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%I #6 Mar 24 2017 00:47:52
%S 0,4,64,2880,208896,23193600,3640688640,768126320640,209688566169600,
%T 71921062285148160,30278182590480384000,15350836256712740044800,
%U 9225766813653105691852800,6485670333458406942179328000,5272823572160895949091320627200
%N Sequence a(n) gives the number of ways to seat 2n people around a circular table so that person i does not sit across from person n+i for any 1 <= i <= n.
%F a(n) = (n!)^2/(2*n)*sum{k = 0..n+1}((-1)^k/k!*binomial(2*n-2*k, n-k)*2^k).
%e When n=2, there are four people seated around a circular table. Person 1 can sit across from either person 2 or person 4, and person 3 can sit either to the left or to the right of person 1. Thus a(2) = 2*2=4.
%o (PARI) a(n) = n!^2/(2*n)*sum(k = 0,n+1,(-1)^k/k!*binomial(2*n-2*k, n-k)*2^k) \\ _Michel Marcus_, Jul 11 2013
%K nonn
%O 1,2
%A Steven Klee (klees(AT)math.washington.edu), Nov 03 2009
%E More terms from _Michel Marcus_, Jul 11 2013
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