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%I #7 Mar 12 2022 13:20:11
%S 1,0,1,0,1,1,2,0,2,1,2,4,1,3,1,4,8,6,3,4,1,12,12,18,9,6,5,1,24,36,30,
%T 32,14,10,6,1,54,84,78,64,51,22,15,7,1,130,184,204,152,120,77,34,21,8,
%U 1,300,452,462,416,280,205,113,51,28,9,1,706,1084,1130,1000,770,492,328,163,74,36,10,1
%N Triangle read by rows: T(n,k) is the number of weighted lattice paths in L_n having k (1,0)-steps 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)=A182894(n).
%C Sum(k*T(n,k), k=0..n)=A182895(n).
%D M. Bona and A. Knopfmacher, On the probability that certain compositions have the same number of parts, Ann. Comb., 14 (2010), 291-306.
%D E. Munarini, N. Zagaglia Salvi, On the rank polynomial of the lattice of order ideals of fences and crowns, Discrete Mathematics 259 (2002), 163-177.
%F G.f.: G(t,z) =1/[(1-t)z(1+z)+sqrt((1+z+z^2)(1-3z+z^2))].
%e T(3,2)=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 two (1,0)-steps at level 0.
%e Triangle starts:
%e 1;
%e 0,1;
%e 0,1,1;
%e 2,0,2,1;
%e 2,4,1,3,1;
%e 4,8,6,3,4,1.
%p G:=1/((1-t)*z*(1+z)+sqrt((1+z+z^2)*(1-3*z+z^2))): Gser:=simplify(series(G,z=0,15)): for n from 0 to 11 do P[n]:=sort(coeff(Gser,z,n)) od: for n from 0 to 11 do seq(coeff(P[n],t,k),k=0..n) od; # yields sequence in triangular form
%Y Cf. A051286, A182894, A182895.
%K nonn,tabl
%O 0,7
%A _Emeric Deutsch_, Dec 12 2010
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