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%I #22 Oct 05 2016 15:20:21
%S 1,1,1,1,1,2,1,1,2,3,4,3,2,1,1,8,36,120,322,728,1428,2472,3823,5328,
%T 6728,7728,8092,7728,6728,5328,3823,2472,1428,728,322,120,36,8,1,1,8,
%U 84,504,3178,15512,74788,311144,1252819,4577328,16087512,52691408,165911284
%N Irregular triangle read by rows: T(n,k) = number of size k subsets of S_n that remain unchanged by a rotation of 180 degrees.
%C A permutation, p, can be thought of as a set of points (i, p(i)). If you plot all the points and rotate the picture by 180 degrees then you get a permutation back.
%C T(n,k) is the number of size k subsets of S_n that remain unchanged by a rotation of 180 degrees.
%F T(n,k) = Sum_( binomial( n! - R(n), i ) * binomial( R(n), k-2*i ) for i in [0..floor(k/2)] ) where R(n) = A037223(n).
%e For n = 3 and k = 3, the subsets unchanged by rotating 180 degrees are {213,132,123}, {231,312,123}, {321,132,213} and {321,231,312} so T(3,3) = 4.
%e Triangle starts:
%e 1, 1;
%e 1, 1;
%e 1, 2, 1;
%e 1, 2, 3, 4, 3, 2, 1;
%Y Row lengths give A038507.
%Y Cf. A037223.
%K nonn,tabf
%O 0,6
%A _Christian Bean_, Sep 28 2016
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