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%I #17 Dec 27 2021 08:26:38
%S 1,1,1,2,1,1,3,2,1,3,1,2,3,1,3,4,3,1,3,2,3,5,1,2,5,1,3,4,1,1,7,6,1,3,
%T 1,4,5,3,1,4,1,7,3,4,5,7,3,2,7,1,1,8,1,3,3,4,3,7,5,2,5,3,9,10,1,5,7,2,
%U 1,3,3,6,5,1,5,8,7,3,3,4,1,9,1,2,11
%N Number of cycles and fixed points in the permutation (n, n-2, n-4, ..., 1, ..., n-3, n-1).
%C The above permutation (see A130517) can be generated by taking S_n: (1, 2, ..., n) and reversing the first two, first three and so on till first n, elements in sequence. Interestingly this permutation orbit has length given by A003558.
%H T. D. Noe, <a href="/A141110/b141110.txt">Table of n, a(n) for n = 1..10000</a>
%e a(20) = 2, since (20, 18, 16, 14, 12, 10, 8, 6, 4, 2, 1, 3, 5, 7, 9, 11, 13, 15, 17, 19) has two cycles (1, 20, 19, 17, 13, 5, 12, 3, 16, 11) and (2, 18, 15, 9, 4, 14, 7, 8, 6, 10).
%o (Python)
%o from sympy.combinatorics import Permutation
%o def a(n):
%o p = list(range(n, 0, -2)) + list(range(1+(n%2), n, 2))
%o return Permutation([pi-1 for pi in p]).cycles
%o print([a(n) for n in range(1, 86)]) # _Michael S. Branicky_, Dec 27 2021
%Y Cf. A130517 (permutations), A003558 (order).
%K easy,nonn
%O 1,4
%A _Ramasamy Chandramouli_, Jun 05 2008
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