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A007060
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Number of ways n married couples can sit in a row without any spouses next to each other.
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13
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1, 0, 8, 240, 13824, 1263360, 168422400, 30865121280, 7445355724800, 2287168006717440, 871804170613555200, 403779880746418176000, 223346806774106790297600, 145427383048755178635264000, 110105698060190464791596236800, 95914116314126658718742347776000, 95252504853751428295192341381120000
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OFFSET
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0,3
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COMMENTS
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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
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LINKS
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FORMULA
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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) = n!*hypergeom([-n, n+1],[],1/2)*(-2)^n. - Mark van Hoeij, Nov 13 2009
a(n) = 2*n((2*n-1)*a(n-1) + (2*n-2)*a(n-2)), n > 1. - Aaron Meyerowitz, May 14 2014
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
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EXAMPLE
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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}.
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MAPLE
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seq(add((-1)^i*binomial(n, i)*2^i*(2*n-i)!, i=0..n), n=0..20);
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MATHEMATICA
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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}]
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn,easy,nice
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AUTHOR
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David Roberts Keeney (David.Roberts.Keeney(AT)directory.Reed.edu)
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EXTENSIONS
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STATUS
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approved
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