|
%I #20 Oct 05 2017 16:00:00
%S 1,1,36,3654,766480,275234400,151111164204,117774526188844,
%T 123672890985095232,168324948170849366820,288216356245328994082600,
%U 606320062786763763996747618,1537230010624231669678572481296,4622745700243196227504110670860680
%N Number of ways to put n indistinguishable balls into n^3 distinguishable boxes.
%C See A165817 for the case n indistinguishable balls into 2*n distinguishable boxes.
%C See A054688 for the case n indistinguishable balls into n^2 distinguishable boxes.
%C a(n) is the number of (weak) compositions of n into n^3 parts. - _Joerg Arndt_, Oct 04 2017
%F a(n) = binomial(n^3+n-1, n).
%F Let denote P(n) = the number of integer partitions of n,
%F p(i) = the number of parts of the i-th partition of n,
%F d(i) = the number of different parts of the i-th partition of n,
%F m(i,j) = multiplicity of the j-th part of the i-th partition of n.
%F Then one has:
%F a(n) = Sum_{i=1..P(n)} (n^3)!/((n^3-p(i))!*(Product_{j=1..d(i)} m(i,j)!)).
%F a(n) = [x^n] 1/(1 - x)^(n^3). - _Ilya Gutkovskiy_, Oct 03 2017
%e For n = 2 the a(2) = 36 solutions are
%e [0, 0, 0, 0, 0, 0, 0, 2]
%e [0, 0, 0, 0, 0, 0, 1, 1]
%e [0, 0, 0, 0, 0, 0, 2, 0]
%e [0, 0, 0, 0, 0, 1, 0, 1]
%e [0, 0, 0, 0, 0, 1, 1, 0]
%e [0, 0, 0, 0, 0, 2, 0, 0]
%e [0, 0, 0, 0, 1, 0, 0, 1]
%e [0, 0, 0, 0, 1, 0, 1, 0]
%e [0, 0, 0, 0, 1, 1, 0, 0]
%e [0, 0, 0, 0, 2, 0, 0, 0]
%e [0, 0, 0, 1, 0, 0, 0, 1]
%e [0, 0, 0, 1, 0, 0, 1, 0]
%e [0, 0, 0, 1, 0, 1, 0, 0]
%e [0, 0, 0, 1, 1, 0, 0, 0]
%e [0, 0, 0, 2, 0, 0, 0, 0]
%e [0, 0, 1, 0, 0, 0, 0, 1]
%e [0, 0, 1, 0, 0, 0, 1, 0]
%e [0, 0, 1, 0, 0, 1, 0, 0]
%e [0, 0, 1, 0, 1, 0, 0, 0]
%e [0, 0, 1, 1, 0, 0, 0, 0]
%e [0, 0, 2, 0, 0, 0, 0, 0]
%e [0, 1, 0, 0, 0, 0, 0, 1]
%e [0, 1, 0, 0, 0, 0, 1, 0]
%e [0, 1, 0, 0, 0, 1, 0, 0]
%e [0, 1, 0, 0, 1, 0, 0, 0]
%e [0, 1, 0, 1, 0, 0, 0, 0]
%e [0, 1, 1, 0, 0, 0, 0, 0]
%e [0, 2, 0, 0, 0, 0, 0, 0]
%e [1, 0, 0, 0, 0, 0, 0, 1]
%e [1, 0, 0, 0, 0, 0, 1, 0]
%e [1, 0, 0, 0, 0, 1, 0, 0]
%e [1, 0, 0, 0, 1, 0, 0, 0]
%e [1, 0, 0, 1, 0, 0, 0, 0]
%e [1, 0, 1, 0, 0, 0, 0, 0]
%e [1, 1, 0, 0, 0, 0, 0, 0]
%e [2, 0, 0, 0, 0, 0, 0, 0]
%p a:= n-> binomial(n^3+n-1, n): seq(a(n), n=0..16);
%t Table[Binomial[n^3 + n - 1, n], {n, 0, 13}] (* _Michael De Vlieger_, Oct 05 2017 *)
%o (PARI) a(n) = binomial(n^3+n-1, n); \\ _Altug Alkan_, Oct 03 2017
%Y Cf. A001700, A054688, A060690, A165817.
%K nonn
%O 0,3
%A _Thomas Wieder_, Oct 03 2009
|
