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%I #41 Jun 05 2021 06:29:28
%S 1,2,1,5,3,1,14,9,4,1,42,28,14,5,1,132,90,48,20,6,1,429,297,165,75,27,
%T 7,1,1430,1001,572,275,110,35,8,1,4862,3432,2002,1001,429,154,44,9,1,
%U 16796,11934,7072,3640,1638,637,208,54,10,1,58786,41990,25194,13260
%N Triangle read by rows giving partial row sums of triangle A033184(n,m), n >= m >= 1 (Catalan triangle).
%C In the language of the Shapiro et al. reference (given in A053121) such a lower triangular (ordinary) convolution array, considered as a matrix, belongs to the Riordan-group. The g.f. for the row polynomials p(n,x) (increasing powers of x) is (c(z)^2)/(1-x*z*c(z)) with c(z) = g.f. A000108 (Catalan numbers).
%C This coincides with the lower triangular Catalan convolution matrix A033184 with first row and first column deleted: a(n,m)= A033184(n+2,m+2), n >= m >= 0, a(n,m) := 0 if n<m.
%C The Catalan convolution matrix R(n,m) = A033184(n+1,m+1), n >= m >= 0, is the only Riordan-type matrix with R(0,0)=1 whose partial row sums (prs) matrix satisfies (prs(R))(n,m)= R(n+1,m+1), n >= m >= 0.
%C Riordan array (c(x)^2,x*c(x)) where c(x)is the g.f. of A000108. - _Philippe Deléham_, Nov 11 2009
%F T(n, m) = Sum_{k=m..n} A033184(n+1, k+1), (partial row sums in columns m).
%F Column m recursion: a(n, m)= sum(a(j-1, m)*A033184(n-j+1, 1), j=m..n) + A033184(n+1, m+1) if n >= m >= 0, a(n, m) := 0 if n<m.
%F G.f. for column m: (c(x)^2)*(x*c(x))^m, m >= 0, with c(x) = g.f. A000108.
%F From _Gary W. Adamson_, Jan 19 2012: (Start)
%F n-th row of the triangle = top row of M^n, where M is the following infinite square production matrix:
%F 2, 1, 0, 0, 0, ...
%F 1, 1, 1, 0, 0, ...
%F 1, 1, 1, 1, 0, ...
%F 1, 1, 1, 1, 1, ...
%F ...
%F (End)
%F G.f.: (((2-2*x)*y)/(2*y+x*sqrt(1-4*y)-x)-1)/(x*y). - _Vladimir Kruchinin_, Apr 13 2015
%F T(n, m) = (m+1) * binomial(2*n - m, n) / (n+1) if n>=m>=1. - _Michael Somos_, Oct 01 2018
%e Triangle starts:
%e 1;
%e 2, 1;
%e 5, 3, 1;
%e 14, 9, 4, 1;
%e 42, 28, 14, 5, 1;
%e 132, 90, 48, 20, 6, 1;
%e ...
%e Fourth row polynomial (n=3): p(3,x)= 14 + 9*x + 4*x^2 + x^3.
%e Top row of M^3 = [14, 9, 4, 1, 0, 0, 0, ...].
%t T[n_, k_] := SeriesCoefficient[((2-2*x)*y)/(2*y+x*Sqrt[1-4*y]-x), {x, 0, n}, {y, 0, k}]; Table[T[n-k+2, k], {n, 0, 10}, {k, n+1, 1, -1}] // Flatten (* _Jean-François Alcover_, Apr 13 2015, after _Vladimir Kruchinin_ *)
%t T[ n_, k_] := (k + 1) Binomial[2 n - k, n] / (n + 1); (* _Michael Somos_, Oct 01 2018 *)
%o (PARI)
%o tabl(nn) = {
%o default(seriesprecision, nn+1);
%o my( gf = ((2-2*x)*y)/(2*y+x*sqrt(1-4*y)-x) + O(x^nn) );
%o for (n=0, nn-1, my( P = polcoeff(gf, n, x) );
%o for (k=0, nn-1, print1(polcoeff(P, k, y), ", "); );
%o print(); );
%o } \\ _Michel Marcus_, Apr 13 2015
%Y Cf. A033184, A000108. Row sums: a(n+1, 1).
%K easy,nonn,tabl
%O 0,2
%A _Wolfdieter Lang_, Apr 27 2000 and May 08 2000
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