A007060 Number of ways n married couples can sit in a row without any spouses next to each other.

1, 0, 8, 240, 13824, 1263360, 168422400, 30865121280, 7445355724800, 2287168006717440, 871804170613555200, 403779880746418176000, 223346806774106790297600, 145427383048755178635264000, 110105698060190464791596236800, 95914116314126658718742347776000, 95252504853751428295192341381120000

Offset: 0

Author

David Roberts Keeney (David.Roberts.Keeney(AT)directory.Reed.edu)

Keywords

• nonn
• easy
• nice

Comments

Limit_{n->oo} a(n)/(2n)! = 1/e.

Also the number of (directed) Hamiltonian paths of the n-cocktail party graph. - Eric W. Weisstein, Dec 16 2013

Also the number of ways to label the cells of a 2 X n grid such that no vertically adjacent cells have adjacent labels. - Sela Fried, May 29 2023

Examples

			For n = 2, the a(2) = 8 solutions for the couples {1,2} and {3,4} are {1324, 1423, 2314, 2413, 3142, 3241, 4132, 4231}.
		

Links

• Andrew Woods, Table of n, a(n) for n = 0..100
• G. Almkvist, Letter to N. J. A. Sloane, Apr. 1992.
• Eric Weisstein's World of Mathematics, Cocktail Party Graph.
• Eric Weisstein's World of Mathematics, Hamiltonian Path.

Crossrefs

Cf. A000166, A000806, A114938, A177840, A053983, A053984, A193624, A229430, A330266.

Programs

Maple

seq(add((-1)^i*binomial(n, i)*2^i*(2*n-i)!, i=0..n),n=0..20);

Mathematica

Table[Sum[(-1)^i Binomial[n,i] (2 n - i)! 2^i, {i, 0, n}], {n, 0, 20}]
Table[(2 n)! Hypergeometric1F1[-n, -2 n, -2], {n, 0, 20}]

PARI

a(n)=sum(k=0, n, binomial(n, k)*(-1)^(n-k)*(n+k)!*2^(n-k)) \\ Charles R Greathouse IV, May 11 2016

Python

from sympy import binomial, subfactorial
def a(n): return sum([(-1)**(n - k)*binomial(n, k)*subfactorial(2*k) for k in range(n + 1)]) # Indranil Ghosh, Apr 28 2017

Formula

a(n) = (Pi*BesselI(n+1/2,1)*(-1)^n+BesselK(n+1/2,1))*exp(-1)*(2/Pi)^(1/2)*2^n*n!. - Mark van Hoeij, Nov 12 2009

a(n) = (-1)^n*2^n*n!*A000806(n), n>0. - Vladeta Jovovic, Nov 19 2009

a(n) = n!*hypergeom([-n, n+1],[],1/2)*(-2)^n. - Mark van Hoeij, Nov 13 2009

a(n) = 2^n * A114938(n). - Toby Gottfried, Nov 22 2010

a(n) = 2*n((2*n-1)*a(n-1) + (2*n-2)*a(n-2)), n > 1. - Aaron Meyerowitz, May 14 2014

From Peter Bala, Mar 06 2015: (Start)

a(n) = Sum_{k = 0..n} (-1)^(n-k)*binomial(n,k)*A000166(2*k).

For n >= 1, Integral_{x = 0..1} (x^2 - 1)^n*exp(x) dx = a(n)*e - A177840(n). Hence lim_{n->oo} A177840(n)/a(n) = e. (End)

a(n) ~ sqrt(Pi) * 2^(2*n+1) * n^(2*n + 1/2) / exp(2*n+1). - Vaclav Kotesovec, Mar 09 2016

Extensions

More terms from Michel ten Voorde, Apr 11 2001