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.
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 (conjectured).
LINKS
Witold Tatkiewicz, Rows n = 0..17 of triangle, flattened
Witold Tatkiewicz, link for java program
FORMULA
T(n,n) = n for n>2.
T(n,n-1) = 0 for n>1.
T(n,n-3) = 1/2*n^3 + 3/4*n^2 - 2 (conjectured);
T(n,n-3) = (2/3)*n^4 + 3*n^3 + (1/3)*n^2 - 7*n + 3 for n > 4 (conjectured);
T(n,n-4) = (25/24)*n^5 + (73/12)*n^4 + (5/8)*n^3 - (253/12)*n^2 + (76/3)*n - 12 for n > 5 (conjectured);
T(n,n-5) = (26/15)*n^6 + (77/6)*n^5 + 7*n^4 - (97/3)*n^3 + (2314/15)*n^2 - 273/2*n + 65 for n > 5 (conjectured);
T(n,n-6) = (707/240)*n^7 + (2093/80)*n^6 + (2009/80)*n^5 - (245/16)*n^4 + (78269/120)*n^3 - (18477/20)*n^2 + (10647/0)*n - 342 for n > 6 (conjectured).
EXAMPLE
Assuming the initial order was {1,2,3,4,5} (therefore 1 and 5 form a 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) but not reflection ({1,2,3,5,4} and {4,5,3,2,1} are counted as one sequence) the total number of ways is 5*5 and therefore T(5,3)=25.
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 3
4 0 0 8 0 4
5 5 0 25 25 0 5
6 18 72 90 120 54 0 6
7 161 490 784 637 343 98 0 7
PROG
(Java) See Links section.
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Witold Tatkiewicz, Aug 01 2019
STATUS
approved