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FORMULA
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a(n)=n!*sum(m=0,n-1,(-1^m/m!)sum(j=0,n-m,C(n-m)/j!)); (n-1)a(n)=n(2n-3)a(n-1)-n(n-1)(n-4)a(n-2)-n(n-1)(n-2)a(n-3), a(1)=1 and a(n)= 0 if n<1
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EXAMPLE
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a(3) = 12 because there are exactly 12 partial bijections (on a 3-element set) with exactly 1 fixed point, namely: (1)->(1), (2)->(2), (3)->(3), (1,2)->(1,3), (1,2)->(3,2), (1,3)->(1,2), (1,3)->(2,3), (2,3)->(2,1), (2,3)->(1,3), (1,2,3)->(1,3,2), (1,2,3)->(3,2,1), (1,2,3)->(2,1,3) - the mappings are coordinate-wise.
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