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%I #13 Aug 15 2013 04:39:26
%S 0,1,3,12,40,150,546,2128,8352,34380,144100,626736,2784288,12753832,
%T 59692920,286857600,1407536896,7069630608,36217682352,189489626560,
%U 1010037302400,5488251406176,30348031302688,170812160339712
%N Number of 2-cycles in all involutions of {1,2,...,n}.
%H Alois P. Heinz, <a href="/A162970/b162970.txt">Table of n, a(n) for n = 1..300</a>
%F a(n) = (1/2)*n*(n-1)*I(n-2) for n>=2, where I(n)=A000085(n) is the number of involutions of {1,2,...,n}.
%F Rec. rel.: a(n) = [n/(n-2)][a(n-1) + (n-1)a(n-2)], a(1)=0, a(2)=1.
%F E.g.f.: x^2/2 * exp(x+x^2/2).
%F a(n) ~ sqrt(2)/4 * n^(n/2+1)*exp(sqrt(n)-n/2-1/4) * (1-17/(24*sqrt(n))). - _Vaclav Kotesovec_, Aug 15 2013
%e a(3) = 3 because in (1)(2)(3), (1)(23), (12)(3), (13)(2) we have three 2-cycles.
%p a[1] := 0: a[2] := 1: for n from 3 to 27 do a[n] := n*(a[n-1]+(n-1)*a[n-2])/(n-2) end do: seq(a[n], n = 1 .. 27);
%t Range[0, 20]! CoefficientList[ Series[x^2/2 Exp[x+x^2/2], {x, 0, 20}], x] // Rest
%Y Cf. A000085, A013989.
%K nonn
%O 1,3
%A _Emeric Deutsch_, Jul 22 2009
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