A151254 Number of walks within N^3 (the first octant of Z^3) starting at (0,0,0) and consisting of n steps taken from {(-1, 0, 0), (1, 0, 0), (1, 0, 1), (1, 1, 0), (1, 1, 1)}.

1, 4, 20, 96, 480, 2368, 11840, 58880, 294400, 1468416, 7342080, 36667392, 183336960, 916144128, 4580720640, 22896574464, 114482872320, 572320645120, 2861603225600, 14306741583872, 71533707919360, 357650927714304, 1788254638571520, 8941026626502656, 44705133132513280, 223522175800311808

Offset: 0

Author

Manuel Kauers, Nov 18 2008

Keywords

• nonn
• walk

Comments

Hankel transform is 4^binomial(n+1,2). - Philippe Deléham, Feb 01 2009

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• Paul Barry, A Note on a One-Parameter Family of Catalan-Like Numbers, JIS 12 (2009) 09.5.4
• Alin Bostan and Manuel Kauers, Automatic Classification of Restricted Lattice Walks, arXiv:0811.2899 [math.CO], 2009.

Crossrefs

Cf. A000108, A001405, A120730, A151162, A151281, A151403, A156058.

Programs

Magma

[n le 3 select Factorial(n+2)/6 else (5*n*Self(n-1) + 16*(n-3)*Self(n-2) - 80*(n-3)*Self(n-3))/n: n in [1..30]]; // G. C. Greubel, Nov 09 2022

Mathematica

aux[i_, j_, k_, n_]:= Which[Min[i, j, k, n]<0 || Max[i, j, k]>n, 0, n==0, KroneckerDelta[i, j, k, n], True, aux[i, j, k, n] = aux[-1+i, -1+j, -1+k, -1+n] + aux[-1+i, -1+j, k, -1+n] + aux[-1+i, j, -1+k, -1+n] + aux[-1+i, j, k, -1 + n] + aux[1+i, j, k, -1+n]]; Table[Sum[aux[i,j,k,n], {i,0,n}, {j,0,n}, {k,0,n}], {n, 0, 30}]
a[n_]:= a[n]= If[n<3, (n+3)!/3!, (5*(n+1)*a[n-1] +16*(n-2)*a[n-2] -80*(n-2)*a[n- 3])/(n+1)]; Table[a[n], {n, 0, 30}] (* G. C. Greubel, Nov 09 2022 *)

SageMath

def a(n): # a = A151254
    if (n==0): return 1
    elif (n%2==1): return 5*a(n-1) - 4^((n-1)/2)*catalan_number((n-1)/2)
    else: return 5*a(n-1)
[a(n) for n in (0..30)] # G. C. Greubel, Nov 09 2022

Formula

a(n) = Sum_{k=0..n} A120730(n,k)*4^k. - Philippe Deléham, Feb 01 2009

From Philippe Deléham, Feb 02 2009: (Start)

a(2n+2) = 5*a(2n+1), a(2n+1) = 5*a(2n) - 4^n*A000108(n) = 5*a(2n) - A151403(n).

G.f.: (sqrt(1-16*x^2) + 8*x - 1)/(8*x*(1-5*x)). (End)

a(n) = (5*(n+1)*a(n-1) + 16*(n-2)*a(n-2) - 80*(n-2)*a(n-3))/(n+1). - G. C. Greubel, Nov 09 2022