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A064036 Number of walks of length n on cubic lattice, starting at origin, staying in first (nonnegative) octant.

%I A064036 #20 Jun 10 2019 03:11:58
%S A064036 1,3,12,51,234,1110,5460,27405,140490,729918,3845016,20447658,
%T A064036 109801692,593806356,3234529584,17715445605,97567971930,539701180590,
%U A064036 2998595422680,16719506691030,93559970043540,525093580540620,2955822168597480,16680150247605390,94365481922990460
%N A064036 Number of walks of length n on cubic lattice, starting at origin, staying in first (nonnegative) octant.
%H A064036 Robert Israel, <a href="/A064036/b064036.txt">Table of n, a(n) for n = 0..500</a>
%H A064036 R. K. Guy, <a href="http://www.cs.uwaterloo.ca/journals/JIS/VOL3/GUY/catwalks.html">Catwalks, Sandsteps and Pascal Pyramids</a>, J. Integer Seqs., Vol. 3 (2000), #00.1.6.
%F A064036 a(n) = sum_j[C(n, j)B(j)B(j+1)B(n-j)] where B(k)=C(k, [k/2])=A001405(k)
%F A064036 E.g.f.: (BesselI(0, 2*x)+BesselI(1, 2*x))^3. - _Vladeta Jovovic_, Apr 28 2003
%F A064036 From _Vaclav Kotesovec_, Jun 10 2019: (Start)
%F A064036 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).
%F A064036 a(n) ~ 6^(n + 3/2) / (Pi*n)^(3/2). (End)
%e A064036 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.
%p A064036 S:= series((BesselI(0,2*x)+BesselI(1,2*x))^3, x, 101):
%p A064036 seq(simplify(coeff(S,x,n))*n!, n=0..100); # _Robert Israel_, Oct 10 2016
%Y A064036 Cf. A064037. The two- and one-dimensional equivalents are A005566 and A001405. With no restriction on the walks, the number is 6^n, i.e. A000400.
%K A064036 nonn
%O A064036 0,2
%A A064036 _Henry Bottomley_, Aug 23 2001