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Triangle of C(n+1,k)*C(2*n-3*k,n-3*k)/(n+1) by rows.
1

%I #11 May 30 2018 13:51:50

%S 1,1,2,5,1,14,5,42,21,132,84,3,429,330,28,1430,1287,180,4862,5005,990,

%T 12,16796,19448,5005,165,58786,75582,24024,1430,208012,293930,111384,

%U 10010,55,742900,1144066,503880,61880,1001,2674440,4457400,2238390

%N Triangle of C(n+1,k)*C(2*n-3*k,n-3*k)/(n+1) by rows.

%H G. C. Greubel, <a href="/A073187/b073187.txt">Rows n=0..150 of triangle, flattened</a>

%e 1;

%e 1;

%e 2;

%e 5, 1;

%e 14, 5;

%e 42, 21;

%e 132, 84, 3;

%e 429, 330, 28;

%e 1430, 1287, 180; ...

%t T[n_, k_]:= If[k < 0 || k > n/3, 0, Binomial[n + 1, k]*Binomial[2*n - 3*k, n - 3*k]/(n + 1)]; Table[T[n, k], {n,0,10}, {k,0,Floor[n/3]}]//Flatten (* _G. C. Greubel_, May 29 2018 *)

%o (PARI) alias(C,binomial); T(n,k)=if(k<0 || k>n/3,0,C(n+1,k)*C(2*n-3*k,n-3*k)/(n+1))

%Y Row sums give A071969.

%K nonn,tabf

%O 1,3

%A _Michael Somos_, Jul 19 2002