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 1-3 of 3 results.

A059375 Number of seating arrangements for the ménage problem.

Original entry on oeis.org

1, 0, 0, 12, 96, 3120, 115200, 5836320, 382072320, 31488549120, 3191834419200, 390445460697600, 56729732529254400, 9659308746908620800, 1905270127543015833600, 431026303509734220288000, 110865322076320374571008000, 32172949121885378686623744000
Offset: 0

Views

Author

N. J. A. Sloane, Jan 28 2001

Keywords

Comments

The "probleme des menages" asks for the number of gender-alternating seating arrangements for n couples around a circular table with the condition that no two spouses are seated adjacently. - Paul C. Kainen and Michael Somos, Mar 11 2011

Examples

			a(3) = 12 because there is a unique seating arrangement up to circular and clockwise / counterclockwise symmetry. - _Paul C. Kainen_ and _Michael Somos_, Mar 11 2011

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 184, mu*(n).
  • H. J. Ryser, Combinatorial Mathematics. Mathematical Association of America, Carus Mathematical Monograph 14, 1963, p. 32. equation (2.3).

Links

Crossrefs

Cf. A000179, A258338, A258664, A258665, A258666, A258667, A258673, A259212.

Programs

  • Mathematica
    a[0] = 1; a[1] = 0; a[n_] := 4n n! Sum[(-1)^k Binomial[2n-k, k] (n-k)! / (2n-k), {k, 0, n}]; Table[a[n], {n, 0, 17}] (* Jean-François Alcover, Jun 19 2017, from 1st formula *)
  • PARI
    {a(n) = local(A); if( n<3, n==0, A = vector(n); A[3] = 1; for(k=4, n, A[k] = (k * (k - 2) * A[k-1] + k * A[k-2] - 4 * (-1)^k) / (k-2)); 2 * n! * A[n])} /* Michael Somos, Mar 11 2011 */

Formula

a(n) = A000179(n) * 2 * n!.
a(n) = A094047(n) * 2 * n.

A114939 Number of essentially different seating arrangements for n couples around a circular table with 2n seats avoiding spouses being neighbors and avoiding clusters of 3 persons with equal gender.

Original entry on oeis.org

0, 1, 7, 216, 10956, 803400, 83003040, 11579823360, 2080493573760, 469031859192960, 129727461014726400, 43176116371928601600, 17025803126147196057600, 7850538273249476117913600
Offset: 1

Views

Author

Hugo Pfoertner, Jan 08 2006

Keywords

Comments

Arrangements that differ only by rotation or reflection are excluded by the following conditions: Seat number 1 is assigned to person (a). Person (a)'s spouse (A) can only take seats with numbers <=(n+1). If (A) gets seat n+1 (i.e. sits exactly opposite to her/his spouse) then person (B) can only take seats with numbers <= n.

Examples

			a(2)=1 because the only valid arrangement is aBAb.
a(3)=7 because the only valid arrangements under the given conditions are: abAcBC, aBAcbC, aBcAbC, aBcACb, acAbCB, acBAbC, aCAbcB.

Links

  • M. A. Alekseyev, Weighted de Bruijn Graphs for the Menage Problem and Its Generalizations. Lecture Notes in Computer Science 9843 (2016), 151-162. doi:10.1007/978-3-319-44543-4_12; arXiv:1510.07926 [math.CO], 2015-2016.

Crossrefs

Cf. A114938, A137729, A137730, A137737, A137749, A258338.

Programs

  • Mathematica
    a[1] = 0;
a[n_] := (n-1)!/4 Sum[(-1)^j(n-j)! SeriesCoefficient[ SeriesCoefficient[Tr[ MatrixPower[{{0, 1, 0, y^2, 0, 0}, {z y^2, 0, 1, 0, y^2, 0}, {z y^2, 0, 0, 0, y^2, 0}, {0, 1, 0, 0, 0, z}, {0, 1, 0, y^2, 0, z}, {0, 0, 1, 0, y^2, 0}}, 2n]], {y, 0, 2n}] , {z, 0, j}], {j, 0, n}];
Array[a, 14] (* Jean-François Alcover, Dec 03 2018, from PARI *)
  • PARI
    { a(n) = if(n<=1, 0, (-1)^n*(n-1)!*2^(n-1) + n! * polcoeff( polcoeff( [0, 2*y*z^3 + z^2, -3*y*z^5 - 4*z^4 + ((-2*y^2 - 1)/y)*z^3, 6*y*z^7 + (4*y^2 + 11)*z^6 + ((8*y^2 + 4)/y)*z^5 + 3*z^4] * sum(j=0,n-1, j! * [0, 0, 0, -z^6 + z^4; 1, 0, 0, ((y^2 + 1)/y)*z^5 - 2*z^4 + ((-y^2 - 1)/y)*z^3; 0, 1, 0, ((2*y^2 + 2)/y)*z^3 + z^2; 0, 0, 1, -2*z^2]^(n+j) ) * [1,0,0,0]~, 2*n,z), 0,y) / 2 ); }

Formula

See Alekseyev (2016) and the PARI code for the formula.
a(n) = A258338(n) / (4*n).

Extensions

a(4)-a(7) corrected, formula and further term provided by Max Alekseyev, Feb 15 2008

A264801 Number of essentially different seating arrangements for 2n couples around a circular table with 4n seats such that no spouses are neighbors, the neighbors of each person have opposite gender and no person's neighbors belong to the same couple.

Original entry on oeis.org

0, 6, 2400, 6375600, 45927907200, 713518388352000, 21216194909362252800, 1105729617210350356224000, 94398452626533646953922560000, 12514511465855205467497303154688000, 2467490887755897725667792936979169280000, 698323914872709997998407130752506728284160000
Offset: 1

Views

Author

Hugo Pfoertner, Nov 25 2015

Keywords

Comments

This might be called the "maximum diversity" menage problem. Arrangements that differ only by rotation or reflection are excluded by the following conditions: Seat number 1 is assigned to person A. Seat number 2 can only be taken by a person of the same gender as A. The second condition forces an mmffmmff... pattern.

Examples

			a(1)=0 because with 2 couples it is impossible to satisfy all three conditions.
a(2)=6 because only the following arrangements are possible with 4 couples: ABdaCDbc, ABcaDCbd, ACdaBDcb, ACbaDBcd, ADcaBCdb, ADbaCBdc. This corresponds to the (2*2-1)! possibilities for persons B, C and D to choose a seat. After the positions of A, B, C and D are fixed, only A000183(2*2)=1 possibility remains to arrange their spouses a, b, c  and d.

Crossrefs

Cf. A000183, A007060, A094047, A114939, A258338.

Programs

  • PARI
    a000183(N)={my(a0=[0,0,0,1,2,20],a=vector(N),
f(x)=fibonacci(x-1)+fibonacci(x+1)+2;);
if(N<7,a=a0[1..N],for(k=1,6,a[k]=a0[k]);
for(n=7,N,a[n] = (-1)^n*(4*n+f(n)) +
 (n/(n-1))*((n+1)*a[n-1] + 2*(-1)^n*f(n-1))
  - ((2*n)/(n-2))*((n-3)*a[n-2] + (-1)^n*f(n-2))
  + (n/(n-3))*((n-5)*a[n-3] + 2*(-1)^(n-1)*f(n-3))
  + (n/(n-4))*(a[n-4] + (-1)^(n-1)*f(n-4))));a};
a264901(limit)={my(a183=a000183(2*limit)); for(n=1,limit,print1((2*n-1)!*a183[2*n],", "))};
a264901(12) \\ Hugo Pfoertner, Sep 05 2020

Formula

a(n) = (2*n-1)! * A000183(2*n).
Showing 1-3 of 3 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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