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A144087
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a(n) is the number of partial bijections (or subpermutations) of an n-element set with exactly 2 fixed points.
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4
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0, 0, 1, 3, 24, 180, 1620, 16380, 184800, 2298240, 31222800, 459874800, 7296791040, 124047443520, 2248897210560, 43301275617600, 882304501478400, 18964350332928000, 428768570841811200, 10170992126597702400, 252555415474602240000, 6550785133563775104000, 177151172210521513804800
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OFFSET
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0,4
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LINKS
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FORMULA
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E.g.f.: (x^k/k!)*exp(x^2/(1-x))/(1-x) where k=2. - Joerg Arndt, Jul 11 2011
a(n) = (n!/2)*Sum_{m=0..n-2} (-1^m/m!)*Sum_{j=0..n-m} C(n-m,j)/j!;
(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.
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
a(n) = (n!/2) * Sum_{k=0..n-2} binomial(k,n-2-k)/(n-2-k)!. - Seiichi Manyama, Aug 06 2024
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EXAMPLE
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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.
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PROG
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(PARI) x='x+O('x^66); /* that many terms */
k=2; egf=x^k/k!*exp(x^2/(1-x))/(1-x);
Vec(serlaplace(egf)) /* show terms, starting with 1 */
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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