A182879 The sum of the lengths of all weighted lattice paths in L_n.

0, 1, 3, 11, 33, 96, 278, 787, 2205, 6133, 16941, 46554, 127390, 347331, 944121, 2559607, 6923529, 18690138, 50364988, 135506485, 364063815, 976880631, 2618206923, 7009868646, 18749876418, 50107633501, 133800148323, 357012426677, 951936494055

Offset: 0

Author

Emeric Deutsch, Dec 11 2010

Keywords

• nonn

Comments

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

Examples

			a(3)=11. Indeed, denoting by h (H) the (1,0)-step of weight 1 (2), and u=(1,1), d=(1,-1), there are five paths of weight 3: hhh, hH, Hh, ud, and du; their lengths are 3, 2, 2, 2,and 2, respectively.
		

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• Jean-Luc Baril, Nathanaël Hassler, Sergey Kirgizov, and José L. Ramírez, Grand zigzag knight's paths, arXiv:2402.04851 [math.CO], 2024.
• Miklós Bóna and Arnold Knopfmacher, On the probability that certain compositions have the same number of parts, Ann. Comb., 14 (2010), 291-306.
• Emanuele Munarini and Norma Zagaglia Salvi, On the Rank Polynomial of the Lattice of Order Ideals of Fences and Crowns, Discrete Mathematics 259 (2002), 163-177.

Crossrefs

Cf. A182878.

Programs

Maple

a:=n->sum(k*binomial(k,n-k)^2,k=0..n): seq(a(n),n=0..28);

Mathematica

CoefficientList[Series[x*(1+2*x^2-x^3)/((1-3*x+x^2)*(1+x+x^2))^(3/2), {x, 0, 20}], x] (* Vaclav Kotesovec, Mar 06 2016 *)

PARI

z='z+O('z^50); concat([0], Vec(z*(1+2*z^2-z^3)/((1-3*z+z^2)*(1+z+z^2))^(3/2))) \\ G. C. Greubel, Mar 25 2017

Formula

a(n) = Sum_{k=0..n} k*binomial(k,n-k)^2.

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

a(n) ~ sqrt(20 + 9*sqrt(5)) * ((3 + sqrt(5))/2)^n * sqrt(n) / (10*sqrt(Pi)). - Vaclav Kotesovec, Mar 06 2016

Conjecture: (n-1)*(310*n-781)*a(n) + (-882*n^2+3845*n-3679)*a(n-1) + (214*n^2-2903*n+3751)*a(n-2) + (-358*n^2-459*n+1955)*a(n-3) + (834*n^2-3631*n+4065)*a(n-4) - (262*n-663)*(n-3)*a(n-5) = 0. - R. J. Mathar, Jun 14 2016

Recurrence (of order 4): (n-1)*(2*n-5)*(2*n^2 - 15*n + 24)*a(n) = 2*(n-3)*(4*n^3 - 30*n^2 + 52*n - 13)*a(n-1) + (4*n^4 - 40*n^3 + 133*n^2 - 201*n + 96)*a(n-2) + 2*(4*n^4 - 38*n^3 + 110*n^2 - 126*n + 51)*a(n-3) - (n-2)*(2*n-3)*(2*n^2 - 11*n + 11)*a(n-4). - Vaclav Kotesovec, Sep 23 2017

a(n) ~ phi^(2*n + 3) * sqrt(n) / (2 * 5^(3/4) * sqrt(Pi)), where phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Sep 23 2017