A141110 Number of cycles and fixed points in the permutation (n, n-2, n-4, ..., 1, ..., n-3, n-1).

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, 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, 1, 3, 3, 6, 5, 1, 5, 8, 7, 3, 3, 4, 1, 9, 1, 2, 11

Offset: 1

Author

Ramasamy Chandramouli, Jun 05 2008

Keywords

• easy
• nonn

Comments

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.

Examples

			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).
		

Links

• T. D. Noe, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A130517 (permutations), A003558 (order).

Programs

Python

from sympy.combinatorics import Permutation
def a(n):
    p = list(range(n, 0, -2)) + list(range(1+(n%2), n, 2))
    return Permutation([pi-1 for pi in p]).cycles
print([a(n) for n in range(1, 86)]) # Michael S. Branicky, Dec 27 2021