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A200472 Triangle T(n,k) is the number of ways to assign n people to k unlabeled groups of equal size.

%I A200472 #21 Jan 28 2016 08:27:58
%S A200472 1,1,1,1,0,1,1,3,0,1,1,0,0,0,1,1,10,15,0,0,1,1,0,0,0,0,0,1,1,35,0,105,
%T A200472 0,0,0,1,1,0,280,0,0,0,0,0,1,1,126,0,0,945,0,0,0,0,1,1,0,0,0,0,0,0,0,
%U A200472 0,0,1,1,462,5775,15400,0,10395,0,0,0,0,0,1
%N A200472 Triangle T(n,k) is the number of ways to assign n people to k unlabeled groups of equal size.
%C A200472 If k is not a factor of n, T(n,k) = 0. If k is a factor of n, T(n,k) = (n!/k!)/(n/k)!^k. If n is a multiple of k, we may obtain T(n,k) by arranging all n people in an ordered line, which can be done in n! ways. Peel off the first n/k people for "group 1", the next n/k people for "group 2", ..., and the last n/k people for "group k". Since the k groups are actually unlabeled, we must divide n! by k! Also, since the ordering of the n/k people within each of the k groups is not of importance, we must now divide by (n/k)!^k. Therefore, T(n,k) = (n!/k!)/(n/k)!^k.
%C A200472 Also, T(2n,n) provide the sequence consisting of the products of consecutive odd integers.
%H A200472 Dennis P. Walsh, <a href="http://frank.mtsu.edu/~dwalsh/GROUPCT3.pdf"> Note on assigning n people into k unlabeled groups of equal size</a>
%F A200472 For k that divide n, T(n,k) = (n!/k!)/((n/k)!)^k; otherwise, T(n,k) = 0.
%F A200472 E.g.f. when k is fixed: (1/k!) sum(j>=1, (x^j/j!)^k ).
%F A200472 E.g.f. for T(n*r,n): exp(x^r/r!).
%F A200472 T(2n,n) = (2n-1)!! = (2n-1)(2n-3)...(3)(1).
%e A200472 Triangle T(n,k) begins
%e A200472 1;
%e A200472 1,   1;
%e A200472 1,   0,    1;
%e A200472 1,   3,    0,     1;
%e A200472 1,   0,    0,     0,   1;
%e A200472 1,  10,   15,     0,   0,     1;
%e A200472 1,   0,    0,     0,   0,     0,  1;
%e A200472 1,  35,    0,   105,   0,     0,  0,  1;
%e A200472 1,   0,  280,     0,   0,     0,  0,  0,  1;
%e A200472 1, 126,    0,     0, 945,     0,  0,  0,  0,  1;
%e A200472 1,   0,    0,     0,   0,     0,  0,  0,  0,  0,  1;
%e A200472 1, 462, 5775, 15400,   0, 10395,  0,  0,  0,  0,  0,  1;
%e A200472 ...
%e A200472 T(6,2) = 10 since there are 10 ways to assign 6 people (A,B,C,D,E,F) into 2 groups of size 3. The assignments are {A,B,C}|{D,E,F}, {A,B,D}|{C,E,F}, {A,B,E}|{C,D,F}, {A,B,F}|{C,D,E}, {A,C,D}|{B,E,F}, {A,C,E}|{B,D,F}, {A,C,F}|{B,D,E}, {B,C,D}|{A,E,F}, {B,C,E}|{A,D,F}, and {B,C,F}|{A,D,E}.
%p A200472 T:= (n, k)-> `if`(modp(n, k)=0, n!/(k!*((n/k)!)^k), 0):
%p A200472 seq(seq(T(n, k), k=1..n), n=1..20);
%t A200472 nn=11;s=Sum[Exp[y x^i/i!]-1,{i,1,nn}];Range[0,nn]!CoefficientList[Series[s,{x,0,nn}],{x,y}]//Grid  (* _Geoffrey Critzer_, Sep 15 2012 *)
%o A200472 (PARI)
%o A200472 T(n,k) = if(n%k!=0, 0, (n!/k!)/((n/k)!)^k );
%o A200472 for (n=1,15, for (k=1,n, print1(T(n,k),", "));print());
%o A200472 /* _Joerg Arndt_, Sep 16 2012 */
%Y A200472 T(2n,n) is A001147(n).
%Y A200472 T(3n,n) is A025035(n).
%Y A200472 T(4n,n) is A025036(n).
%Y A200472 Row sums of T(n,k) provide A038041(n).
%Y A200472 A200473 is A200472 with zeros removed.
%K A200472 nonn,easy,tabl
%O A200472 1,8
%A A200472 _Dennis P. Walsh_, Nov 18 2011