cp's OEIS Frontend

A277080 Irregular triangle read by rows: T(n,k) = number of size k subsets of S_n that remain unchanged by reverse.

%I A277080 #32 Oct 03 2016 16:05:53
%S A277080 1,1,1,1,1,0,1,1,0,3,0,3,0,1,1,0,12,0,66,0,220,0,495,0,792,0,924,0,
%T A277080 792,0,495,0,220,0,66,0,12,0,1,1,0,60,0,1770,0,34220,0,487635,0,
%U A277080 5461512,0,50063860,0,386206920,0,2558620845,0,14783142660,0,75394027566
%N A277080 Irregular triangle read by rows: T(n,k) = number of size k subsets of S_n that remain unchanged by reverse.
%C A277080 The reverse of a permutation is the reverse in one line notation. For example the reverse of 43521 is 12534.
%C A277080 T(n,k) is the number of size k bases of S_n which remain unchanged by reverse.
%F A277080 T(n,k) = C(n!/2, k/2) if k is even and T(n,k) = 0 if k is odd.
%e A277080 For n = 4 and k = 2 the subsets that remain unchanged by reverse are {4321, 1234}, {1243, 3421}, {4231, 1324}, {1342, 2431}, {1423, 3241}, {1432, 2341}, {2134, 4312}, {3412, 2143}, {2314, 4132}, {3142, 2413}, {4213, 3124} and {4123, 3214} so T(4,2) = 12.
%e A277080 For n = 3 and k = 4 the subsets that remain unchanged by reverse are {231, 321, 132, 123}, {321, 213, 312, 123} and {231, 132, 312, 213} so T(3,4) = 3.
%e A277080 The triangle starts:
%e A277080 1, 1;
%e A277080 1, 1;
%e A277080 1, 0, 1;
%e A277080 1, 0, 3, 0, 3, 0, 1;
%o A277080 (Sage) def T(n,k):
%o A277080     if k % 2 == 1:
%o A277080         return 0
%o A277080     return binomial( factorial(n)/2, k/2 )
%Y A277080 Row lengths give A038507.
%K A277080 nonn,tabf
%O A277080 0,10
%A A277080 _Christian Bean_, Sep 28 2016