OFFSET
1,8
COMMENTS
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.
Also, T(2n,n) provide the sequence consisting of the products of consecutive odd integers.
LINKS
FORMULA
For k that divide n, T(n,k) = (n!/k!)/((n/k)!)^k; otherwise, T(n,k) = 0.
E.g.f. when k is fixed: (1/k!) sum(j>=1, (x^j/j!)^k ).
E.g.f. for T(n*r,n): exp(x^r/r!).
T(2n,n) = (2n-1)!! = (2n-1)(2n-3)...(3)(1).
EXAMPLE
Triangle T(n,k) begins
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, 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, 0, 0, 1;
1, 462, 5775, 15400, 0, 10395, 0, 0, 0, 0, 0, 1;
...
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}.
MAPLE
T:= (n, k)-> `if`(modp(n, k)=0, n!/(k!*((n/k)!)^k), 0):
seq(seq(T(n, k), k=1..n), n=1..20);
MATHEMATICA
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 *)
PROG
(PARI)
T(n, k) = if(n%k!=0, 0, (n!/k!)/((n/k)!)^k );
for (n=1, 15, for (k=1, n, print1(T(n, k), ", ")); print());
/* Joerg Arndt, Sep 16 2012 */
CROSSREFS
KEYWORD
AUTHOR
Dennis P. Walsh, Nov 18 2011
STATUS
approved