A200473 -id:A200473 - cp's OEIS frontend

A038041 Number of ways to partition an n-set into subsets of equal size.

1, 2, 2, 5, 2, 27, 2, 142, 282, 1073, 2, 32034, 2, 136853, 1527528, 4661087, 2, 227932993, 2, 3689854456, 36278688162, 13749663293, 2, 14084955889019, 5194672859378, 7905858780927, 2977584150505252, 13422745388226152, 2, 1349877580746537123, 2

Offset: 1

Christian G. Bower

a(n) = 2 iff n is prime with a(p) = card{ 1|2|3|...|p-1|p, 123...p } = 2. - Bernard Schott, May 16 2019

			a(4) = card{ 1|2|3|4, 12|34, 14|23, 13|24, 1234 } = 5.
From _Gus Wiseman_, Jul 12 2019: (Start)
The a(6) = 27 set partitions:
  {{1}{2}{3}{4}{5}{6}}  {{12}{34}{56}}  {{123}{456}}  {{123456}}
                        {{12}{35}{46}}  {{124}{356}}
                        {{12}{36}{45}}  {{125}{346}}
                        {{13}{24}{56}}  {{126}{345}}
                        {{13}{25}{46}}  {{134}{256}}
                        {{13}{26}{45}}  {{135}{246}}
                        {{14}{23}{56}}  {{136}{245}}
                        {{14}{25}{36}}  {{145}{236}}
                        {{14}{26}{35}}  {{146}{235}}
                        {{15}{23}{46}}  {{156}{234}}
                        {{15}{24}{36}}
                        {{15}{26}{34}}
                        {{16}{23}{45}}
                        {{16}{24}{35}}
                        {{16}{25}{34}}
(End)

Alois P. Heinz, Table of n, a(n) for n = 1..250
Gus Wiseman, Sequences counting and ranking multiset partitions whose part lengths, sums, or averages are constant or strict.

Cf. A061095 (same but with labeled boxes), A005225, A236696, A055225, A262280, A262320.
Column k=1 of A208437.
Row sums of A200472 and A200473.
Cf. A000110, A007837 (different lengths), A035470 (equal sums), A275780, A317583, A320324, A322794, A326512 (equal averages), A326513.

A038041 := proc(n) local d;
add(n!/(d!*(n/d)!^d), d = numtheory[divisors](n)) end:
seq(A038041(n),n = 1..29); # Peter Luschny, Apr 16 2011

a[n_] := Block[{d = Divisors@ n}, Plus @@ (n!/(#! (n/#)!^#) & /@ d)]; Array[a, 29] (* Robert G. Wilson v, Apr 16 2011 *)
Table[Sum[n!/((n/d)!*(d!)^(n/d)), {d, Divisors[n]}], {n, 1, 31}] (* Emanuele Munarini, Jan 30 2014 *)
sps[{}]:={{}};sps[set:{i_,_}]:=Join@@Function[s,Prepend[#,s]&/@sps[Complement[set,s]]]/@Cases[Subsets[set],{i,_}];
Table[Length[Select[sps[Range[n]],SameQ@@Length/@#&]],{n,0,8}] (* Gus Wiseman, Jul 12 2019 *)

a(n):= lsum(n!/((n/d)!*(d!)^(n/d)),d,listify(divisors(n)));
makelist(a(n),n,1,40); /* Emanuele Munarini, Feb 03 2014 */

/* compare to A061095 */
mnom(v)=
/* Multinomial coefficient s! / prod(j=1, n, v[j]!) where
  s= sum(j=1, n, v[j]) and n is the number of elements in v[]. */
sum(j=1, #v, v[j])! / prod(j=1, #v, v[j]!)
A038041(n)={local(r=0);fordiv(n,d,r+=mnom(vector(d,j,n/d))/d!);return(r);}
vector(33,n,A038041(n)) /* Joerg Arndt, Apr 16 2011 */

import math
def a(n):
    count = 0
    for k in range(1, n + 1):
        if n % k == 0:
            count += math.factorial(n) // (math.factorial(k) ** (n // k) * math.factorial(n // k))
    return count # Paul Muljadi, Sep 25 2024

a(n) = Sum_{d divides n} (n!/(d!*((n/d)!)^d)).

E.g.f.: Sum_{k >= 1} (exp(x^k/k!)-1).

More terms from Erich Friedman

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

Original entry on oeis.org

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

Offset: 1

Dennis P. Walsh, Nov 18 2011

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.

			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}.

Dennis P. Walsh, Note on assigning n people into k unlabeled groups of equal size

T(2n,n) is A001147(n).
T(3n,n) is A025035(n).
T(4n,n) is A025036(n).
Row sums of T(n,k) provide A038041(n).
A200473 is A200472 with zeros removed.

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);

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 *)

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 */

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).

cp's OEIS Frontend

A038041 Number of ways to partition an n-set into subsets of equal size.

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