%I #43 Nov 12 2023 12:07:18
%S 6,180,5040,143640,4199580,125621496,3830266440,118655943120,
%T 3724872182460,118248726796200,3789926661961440,122473276342326000,
%U 3986235855826497000,130561182081992667600,4300094066688571550400
%N Number of walks of length 2n+3 from origin to (1,1,1) on a cubic lattice.
%C a(n) is the number of walks of length 2*n+3 in a cubic lattice that begin at the origin and end at (1,1,1) using steps (1,0,0), (-1,0,0), (0,1,0), (0,-1,0), (0,0,1), (0,0,-1).
%H G. C. Greubel, <a href="/A135395/b135395.txt">Table of n, a(n) for n = 0..250</a>
%H S. Hollos and R. Hollos, <a href="http://www.exstrom.com/math/lattice/latpath.html">Lattice Paths and Walks</a>.
%F a(n) = binomial(2n+3,n) * Sum_{k=0..n} (binomial(n,k) * binomial(n+3,k+2) * binomial(2k+2,k+1)).
%F G.f.: ((12*(4*x-1)*(36*x-1)/x)*g'' + (12*(288*x^2-60*x+1)/x^2)*g' + (72*(6*x-1)/x^2)*g)/288 where g is the o.g.f. of A002896. - _Mark van Hoeij_, Nov 12 2011
%F From _Vaclav Kotesovec_, Nov 27 2017: (Start)
%F Recurrence: n*(n+2)*(n+3)*a(n) = 4*(2*n + 3)*(5*n^2 + 10*n + 3)*a(n-1) - 36*n*(2*n + 1)*(2*n + 3)*a(n-2).
%F a(n) ~ 2^(2*n + 1) * 3^(2*n + 9/2) / (Pi*n)^(3/2). (End)
%F a(n) = (2*n+1)*(2*n+3)*binomial(2*n,n)*((n+3)*A005802(n+1)-(n+1)*A005802(n)). - _Mark van Hoeij_, Nov 12 2023
%p sq := (1-40*x+144*x^2)^(1/2); pb := 54*x*(108*x^2-27*x+1+(9*x-1)*sq);
%p H1 := hypergeom([7/6,1/3],[1],pb); H2 := hypergeom([1/6,4/3],[1],pb);
%p fa := (10-72*x-6*sq)^(1/2)/(432*x^3);
%p ogf := fa*((648*x^2-162*x+(54*x+3)*sq+5)*H1^2 - (648*x^2-342*x+(54*x+6)*sq+10)*H1*H2 - (180*x-5-3*sq)*H2^2);
%p series(ogf,x=0,20) # _Mark van Hoeij_, Nov 12 2011
%t Table[Binomial[2n+3,n]Sum[Binomial[n,k]Binomial[n+3,k+2]Binomial[2k+2,k+1],{k,0,n}],{n,0,20}] (* _Harvey P. Dale_, Mar 20 2012 *)
%o (Maxima) a(n) = binomial(2n+3,n) * sum( binomial(n,k) * binomial(n+3,k+2) * binomial(2k+2,k+1), k, 0, n )
%o (PARI) a(n) = binomial(2*n+3,n) * sum(k=0,n, binomial(n,k) * binomial(n+3,k+2) * binomial(2*k+2,k+1)) \\ _Charles R Greathouse IV_, Oct 12 2016
%Y Cf. A002896.
%K easy,nonn
%O 0,1
%A Stefan Hollos (stefan(AT)exstrom.com), Dec 11 2007
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