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.

A151162 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)}.

Original entry on oeis.org

1, 3, 12, 45, 180, 702, 2808, 11097, 44388, 176418, 705672, 2812482, 11249928, 44903484, 179613936, 717517521, 2870070084, 11470898106, 45883592424, 183438670950, 733754683800, 2934026948196, 11736107792784, 46934017407594, 187736069630376, 750833732416212, 3003334929664848
Offset: 0

Views

Author

Manuel Kauers, Nov 18 2008

Keywords

Comments

Hankel transform is 3^C(n+1,2). - Philippe Deléham, Feb 01 2009
Inverse binomial transform of A151253. - Philippe Deléham, Feb 03 2009

Crossrefs

Programs

  • Magma
    [n le 3 select Factorial(n+1)/2 else (4*n*Self(n-1) + 12*(n-3)*Self(n-2) - 48*(n-3)*Self(n-3))/n: n in [1..41]]; // 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, 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+2)!/2, (4*(n+1)*a[n-1] +12*(n-2)*a[n-2] -48*(n-2)*a[n- 3])/(n+1)]; Table[a[n], {n,0,40}] (* G. C. Greubel, Nov 09 2022 *)
  • SageMath
    def a(n): # a = A151162
        if (n==0): return 1
        elif (n%2==1): return 4*a(n-1) - 3^((n-1)/2)*catalan_number((n-1)/2)
        else: return 4*a(n-1)
    [a(n) for n in (0..40)] # G. C. Greubel, Nov 09 2022

Formula

a(n) = Sum_{k=0..n} A120730(n,k)*3^k. - Philippe Deléham, Feb 01 2009
From Philippe Deléham, Feb 02 2009: (Start)
a(2*n+2) = 4*a(2*n+1), a(2*n+1) = 4*a(2*n) - 3^n*A000108(n)
a(2*n+1) = 4*a(2*n) - A005159(n).
G.f.: (sqrt(1 - 12*x^2) + 6*x - 1)/(6*x*(1 - 4*x)). (End)
a(n) ~ 2^(2*n+1)/3. - Vaclav Kotesovec, Oct 31 2017
a(n) = (4*(n+1)*a(n-1) + 12*(n-2)*a(n-2) - 48*(n-2)*a(n-3))/(n+1). - G. C. Greubel, Nov 09 2022