%I
%S 1,1,0,1,0,1,1,0,1,1,1,0,2,1,2,1,0,3,2,2,2,1,0,6,3,4,2,4,1,0,10,6,6,4,
%T 4,4,1,0,20,10,12,6,8,4,9,1,0,35,20,20,12,12,8,9,9,1,0,70,35,40,20,24,
%U 12,18,9,23,1,0,126,70,70,40,40,24,27,18,23,23,1,0,252,126,140,70,80,40,54,27,46,23,65
%N Triangle read by rows: T(n,k) is the number of dispersed Dyck paths (i.e., Motzkin paths with no (1,0) steps at positive heights) of length n having abscissa of the first return to the horizontal axis equal to k (assumed to be 0 if there are no such returns).
%C Sum of entries in row n is binomial(n, floor(n/2)) = A001405(n).
%C Sum_{k>=0} k*T(n,k) = A093387(n+1).
%F T(n,0)=1; T(n,1)=0;
%F T(n,k) = binomial(nk, floor((nk)/2))*Sum_{j=0..floor(k/2)1} c(j), where 2<=k<=n and c(j) = binomial(2*j,j)/(j+1) are the Catalan numbers.
%F G.f.: G(t,z) = 1/(1z)+(1sqrt(14*t^2*z^2))/((1t*z)*(12*z+sqrt(14*z^2)).
%F For k>=1, g.f. of column 2k is b_{k1}*z^{2k}*g and of column 2k+1 is b_{k1}*z^{2*k+1}*g, where g = 2/(12*z+sqrt(14*z^2)) and b(k) = Sum_{j=0..k1} c(j) with c(j) = binomial(2*j,j)/(j+1) = A000108(j) (the Catalan numbers).
%e T(5,3)=2 because we have HUDHH and HUDUD, where U=(1,1), D=(1,1), H=(1,0).
%e Triangle starts:
%e 1;
%e 1, 0;
%e 1, 0, 1;
%e 1, 0, 1, 1;
%e 1, 0, 2, 1, 2;
%e 1, 0, 3, 2, 2, 2;
%e 1, 0, 6, 3, 4, 2, 4;
%p c := proc (j) options operator, arrow: binomial(2*j, j)/(j+1) end proc: T := proc (n, k) if n < k then 0 elif k = 0 then 1 elif k = 1 then 0 else binomial(nk, floor((1/2)*n(1/2)*k))*(sum(c(j), j = 0 .. floor((1/2)*k)1)) end if end proc: for n from 0 to 12 do seq(T(n, k), k = 0 .. n) end do; # yields sequence in triangular form
%p G := (1t*z+t^2*z^2*g*Ct^2*z^3*g*C)/((1z)*(1t*z)): g := 2/(12*z+sqrt(14*z^2)): C := ((1sqrt(14*t^2*z^2))*1/2)/(t^2*z^2): Gser := simplify(series(G, z = 0, 15)): for n from 0 to 12 do P[n] := sort(coeff(Gser, z, n)) end do: for n from 0 to 12 do seq(coeff(P[n], t, k), k = 0 .. n) end do; # yields sequence in triangular form
%Y Cf. A001405, A191313.
%K nonn,tabl
%O 0,13
%A _Emeric Deutsch_, May 30 2011
