%I #9 Jan 25 2020 18:12:03
%S 2,4,2,12,4,4,40,12,8,10,140,40,24,20,28,504,140,80,60,56,84,1848,504,
%T 280,200,168,168,264,6864,1848,1008,700,560,504,528,858,25740,6864,
%U 3696,2520,1960,1680,1584,1716,2860,97240,25740,13728,9240,7056,5880,5280
%N Triangle read by rows: T(n,k) is the number of Grand Dyck paths of semilength n having first return to the x-axis at (2k,0) (n,k >= 1). (A Grand Dyck path of semilength n is a path in the half-plane x >= 0, starting at (0,0), ending at (2n,0) and consisting of steps u=(1,1) and d=(1,-1)).
%C Row sums are the central binomial coefficients (A000984).
%C T(n,0) = 2*A028329(n-1).
%C Sum_{k>=1} k*T(n,k) = 2^(2n-1) (A004171).
%C For returns to the x-axis arriving from above, see A039599.
%F T(n,k) = 2*binomial(2k-2,k-1)*binomial(2n-2k,n-k)/k.
%F G.f. = G(t,z) = (1-sqrt(1-4tz))/sqrt(1-4z).
%F T(n+1,k+1) = 2*(n-k+1)*A078391(n,k), n >= 0, k >= 0. - _Philippe Deléham_, Dec 13 2006
%e T(3,2)=4 because we have uudd|ud, uudd|du, dduu|ud and dduu|du (first return to the x-axis shown by | ).
%e Triangle starts:
%e 2;
%e 4, 2;
%e 12, 4, 4;
%e 40, 12, 8, 10;
%e 140, 40, 24, 20, 28;
%p T:=(n,k)->2*binomial(2*k-2,k-1)*binomial(2*n-2*k,n-k)/k: for n from 1 to 10 do seq(T(n,k),k=1..n) od; # yields sequence in triangular form
%Y Cf. A000984, A028329, A004171, A039599.
%K nonn,tabl
%O 1,1
%A _Emeric Deutsch_, May 06 2006