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%I #31 Feb 03 2021 21:46:53
%S 1,1,1,1,1,2,1,1,4,7,8,7,4,1,1,10,52,190,546,1302,2660,4754,7535,
%T 10692,13672,15820,16604,15820,13672,10692,7535,4754,2660,1302,546,
%U 190,52,10,1,1,26,372,3822,31306,216086,1300420,6981650,33992275,151945820
%N Irregular triangle read by rows: T(n,k) = number of size k subsets of S_n that remain unchanged under the operation of replacing a permutation with its inverse.
%C T(n,k) is the number of size k subsets of S_n that remain unchanged under the operation of replacing a permutation with its inverse.
%H Andrew Howroyd, <a href="/A277081/b277081.txt">Table of n, a(n) for n = 0..880</a> (rows 0..6)
%F T(n,k) = Sum( C((n! - I(n))/2, i)*C(I(n), k - 2*i) for i in [0..floor(k/2)]) where I(n) = A000085(n).
%e For n = 3 and k = 3 the subsets unchanged by inverse are {213,132,123}, {321,132,123}, {321,213,123}, {231,312,123}, {321,132,213}, {132,312,231},{213,312,231}, {321,231,312} hence T(3,3) = 8. (Here we are using the one-line notation for permutations, not the product of cycles form.)
%e Triangle starts:
%e 1, 1;
%e 1, 1;
%e 1, 2, 1;
%e 1, 4, 7, 8, 7, 4, 1;
%o (PARI) \\ here b(n) is A000085(n).
%o b(n)={sum(k=0, n\2, n!/((n-2*k)!*2^k*k!))}
%o Row(n)={my(t=b(n)); vector(n!+1, k, k--; sum(i=0, k\2, binomial((n!-t)/2, i)*binomial(t, k-2*i)))}
%o { for(n=0, 4, print(Row(n))) } \\ _Andrew Howroyd_, Feb 03 2021
%Y Row lengths give A038507.
%Y Cf. A000085.
%K nonn,tabf
%O 0,6
%A _Christian Bean_, Sep 28 2016
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