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%I #15 Dec 25 2018 13:16:46
%S 4,23,107,464,1950,8063,33033,134576,546312,2212550,8946454,36134656,
%T 145831270,588199815,2371435125,9557736480,38511326040,155143873170,
%U 624899673690,2516678580000,10134353299980,40805797511622
%N Number of permutations of length n containing 2 occurrences of 132.
%H T. Mansour and A. Vainshtein, <a href="http://arXiv.org/abs/math.CO/0105073">Counting occurrences of 132 in a permutation</a>, arXiv:math/0105073 [math.CO], 2001.
%F a(n) = C(2*n-6,n-2)*(n^3+17*n^2-80*n+80)/(2n(n-1)).
%t Table[Binomial[2n-6,n-2] (n^3+17n^2-80n+80)/(2n(n-1)),{n,4,30}] (* _Harvey P. Dale_, Dec 25 2018 *)
%o (PARI) a(n)=binomial(2*n-6,n-2)*(n^3+17*n^2-80*n+80)/2/n/(n-1)
%Y Column k=2 of A263771.
%K nonn
%O 4,1
%A _Benoit Cloitre_, May 27 2003
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