A182884 Number of (1,0)-steps of weight 1 in all weighted lattice paths in L_n.
0, 1, 2, 5, 16, 44, 122, 341, 940, 2581, 7064, 19258, 52348, 141935, 383962, 1036633, 2793812, 7517698, 20200330, 54209775, 145309380, 389091111, 1040853492, 2781908250, 7429184976, 19824925429, 52866176702, 140883978971, 375216491080
Offset: 0
Keywords
Examples
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.
Links
- Robert Israel, Table of n, a(n) for n = 0..2388
- M. Bona and A. Knopfmacher, On the probability that certain compositions have the same number of parts, Ann. Comb., 14 (2010), 291-306.
- 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.
Crossrefs
Cf. A182882.
Programs
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Maple
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);
Formula
a(n) = Sum_{k>=0} k*A182882(n,k).
G.f.: z*(1-z-z^2)/((1-3*z+z^2)*(1+z+z^2))^(3/2).
(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
Comments