Ramasamy Chandramouli - cp's OEIS frontend

User: Ramasamy Chandramouli

Ramasamy Chandramouli has authored 2 sequences.

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

Ramasamy Chandramouli, Jun 05 2008

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.

			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).

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

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

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

A130320 Given n numbers n>(n-1)>(n-2)>...>2>1, adding the first and last numbers leads to the identity n+1 = (n-1)+2 = (n-2)+3 = ... In case if some positive x_1, x_2, ... are added to n, (n-1) etc, the strict inequality could be retained. This could be repeated finitely many times till it ends in inequality of form M > N where M-N is minimal. This sequence gives the value of M for different n.

Original entry on oeis.org

1, 2, 4, 6, 10, 16, 18, 22, 34, 40, 56, 64, 66, 74, 78, 86, 130, 142, 148, 160, 216, 232, 240, 256, 258, 274, 282, 298, 302, 318, 326, 342, 514, 538, 550, 574, 580, 604, 616, 640, 856, 888, 904, 936, 944, 976, 992, 1024, 1026, 1058, 1074, 1106, 1114, 1146, 1162, 1194, 1198

Offset: 1

Ramasamy Chandramouli, May 23 2007

Apparently contains 2^(2k+1) and 2^k+2. - Ralf Stephan, Nov 10 2013

			a(5) = 10 because we have 5 > 4 > 3 > 2 > 1.
To follow a strict inequality we would have 5 + x > 4 + y > 3 > 2 > 1, where x >= 0, y >= 0.
The next level of inequality gives 1 + 5 + x > 2 + 4 + y > 3. This implies x > y.
Continuing with next level gives 3 + 6 + x > 6 + y. This gives x = 1, y = 0.
Hence 10 > 6 giving a(5) = 10.

Ramasamy Chandramouli, Table of n, a(n) for n = 1..17000

For n of form 2^k, we have a(n) = 4a(n-1) - 2 with a(1) = 2. For n of form 2^k + 2^(k-1), a(n) = 4a(n-1) with a(1) = 4.

cp's OEIS Frontend

User: Ramasamy Chandramouli

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Python