%I #11 Apr 10 2020 08:16:32
%S 1,1,2,3,6,3,12,24,15,4,55,110,75,28,5,273,546,390,168,45,6,1428,2856,
%T 2100,980,315,66,7,7752,15504,11628,5712,2040,528,91,8,43263,86526,
%U 65835,33516,12825,3762,819,120,9,246675,493350,379500,198352,79695,25410,6370,1200,153,10
%N Triangle read by rows: T(n,k) is the number of noncrossing trees with n edges in which the leftmost child of the root has degree k.
%C Row n contains n terms. Column k=0 and row sums yield the ternary numbers (A001764).
%H Andrew Howroyd, <a href="/A101401/b101401.txt">Table of n, a(n) for n = 1..1275</a>
%F T(n, k) = ((k+1)(2k+1)/(3n-k-2)) binomial(3n-k-2, 2n-1).
%F G.f.: zg/(1-tzg^2)^2, where g = 1+zg^3 is the g.f. of the ternary numbers (A001764).
%e T(2,0)=1 and T(2,1)=2 because the noncrossing trees with 2 edges are /\, |_ and _|.
%e Triangle starts:
%e 1;
%e 1, 2;
%e 3, 6, 3;
%e 12, 24, 15, 4;
%e 55, 110, 75, 28, 5;
%e 273, 546, 390, 168, 45, 6;
%e ...
%p T:=proc(n,k) if n=1 and k=1 then 0 elif k<=n then (k+1)*(2*k+1)*binomial(3*n-k-2,2*n-1)/(3*n-k-2) else 0 fi end: for n from 1 to 10 do seq(T(n,k),k=0..n-1) od; # yields sequence in triangular form
%t T[n_, k_] := ((k + 1)(2k + 1)/(3n - k - 2)) Binomial[3n - k - 2, 2n - 1];
%t Table[T[n, k], {n, 1, 10}, {k, 0, n - 1}] // Flatten (* _Jean-François Alcover_, Apr 10 2020 *)
%o (PARI)
%o T(n, k) = if(k<n, (k+1)*(2*k+1)*binomial(3*n-k-2, 2*n-1)/(3*n-k-2));
%o for(n=1, 10, for(k=0, n-1, print1(T(n, k), ", ")); print); \\ _Andrew Howroyd_, Nov 06 2017
%Y Cf. A001764.
%Y Column k=1 is A046646.
%K nonn,tabl
%O 1,3
%A _Emeric Deutsch_, Jan 15 2005