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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)}.
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%I #15 Nov 09 2022 19:17:35

%S 1,4,20,96,480,2368,11840,58880,294400,1468416,7342080,36667392,

%T 183336960,916144128,4580720640,22896574464,114482872320,572320645120,

%U 2861603225600,14306741583872,71533707919360,357650927714304,1788254638571520,8941026626502656,44705133132513280,223522175800311808

%N 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)}.

%C Hankel transform is 4^binomial(n+1,2). - _Philippe Deléham_, Feb 01 2009

%H G. C. Greubel, <a href="/A151254/b151254.txt">Table of n, a(n) for n = 0..1000</a>

%H Paul Barry, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL12/Barry2/barry73.html">A Note on a One-Parameter Family of Catalan-Like Numbers</a>, JIS 12 (2009) 09.5.4

%H Alin Bostan and Manuel Kauers, <a href="https://arxiv.org/abs/0811.2899">Automatic Classification of Restricted Lattice Walks</a>, arXiv:0811.2899 [math.CO], 2009.

%F a(n) = Sum_{k=0..n} A120730(n,k)*4^k. - _Philippe Deléham_, Feb 01 2009

%F From _Philippe Deléham_, Feb 02 2009: (Start)

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

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

%F 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

%t 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}]

%t 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 *)

%o (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

%o (SageMath)

%o def a(n): # a = A151254

%o if (n==0): return 1

%o elif (n%2==1): return 5*a(n-1) - 4^((n-1)/2)*catalan_number((n-1)/2)

%o else: return 5*a(n-1)

%o [a(n) for n in (0..30)] # _G. C. Greubel_, Nov 09 2022

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

%K nonn,walk

%O 0,2

%A _Manuel Kauers_, Nov 18 2008