%I
%S 1,1,1,1,1,2,1,3,2,1,4,5,2,5,9,4,3,6,14,12,8,9,20,25,8,13,14,27,44,28,
%T 31,29,40,70,66,16,49,54,62,104,129,64,109,115,116,159,225,168,32,170,
%U 212,217,250,363,360,144,371,430,445,444,581,681,416,64,581,772,854,820,938,1182,968,320
%N Triangle read by rows: T(n,k) is the number of dispersed Dyck paths of length n (i.e., of Motzkin paths of length n with no (1,0)steps at positive heights) for which the sum of the heights of its base pyramid is k. A base pyramid is a factor of the form U^j D^j (j>0), starting at the horizontal axis. Here U=(1,1) and D=(1,1).
%C Row n has 1+ceiling(n/2) entries.
%C Sum of entries in row n is binomial(n, floor(n/2)) = A001405(n).
%C T(n,0) = A191393(n).
%C Sum_{k>=0} k*T(n,k) = A191397(n).
%F G.f.: G=G(t,z) satisfies G = 1+z*G+z^2*G*(c+t/(1t*z^2)1/(1z^2)), where c = (1sqrt(14*z^2))/(2*z^2) (the Catalan function with argument z^2).
%e T(4,2)=2 because we have UDUD and UUDD, where U=(1,1), D=(1,1), H=(1,0).
%e Triangle starts:
%e 1;
%e 1;
%e 1, 1;
%e 1, 2;
%e 1, 3, 2;
%e 1, 4, 5;
%e 2, 5, 9, 4;
%e 3, 6, 14, 12;
%e 8, 9, 20, 25, 8;
%p eq := G = 1+z*G+z^2*G*(c+t/(1t*z^2)1/(1z^2)): c := ((1sqrt(14*z^2))*1/2)/z^2: g := simplify(solve(eq, G)): gser := simplify(series(g, z = 0, 19)): for n from 0 to 15 do P[n] := sort(expand(coeff(gser, z, n))) end do: for n from 0 to 15 do seq(coeff(P[n], t, k), k = 0 .. floor((1/2)*n)) end do; # yields sequence in triangular form
%Y Cf. A001405, A191393, A191397.
%K nonn,tabf
%O 0,6
%A _Emeric Deutsch_, Jun 04 2011
