A187430 Number of nonnegative walks of n steps with step sizes 1 and 2, starting and ending at 0.
1, 0, 2, 2, 11, 24, 93, 272, 971, 3194, 11293, 39148, 139687, 497756, 1798002, 6517194, 23807731, 87336870, 322082967, 1192381270, 4431889344, 16527495396, 61831374003, 231973133544, 872598922407, 3290312724374, 12434632908623, 47089829065940, 178672856753641
Offset: 0
Examples
The 11 length-4 walks are 0,2,4,2,0; 0,2,3,2,0; 0,2,3,1,0; 0,2,1,2,0; 0,2,0,2,0; 0,2,0,1,0; 0,1,3,2,0; 0,1,3,1,0; 0,1,2,1,0; 0,1,0,2,0; and 0,1,0,1,0.
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..1000
- C. Banderier, Analytic combinatorics of random walks and planar maps, PhD Thesis, 2001 , Example 11.
- C. Banderier, C. Krattenthaler, A. Krinik, D. Kruchinin, V. Kruchinin, D. Nguyen, and M. Wallner, Explicit formulas for enumeration of lattice paths: basketball and the kernel method, arXiv preprint arXiv:1609.06473 [math.CO], 2016.
- Jean-Luc Baril and José L. Ramírez, Knight's paths towards Catalan numbers, Univ. Bourgogne Franche-Comté (2022).
Programs
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Maple
a:= proc(n) option remember; `if`(n<3, (n-1)*(3*n-2)/2, ((n+1)*(115*n^3-137*n^2-10*n+8) *a(n-1) +4*(2*n-1)*(5*n^3+36*n^2-26*n-12) *a(n-2) -36*(n-2)*(2*n-1)*(2*n-3)*(5*n+1) *a(n-3)) / (2*(5*n-4)*(2*n+1)*(n+2)*(n+1))) end: seq(a(n), n=0..30); # Alois P. Heinz, May 16 2013
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Mathematica
a[n_] := (Sum[Binomial[n+1, l]*Sum[Binomial[n-2*i-1, 2*l-1]*Binomial[n-l+1, i], {i, 0, (n-1)/2}], {l, 0, n+1}] + (((-1)^n+1)*Binomial[n+1, n/2])/2)/(n+1); Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Oct 24 2016, after Vladimir Kruchinin *)
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Maxima
a(n):=((sum(binomial(n+1,l)*sum(binomial(n-2*i-1,2*l-1)*binomial(n-l+1,i),i,0,(n-1)/2),l,0,n+1))+(((-1)^n+1)*binomial(n+1,n/2))/2)/(n+1); /* Vladimir Kruchinin, Jun 26 2015 */
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PARI
al(n)={local(r,p); r=vector(n);r[1]=p=1; for(k=2,n,p*=1+x+x^3+x^4;p=(p-polcoeff(p,0)-polcoeff(p,1)*x)/x^2;r[k]=polcoeff(p,0)); r}
Formula
G.f.: 1/(2*x)-(1+(1-4*x)^(1/2))*((2+2*(1-4*x)^(1/2)+12*x)^(1/2)-2)/(8*x^2). - Mark van Hoeij, May 16 2013
a(n) ~ (3/sqrt(5)-1) * 2^(2*n+1) / (sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Sep 09 2014
G.f.: exp( Sum_{n>=1} A092765(n)*x^n/n ), where A092765(n) = Sum_{k=0..n} binomial(n,k)*binomial(n,2*n-3*k). - Paul D. Hanna, May 31 2015
a(n) = ((Sum_{l=0..n+1} (C(n+1,l)*Sum_{i=0..(n-1)/2}(C(n-2*i-1,2*l-1)*C(n-l+1,i))))+(((-1)^n+1)/2*C(n+1,n/2)))/(n+1). - Vladimir Kruchinin, Jun 26 2015
Sum_{n>=0} a(n)*x^(n+1) is the compositional inverse of x*(1-x^2)^2/(1+x^3)^2. - Ira M. Gessel, Sep 19 2017
Conjecture: 1 + Sum_{n>=0} a(n)*(-1)^n x^(n+1)/(1-x)^(2*n+2) = C(x), the g.f. for the Catalan numbers A000108. - Benedict W. J. Irwin, Jan 13 2017
D-finite with recurrence 2*(2*n+1)*(n+2)*(n+1)*a(n) +(n+1)*(n^2-27*n+2)*a(n-1) +2*(-73*n^3+204*n^2-167*n+6)*a(n-2) +12*(n-3)*(2*n-3)*(4*n-7)*a(n-3) +216*(2*n-5)*(n-3)*(2*n-3)*a(n-4)=0. - R. J. Mathar, Sep 29 2020
From Seiichi Manyama, Jan 17 2024: (Start)
G.f.: (1/x) * Series_Reversion( x * (1-x)^2 / (1-x+x^2)^2 ).
a(n) = (1/(n+1)) * Sum_{k=0..floor(n/2)} binomial(2*n+2,k) * binomial(n-k-1,n-2*k). (End)
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