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%I #22 Apr 24 2013 12:22:21
%S 1,1,1,1,2,3,1,3,6,10,1,4,10,20,31,1,5,15,35,65,101,1,6,21,56,120,216,
%T 336,1,7,28,84,203,413,728,1128,1,8,36,120,322,728,1428,2472,3823,1,9,
%U 45,165,486,1206,2598,4950,8451,13051,1,10
%N Triangle of numbers C^(3)(n,k) of combinations with repetitions from n different elements over k for each of them not more than three appearances allowed.
%C The left side of triangle consists of 1's, while the right side is formed by A005725. Further, T(n,0)=1, T(n,1)=n, T(n,2)=A000217(n) for n>1, T(n,3)=A000292(n) for n>=3, T(n,4)=A005718(n) for n>=2, T(n,5)=A005719(n) for n>=5, T(n,6)=A005720(n) for n>=6, T(n,7)=A001919(n) for n>=7, T(n,8)=A064055(n) for n>=5.
%H Peter J. C. Moses, <a href="/A213742/b213742.txt">Rows n = 0..50 of triangle, flattened</a>
%F C^(3)(n,k)=sum{r=0,...,floor(k/4)}(-1)^r*C(n,r)*C(n-4*r+k-1, n-1)
%e Triangle begins
%e n/k.|..0.....1.....2.....3.....4.....5.....6.....7
%e ==================================================
%e .0..|..1
%e .1..|..1.....1
%e .2..|..1.....2.....3
%e .3..|..1.....3.....6....10
%e .4..|..1.....4....10....20....31
%e .5..|..1.....5....15....35....65....101
%e .6..|..1.....6....21....56...120....216...336
%e .7..|..1.....7....28....84...203....413...728....1128
%t Flatten[Table[Sum[(-1)^r Binomial[n,r] Binomial[n-# r+k-1,n-1],{r,0,Floor[k/#]}],{n,0,15},{k,0,n}]/.{0}->{1}]&[4] (* _Peter J. C. Moses_, Apr 16 2013 *)
%Y Cf. A007318, A005725, A111808, A000217, A000292, A005718, A005719, A005720, A001919, A064055.
%K nonn,tabl
%O 0,5
%A _Vladimir Shevelev_ and _Peter J. C. Moses_, Jun 19 2012
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