proposed

approved

editing

proposed

According to rook theory, John Riordan considered a(1) to be -1. - Vladimir Shevelev, Apr 02 2010.

This is also the value that the formulas of Touchard and of Wyman and Moser give and is compatible with many recurrences. - William P. Orrick, Aug 31 2020.

Or, for n >= 3, the number of 3 X n Latin rectangles the second row of which is full cycle with a fixed order of its elements, e.g., the cycle (x_2,x_3,...,x_n,x_1) with x_1 < x_2 < ... < x_n. - Vladimir Shevelev, Mar 22 2010

001*11

1*0011

11001* (1)

11*100

0111*0,

We can indicate a one-to-one correspondence between arrangements of n non-attacking rooks on the 1's of a matrix A_n and arrangements of n married couples around a circular table by the rules of the ménage problem, after the ladies w_1, w_2, ..., w_n have taken the chairs numbered

2*n, 2, 4, ..., 2*n-2 (2)

respectively. Suppose we consider an arrangement of rooks: (1,j_1), (2,j_2), ..., (n,j_n). Then the men m_1, m_2, ..., m_n took chairs with numbers

2*j_i - 3 (mod 2*n), (3)

Nick Hobson, <a href="/A000179/a000179.py.txt">Python program for this sequence</a>.

Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/MarriedCouplesProblem.html">Married Couples Problem</a>.

Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/RooksProblem.html">Rooks Problem</a>.

Wikipedia, <a href="https://en.wikipedia.org/wiki/M%C3%A9nage_problem">Menage problem</a>.

a(2^k+2) == 0 (mod 2^k); for k >= 2, a(2^k) == 2(mod 2^k). - Vladimir Shevelev, Jan 14 2011

a(k+4*p) - 2*a(k+2*p) + a(k) is divisible by p, for any k > 0 and any prime p. - Mark van Hoeij, Jan 11 2022

approved

editing

reviewed

approved

proposed

reviewed

editing

proposed

Peter Kagey, <a href="https://arxiv.org/abs/2210.17021">Ranking and Unranking Restricted Permutations</a>, arXiv:2210.17021 [math.CO], 2022.

approved

editing

reviewed

approved

proposed

reviewed

editing

proposed