A064036 Number of walks of length n on cubic lattice, starting at origin, staying in first (nonnegative) octant.
1, 3, 12, 51, 234, 1110, 5460, 27405, 140490, 729918, 3845016, 20447658, 109801692, 593806356, 3234529584, 17715445605, 97567971930, 539701180590, 2998595422680, 16719506691030, 93559970043540, 525093580540620, 2955822168597480, 16680150247605390, 94365481922990460
Offset: 0
Keywords
Examples
a(2)=12 since a(1) is obviously 3 and from each of these three positions there are four possible steps which remain in the first octant.
Links
- Robert Israel, Table of n, a(n) for n = 0..500
- R. K. Guy, Catwalks, Sandsteps and Pascal Pyramids, J. Integer Seqs., Vol. 3 (2000), #00.1.6.
Crossrefs
Programs
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Maple
S:= series((BesselI(0,2*x)+BesselI(1,2*x))^3, x, 101): seq(simplify(coeff(S,x,n))*n!, n=0..100); # Robert Israel, Oct 10 2016
Formula
a(n) = sum_j[C(n, j)B(j)B(j+1)B(n-j)] where B(k)=C(k, [k/2])=A001405(k)
E.g.f.: (BesselI(0, 2*x)+BesselI(1, 2*x))^3. - Vladeta Jovovic, Apr 28 2003
From Vaclav Kotesovec, Jun 10 2019: (Start)
Recurrence: (n+1)*(n+2)*(n+3)*a(n) = 4*(5*n^2+10*n+3)*a(n-1) + 4*(n-1)*(10*n^2+10*n-9)*a(n-2) - 144*(n-2)*(n-1)*a(n-3) - 144*(n-3)*(n-2)*(n-1)*a(n-4).
a(n) ~ 6^(n + 3/2) / (Pi*n)^(3/2). (End)
Comments