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Number of (1,0)-steps of weight 1 in all weighted lattice paths in L_n.
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%I #20 Oct 13 2021 00:40:37

%S 0,1,2,5,16,44,122,341,940,2581,7064,19258,52348,141935,383962,

%T 1036633,2793812,7517698,20200330,54209775,145309380,389091111,

%U 1040853492,2781908250,7429184976,19824925429,52866176702,140883978971,375216491080

%N Number of (1,0)-steps of weight 1 in all weighted lattice paths in L_n.

%C L_n is the set of lattice paths of weight n that start at (0,0) and end on the horizontal axis and whose steps are of the following four kinds: a (1,0)-step with weight 1; a (1,0)-step with weight 2; a (1,1)-step with weight 2; a (1,-1)-step with weight 1. The weight of a path is the sum of the weights of its steps.

%H Robert Israel, <a href="/A182884/b182884.txt">Table of n, a(n) for n = 0..2388</a>

%H M. Bona and A. Knopfmacher, <a href="http://dx.doi.org/10.1007/s00026-010-0060-7">On the probability that certain compositions have the same number of parts</a>, Ann. Comb., 14 (2010), 291-306.

%H E. Munarini, N. Zagaglia Salvi, <a href="http://dx.doi.org/10.1016/S0012-365X(02)00378-3">On the Rank Polynomial of the Lattice of Order Ideals of Fences and Crowns</a>, Discrete Mathematics 259 (2002), 163-177.

%F a(n) = Sum_{k>=0} k*A182882(n,k).

%F G.f.: z*(1-z-z^2)/((1-3*z+z^2)*(1+z+z^2))^(3/2).

%F (n+3)*a(n)-n*a(n+1)+(-18-4*n)*a(n+2)+(6-n)*a(n+3)+(14+3*n)*a(n+5)+(-5-n)*a(n+6) = 0. - _Robert Israel_, Dec 30 2016

%e a(3)=5. 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; the total number of h steps in them is 0+0+1+1+3=5.

%p G:=z*(1-z-z^2)/((1-3*z+z^2)*(1+z+z^2))^(3/2): Gser:=series(G,z=0,35): seq(coeff(Gser,z,n),n=0..28);

%Y Cf. A182882.

%K nonn

%O 0,3

%A _Emeric Deutsch_, Dec 11 2010