%I
%S 1,1,1,1,3,2,7,3,1,15,8,3,35,21,6,1,83,50,16,4,197,123,45,10,1,473,
%T 308,117,28,5,1145,769,304,83,15,1,2787,1926,798,232,45,6,6819,4843,
%U 2085,636,140,21,1,16759,12204,5433,1744,416,68,7,41345,30813,14154,4749,1200,222,28,1
%N Triangle read by rows: T(n,k) is the number of weighted lattice paths in L_n having k (1,0)steps of weight 2 at level 0. The members of L_n are paths of weight n that start at (0,0) , end on the horizontal axis and whose steps are of the following four kinds: an (1,0)step with weight 1, an (1,0)step with weight 2, a (1,1)step with weight 2, and a (1,1)step with weight 1. The weight of a path is the sum of the weights of its steps.
%C Sum of entries in row n is A051286(n).
%C T(n,0)=A182892(n).
%C Sum(k*T(n,k), k=0..n)=A182890(n1).
%D M. Bona and A. Knopfmacher, On the probability that certain compositions have the same number of parts, Ann. Comb., 14 (2010), 291306.
%D E. Munarini, N. Zagaglia Salvi, On the rank polynomial of the lattice of order ideals of fences and crowns, Discrete Mathematics 259 (2002), 163177.
%F G.f. G(t,z) =1/[z^2tz^2+sqrt((1+z+z^2)(13z+z^2))].
%e T(3,1)=2. Indeed, denoting by h (H) the (1,0)step of weight 1 (2), and u=(1,1), d=(1,1), the five paths of weight 3 are ud, du, hH, Hh, and hhh; two of them, namely hH and Hh, have exactly one Hstep at level 0.
%e Triangle starts:
%e 1;
%e 1;
%e 1,1;
%e 3,2;
%e 7,3,1;
%e 15,8,3;
%e 35,21,6,1;
%p G:=1/(z^2t*z^2+sqrt((1+z+z^2)*(13*z+z^2))): Gser:=simplify(series(G,z=0,18)): for n from 0 to 14 do P[n]:=sort(coeff(Gser,z,n)) od: for n from 0 to 14 do seq(coeff(P[n],t,k),k=0..floor(n/2)) od; # yields sequence in triangular form
%Y Cf. A051286, A182890, A182892.
%K nonn,tabf
%O 0,5
%A _Emeric Deutsch_, Dec 12 2010
