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The number of ways of putting n labeled items into k labeled boxes so that each box receives at least 2 objects.
+10
7
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
2,4
COMMENTS
Equivalently, the number of ordered set partitions of the set [n] into k blocks of size at least two. When the boxes are unlabeled or the set partitions unordered we obtain A008299.
The number of doubly-surjective functions f:[n]->[k], where a doubly-surjective function f has pre-image sets of size at least 2 for each element of the codomain. Also, the number of ways to distribute n different toys to k different children so that each child gets at least two toys. - Dennis P. Walsh, Apr 09 2013
T(n,k) is the number of chains 0 = x_0 < x_1 < ... < x_k = 1 in the Boolean lattice of rank n, such that x_i is not covered by x_(i+1) for all i. - Geoffrey Critzer, Jul 15 2018
REFERENCES
P. Flajolet and R. Sedgewick, Analytic Combinatorics, Cambridge University Press, 2009, page 100-109.
LINKS
Marko R. Riedel, Count by PIE and symbolic combinatorics, Math Stackexchange, March 2022.
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
EXAMPLE
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.
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
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 *)
PROG
(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
CROSSREFS
T(n,2) = A052515(n), T(n,3) = A224541(n), T(n,4) = A224542(n).
Cf. A032032 (row sums).
KEYWORD
nonn,easy,tabf
AUTHOR
Peter Bala, Dec 04 2011
STATUS
approved
Number of doubly-surjective functions f:[n]->[4].
+10
1
2520, 30240, 226800, 1367520, 7271880, 35692800, 165957792, 742822080, 3234711480, 13803744864, 58021888080, 241116750624, 993313349544, 4064913201216, 16549636147968, 67112688842496, 271323921459096, 1094303232174240, 4405390451382960, 17709538489849440
OFFSET
8,1
COMMENTS
Fourth column of A200091: A200091(n,4)=a(n).
Also, a(n) is (i) the number of length-n words on the alphabet A, B, C, and D with each letter of the alphabet occurring at least twice; (ii) the number of ways to distribute n different toys to 4 children so that each child gets at least two toys; (iii) the number of ways to put n numbered balls into 4 labeled boxes so that each box gets at least two balls; (iv) the number of n-digit positive integers consisting of only digits 1,2,3, and 4 with each of these digits appearing at least twice.
A doubly-surjective function f:D->C is such that the pre-image set f^-1(y) has size at least 2 for each y in C.
Triangle A200091 provides the number of doubly-surjective functions f from a set of size n onto a set of size k. Hence a(n) is column 4 of A200091.
FORMULA
a(n) = 4^n-4*3^n-4*n*3^(n-1)+(9*n+3*n^2)*2^(n-1)+6*2^n-4-8*n-4*n^3;
a(n) = sum(n!/(i!j!k!m!) over <i,j,k,m> such that i,j,k, and m are all at least 2 and i+j+k+m=n.
E.g.f.: (exp(x)-x-1)^4.
a(n) = 24*A058844(n). - Alois P. Heinz, Apr 10 2013
G.f.: 24*x^8*(288*x^6-1560*x^5+3500*x^4-4130*x^3+2625*x^2-840*x+105) / ((x-1)^4*(2*x-1)^3*(3*x-1)^2*(4*x-1)). - Colin Barker, Jun 04 2013
EXAMPLE
a(9) = 30240 since there are 30240 ways to distribute 9 different toys to 4 children so that each child gets at least 2 toys. One child must get 3 toys and the other children get 2 toys each. There are 4 ways to pick the lucky kid. There are C(9,3) ways to choose the 3 toys for the lucky kid. There are 6!/(2!)^3 ways to distribute the remaining 6 toys among the 3 kids. We obtain 4*C(9,3)*6!/8=30240.
MAPLE
seq(eval(diff((exp(x)-x-1)^4, x$n), x=0), n=8..40);
MATHEMATICA
nn=27; Drop[Range[0, nn]! CoefficientList[Series[(Exp[x]-x-1)^4, {x, 0, nn}], x], 8] (* Geoffrey Critzer, Sep 28 2013 *)
PROG
(PARI) x='x+O('x^66); Vec(serlaplace((exp(x)-x-1)^4)) /* Joerg Arndt, Apr 10 2013 */
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Dennis P. Walsh, Apr 09 2013
STATUS
approved

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