%I #13 Feb 12 2023 12:34:14
%S 1,1,1,0,1,2,2,5,0,1,3,5,14,14,42,0,1,4,9,28,42,132,132,429,0,1,5,14,
%T 48,90,297,429,1430,1430,4862,0,1,6,20,75,165,572,1001,3432,4862,
%U 16796,16796,58786,0,1,7,27,110,275,1001,2002,7072,11934,41990,58786,208012
%N Slanted Catalan convolution table, read by rows of 2*n+1 terms in row n, where T(n,k) = C(n+2*k-[k/2],k)*(n-[k/2])/(n+2*k-[k/2]).
%C Row sums form A100248. Antidiagonal sums form A100249.
%F T(n, k) = A033184(n-[k/2], k) for n>0 (with A033184 formatted as a square array).
%F G.f. A(x, y) satisfies:
%F A(x^2, y) = ( (1+x)/(2*y - x*(1 - sqrt(1 - 4*x*y))) - (1-x)/(2*y + x*(1 - sqrt(1 + 4*x*y))) )*y/x.
%e Rows begin:
%e [1],
%e [1,1,0],
%e [1,2,2,5,0],
%e [1,3,5,14,14,42,0],
%e [1,4,9,28,42,132,132,429,0],
%e [1,5,14,48,90,297,429,1430,1430,4862,0],
%e [1,6,20,75,165,572,1001,3432,4862,16796,16796,58786,0],...
%e and is derived from the square array of Catalan convolutions (A033184)
%e by shifting each column k down by [k/2] rows.
%o (PARI) {T(n,k) = if(n==k&k==0,1,binomial(n+2*k-(k\2),k)*(n-(k\2))/(n+2*k-(k\2)))}
%o for(n=0,10,for(k=0,2*n, print1(T(n,k),", "));print(""))
%o (PARI) {T(n,k) = polcoeff(((1-sqrt(1-4*z +z*O(z^(2*n))))/(2*z))^(n-k\2),k,z)}
%o for(n=0,10,for(k=0,2*n, print1(T(n,k),", "));print(""))
%Y Cf. A033184, A100248, A100249.
%K nonn,tabf
%O 0,6
%A _Paul D. Hanna_, Nov 09 2004