%I #6 Sep 17 2014 15:48:13
%S 1,1,2,4,8,17,36,1,80,2,180,5,410,13,946,32,2203,80,5173,199,1,12233,
%T 499,3,29108,1255,9,69643,3161,28,167437,7984,81,404311,20206,231,
%U 980125,51228,650,1,2384441,130090,1812,4,5819576,330835,5016,14
%N Triangle read by rows: T(n,k) is the number of weighted lattice paths B(n) having k uudd strings.
%C B(n) is the set of lattice paths of weight n that start in (0,0), end on the horizontal axis and never go below this axis, whose steps are of the following four kinds: h = (1,0) of weight 1, H = (1,0) of weight 2, u = (1,1) of weight 2, and d = (1,1) of weight 1. The weight of a path is the sum of the weights of its steps.
%C Row n contains 1 + floor(n/6) entries.
%C Sum of entries in row n is A004148(n+1) (the 2ndary structure numbers).
%C T(n,0) = A247298(n).
%C Sum(k*T(n,k), k=0..n) =A110320(n5) (n>=6)
%H M. Bona and A. Knopfmacher, <a href="http://dx.doi.org/10.1007/s0002601000607">On the probability that certain compositions have the same number of parts</a>, Ann. Comb., 14 (2010), 291306.
%F G.f. G = G(t,z) satisfies G = 1 + z*G + z^2*G + z^3*G*(G  z^3 + t*z^3).
%e T(6,1)=1 because among the 37 (=A004148(7)) paths in B(6) only uudd contains uudd.
%e T(13,2)=3 because we have huudduudd, uuddhuudd, and uudduuddh.
%e Triangle starts:
%e 1;
%e 1;
%e 2;
%e 4;
%e 8;
%e 17;
%e 36,1;
%e 80,2;
%p eq := G = 1+z*G+z^2*G+z^3*(Gz^3+t*z^3)*G: G := RootOf(eq, G): Gser := simplify(series(G, z = 0, 30)): for n from 0 to 25 do P[n] := sort(coeff(Gser, z, n)) end do: for n from 0 to 25 do seq(coeff(P[n], t, k), k = 0 .. floor((1/6)*n)) end do; # yields sequence in triangular form
%Y Cf. A004148, A110320, A247298.
%K nonn,tabf
%O 0,3
%A _Emeric Deutsch_, Sep 17 2014
