cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A135395 Number of walks of length 2n+3 from origin to (1,1,1) on a cubic lattice.

Original entry on oeis.org

6, 180, 5040, 143640, 4199580, 125621496, 3830266440, 118655943120, 3724872182460, 118248726796200, 3789926661961440, 122473276342326000, 3986235855826497000, 130561182081992667600, 4300094066688571550400
Offset: 0

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Author

Stefan Hollos (stefan(AT)exstrom.com), Dec 11 2007

Keywords

Comments

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).

Crossrefs

Cf. A002896.

Programs

  • Maple
    sq := (1-40*x+144*x^2)^(1/2); pb := 54*x*(108*x^2-27*x+1+(9*x-1)*sq);
    H1 := hypergeom([7/6,1/3],[1],pb); H2 := hypergeom([1/6,4/3],[1],pb);
    fa := (10-72*x-6*sq)^(1/2)/(432*x^3);
    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);
    series(ogf,x=0,20) # Mark van Hoeij, Nov 12 2011
  • Mathematica
    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 *)
  • Maxima
    a(n) = binomial(2n+3,n) * sum( binomial(n,k) * binomial(n+3,k+2) * binomial(2k+2,k+1), k, 0, n )
    
  • 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

Formula

a(n) = binomial(2n+3,n) * Sum_{k=0..n} (binomial(n,k) * binomial(n+3,k+2) * binomial(2k+2,k+1)).
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
From Vaclav Kotesovec, Nov 27 2017: (Start)
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).
a(n) ~ 2^(2*n + 1) * 3^(2*n + 9/2) / (Pi*n)^(3/2). (End)
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