A200091 The number of ways of putting n labeled items into k labeled boxes so that each box receives at least 2 objects.
1, 1, 1, 6, 1, 20, 1, 50, 90, 1, 112, 630, 1, 238, 2940, 2520, 1, 492, 11508, 30240, 1, 1002, 40950, 226800, 113400, 1, 2024, 137610, 1367520, 2079000, 1, 4070, 445896, 7271880, 22869000, 7484400, 1, 8164, 1410552, 35692800, 196396200, 194594400, 1, 16354
Offset: 2
Examples
Table begins
n |k=1 2 3 4
----+-------------------
2 | 1
3 | 1
4 | 1 6
5 | 1 20
6 | 1 50 90
7 | 1 112 630
8 | 1 238 2940 2520
9 | 1 492 11508 30240
...
T(4,2) = 6: The arrangements of 4 objects into 2 boxes { } and [ ] so that each box contains at least 2 items are {1,2}[3,4], {1,3}[2,4], {2,3}[1,4] and the 3 other possibilities where the contents of a pair of boxes are swapped.
References
- P. Flajolet and R. Sedgewick, Analytic Combinatorics, Cambridge University Press, 2009, page 100-109.
Links
- Muniru A Asiru, Table of n, a(n) for n = 2..626
- Marko R. Riedel, Count by PIE and symbolic combinatorics, Math Stackexchange, March 2022.
- Dennis Walsh, Notes on doubly-surjective finite functions
Crossrefs
Programs
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GAP
Flat(List([2..14],n->List([1..Int(n/2)],k->Sum([0..k],j->(-1)^j*Binomial(k,j)*(Sum([0..j],i->Binomial(j,i)*(Binomial(n,i)*Factorial(i)*(k-j)^(n-i)))))))); # Muniru A Asiru, Jul 17 2018
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Maple
seq(seq(eval(diff((exp(x)-x-1)^k,x$n),x=0),k=1..floor(n/2)),n=2..20); # Dennis P. Walsh, Apr 09 2013 T := proc(n,k) local r; k!* add(binomial(n,r)*(-1)^r*Stirling2(n-r,k-r), r=0..min(n,k)); end; # Marko Riedel, Mar 25 2022
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Mathematica
t[n_, k_] := k! * Sum[ (-1)^i*Binomial[n, i] * Sum[ (-1)^j*(k - i - j)^(n - i) / (j!*(k - i - j)!), {j, 0, k - i}], {i, 0, k}]; Table[ t[n, k], {n, 2, 14}, {k, 1, n/2}] // Flatten (* Jean-François Alcover, Apr 10 2013 *)
Formula
E.g.f. with additional 1: 1/(1-t*(exp(x)-1-x)) = 1 + t*x^2/2! + t*x^3/3! + (t+6*t^2)*x^4/4! + ....
E.g.f. (k fixed): (exp(x)-x-1)^k. - Dennis P. Walsh, Apr 09 2013
Recurrence relation: T(n+1,k) = k*(T(n,k) + n*T(n-1,k-1)). T(n,k) = k!*A008299(n,k).
T(n,k+j) = Sum_{i=0..n} C(n,i)*T(i,k)*T(n-i,j). - Dennis P. Walsh, Apr 09 2013
T(n,k) = Sum_{j=0..k} (-1)^j*C(k,j)*(Sum_{i=0..j} C(j,i)*C(n,i)*i!*(k-j)^(n-i)) for 1 <= k <= n/2 and n >= 2. - Dennis P. Walsh, Apr 10 2013
T(n,k) = k!*Sum_{r=0..min(n,k)} binomial(n,r)*(-1)^r*Stirling2(n-r, k-r). - Marko Riedel, Mar 25 2022
Comments