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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. 4

%I #52 Aug 06 2019 21:50:17

%S 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,

%T 12,322,980,1568,1274,686,196,0,14,2832,8704,11840,10240,4832,1536,

%U 320,0,16,27954,81000,108054,85050,43902,13446,2970,486,0,18,299260,834800,1071700,828400,416200,141520,31000,5200,700,0,20

%N 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.

%C 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.

%C Sum of each row is equal to n!.

%C 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).

%H Witold Tatkiewicz, <a href="/A326390/b326390.txt">Rows n = 0..17 of triangle, flattened</a>

%H Witold Tatkiewicz, <a href="https://pastebin.com/q8ZxekrZ">Link for Java program</a>

%F T(n,n) = 2*n for n > 2;

%F T(n,n-1) = 0 for n > 1;

%F T(n,n-2) = n^2*(n-3) for n > 3 (conjectured);

%F T(n,n-3) = (3/4)*n^4 + 6*n^3 + (2/3)*n^2 - 14*n + 6 for n > 4 (conjectured);

%F 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);

%F 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);

%F 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).

%e 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.

%e Unfolded table with n individuals (rows) forming k pairs (columns):

%e 0 1 2 3 4 5 6 7

%e 0 1

%e 1 0 1

%e 2 0 0 2

%e 3 0 0 0 6

%e 4 0 0 16 0 8

%e 5 10 0 50 50 0 10

%e 6 36 144 180 240 108 0 12

%e 7 322 980 1568 1274 686 196 0 14

%o (Java) See Links section

%Y Cf. A089222 (column k=0).

%Y Cf. A000142 sum of each row.

%Y Cf. A326397 (disregards reflection symmetry), A326404 (disregards circular symmetry), A326411 (disregards both circular and reflection symmetry).

%K nonn,tabl

%O 0,6

%A _Witold Tatkiewicz_, Jul 03 2019

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