A165984 Number of ways to put n indistinguishable balls into n^3 distinguishable boxes.

1, 1, 36, 3654, 766480, 275234400, 151111164204, 117774526188844, 123672890985095232, 168324948170849366820, 288216356245328994082600, 606320062786763763996747618, 1537230010624231669678572481296, 4622745700243196227504110670860680

Offset: 0

Author

Thomas Wieder, Oct 03 2009

Keywords

• nonn

Comments

See A165817 for the case n indistinguishable balls into 2*n distinguishable boxes.

See A054688 for the case n indistinguishable balls into n^2 distinguishable boxes.

a(n) is the number of (weak) compositions of n into n^3 parts. - Joerg Arndt, Oct 04 2017

Examples

			For n = 2 the a(2) = 36 solutions are
[0, 0, 0, 0, 0, 0, 0, 2]
[0, 0, 0, 0, 0, 0, 1, 1]
[0, 0, 0, 0, 0, 0, 2, 0]
[0, 0, 0, 0, 0, 1, 0, 1]
[0, 0, 0, 0, 0, 1, 1, 0]
[0, 0, 0, 0, 0, 2, 0, 0]
[0, 0, 0, 0, 1, 0, 0, 1]
[0, 0, 0, 0, 1, 0, 1, 0]
[0, 0, 0, 0, 1, 1, 0, 0]
[0, 0, 0, 0, 2, 0, 0, 0]
[0, 0, 0, 1, 0, 0, 0, 1]
[0, 0, 0, 1, 0, 0, 1, 0]
[0, 0, 0, 1, 0, 1, 0, 0]
[0, 0, 0, 1, 1, 0, 0, 0]
[0, 0, 0, 2, 0, 0, 0, 0]
[0, 0, 1, 0, 0, 0, 0, 1]
[0, 0, 1, 0, 0, 0, 1, 0]
[0, 0, 1, 0, 0, 1, 0, 0]
[0, 0, 1, 0, 1, 0, 0, 0]
[0, 0, 1, 1, 0, 0, 0, 0]
[0, 0, 2, 0, 0, 0, 0, 0]
[0, 1, 0, 0, 0, 0, 0, 1]
[0, 1, 0, 0, 0, 0, 1, 0]
[0, 1, 0, 0, 0, 1, 0, 0]
[0, 1, 0, 0, 1, 0, 0, 0]
[0, 1, 0, 1, 0, 0, 0, 0]
[0, 1, 1, 0, 0, 0, 0, 0]
[0, 2, 0, 0, 0, 0, 0, 0]
[1, 0, 0, 0, 0, 0, 0, 1]
[1, 0, 0, 0, 0, 0, 1, 0]
[1, 0, 0, 0, 0, 1, 0, 0]
[1, 0, 0, 0, 1, 0, 0, 0]
[1, 0, 0, 1, 0, 0, 0, 0]
[1, 0, 1, 0, 0, 0, 0, 0]
[1, 1, 0, 0, 0, 0, 0, 0]
[2, 0, 0, 0, 0, 0, 0, 0]
		

Crossrefs

Cf. A001700, A054688, A060690, A165817.

Programs

Maple

a:= n-> binomial(n^3+n-1, n): seq(a(n), n=0..16);

Mathematica

Table[Binomial[n^3 + n - 1, n], {n, 0, 13}] (* Michael De Vlieger, Oct 05 2017 *)

PARI

a(n) = binomial(n^3+n-1, n); \\ Altug Alkan, Oct 03 2017

Formula

a(n) = binomial(n^3+n-1, n).

Let denote P(n) = the number of integer partitions of n,

p(i) = the number of parts of the i-th partition of n,

d(i) = the number of different parts of the i-th partition of n,

m(i,j) = multiplicity of the j-th part of the i-th partition of n.

Then one has:

a(n) = Sum_{i=1..P(n)} (n^3)!/((n^3-p(i))!*(Product_{j=1..d(i)} m(i,j)!)).

a(n) = [x^n] 1/(1 - x)^(n^3). - Ilya Gutkovskiy, Oct 03 2017