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A144088
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T(n,k) is the number of partial bijections (or subpermutations) of an n-element set with exactly k fixed points.
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4
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1, 1, 1, 4, 2, 1, 18, 12, 3, 1, 108, 72, 24, 4, 1, 780, 540, 180, 40, 5, 1, 6600, 4680, 1620, 360, 60, 6, 1, 63840, 46200, 16380, 3780, 630, 84, 7, 1, 693840, 510720, 184800, 43680, 7560, 1008, 112, 8, 1, 8361360, 6244560, 2298240, 554400, 98280, 13608, 1512, 144, 9, 1
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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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T(n,k) = C(n,k)*(n-k)! * Sum_{m=0..n-k} (-1^m/m!)*Sum_{j=0..n-m} C(n-m,j)/j!.
(n-k)*T(n,k) = n*(2n-2k-1)*T(n-1,k) - n*(n-1)*(n-k-3)*T(n-2,k) - n*(n-1)*(n-2)*T(n-3,k), T(k,k)=1 and T(n,k)=0 if n < k.
E.g.f.: exp(log(1/(1-x)) - x + y*x)*exp(x/(1-x)). - Geoffrey Critzer, Nov 29 2021
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EXAMPLE
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Triangle begins:
1;
1, 1;
4, 2, 1;
18, 12, 3, 1;
108, 72, 24, 4, 1;
780, 540, 180, 40, 5, 1;
6600, 4680, 1620, 360, 60, 6, 1;
63840, 46200, 16380, 3780, 630, 84, 7, 1;
...
T(3,1) = 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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MATHEMATICA
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max = 7; f[x_, k_] := (x^k/k!)*(Exp[x^2/(1-x)]/(1-x)); t[n_, k_] := n!*SeriesCoefficient[ Series[ f[x, k], {x, 0, max}], n]; Flatten[ Table[ t[n, k], {n, 0, max}, {k, 0, n}]](* Jean-François Alcover, Mar 12 2012, from e.g.f. by Joerg Arndt *)
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PROG
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(PARI)
T(n) = {my(egf=exp(log(1/(1-x) + O(x*x^n)) - x + y*x + x/(1-x))); Vec([Vecrev(p) | p<-Vec(serlaplace(egf))])}
{ my(A=T(10)); for(n=1, #A, print(A[n])) } \\ Andrew Howroyd, Nov 29 2021
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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