%I #8 Aug 30 2017 09:37:59
%S 1,1,1,4,5,1,25,33,9,1,190,256,78,13,1,1606,2186,703,139,17,1,14506,
%T 19863,6591,1430,216,21,1,137089,188449,63813,14669,2501,309,25,1,
%U 1338790,1845416,633808,151532,27940,3980,418,29,1,13403950,18513822
%N Riordan array (1/(1-xc(3x)), xc(3x)/(1-xc(3x))), c(x) the g.f. of A000108.
%C Product of (1, xc(3x)) and (1/(1-x), x/(1-x)) (A110518 and A007318). The binomial transform of the inverse of this triangle has general element (-3)^(n-k)*C(k,n-k), that is, it is the Riordan array (1, x(1-3x)) [A110517]. Row sums are A110520. Diagonal sums are A110521.
%H G. C. Greubel, <a href="/A110519/b110519.txt">Table of n, a(n) for the first 50 rows, flattened</a>
%F Number triangle T(0,k) = 0^k, T(n,k) = Sum_{j=0..n} j*C(2n-j-1, n-j)* C(j, k)3^(n-j)/n, n > 0, k > 0. Deleham triangle Delta(0^n, 3-2*0^n) [see construction in A084938].
%e Rows begin
%e 1;
%e 1, 1;
%e 4, 5, 1;
%e 25, 33, 9, 1;
%e 190, 256, 78, 13, 1;
%e 1606, 2186, 703, 139, 17, 1;
%t T[0, 0] := 1; T[0, k_] := 0; T[n_, k_] := Sum[j*3^(n - j)*Binomial[2*n - j - 1, n - j]*Binomial[j, k]/n, {j, 0, n}]; Table[T[n, k], {n, 0, 20}, {k, 0, n}] // Flatten ( _G. C. Greubel_, Aug 29 2017 *)
%o (PARI) concat([1], for(n=1, 10, for(k=0,n, print1(sum(j=0,n, j*binomial(2*n-j-1,n-j)*binomial(j,k)*3^(n-j)/n), ", ")))) \\ _G. C. Greubel_, Aug 29 2017
%K easy,nonn,tabl
%O 0,4
%A _Paul Barry_, Jul 24 2005