A182881 Number of (1,1)-steps in all weighted lattice paths in L_n.
0, 0, 0, 2, 6, 18, 56, 162, 462, 1306, 3648, 10116, 27892, 76524, 209112, 569506, 1546542, 4189314, 11323480, 30548190, 82272330, 221240070, 594131160, 1593553452, 4269391596, 11426761548, 30554523096, 81631135502, 217918012002
Offset: 0
Keywords
Examples
a(3)=2. 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, containing a total of 1+1+0+0+0=2 u steps.
Links
- G. C. Greubel, Table of n, a(n) for n = 0..1000
- 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. A182880.
Programs
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Maple
g:=2*z^3/((1-3*z+z^2)*(1+z+z^2))^(3/2): gser:=series(g,z=0,32): seq(coeff(gser,z,n),n=0..28);
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Mathematica
CoefficientList[Series[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([0,0,0], Vec(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} k*A182880(n,k).
G.f.: 2*z^3/[(1-3*z+z^2)*(1+z+z^2)]^(3/2).
a(n) ~ ((3 + sqrt(5))/2)^n * sqrt(n) / (2*sqrt(Pi)*5^(3/4)). - Vaclav Kotesovec, Mar 06 2016
Conjecture: (-n+3)*a(n) +(2*n-5)*a(n-1) +(n-2)*a(n-2) +(2*n-3)*a(n-3) +(-n+1)*a(n-4)=0. - R. J. Mathar, Jun 14 2016
Comments