%I #2 Mar 30 2012 18:37:08
%S 1,1,1,1,1,4,4,4,4,3,3,2,2,1,1,28,28,28,28,24,24,20,20,16,16,12,9,9,6,
%T 4,4,2,1,1,280,280,280,280,252,252,224,224,196,196,168,144,144,120,
%U 100,100,80,64,64,48,36,27,27,18,12,8,8,4,2,1,1,3640,3640,3640,3640,3360
%N Triple factorial triangle, read by rows of 3n(n+1)/2+1 terms, where row n+1 is generated from row n by first inserting zeros in row n at positions {[m*(m+5)/6], m=0..3n-1} and then taking partial sums, starting with a '1' in row 0.
%C Square array A136212 is generated by a complementary process. This is the triple factorial variant of triangles A135877 (double factorials) and A127452 (factorials).
%F Column 0 forms the triple factorials A007559.
%e Triangle begins:
%e 1;
%e 1,1,1,1;
%e 4,4,4,4,3,3,2,2,1,1;
%e 28,28,28,28,24,24,20,20,16,16,12,9,9,6,4,4,2,1,1;
%e 280,280,280,280,252,252,224,224,196,196,168,144,144,120,100,100,80,64,64,48,36,27,27,18,12,8,8,4,2,1,1;
%e 3640,3640,3640,3640,3360,3360,3080,3080,2800,2800,2520,2268,2268,2016,1792,1792,1568,1372,1372,1176,1008,864,864,720,600,500,500,400,320,256,256,192,144,108,81,81,54,36,24,16,16,8,4,2,1,1;
%e ...
%e To generate row 3, start with row 2:
%e [4,4,4,4,3,3,2,2,1,1];
%e insert zeros at positions [0,1,2,4,6,8,11,14,17] to get:
%e [0,0,0,4,0,4,0,4,0,4,3,0,3,2,0,2,1,0,1],
%e then take reverse partial sums (from right to left) to obtain row 3:
%e [28,28,28,28,24,24,20,20,16,16,12,9,9,6,4,4,2,1,1].
%e Continuing in this way will generate all the rows of this triangle.
%o (PARI) {T(n,k)=local(A=[1],B);if(n>0,for(i=1,n,m=1;B=[0,0]; for(j=1,#A,if(j+m-1==(m*(m+7))\6,m+=1;B=concat(B,0));B=concat(B,A[j])); A=Vec(Polrev(Vec(Pol(B)/(1-x+O(x^#B)))))));if(k+1>#A,0,A[k+1])}
%Y Cf. A007559; related tables: A136212, A136218, A136214, A135877.
%K nonn,tabl
%O 0,6
%A _Paul D. Hanna_, Dec 22 2007