%I
%S 1,1,2,3,5,4,12,19,16,8,55,85,73,44,16,273,416,361,234,112,32,1428,
%T 2156,1883,1269,680,272,64,7752,11628,10200,7043,4016,1856,640,128,
%U 43263,64581,56829,39897,23665,11864,4848,1472,256,246675,366850,323587,229936,140161,74050,33360,12256,3328,512
%N Triangle read by rows: T(n,k) is the number of noncrossing trees with n edges in which the leftmost leaf is at level k.
%C T(n,k) is also the number of diagonally convex directed polyominoes with n diagonals and having k diagonals of length 1. Proof: the two triangles have the same g.f.
%C Row n has n terms. Column 1 and row sums yield the ternary numbers (A001764).
%H Andrew Howroyd, <a href="/A101409/b101409.txt">Table of n, a(n) for n = 1..1275</a>
%H M. BousquetMélou, <a href="http://dx.doi.org/10.1006/eujc.1996.0029">Percolation models and animals</a>, Europ. J. Combinatorics, 17, 1996, 343369 (Prop. 2.4).
%H E. Deutsch, S. Feretic and M. Noy, <a href="http://dx.doi.org/10.1016/S0012365X(02)003400">Diagonally convex directed polyominoes and even trees: a bijection and related issues</a>, Discrete Math., 256 (2002), 645654.
%F T(n,k) = Sum_{i=0..k1}((k+i)/(2*nk+i)) binomial(k1, i) binomial(3n2k+i1, nk).
%F G.f. = (1tzg^2)/(1tzgtzg^2), where g=1+zg^3 is the g.f. of the ternary numbers (A001764). (An explicit expression for g is given in the Maple program.)
%e T(2,1)=1 and T(2,2)=2 because the noncrossing trees with 2 edges are /\, /_ and _\.
%e Or, T(2,2)=2 because the horizontal domino and the vertical domino have 2 diagonals of length 1 each.
%e Triangle begins:
%e 1;
%e 1, 2;
%e 3, 5, 4;
%e 12, 19, 16, 8;
%e 55, 85, 73, 44, 16;
%p G:=t*z*g/(1t*z*gt*z*g^2): g:=2*sin(arcsin(3*sqrt(3*z)/2)/3)/sqrt(3*z): Gser:=simplify(series(G,z=0,12)): Gser:=simplify(series(G,z=0,14)): for n from 1 to 10 do P[n]:=sort(coeff(Gser,z^n)) od: for n from 1 to 10 do seq(coeff(P[n],t^k),k=1..n) od;
%p T:=proc(n,k) if k=1 then binomial(3*n3,n1)/(2*n1) elif k<=n then sum(((k+i)/(2*nk+i))*binomial(k1,i)*binomial(3*n2*k+i1,nk),i=0..k1) else 0 fi end: for n from 1 to 10 do seq(T(n,k),k=1..n) od; # yields sequence in triangular form
%t T[n_, k_] := Sum[(k+i)/(2nk+i) Binomial[k1, i] Binomial[3n2k+i1, nk], {i, 0, k1}]; Table[T[n, k], {n, 1, 10}, {k, 1, n}] // Flatten (* _JeanFrançois Alcover_, Mar 18 2017 *)
%o (PARI) T(n,k)={sum(i=0, k1, ((k+i)/(2*nk+i))*binomial(k1, i)*binomial(3*n2*k+i1, nk))} \\ _Andrew Howroyd_, Nov 17 2017
%Y Cf. A001764.
%K nonn,tabl
%O 1,3
%A _Emeric Deutsch_, Jan 15 2005 and Jan 17 2005
%E Edited by _N. J. A. Sloane_, Jan 25 2009 at the suggestion of _R. J. Mathar_ and _Max Alekseyev_
