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%I #13 Sep 24 2019 09:18:17
%S 1,0,1,0,0,1,0,0,0,3,0,0,8,0,4,5,0,25,25,0,5,18,72,90,120,54,0,6,161,
%T 490,784,637,343,98,0,7,1416,4352,5920,5120,2416,768,160,0,8,13977,
%U 40500,54027,42525,21951,6723,1485,243,0,9,149630,417400,535850,414200,208100,70760,15500,2600,350,0,10,1737241,4691654
%N Triangle T(n,k) read by rows: T(n,k) = number of ways of seating n people around a table for the second time so that k pairs are maintained. Reflected sequences are counted as one.
%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 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).
%H Witold Tatkiewicz, <a href="/A326397/b326397.txt">Rows n = 0..17 of triangle, flattened</a>
%H Witold Tatkiewicz, <a href="https://pastebin.com/VGC8SBUq">link for java program</a>
%F T(n,n) = n for n>2.
%F T(n,n-1) = 0 for n>1.
%F T(n,n-3) = 1/2*n^3 + 3/4*n^2 - 2 (conjectured);
%F T(n,n-3) = (2/3)*n^4 + 3*n^3 + (1/3)*n^2 - 7*n + 3 for n > 4 (conjectured);
%F 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);
%F 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);
%F 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).
%e 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.
%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 1
%e 3 0 0 0 3
%e 4 0 0 8 0 4
%e 5 5 0 25 25 0 5
%e 6 18 72 90 120 54 0 6
%e 7 161 490 784 637 343 98 0 7
%o (Java) See Links section.
%Y Cf. A001710 sum of each row.
%Y Cf. A326390 (with reflection symmetry), A326404 (with reflection symmetry, but disregards circular symmetry), A326411 (disregards both circular and reflection symmetry).
%K nonn,tabl
%O 0,10
%A _Witold Tatkiewicz_, Aug 01 2019
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