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%I #32 Dec 06 2022 17:17:59
%S 1,1,1,1,2,2,1,3,6,4,1,4,12,16,9,1,5,20,40,45,21,1,6,30,80,135,126,51,
%T 1,7,42,140,315,441,357,127,1,8,56,224,630,1176,1428,1016,323,1,9,72,
%U 336,1134,2646,4284,4572,2907,835,1,10,90,480,1890,5292,10710,15240,14535,8350,2188
%N Triangle read by rows: T(n,k) is the number of ordered trees with n edges and k branches.
%C Row sums are the Catalan numbers A000108. Diagonal entries are the Motzkin numbers A001006.
%C Equals binomial transform of an infinite lower triangular matrix with A001006 as the main diagonal and the rest zeros. [_Gary W. Adamson_, Dec 31 2008] [Corrected by _Paul Barry_, Mar 06 2011]
%C Reversal of A091869. Diagonal sums are A026418(n+2). [_Paul Barry_, Mar 06 2011]
%H Michael De Vlieger, <a href="/A091187/b091187.txt">Table of n, a(n) for n = 1..11325</a> (rows 1..150, flattened)
%H J.-L. Baril, S. Kirgizov, <a href="http://jl.baril.u-bourgogne.fr/Stirling.pdf">The pure descent statistic on permutations</a>, Preprint, 2016.
%H J. Riordan, <a href="http://dx.doi.org/10.1016/S0097-3165(75)80010-0">Enumeration of plane trees by branches and endpoints</a>, J. Combinat. Theory, Ser A, 19, 214-222, 1975.
%H Lin Yang and Shengliang Yang, <a href="https://doi.org/10.4208/jms.v56n1.23.01">Protected Branches in Ordered Trees</a>, J. Math. Study (2023) Vol. 56, No. 1, 1-17.
%F T(n,k) = M(k-1)*binomial(n-1, k-1), where M(k) = A001006(k) = (Sum_{q=0..ceiling((k+1)/2)} binomial(k+1, q)*binomial(k+1-q, q-1))/(k+1) is a Motzkin number.
%F G.f.: G = G(t,z) satisfies t*z*G^2 -(1 - z + t*z)*G + 1- z + t*z = 0.
%F From _Paul Barry_, Mar 06 2011: (Start)
%F G.f.: 1/(1-x-xy-x^2y^2/(1-x-xy-x^2y^2/(1-x-xy-x^2y^2/(1-... (continued fraction).
%F G.f.: (1-x(1+y)-sqrt(1-2x(1+y)+x^2(1+2y-3y^2)))/(2x^2*y^2).
%F E.g.f.: exp(x(1+y))*Bessel_I(1,2*x*y)/(x*y). (End)
%e Triangle begins:
%e 1;
%e 1, 1;
%e 1, 2, 2;
%e 1, 3, 6, 4;
%e 1, 4, 12, 16, 9;
%e 1, 5, 20, 40, 45, 21;
%e 1, 6, 30, 80, 135, 126, 51;
%e 1, 7, 42, 140, 315, 441, 357, 127;
%p M := n->sum(binomial(n+1,q)*binomial(n+1-q,q-1),q=0..ceil((n+1)/2))/(n+1): T := (n,k)->binomial(n-1,k-1)*M(k-1): seq(seq(T(n,k),k=1..n),n=1..13);
%t (* m = MotzkinNumber *) m[0] = 1; m[n_] := m[n] = m[n - 1] + Sum[m[k]*m[n - 2 - k], {k, 0, n - 2}]; t[n_, k_] := m[k - 1]*Binomial[n - 1, k - 1]; Table[t[n, k], {n, 1, 11}, {k, 1, n}] // Flatten (* _Jean-François Alcover_, Jul 10 2013 *)
%Y Cf. A001006, A000108.
%Y Cf. A007476. [_Gary W. Adamson_, Dec 31 2008]
%K nonn,tabl
%O 1,5
%A _Emeric Deutsch_, Feb 23 2004
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