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%I #5 Mar 30 2012 17:36:25
%S 1,1,2,4,1,9,2,21,5,48,14,1,112,38,3,263,104,9,623,276,31,1,1484,730,
%T 99,4,3550,1921,309,14,8525,5034,929,56,1,20537,13145,2739,205,5,
%U 49612,34208,7956,716,20,120136,88780,22804,2394,90,1,291519,229860,64650
%N Triangle read by rows: T(n,k) is the number of weighted lattice paths in L_n having k peaks.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. A peak is a (1,1)-step followed by a (1,-1)-step.
%C Number of entries in row n is 1+floor(n/3).
%C Sum of entries in row n is A051286(n).
%C T(n,0)= A182904(n).
%C Sum(k*T(n,k), k>=0)=A182884(n-2).
%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 Let F=F(t,s,x,y,z) be the 5-variate g.f. of the considered weighted lattice paths, where z marks weight, t (s) marks number of peaks (valleys), x (y) indicates that the path starts with a (1,1)-step ((1,-1)-step). Then F(t,s,x,y,z)=1+z(1+z)F(t,s,1,1,z)+xz^3[t+H(t,s,z)-1]F(t,s,s,1,z)+yz^3[s+H(s,t,z)-1]F(t,s,1,t,z), where H=H(t,s,z) is given by H=1+zH+z^2*H+z^3*(t-1+H)[s(H-1-zH-z^2*H)+1+zH+z^2*H] (see A182900).
%e T(7,2)=3. Indeed, denoting by h (H) the (1,0)-step of weight 1 (2), and U=(1,1), D=(1,-1), we have hUDUD, UDhUD, UDUDh.
%e Triangle starts:
%e 1;
%e 1;
%e 2;
%e 4,1;
%e 9,2;
%e 21,5;
%e 48,14,1;
%Y Cf. A051286, A182904, A182884, A182900.
%K nonn,tabf
%O 0,3
%A _Emeric Deutsch_, Dec 16 2010
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