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Number of (1,0)-steps at level 0 in all weighted lattice paths in L_n.
3

%I #24 Jul 26 2022 13:28:21

%S 0,1,3,7,19,50,130,341,893,2337,6119,16020,41940,109801,287463,752587,

%T 1970299,5158310,13504630,35355581,92562113,242330757,634430159,

%U 1660959720,4348449000,11384387281,29804712843,78029751247,204284540899

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

%C The members of L_n are 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, and a (1,-1)-step with weight 1. The weight of a path is the sum of the weights of its steps.

%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.

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (2,1,2,-1)

%F a(n) = Sum_{k>=0} k*A182893(n,k).

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

%F a(n) = (A000032(2n+1) - A010892(2n))/4. - _John M. Campbell_, Dec 30 2016

%F 4*a(n) = -A057078(n) +A002878(n). - _R. J. Mathar_, Jul 26 2022

%e a(3) = 7. 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; they contain 0+0+2+2+3=7 (1,0)-steps at level 0.

%p G:=z*(1+z)/(1+z+z^2)/(1-3*z+z^2): Gser:=series(G,z=0,32): seq(coeff(Gser,z,n),n=0..28);

%t LinearRecurrence[{2,1,2,-1},{0,1,3,7},30] (* _Harvey P. Dale_, Jan 05 2022 *)

%Y Cf. A182893.

%K nonn

%O 0,3

%A _Emeric Deutsch_, Dec 12 2010