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%I #21 Aug 07 2024 02:26:50
%S 0,0,1,3,24,180,1620,16380,184800,2298240,31222800,459874800,
%T 7296791040,124047443520,2248897210560,43301275617600,882304501478400,
%U 18964350332928000,428768570841811200,10170992126597702400,252555415474602240000,6550785133563775104000,177151172210521513804800
%N a(n) is the number of partial bijections (or subpermutations) of an n-element set with exactly 2 fixed points.
%H Vaclav Kotesovec, <a href="/A144087/b144087.txt">Table of n, a(n) for n = 0..440</a>
%H A. Laradji and A. Umar, <a href="http://dx.doi.org/10.1007/s00233-007-0732-8">Combinatorial results for the symmetric inverse semigroup</a>, Semigroup Forum 75, (2007), 221-236.
%F a(n) = (n*(n-1)/2)*A144085(n-2).
%F E.g.f.: (x^k/k!)*exp(x^2/(1-x))/(1-x) where k=2. - _Joerg Arndt_, Jul 11 2011
%F a(n) = (n!/2)*Sum_{m=0..n-2} (-1^m/m!)*Sum_{j=0..n-m} C(n-m,j)/j!;
%F (n-2)*a(n) = n*(2*n-5)*a(n-1) - n*(n-1)*(n-5)*a(n-2) - n*(n-1)*(n-2)*a(n-3), a(2)=1 and a(n)=0 if n < 2.
%F a(n) ~ n^(n + 1/4) * exp(2*sqrt(n) - 3/2 - n) / 2^(3/2) * (1 - 17/(48*sqrt(n))). - _Vaclav Kotesovec_, Dec 01 2021
%F a(n) = (n!/2) * Sum_{k=0..n-2} binomial(k,n-2-k)/(n-2-k)!. - _Seiichi Manyama_, Aug 06 2024
%e a(3) = 3 because there are exactly 3 partial bijections (on a 3-element set) with exactly 2 fixed points, namely: (1,2)->(1,2), (1,3)->(1,3), (2,3)->(2,3) - the mappings are coordinate-wise.
%o (PARI) x='x+O('x^66); /* that many terms */
%o k=2; egf=x^k/k!*exp(x^2/(1-x))/(1-x);
%o Vec(serlaplace(egf)) /* show terms, starting with 1 */
%o /* _Joerg Arndt_, Jul 11 2011 */
%Y Column k=2 of A144088.
%Y Cf. A144085.
%K nonn
%O 0,4
%A _Abdullahi Umar_, Sep 11 2008
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