cp's OEIS Frontend

A192012 Number of ways to use the elements of set {1,..,k}, 0<=k<=3*n, once each to form a collection of n (possibly empty) sets, each with at most 3 elements.

%I A192012 #30 Oct 21 2024 09:02:46
%S A192012 1,4,35,877,46173,4108044,550917287,103674052788,26046619272535,
%T A192012 8420151470990221,3404266960229749907,1682802564587905472500,
%U A192012 998472258682783813839141,700281698972322460184258208,573086115189070229131370358179,541208343386984031504989621465925
%N A192012 Number of ways to use the elements of set {1,..,k}, 0<=k<=3*n, once each to form a collection of n (possibly empty) sets, each with at most 3 elements.
%C A192012 Number of partitions of the set {0,1,...,3*n} into n parts of size <=3.
%H A192012 Alois P. Heinz, <a href="/A192012/b192012.txt">Table of n, a(n) for n = 0..100</a>
%H A192012 R. A. Proctor, <a href="http://arxiv.org/abs/math/0606404">Let's Expand Rota's Twelvefold Way for Counting Partitions!</a>, arXiv:math/0606404 [math.CO], 2006-2007.
%F A192012 a(n) = Sum_{k=0..3*n} Sum_{i=0..n} Sum_{j=0..3*i-k} k! *C(i,j) *C(j,k+2*j-3*i) / (i! * 3^(i-j) * 2^(k+j-2*i) ).
%e A192012 a(0) = 1 = card({[e]}) where e denotes the empty set.
%e A192012 a(1) = 4 = card({[e],[1],[12],[123]}).
%e A192012 a(2) = 35 = card({ [e,e],[e,1],[e,12],[1,2],[e,123],[1,23],[2,13],[3,12],
%e A192012 [1,234],[2,134],[3,124],[4,123],[12,34],[13,24],[14,23],[12,345],[13,245],[14,235],[15,324],[23,145],[24,135],[25,134],[34,125],[35,124],[45,123],
%e A192012 [123,456],[124,356],[125,346],[126,345],[134,256],[135,246],[136,245],[145,236],[146,235],[156,234] }).
%t A192012 Table[Sum[k!/(i!3^(i - j)2^(k + j - 2i))Binomial[i, j] Binomial[j,k + 2j - 3i], {k, 0, 3n}, {i, 0, n}, {j, 0, 3i - k}], {n, 0, 15}]
%Y A192012 Partial sums of A144416.
%K A192012 nonn
%O A192012 0,2
%A A192012 _Adi Dani_, Jun 22 2011