A326390 The number of ways of seating n people around a table for the second time so that k pairs are maintained. T(n,k) read by rows.
1, 0, 1, 0, 0, 2, 0, 0, 0, 6, 0, 0, 16, 0, 8, 10, 0, 50, 50, 0, 10, 36, 144, 180, 240, 108, 0, 12, 322, 980, 1568, 1274, 686, 196, 0, 14, 2832, 8704, 11840, 10240, 4832, 1536, 320, 0, 16, 27954, 81000, 108054, 85050, 43902, 13446, 2970, 486, 0, 18, 299260, 834800, 1071700, 828400, 416200, 141520, 31000, 5200, 700, 0, 20
Offset: 0
Examples
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
Links
- Witold Tatkiewicz, Rows n = 0..17 of triangle, flattened
- Witold Tatkiewicz, Link for Java program
Crossrefs
Programs
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Java
See Links section
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).
Comments