A264801 - OEIS

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.

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

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,2

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.

LINKS

Table of n, a(n) for n=1..12.

FORMULA

a(n) = (2*n-1)! * A000183(2*n).

EXAMPLE

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.

PROG

(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

CROSSREFS