OFFSET
0,6
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 each row is equal to n!.
Weighted average of each row using k as weights converges to 2 for large n and is given with following formula: (Sum_{k} T(n,k)*k)/n! = 2/(n-1) + 2 (conjectured).
LINKS
Witold Tatkiewicz, Rows n = 0..17 of triangle, flattened
Witold Tatkiewicz, Link for Java program
FORMULA
T(n,n) = 2*n for n > 2;
T(n,n-1) = 0 for n > 1;
T(n,n-2) = n^2*(n-3) for n > 3 (conjectured);
T(n,n-3) = (3/4)*n^4 + 6*n^3 + (2/3)*n^2 - 14*n + 6 for n > 4 (conjectured);
T(n,n-4) = (25/12)*n^5 + (73/6)*n^4 + (5/4)*n^3 - (253/6)*n^2 + (152/3)*n - 24 for n > 5 (conjectured);
T(n,n-5) = (52/15)*n^6 + (77/3)*n^5 + 14*n^4 - (194/3)*n^3 + (4628/15)*n^2 - 273*n + 130 for n > 5 (conjectured);
T(n,n-6) = (707/120)*n^7 + (2093/40)*n^6 + (2009/40)*n^5 - (245/8)*n^4 + (78269/60)*n^3 - (18477/10)*n^2 + (21294/10)*n - 684 for n > 6 (conjectured).
EXAMPLE
Assuming initial order was {1,2,3,4,5} (therefore 1 and 5 forms pair as first and last person are neighbors in case of 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 allow for rotation ({1,2,3,5,4} and {2,3,5,4,1} are different) and reflection ({1,2,3,5,4} and {4,5,3,2,1} are also different) the total number of ways is 5*2*5 and therefore T(5,3)=50.
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 2
3 0 0 0 6
4 0 0 16 0 8
5 10 0 50 50 0 10
6 36 144 180 240 108 0 12
7 322 980 1568 1274 686 196 0 14
PROG
(Java) See Links section
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Witold Tatkiewicz, Jul 03 2019
STATUS
approved