OFFSET
0,10
COMMENTS
Definition requires "pairs" and for n=0 it is assumed that there is 1 way of seating 0 people around a table for the second time so that 0 pairs are maintained and 1 person forms only one pair with him/herself. Therefore T(0,0)=1, T(1,0)=0 and T(1,1)=1.
Sum of row n is equal to (n-1)! for n > 1.
Conjecture: The weighted average of each row using k as weights converges to 2 for large n and is given by the following formula: (Sum_{k} T(n,k)*k)/(Sum_{k} T(n,k)) = 2/(n-1) + 2.
LINKS
Witold Tatkiewicz, Rows n = 0..17 of triangle, flattened
Witold Tatkiewicz, Link for Java program.
FORMULA
T(n,n) = 2 for n > 2;
T(n,n-1) = 0 for n > 1.
Conjectures:
T(n,n-2) = n^2 + n - 2 for n > 3;
T(n,n-3) = (4/3)*n^3 + 2*n^2 - (16/3)*n + 2 for n > 4;
T(n,n-4) = (25/12)*n^4 + (23/6)*n^3 - (169/12)*n^2 + (85/6)*n - 6 for n > 5;
T(n,n-5) = (52/15)*n^5 + (25/3)*n^4 - (83/3)*n^3 + (221/3)*n^2 - (299/5)*n + 26 for n > 5;
T(n,n-6) = (707/120)*n^6 + (2037/120)*n^5 - (413/8)*n^4 + (2233/8)*n^3 - (5554/15)*n^2 + (3739/10)*n - 114 for n > 6.
EXAMPLE
Assuming the initial order was {1,2,3,4,5} (therefore 1 and 5 form a pair as the first and last persons are neighbors in the case of a round table) there are 5 sets of ways of seating them again so that 3 pairs are conserved: {1,2,3,5,4}, {2,3,4,1,5}, {3,4,5,2,1}, {4,5,1,3,2}, {5,1,2,4,3}. Since within each set we do not allow for circular symmetry (e.g., {1,2,3,5,4} and its rotation to form {2,3,5,4,1} are counted as one) but we allow reflection ({1,2,3,5,4} and {4,5,3,2,1} are considered distinct), the total number of ways is 5*2 and therefore T(5,3)=10.
Unfolded table with n individuals (rows) forming k pairs (columns):
0 1 2 3 4 5 6 7
0 1
1 0 1
2 0 0 1
3 0 0 0 2
4 0 0 4 0 2
5 2 0 10 10 0 2
6 6 24 30 40 18 0 2
7 46 140 224 182 98 28 0 2
PROG
(Java) See Links section
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Witold Tatkiewicz, Aug 01 2019
STATUS
approved