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 1-10 of 11 results. Next

A000029 Number of necklaces with n beads of 2 colors, allowing turning over (these are also called bracelets).

Original entry on oeis.org

1, 2, 3, 4, 6, 8, 13, 18, 30, 46, 78, 126, 224, 380, 687, 1224, 2250, 4112, 7685, 14310, 27012, 50964, 96909, 184410, 352698, 675188, 1296858, 2493726, 4806078, 9272780, 17920860, 34669602, 67159050, 130216124, 252745368, 490984488, 954637558, 1857545300
Offset: 0

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Author

Keywords

Comments

"Necklaces with turning over allowed" are usually called bracelets. - Joerg Arndt, Jun 10 2016

Examples

			For n=2, the three bracelets are AA, AB, and BB. For n=3, the four bracelets are AAA, AAB, ABB, and BBB. - _Robert A. Russell_, Sep 24 2018
		

References

  • J. L. Fisher, Application-Oriented Algebra (1977), ISBN 0-7002-2504-8, circa p. 215.
  • Martin Gardner, "New Mathematical Diversions from Scientific American" (Simon and Schuster, New York, 1966), pages 245-246.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • N. Zagaglia Salvi, Ordered partitions and colourings of cycles and necklaces, Bull. Inst. Combin. Appl., 27 (1999), 37-40.

Crossrefs

Row sums of triangle in A052307, second column of A081720, column 2 of A051137.
Cf. A000011, A000013, A000031 (turning over not allowed), A001371 (primitive necklaces), A059076, A164090.

Programs

  • Maple
    with(numtheory): A000029 := proc(n) local d,s; if n = 0 then return 1 else if n mod 2 = 1 then s := 2^((n-1)/2) else s := 2^(n/2-2)+2^(n/2-1) fi; for d in divisors(n) do s := s+phi(d)*2^(n/d)/(2*n) od; return s; fi end:
  • Mathematica
    a[0] := 1; a[n_] := Fold[#1 + EulerPhi[#2]2^(n/#2)/(2n) &, If[OddQ[n], 2^((n - 1)/2), 2^(n/2 - 1) + 2^(n/2 - 2)], Divisors[n]]
    mx=40;CoefficientList[Series[(1-Sum[ EulerPhi[n]*Log[1-2*x^n]/n,{n,mx}]+(1+x)^2/(1-2*x^2))/2,{x,0,mx}],x] (* Herbert Kociemba, Nov 02 2016 *)
    a[0] = 1; a[n_] := (1/4)*(Mod[n, 2] + 3)*2^Quotient[n, 2] + DivisorSum[n, EulerPhi[#]*2^(n/#)&]/(2*n); Array[a, 36, 0] (* Jean-François Alcover, Nov 05 2017 *)
  • PARI
    a(n)=if(n<1,!n,(n%2+3)/4*2^(n\2)+sumdiv(n,d,eulerphi(n/d)*2^d)/2/n)
    
  • Python
    from sympy import divisors, totient
    def a(n):
        return 1 if n<1 else ((2**(n//2+1) if n%2 else 3*2**(n//2-1)) + sum(totient(n//d)*2**d for d in divisors(n))//n)//2
    print([a(n) for n in range(51)]) # Indranil Ghosh, Apr 23 2017

Formula

a(n) = Sum_{d divides n} phi(d)*2^(n/d)/(2*n) + either 2^((n - 1)/2) if n odd or 2^(n/2 - 1) + 2^(n/2 - 2) if n even.
G.f.: (1 - Sum_{n>=1} phi(n)*log(1 - 2*x^n)/n + (1 + x)^2/(1 - 2*x^2))/2. - Herbert Kociemba, Nov 02 2016
Equals (A000031 + A164090) / 2 = A000031 - A059076 = A059076 + A164090. - Robert A. Russell, Sep 24 2018
From Richard L. Ollerton, May 04 2021: (Start)
a(0) = 1; a(n) = Sum_{k=1..n} 2^gcd(n,k)/(2*n) + either 2^((n - 1)/2) if n odd or 2^(n/2 - 1) + 2^(n/2 - 2) if n even.
a(0) = 1; a(n) = A000031(n)/2 + (2^floor((n+1)/2) + 2^ceiling((n+1)/2))/4. See A051137. (End)

Extensions

More terms from Christian G. Bower

A293496 Array read by antidiagonals: T(n,k) = number of chiral pairs of necklaces with n beads using a maximum of k colors.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 4, 3, 0, 0, 0, 0, 10, 15, 12, 1, 0, 0, 0, 20, 45, 72, 38, 2, 0, 0, 0, 35, 105, 252, 270, 117, 6, 0, 0, 0, 56, 210, 672, 1130, 1044, 336, 14, 0, 0, 0, 84, 378, 1512, 3535, 5270, 3795, 976, 30, 0
Offset: 1

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Author

Andrew Howroyd, Oct 10 2017

Keywords

Comments

An orientable necklace when turned over does not leave it unchanged. Only one necklace in each pair is included in the count.
The number of chiral bracelets. An achiral bracelet is the same as its reverse, while a chiral bracelet is equivalent to its reverse. - Robert A. Russell, Sep 28 2018

Examples

			Array begins:
  ==========================================================
  n\k | 1  2    3     4      5       6        7        8
  ----+-----------------------------------------------------
   1  | 0  0    0     0      0       0        0        0 ...
   2  | 0  0    0     0      0       0        0        0 ...
   3  | 0  0    1     4     10      20       35       56 ...
   4  | 0  0    3    15     45     105      210      378 ...
   5  | 0  0   12    72    252     672     1512     3024 ...
   6  | 0  1   38   270   1130    3535     9156    20748 ...
   7  | 0  2  117  1044   5270   19350    57627   147752 ...
   8  | 0  6  336  3795  23520  102795   355656  1039626 ...
   9  | 0 14  976 14060 106960  556010  2233504  7440216 ...
  10  | 0 30 2724 51204 483756 3010098 14091000 53615016 ...
  ...
For T(3,4)=4, the chiral pairs are ABC-ACB, ABD-ADB, ACD-ADC, and BCD-BDC.
For T(4,3)=3, the chiral pairs are AABC-AACB, ABBC-ACBB, and ABCC-ACCB. - _Robert A. Russell_, Sep 28 2018
		

Crossrefs

Programs

  • Mathematica
    b[n_, k_] := (1/n)*DivisorSum[n, EulerPhi[#]*k^(n/#) &];
    c[n_, k_] := If[EvenQ[n], (k^(n/2) + k^(n/2 + 1))/2, k^((n + 1)/2)];
    T[, 1] = T[1, ] = 0; T[n_, k_] := (b[n, k] - c[n, k])/2;
    Table[T[n - k + 1, k], {n, 1, 11}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Oct 11 2017, translated from PARI *)
  • PARI
    \\ here b(n,k) is A075195 and c(n,k) is A284855
    b(n, k) = (1/n) * sumdiv(n, d, eulerphi(d)*k^(n/d));
    c(n, k) = if(n % 2 == 0, (k^(n/2) + k^(n/2+1))/2, k^((n+1)/2));
    T(n, k) = (b(n, k) - c(n, k)) / 2;

Formula

T(n,k) = (A075195(n,k) - A284855(n,k)) / 2.
From Robert A. Russell, Sep 28 2018: (Start)
T(n, k) = -(k^floor((n+1)/2) + k^ceiling((n+1)/2)) / 4 + (1/2n) * Sum_{d|n} phi(d) * k^(n/d)
G.f. for column k: -(kx/4)*(kx+x+2)/(1-kx^2) - Sum_{d>0} phi(d)*log(1-kx^d)/2d. (End)

A305541 Triangle read by rows: T(n,k) is the number of chiral pairs of color loops of length n with exactly k different colors.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 0, 0, 3, 3, 0, 0, 12, 24, 12, 0, 1, 35, 124, 150, 60, 0, 2, 111, 588, 1200, 1080, 360, 0, 6, 318, 2487, 7845, 11970, 8820, 2520, 0, 14, 934, 10240, 46280, 105840, 129360, 80640, 20160, 0, 30, 2634, 40488, 254676, 821592, 1481760, 1512000, 816480, 181440, 0, 62, 7503, 158220, 1344900, 5873760, 14658840, 21772800, 19051200, 9072000, 1814400
Offset: 1

Views

Author

Robert A. Russell, Jun 04 2018

Keywords

Comments

In other words, the number of n-bead bracelets with beads of exactly k different colors that when turned over are different from themselves. - Andrew Howroyd, Sep 13 2019

Examples

			Triangle T(n,k) begins:
  0;
  0,  0;
  0,  0,    1;
  0,  0,    3,     3;
  0,  0,   12,    24,     12;
  0,  1,   35,   124,    150,     60;
  0,  2,  111,   588,   1200,   1080,     360;
  0,  6,  318,  2487,   7845,  11970,    8820,    2520;
  0, 14,  934, 10240,  46280, 105840,  129360,   80640,  20160;
  0, 30, 2634, 40488, 254676, 821592, 1481760, 1512000, 816480, 181440;
  ...
For T(4,3)=3, the chiral pairs are AABC-AACB, ABBC-ACBB, and ABCC-ACCB.
For T(4,4)=3, the chiral pairs are ABCD-ADCB, ABDC-ACDB, and ACBD-ADBC.
		

Crossrefs

Columns 2-6 are A059076, A305542, A305543, A305544, and A305545.
Row sums are A326895.

Programs

  • Mathematica
    Table[(k!/(2n)) DivisorSum[n, EulerPhi[#] StirlingS2[n/#, k] &] - (k!/4) (StirlingS2[Floor[(n+1)/2], k] + StirlingS2[Ceiling[(n+1)/2], k]), {n, 1, 15}, {k, 1, n}] // Flatten
  • PARI
    T(n,k) = {-k!*(stirling((n+1)\2,k,2) + stirling(n\2+1,k,2))/4 + k!*sumdiv(n,d, eulerphi(d)*stirling(n/d,k,2))/(2*n)} \\ Andrew Howroyd, Sep 13 2019

Formula

T(n,k) = -(k!/4)*(S2(floor((n+1)/2),k) + S2(ceiling((n+1)/2),k)) + (k!/(2 n))*Sum_{d|n} phi(d)*S2(n/d,k), where S2(n,k) is the Stirling subset number A008277.
T(n,k) = A087854(n,k) - A273891(n,k).
T(n,k) = (A087854(n,k) - A305540(n,k)) / 2.
T(n, k) = Sum_{i=0..k} (-1)^(k-i)*binomial(k,i)*A293496(n, i). - Andrew Howroyd, Sep 13 2019

A056342 Number of bracelets of length n using exactly two different colored beads.

Original entry on oeis.org

0, 1, 2, 4, 6, 11, 16, 28, 44, 76, 124, 222, 378, 685, 1222, 2248, 4110, 7683, 14308, 27010, 50962, 96907, 184408, 352696, 675186, 1296856, 2493724, 4806076, 9272778, 17920858, 34669600, 67159048, 130216122, 252745366, 490984486, 954637556, 1857545298, 3617214679, 7048675958, 13744694926, 26818405350
Offset: 1

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Author

Keywords

Comments

Turning over will not create a new bracelet.

Examples

			For a(6)=11, the arrangements are AAAAAB, AAAABB, AAABAB, AAABBB, AABAAB, AABBBB, ABABAB, ABABBB, ABBABB, ABBBBB, and AABABB, the last being chiral. Its reverse is AABBAB. - _Robert A. Russell_, Sep 26 2018
		

References

  • M. R. Nester (1999). Mathematical investigations of some plant interaction designs. PhD Thesis. University of Queensland, Brisbane, Australia. [See A056391 for pdf file of Chap. 2]

Crossrefs

Column 2 of A273891.
Equals A052823 - A059076.

Programs

  • Mathematica
    a[n_] := (1/4)*(Mod[n, 2] + 3)*2^Quotient[n, 2] + DivisorSum[n, EulerPhi[#]*2^(n/#)&]/(2*n) - 2; Array[a, 41] (* Jean-François Alcover, Nov 05 2017 *)
    k=2; Table[k! DivisorSum[n, EulerPhi[#] StirlingS2[n/#,k]&]/(2n) + k!(StirlingS2[Floor[(n+1)/2], k] + StirlingS2[Ceiling[(n+1)/2], k])/4, {n,1,30}] (* Robert A. Russell, Sep 26 2018 *)
  • PARI
    a(n) = my(k=2); (k!/4)*(stirling(floor((n+1)/2),k,2) + stirling(ceil((n+1)/2),k,2)) + (k!/(2*n))*sumdiv(n,d,eulerphi(d)*stirling(n/d,k,2)); \\ Michel Marcus, Sep 28 2018

Formula

a(n) = A000029(n) - 2.
From Robert A. Russell, Sep 26 2018: (Start)
a(n) = (A052823(n) + A027383(n-2)) / 2 = A059076(n) + A027383(n-2).
a(n) = (k!/4) * (S2(floor((n+1)/2),k) + S2(ceiling((n+1)/2),k)) + (k!/2n) * Sum_{d|n} phi(d) * S2(n/d,k), where k=2 is the number of colors and S2 is the Stirling subset number A008277.
G.f.: (k!/4) * x^(2k-2) * (1+x)^2 / Product_{i=1..k} (1-i x^2) - Sum_{d>0} (phi(d)/2d) * Sum_{j} (-1)^(k-j) * C(k,j) * log(1-j x^d), where k=2 is the number of colors. (End)

Extensions

More terms from Joerg Arndt, Jun 10 2016

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

A278639 Number of pairs of orientable necklaces with n beads and up to 3 colors; i.e., turning the necklace over does not leave it unchanged. The turned-over necklace is not included in the count.

Original entry on oeis.org

0, 0, 0, 1, 3, 12, 38, 117, 336, 976, 2724, 7689, 21455, 60228, 168714, 475037, 1338861, 3788400, 10742588, 30556305, 87112059, 248967564, 713032782, 2046325125, 5883428618, 16944975048, 48880471500, 141212377489, 408509453511, 1183275193908, 3431504760514
Offset: 0

Views

Author

Herbert Kociemba, Nov 24 2016

Keywords

Comments

Number of chiral bracelets of n beads using up to three different colors.

Examples

			Example: The 3 orientable necklaces with 4 beads and the colors A, B and C are AABC, BBAC and CCAB. The turned-over necklaces AACB, BBCA and CCBA are not included in the count.
For n=6, the three chiral pairs using just two colors are AABABB-AABBAB, AACACC-AACCAC, and BBCBCC-BBCCBC.  The other 35 use three colors. - _Robert A. Russell_, Sep 24 2018
		

Crossrefs

Column 3 of A293496.
Cf. A059076 (2 colors).
a(n) = (A001867(n) - A182751(n-1)) / 2.
Equals A001867 - A027671.
a(n) = A027671(n) - A182751(n-1).

Programs

  • Mathematica
    mx=40;f[x_,k_]:=(1-Sum[EulerPhi[n]*Log[1-k*x^n]/n,{n,1,mx}]-Sum[Binomial[k,i]*x^i,{i,0,2}]/(1-k*x^2))/2;CoefficientList[Series[f[x,3],{x,0,mx}],x]
    k=3; Prepend[Table[DivisorSum[n, EulerPhi[#] k^(n/#) &]/(2n) -(k^Floor[(n+1)/2] + k^Ceiling[(n+1)/2])/4, {n, 1, 30}], 0] (* Robert A. Russell, Sep 24 2018 *)

Formula

G.f.: k=3, (1 - Sum_{n>=1} phi(n)*log(1 - k*x^n)/n - Sum_{i=0..2} binomial(k,i)*x^i / (1 - k*x^2))/2.
For n > 0, a(n) = -(k^floor((n+1)/2) + k^ceiling((n+1)/2))/4 + (1/2n)* Sum_{d|n} phi(d)*k^(n/d), where k=3 is the maximum number of colors. - Robert A. Russell, Sep 24 2018

A278640 Number of pairs of orientable necklaces with n beads and up to 4 colors; i.e., turning the necklace over does not leave it unchanged. The turned-over necklace is not included in the count.

Original entry on oeis.org

0, 0, 0, 4, 15, 72, 270, 1044, 3795, 14060, 51204, 188604, 694130, 2572920, 9567090, 35758704, 134137875, 505159200, 1908554190, 7233104844, 27486506268, 104713296760, 399817262550, 1529746919604, 5864041395730, 22517964582504, 86607602546220, 333599838189804, 1286742419927070, 4969488707124120, 19215357085867800
Offset: 0

Views

Author

Herbert Kociemba, Nov 24 2016

Keywords

Comments

Number of chiral bracelets of n beads using up to four different colors.

Examples

			Example: For 3 beads and the colors A, B, C and D the 4 orientable necklaces are ABC, ABD, ACD and BCD. The turned-over necklaces ACB, ADB, ADC and BDC are not included in the count.
		

Crossrefs

Column 4 of A293496.
Cf. A059076 (2 colors), A278639 (3 colors).
Equals (A001868 - A056486) / 2 = A001868 - A032275 = A032275 - A056486.

Programs

  • Mathematica
    mx=40;f[x_,k_]:=(1-Sum[EulerPhi[n]*Log[1-k*x^n]/n,{n,1,mx}]-Sum[Binomial[k,i]*x^i,{i,0,2}]/(1-k*x^2))/2;CoefficientList[Series[f[x,4],{x,0,mx}],x]
    k=4; Prepend[Table[DivisorSum[n, EulerPhi[#] k^(n/#) &]/(2n) - (k^Floor[(n+1)/2] + k^Ceiling[(n+1)/2])/4, {n, 1, 30}], 0] (* Robert A. Russell, Sep 24 2018 *)

Formula

G.f.: k=4, (1 - Sum_{n>=1} phi(n)*log(1 - k*x^n)/n - Sum_{i=0..2} Binomial[k,i]*x^i / ( 1-k*x^2) )/2.
For n > 0, a(n) = -(k^floor((n+1)/2) + k^ceiling((n+1)/2))/4 + (1/2n)* Sum_{d|n} phi(d)*k^(n/d), where k=4 is the maximum number of colors. - Robert A. Russell, Sep 24 2018

A278641 Number of pairs of orientable necklaces with n beads and up to 5 colors; i.e., turning the necklace over does not leave it unchanged. The turned-over necklace is not included in the count.

Original entry on oeis.org

0, 0, 0, 10, 45, 252, 1130, 5270, 23520, 106960, 483756, 2211650, 10149805, 46911060, 217868310, 1017057518, 4767797895, 22438419120, 105960938380, 501928967930, 2384171386941, 11353241261180, 54185968572450, 259150507387910, 1241763071712930, 5960463867187752, 28656077411358180, 137973711706163210
Offset: 0

Views

Author

Herbert Kociemba, Nov 24 2016

Keywords

Comments

Number of chiral bracelets of n beads using up to five different colors.

Crossrefs

Column 5 of A293496.
Cf. A059076 (2 colors), A278639 (3 colors), A278640 (4 colors).
a(n) = (A001869(n) - A056487(n+1)) / 2 = A032276(n) - A056487(n+1).
Equals A001869 - A032276.

Programs

  • Mathematica
    mx=40;f[x_,k_]:=(1-Sum[EulerPhi[n]*Log[1-k*x^n]/n,{n,1,mx}]-Sum[Binomial[k,i]*x^i,{i,0,2}]/(1-k*x^2))/2;CoefficientList[Series[f[x,5],{x,0,mx}],x]
    k=5; Prepend[Table[DivisorSum[n, EulerPhi[#] k^(n/#) &]/(2n) - (k^Floor[(n+1)/2] + k^Ceiling[(n+1)/2])/4, {n, 1, 30}], 0] (* Robert A. Russell, Sep 24 2018 *)

Formula

G.f.: k=5, (1 - Sum_{n>=1} phi(n)*log(1 - k*x^n)/n - Sum_{i=0..2} Binomial[k,i]*x^i / ( 1-k*x^2) )/2.
For n>0, a(n) = -(k^floor((n+1)/2) + k^ceiling((n+1)/2))/4 + (1/2n)* Sum_{d|n} phi(d)*k^(n/d), where k=5 is the maximum number of colors. - Robert A. Russell, Sep 24 2018

A278642 Number of pairs of orientable necklaces with n beads and up to 6 colors; i.e., turning the necklace over does not leave it unchanged. The turned-over necklace is not included in the count.

Original entry on oeis.org

0, 0, 0, 20, 105, 672, 3535, 19350, 102795, 556010, 3010098, 16467450, 90619690, 502194420, 2798240265, 15671993560, 88156797855, 497837886000, 2821092554035, 16035752398770, 91403856697944, 522308167195260, 2991401733402075, 17168047238861070, 98716274117752900, 568605754068247644, 3280417827002225910, 18953525314104758810
Offset: 0

Views

Author

Herbert Kociemba, Nov 24 2016

Keywords

Comments

Number of chiral bracelets of n beads using up to six different colors.

Crossrefs

Column 6 of A293496.
Cf. A059076 (2 colors), A278639 (3 colors), A278640 (4 colors), A278641 (5 colors).

Programs

  • Mathematica
    mx = 40; f[x_, k_] := (1 - Sum[EulerPhi[n] * Log[1 - k * x^n]/n,{n, mx}] - Sum[Binomial[k, i] * x^i, {i, 0, 2}]/(1 - k * x^2))/2; CoefficientList[Series[f[x, 6], {x, 0, mx}], x]
    k = 6; Prepend[Table[DivisorSum[n, EulerPhi[#] k^(n/#) &]/(2n) - (k^Floor[(n + 1)/2] + k^Ceiling[(n + 1)/2])/4, {n, 30}], 0] (* Robert A. Russell, Sep 24 2018 *)

Formula

Equals (A054625(n) - A056488(n)) / 2 = A054625(n) - A056341(n) = A056341(n) - A056488(n), for n >= 1.
G.f.: k = 6, (1 - Sum_{n >= 1} phi(n)*log(1 - k*x^n)/n - Sum_{i = 0..2} Binomial[k, i]*x^i / ( 1 - k*x^2) )/2.
For n > 0, a(n) = -(k^floor((n+1)/2) + k^ceiling((n+1)/2))/4 + (1/2n)* Sum_{d|n} phi(d)*k^(n/d), where k = 6 is the maximum number of colors. - Robert A. Russell, Sep 24 2018

A059078 Number of orientable necklaces with 2n beads and two colors which when turned over produce their own color complement.

Original entry on oeis.org

0, 0, 0, 1, 2, 6, 12, 27, 54, 113, 228, 465, 934, 1890, 3798, 7644, 15350, 30840, 61878, 124173, 249008, 499318, 1000866, 2005971, 4019446, 8053062, 16131780, 32311665, 64711820, 129589530, 259487040, 519552495, 1040186358, 2082408354
Offset: 0

Views

Author

Henry Bottomley, Dec 22 2000

Keywords

Comments

Clearly in each necklace the number of beads of each of the two colors must be equal and so the number of beads must be even, if a(n) is to be positive.

Programs

  • Mathematica
    a13[n_] := DivisorSum[n, EulerPhi@(2*#)*2^(n/#)&]/(2*n);
    a29[n_] := (1/4)*(Mod[n, 2] + 3)*2^Quotient[n, 2] + DivisorSum[n, EulerPhi[#]*2^(n/#)&]/(2*n);
    a[0] = 0; a[n_] := a29[2*n] - a13[2*n] - 2^(n - 1);
    Array[a, 34, 0] (* Jean-François Alcover, Nov 05 2017 *)

Formula

a(n) = A059076(2*n) - 2*A059053(2*n).
a(n) = A000029(2*n) - A000013(2*n) - A000079(n-1).

Extensions

More terms from Vladeta Jovovic, Mar 06 2001
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