%I
%S 1,1,0,1,1,0,1,2,2,0,1,3,5,5,0,1,4,9,14,14,0,1,5,14,28,42,42,0,1,6,20,
%T 48,90,132,132,0,1,7,27,75,165,297,429,429,0,1,8,35,110,275,572,1001,
%U 1430,1430,0,1,9,44,154,429,1001,2002,3432,4862,4862,0
%N Triangle T(n,k), 0<=k<=n, read by rows given by [1,0,0,0,0,0,0,...] DELTA [0,1,1,1,1,1,1,...] where DELTA is the operator defined in A084938 .
%C Reflected version of A106566.
%F Sum_{k, 0<=k<=n}T(n,k)=A000108(n).
%e Triangle begins:
%e 1;
%e 1, 0;
%e 1, 1, 0;
%e 1, 2, 2, 0;
%e 1, 3, 5, 5, 0;
%e 1, 4, 9, 14, 14, 0;
%e 1, 5, 14, 28, 42, 42, 0;
%e 1, 6, 20, 48, 90, 132, 132, 0;
%e 1, 7, 27, 75, 165, 297, 429, 429, 0;
%e 1, 8, 35, 110, 275, 572, 1001, 1430, 1430, 0;
%e 1, 9, 44, 154, 429, 1001, 2002, 3432, 4862, 4862, 0 ;
%e ...
%t T[n_, k_] := (n  k) Binomial[n + k  1, k]/n; T[0, 0] = 1;
%t Table[T[n, k], {n, 0, 10}, {k, 0, n}] (* _JeanFrançois Alcover_, Jun 14 2019 *)
%o (Sage)
%o @CachedFunction
%o def A130020(n, k):
%o if n==k: return add((1)^j*binomial(n, j) for j in (0..n))
%o return add(A130020(n1, j) for j in (0..k))
%o for n in (0..10) :
%o [A130020(n, k) for k in (0..n)] # _Peter Luschny_, Nov 14 2012
%Y The following are all versions of (essentially) the same Catalan triangle: A009766, A030237, A033184, A059365, A099039, A106566, A130020, A047072.
%Y Diagonals give A000108 A000245 A002057 A000344 A003517 A000588 A003518 A003519 A001392, ...
%K nonn,tabl
%O 0,8
%A _Philippe Deléham_, Jun 16 2007
