%I #17 Aug 26 2021 18:20:50

%S 1,6,12,8,0,0,0,0,6,24,24,0,0,0,0,0,12,24,0,0,0,0,0,0,8,0,0,6,24,24,0,

%T 0,0,0,0,24,48,0,0,0,0,0,0,24,0,0,0,0,0,0,0,0,0,0,12,24,0,0,0,0,0,0,

%U 24,0,6,24,24,0,0,0,0,0,24,48,0,0,0,0,0,0,24,8,0,0,0,0,0,0,0,0,0,24,48,0,0,0

%N Number of integer triples (x,y,z) satisfying |x|^3 + |y|^3 + |z|^3 = n, -n <= x,y,z <= n.

%C A three-dimensional variant of A175362.

%H Robert Israel, <a href="/A175365/b175365.txt">Table of n, a(n) for n = 0..10000</a>

%F G.f.: ( 1 + 2*Sum_{j>=1} x^(j^3) )^3.

%F a(n) = A175362(n) + 2*Sum_{k=1..floor(n^(1/3))} A175362(n - k^3). - _Daniel Suteu_, Aug 15 2021

%e a(2) = 12 counts (x,y,z) = (-1,-1,0), (-1,0,-1), (-1,0,1), (-1,1,0), (0,-1,-1), (0,-1,1), (0,1,-1), (0,1,1), (1,-1,0), (1,0,-1), (1,0,1) and (1,1,0).

%p N:= 100: # to get a(0) to a(N)

%p G:= (1+2*add(x^(j^3),j=1..floor(N^(1/3))))^3:

%p S:= series(G,x,N+1):

%p seq(coeff(S,x,j),j=0..N); # _Robert Israel_, Apr 08 2016

%t CoefficientList[(1 + 2 Sum[x^(j^3), {j, 4}])^3, x] (* _Michael De Vlieger_, Apr 08 2016 *)

%o (PARI) a(n, k=3) = if(n==0, return(1)); if(k <= 0, return(0)); if(k == 1, return(ispower(n, 3))); my(count = 0); for(v = 0, sqrtnint(n, 3), count += (2 - (v == 0))*if(k > 2, a(n - v^3, k-1), if(ispower(n - v^3, 3), 2 - (n - v^3 == 0), 0))); count; \\ _Daniel Suteu_, Aug 15 2021

%K nonn

%O 0,2

%A _R. J. Mathar_, Apr 24 2010