathar, Nov 08 2008
a(n) is the number of partitions of 2*n into powers of 2 less than or equal to 2^3. First differs from A000123 at n=8. - Alois P. Heinz, Apr 02 2012
a(n) is the number of bracelets with 4 black beads and n+3 white beads which have no reflection symmetry. For n=1 we have for example 2 such bracelets with 4 black beads and 4 white beads: BBBWBWWW and BBWBWBWW. - Herbert Kociemba, Nov 27 2016
a(n) is the also number of aperiodic bracelets with 4 black beads and n+3 white beads which have no reflection symmetry. This is equivalent to saying that a(n) is the (n+7)th element of the DHK[4] (bracelet, identity, unlabeled, 4 parts) transform of 1, 1, 1, ... (see Bower's link about transforms). Thus, for n >= 1 , a(n) = (DHK[4] c){n+7}, where c = (1 : n >= 1). This is because every bracelet with 4 black beads and n+3 white beads which has no reflection symmetry must also be aperiodic. This statement is not true anymore if we have k black beads where k is even >= 6. - _Petros Hadjicostas, Feb 24 2019

Examples

			G.f. = 1 + 2*x + 4*x^2 + 6*x^3 + 10*x^4 + 14*x^5 + 20*x^6 + 26*x^7 + 35*x^8 + ...
There are 10 asymmetric nonnegative integer 2 X 2 matrices with sum of elements equal to 7 under action of D_4:
[0 0] [0 0] [0 0] [0 1] [0 1] [0 1] [0 1] [0 2] [0 2] [1 1]
[1 6] [2 5] [3 4] [2 4] [3 3] [4 2] [5 1] [3 2] [4 1] [2 3]
		

Crossrefs

Column k=3 of A181322. Column k = 4 of A180472 (but with different offset).

Programs

  • GAP
    a:=[1,2,4,6,10,14,20,26];; for n in [9..60] do a[n]:=2*a[n-1] -2*a[n-3]+2*a[n-4]-2*a[n-5]+2*a[n-7]-a[n-8]; od; a; # G. C. Greubel, Sep 12 2019
  • Magma
    R:=PowerSeriesRing(Integers(), 60); Coefficients(R!( 1/((1-x)^2*(1-x^2)*(1-x^4)) )); // G. C. Greubel, Sep 12 2019
    
  • Maple
    seq(coeff(series(1/((1-x)^2*(1-x^2)*(1-x^4)), x, n+1), x, n), n = 0..60); # G. C. Greubel, Sep 12 2019
  • Mathematica
    LinearRecurrence[{2,0,-2,2,-2,0,2,-1}, {1,2,4,6,10,14,20,26}, 60] (* Vladimir Joseph Stephan Orlovsky, Feb 23 2012 *)
    gf[x_,k_]:=x^k/2 (1/k Plus@@(EulerPhi[#] (1-x^#)^(-(k/#))&/@Divisors[k])-(1+x)/(1-x^2)^Floor[k/2+1]); CoefficientList[Series[gf[x,4]/x^7,{x,0,60}],x] (* Herbert Kociemba, Nov 27 2016 *)
    Table[(84 +12*(-1)^n +85*n +3*(-1)^n*n +24*n^2 +2*n^3 +12*Sin[n Pi/2])/96, {n,0,60}] (* Eric W. Weisstein, Oct 12 2017 *)
    CoefficientList[Series[1/((1-x)^4*(1+x)^2*(1+x^2)), {x,0,60}], x] (* Eric W. Weisstein, Oct 12 2017 *)
  • PARI
    a(n)=(84+12*(-1)^n+6*I*((-I)^n-I^n)+(85+3*(-1)^n)*n+24*n^2 +2*n^3)/96 \\ Jaume Oliver Lafont, Sep 20 2009
    
  • PARI
    {a(n) = my(s = 1); if( n<-7, n = -8 - n; s = -1); if( n<0, 0, s * polcoeff( 1 / ((1 - x)^2 * (1 - x^2) * (1 - x^4)) + x * O(x^n), n))}; /* Michael Somos, Feb 02 2011 */
    
  • Sage
    def A008804_list(prec):
        P. = PowerSeriesRing(ZZ, prec)
        return P(1/((1-x)^2*(1-x^2)*(1-x^4))).list()
    A008804_list(60) # G. C. Greubel, Sep 12 2019
    

Formula

For a formula for a(n) see A014557.
a(n) = (84 +85*n +24*n^2 +2*n^3 +12*A056594(n+3) +3*(-1)^n*(n+4))/96. - R. J. Mathar, Nov 08 2008
a(n) = 2*(Sum_{k=0..floor(n/2)} A002620(k+2)) - A002620(n/2+2)*(1+(-1)^n)/2. - Paul Barry, Mar 05 2009
G.f.: 1/((1-x)^4*(1+x)^2*(1+x^2)). - Jaume Oliver Lafont, Sep 20 2009
Euler transform of length 4 sequence [2, 1, 0, 1]. - Michael Somos, Feb 05 2011
a(n) = -a(-8 - n) for all n in Z. - Michael Somos, Feb 05 2011
From Herbert Kociemba, Nov 27 2016: (Start)
More generally gf(k) is the g.f. for the number of bracelets without reflection symmetry with k black beads and n-k white beads.
gf(k): x^k/2 * ( (1/k)*Sum_{n|k} phi(n)/(1 - x^n)^(k/n) - (1 + x)/(1 -x^2)^floor(k/2 + 1) ). The g.f. here is gf(4)/x^7 because of the different offset. (End)
E.g.f.: ((48 + 54*x + 15*x^2 + x^3)*cosh(x) + 6*sin(x) + (36 + 57*x + 15*x^2 + x^3)*sinh(x))/48. - Stefano Spezia, May 15 2023
a(n) = A001400(n) + A001400(n-1) + A001400(n-2). - David GarcĂ­a Herrero, Aug 26 2024
a(n) = floor((2*n^3 + 24*n^2 + n*(85+3*(-1)^n) + 96) / 96). - Hoang Xuan Thanh, May 24 2025

A032246 "DHK[ 5 ]" (bracelet, identity, unlabeled, 5 parts) transform of 1,1,1,1,...

Original entry on oeis.org

2, 4, 10, 16, 28, 42, 64, 90, 126, 168, 224, 288, 370, 462, 576, 704, 858, 1030, 1232, 1456, 1716, 2002, 2330, 2688, 3094, 3536, 4032, 4570, 5168, 5814, 6528, 7296, 8140, 9044, 10032, 11088, 12236, 13460, 14784, 16192, 17710, 19320, 21050, 22880, 24840, 26910
Offset: 8

Views

Author

Keywords

Comments

a(n) is the number of bracelets with k = 5 black beads and n-k white beads which have no reflection symmetry. - Herbert Kociemba, Nov 27 2016
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 = 5 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 below 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 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)

Examples

			G.f. = 2*x^8 + 4*x^9 + 10*x^10 + 16*x^11 + 28*x^12 + 42*x^13 + 64*x^14 + ...
		

Crossrefs

Cf. A001399, A008804, A032248. Column k = 5 of A180472.

Programs

  • Magma
    m:=50; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( 2*x^8/((1-x)^2*(1-x^2)^2*(1-x^5)) )); // G. C. Greubel, Feb 25 2019
    
  • Mathematica
    gf[x_,k_]:=x^k/2 (1/k Plus@@(EulerPhi[#] (1-x^#)^(-(k/#))&/@Divisors[k])-(1+x)/(1-x^2)^Floor[k/2+1]); CoefficientList[Series[gf[x,5],{x,0,50}],x] (* Herbert Kociemba, Nov 27 2016 *)
    Drop[CoefficientList[Series[2*x^8/((1-x)^2*(1-x^2)^2*(1-x^5)), {x,0,50}], x], 8] (* G. C. Greubel, Feb 25 2019 *)
  • PARI
    {a(n) = if( n<0, n=5-n); polcoeff( 2 * x^8 / ((1-x)^2*(1-x^2)^2*(1-x^5)) + x * O(x^n), n)}; /* Michael Somos, Nov 28 2016 */
    
  • PARI
    Vec(2*x^8/((1-x)^2*(1-x^2)^2*(1-x^5)) + O(x^40)) \\ Colin Barker, Mar 13 2019
    
  • Sage
    a=(2*x^8/((1-x)^2*(1-x^2)^2*(1-x^5))).series(x, 50).coefficients(x, sparse=False); a[8:] # G. C. Greubel, Feb 25 2019

Formula

G.f.: 2*x^8/((1-x)^2*(1-x^2)^2*(1-x^5)).
From Herbert Kociemba, Nov 27 2016: (Start)
More generally gf(k) is the g.f. for the number of bracelets without reflection symmetry with k black beads and n-k white beads.
gf(k): x^k/2 * ((1/k)*Sum_{n|k} phi(n)/(1 - x^n)^(k/n) - (1 + x)/(1-x^2)^floor(k/2 + 1)). (End)
a(n) = a(5-n) for all n in Z. - Michael Somos, Nov 28 2016
0 = a(n) - 2*a(n+1) - a(n+2) + 4*a(n+3) - a(n+4) - 3*a(n+5) + 3*a(n+6) + a(n+7) - 4*a(n+8) + a(n+9) + 2*a(n+10) - a(n+11) for all n in Z. - Michael Somos, Nov 28 2016

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

Views

Author

Keywords

Comments

Here, a(n) is the number of aperiodic bracelets with k = 6 black beads and n-k = n-6 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: (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)

A180472 Triangle T(n, k) = OC(n, k; not -1), read by rows, where OC(n, k; not -1) is the number of k-subsets of Z_n without -1 as a multiplier, up to congruency.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 2, 2, 2, 0, 0, 0, 0, 0, 0, 3, 4, 4, 3, 0, 0, 0, 0, 0, 0, 4, 6, 10, 6, 4, 0, 0, 0, 0, 0, 0, 5, 10, 16, 16, 10, 5, 0, 0, 0, 0, 0, 0, 7, 14, 28, 30, 28, 14, 7, 0, 0, 0, 0, 0, 0, 8, 20, 42, 56, 56, 42, 20, 8, 0, 0, 0, 0, 0, 0, 10, 26, 64, 91, 113, 91, 64, 26, 10, 0, 0, 0, 0, 0, 0, 12, 35, 90, 150, 197, 197, 150, 90, 35, 12, 0, 0, 0, 0, 0, 0, 14, 44, 126, 224, 340, 370, 340, 224, 126, 44, 14, 0, 0, 0, 0, 0, 0, 16, 56, 168, 336, 544, 680, 680, 544, 336, 168, 56, 16, 0, 0, 0
Offset: 0

Views

Author

John P. McSorley, Sep 06 2010

Keywords

Comments

Let Z_n = {0,1,...,n-1} denote the integers mod n.
Let S be a k-subset of Z_n.
Then S has multiplier -1 iff there is a z in Z_n for which S = -S + z. Otherwise, S doesn't have multiplier -1.
For example in Z_7 the set S = {0,1,2} has multiplier -1 since -S = {0,-1,-2} = {0,5,6} and then {0,1,2} = {0,5,6} + 2, so S = -S + 2. But S={0,1,3} doesn't have multiplier -1.
Let S and S' be two k-subsets of Z_n.
Define an equivalence relation on the set of k-subsets as follows: S is congruent to S' iff S=S'+z or S = -S' + z for some z in Z_n.
Then define OC(n, k) to be the number of such congruence classes.
And define OC(n, k; not -1) to be the number of such congruence classes in which the representative doesn't have -1 as a multiplier.
Then this sequence is the 'OC(n,k; not -1)' triangle read by rows.
For convenience we start the triangle at n = 0, and we have 0 <= k <= n.
See the McSorley and Schoen (2013) reference below for equivalent definitions of this sequence in terms of (n,k)-Ovals and k-compositions of n.
From Petros Hadjicostas, May 29 2019: (Start)
Here, T(n, k) is the number of bracelets (turnover necklaces) of length n that have no reflection symmetry and consist of k white beads and n - k black beads. (Bracelets that have no reflection symmetry are also known as chiral bracelets.)
It is also the number of dihedral compositions of n into k parts with no reflection symmetry. It is also the number of dihedral compositions of n into n - k parts with no reflection symmetry. (For a definition of a dihedral composition, see Knopfmacher and Robbins (2013) in the references.)
For MacMahon's method for transforming a cyclic composition into a necklace and vice versa, see the comments for sequence A308401. See also p. 273 in Sommerville (1909).
(End)

Examples

			The triangle begins (with rows for n >= 0 and columns for k >= 0) as follows:
  0
  0  0
  0  0  0
  0  0  0  0
  0  0  0  0  0
  0  0  0  0  0  0
  0  0  0  1  0  0  0
  0  0  0  1  1  0  0   0
  0  0  0  2  2  2  0   0  0
  0  0  0  3  4  4  3   0  0  0
  0  0  0  4  6 10  6   4  0  0  0
  0  0  0  5 10 16 16  10  5  0  0  0
  0  0  0  7 14 28 30  28 14  7  0  0  0
  0  0  0  8 20 42 56  56 42 20  8  0  0  0
  0  0  0 10 26 64 91 113 91 64 26 10  0  0  0  0
  ...
For example the row which corresponds to Z_7 is: 0 0 0 1 1 0 0 0.
The first '1' here corresponds to the 3-subsets of Z_7.
There are 4 congruence classes of the 3-subsets of Z_7, their representatives are {0,1,2}, {0,2,4}, {0,1,4} and {0,1,3}. The first 3 representatives have multiplier -1, but the last doesn't. Hence there is just one 3-subset of Z_7 without multiplier -1, up to congruency.
		

Crossrefs

This sequence is A052307-A119963. The sequence A052307 is formed from the triangle whose (n, k)-term is the number of k-subsets of Z_n up to congruence, and the sequence A119963 is formed from the triangle whose (n, k)-term is the number of k-subsets of Z_n with multiplier -1 up to congruence.
The row sums of the 'OC(n, k, not -1)' triangle above give sequence A059076.
Cf. A001399 (column k = 3 with different offset), A008804 (column k = 4 with different offset), A032246 (column k = 5), A308401 (column k = 6), A032248 (column k = 7).

Programs

  • PARI
    T(n,k) = if ((n==0) && (k==0), 0, -binomial(floor(n/2) - (k % 2) * (1 - n % 2), floor(k/2)) / 2 + sumdiv(gcd(n,k), d, (eulerphi(d)*binomial(n/d, k/d))) / (2*n));tabl(nn) = for (n=0, nn, for (k=0, n, print1(T(n,k), ", ")); print); \\ Michel Marcus, May 30 2019

Formula

From Petros Hadjicostas, May 29 2019: (Start)
T(n,k) = -binomial(floor(n/2) - (k mod 2) * (1 - (n mod 2)), floor(k/2)) / 2 + Sum_{d|n, d|k} (phi(d)*binomial(n/d, k/d)) / (2*n) for n >= 1 and 0 <= k <= n. (This is a modification of formulas due to Gupta (1979), Shevelev (2004), and W. Bomfim in sequence A052307.)
T(n, k) = A052307(n, k) - A119963(n,k) for 0 <= k <= n. (See the comments in CROSSREFS by J. P. McSorley.)
T(n, k) = T(n, n - k) for 0 <= k <= n.
G.f. for column k >= 1: (x^k/2) * (-(1 + x)/(1 - x^2)^floor((k/2) + 1) + (1/k) * Sum_{m|k} phi(m)/(1 - x^m)^(k/m)). (This formula is due to Herbert Kociemba.)
(End)
Bivariate g.f.: Sum_{n,k >= 0} T(n, k)*x^n*y^k = (1/2) * (1 - (1 + x) * (1 + x*y) / (1 - x^2 * (1 + y^2)) - Sum_{d >= 1} (phi(d) / d) * log(1 - x^d * (1 + y^d))). - Petros Hadjicostas, Jun 15 2019

Extensions

Name edited by Petros Hadjicostas, May 29 2019
Offset corrected by Andrew Howroyd, Sep 27 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 1-8 of 8 results.