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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.
1
0, 6, 2400, 6375600, 45927907200, 713518388352000, 21216194909362252800, 1105729617210350356224000, 94398452626533646953922560000, 12514511465855205467497303154688000, 2467490887755897725667792936979169280000, 698323914872709997998407130752506728284160000
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.
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
KEYWORD
nonn
AUTHOR
Hugo Pfoertner, Nov 25 2015
STATUS
approved