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.

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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

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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

A056346 Number of bracelets of length n using exactly six different colored beads.

Original entry on oeis.org

0, 0, 0, 0, 0, 60, 1080, 11970, 105840, 821952, 5874480, 39713550, 258136200, 1631273220, 10096734312, 61536377700, 370710950400, 2213749658880, 13132080672480, 77509456944318, 455754569692680
Offset: 1

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Author

Keywords

Comments

Turning over will not create a new bracelet.

Examples

			For a(6)=60, pair up the 120 permutations of BCDEF, each with its reverse, such as BCDEF-FEDCB.  Precede the first of each pair with an A, such as ABCDEF.  These are the 60 arrangements, all chiral.  If we precede the second of each pair with an A, such as AFEDCB, we get the chiral partner of each. - _Robert A. Russell_, Sep 27 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 6 of A273891.
Equals (A056286 + A056492) / 2 = A056286 - A305545 = A305545 + A056492.
Cf. A008277.

Programs

  • Mathematica
    t[n_, k_] := (For[t1 = 0; d = 1, d <= n, d++, If[Mod[n, d] == 0, t1 = t1 + EulerPhi[d]*k^(n/d)]]; If[EvenQ[n], (t1 + (n/2)*(1 + k)*k^(n/2))/(2*n), (t1 + n*k^((n + 1)/2))/(2*n)]);
    T[n_, k_] := Sum[(-1)^i*Binomial[k, i]*t[n, k - i], {i, 0, k - 1}];
    a[n_] := T[n, 6];
    Array[a, 21] (* Jean-François Alcover, Nov 05 2017, after Andrew Howroyd *)
    k=6; 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 27 2018 *)
  • PARI
    a(n) = my(k=6); (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 29 2018

Formula

a(n) = A056341(n) - 6*A032276(n) + 15*A032275(n) - 20*A027671(n) + 15*A000029(n) - 6.
From Robert A. Russell, Sep 27 2018: (Start)
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=6 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=6 is the number of colors. (End)
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