%I #17 Aug 28 2019 18:00:57
%S 1,3,1,10,3,1,35,19,3,1,126,65,19,3,1,462,331,92,19,3,1,1716,1190,421,
%T 92,19,3,1,6435,5587,1805,502,92,19,3,1,24310,20613,8771,2075,502,92,
%U 19,3,1,92378,92821,35726,10616,2318,502,92,19,3,1,352716,347930,160205
%N Triangle of numbers obtained from the partition array A134284.
%C This triangle is called s2(3)'.
%H Wolfdieter Lang, <a href="/A134285/a134285.txt">First 10 rows and more. </a>
%F a(n,m) = sum(product(s2(3;j,1)^e(n,m,q,j),j=1..n),k=1..p(n,m)) if n>=m>=1, else 0. Here p(n,m)=A008284(n,m), the number of m parts partitions of n and e(n,m,q,j) is the exponent of j in the q-th m part partition of n. s2(3;n,1) = A035324(n,1) = A001700(n-1) = binomial(2*n-1,n).
%F Row sums = A001700. Triangle A134285 = A001263 * A000012. - _Gary W. Adamson_, Nov 19 2007
%e Triangle starts:
%e 1
%e 3, 1
%e 10, 3, 1
%e 35, 19, 3, 1
%e 126, 65, 19, 3, 1
%e ...
%e a(4,2)=19 because the m=2 parts partitions (1^1,3^1) and (2^2) of n=4 lead to 1^1*10^1 + 3^2 =19, since A001700(n-1)=[1,3,10,...], n>=1.
%Y Row sums A134826. Alternating row sums A134827.
%Y Cf. A001700.
%K nonn,easy,tabl
%O 1,2
%A _Wolfdieter Lang_, Nov 13 2007